Are there categories whose internal hom is somewhat 'exotic'?Is there lore about how endofunctors of Cat interact with the formation of presheaf categories?Internal hom of sheavesEquivalences of Internal CategoriesTwo definitions of modules in monoidal categoryAre dagger categories truly evil?Examples of enriched categories which are (co)powered or (co)tensoredInternal Hom of Deligne' tensor productCombinatorial proof that some model categories are monoidal/enriched?
Are there categories whose internal hom is somewhat 'exotic'?
Is there lore about how endofunctors of Cat interact with the formation of presheaf categories?Internal hom of sheavesEquivalences of Internal CategoriesTwo definitions of modules in monoidal categoryAre dagger categories truly evil?Examples of enriched categories which are (co)powered or (co)tensoredInternal Hom of Deligne' tensor productCombinatorial proof that some model categories are monoidal/enriched?
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The internal hom (or exponential object) is basically a reification of the 'external' hom. It can be defined in any cartesian (or even monoidal, more on this later) category as the right adjoint of the (monoidal) product.
My question is: are there some categories whose internal hom behaves quite unexpectedly? Or even some interesting examples where internalization really deforms the external hom in a non-trivial way.
The question arises from the observaton that the 'external' hom can be assumed to be a set (assume the category to be locally small, or wave your hand hard enough), and thus is already somewhat reified. We expect some 'structure' on it, namely elements and perhaps even subobjects. In a sufficiently rich category (say a topos), this structure is internalizable as well. So it might be the case that some exotic behavior emerge, or some collapse happens.
I expect this to happen mainly for monoidal closed category, because the monoidal product can be quite involved.
On the other hand, Section 3 of the internal hom page of the nLab seems to prove the internal hom shares some strong properties of the 'external' hom, which might hint to the fact they are really 'the same'.
ct.category-theory internalization
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The internal hom (or exponential object) is basically a reification of the 'external' hom. It can be defined in any cartesian (or even monoidal, more on this later) category as the right adjoint of the (monoidal) product.
My question is: are there some categories whose internal hom behaves quite unexpectedly? Or even some interesting examples where internalization really deforms the external hom in a non-trivial way.
The question arises from the observaton that the 'external' hom can be assumed to be a set (assume the category to be locally small, or wave your hand hard enough), and thus is already somewhat reified. We expect some 'structure' on it, namely elements and perhaps even subobjects. In a sufficiently rich category (say a topos), this structure is internalizable as well. So it might be the case that some exotic behavior emerge, or some collapse happens.
I expect this to happen mainly for monoidal closed category, because the monoidal product can be quite involved.
On the other hand, Section 3 of the internal hom page of the nLab seems to prove the internal hom shares some strong properties of the 'external' hom, which might hint to the fact they are really 'the same'.
ct.category-theory internalization
New contributor
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$begingroup$
The internal hom (or exponential object) is basically a reification of the 'external' hom. It can be defined in any cartesian (or even monoidal, more on this later) category as the right adjoint of the (monoidal) product.
My question is: are there some categories whose internal hom behaves quite unexpectedly? Or even some interesting examples where internalization really deforms the external hom in a non-trivial way.
The question arises from the observaton that the 'external' hom can be assumed to be a set (assume the category to be locally small, or wave your hand hard enough), and thus is already somewhat reified. We expect some 'structure' on it, namely elements and perhaps even subobjects. In a sufficiently rich category (say a topos), this structure is internalizable as well. So it might be the case that some exotic behavior emerge, or some collapse happens.
I expect this to happen mainly for monoidal closed category, because the monoidal product can be quite involved.
On the other hand, Section 3 of the internal hom page of the nLab seems to prove the internal hom shares some strong properties of the 'external' hom, which might hint to the fact they are really 'the same'.
ct.category-theory internalization
New contributor
$endgroup$
The internal hom (or exponential object) is basically a reification of the 'external' hom. It can be defined in any cartesian (or even monoidal, more on this later) category as the right adjoint of the (monoidal) product.
My question is: are there some categories whose internal hom behaves quite unexpectedly? Or even some interesting examples where internalization really deforms the external hom in a non-trivial way.
The question arises from the observaton that the 'external' hom can be assumed to be a set (assume the category to be locally small, or wave your hand hard enough), and thus is already somewhat reified. We expect some 'structure' on it, namely elements and perhaps even subobjects. In a sufficiently rich category (say a topos), this structure is internalizable as well. So it might be the case that some exotic behavior emerge, or some collapse happens.
I expect this to happen mainly for monoidal closed category, because the monoidal product can be quite involved.
On the other hand, Section 3 of the internal hom page of the nLab seems to prove the internal hom shares some strong properties of the 'external' hom, which might hint to the fact they are really 'the same'.
ct.category-theory internalization
ct.category-theory internalization
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mattecapumattecapu
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$newcommandHommathrmHomnewcommandSetmathrmSetnewcommandhommathrmhom$In this example one cannot really say that the internal Hom is “exotic”, but at least it is morally different from the external Hom.
Let $G$ be a group and let $Set(G)$ be the category of sets with a $G$-action. Morphisms of this category are mappings of sets which commute with the action. This category has finite products, given by the products of the underlying sets. In fact, it is cartesian closed: The internal Hom $hom(X,Y)$ is the set of all mappings $X to Y$, with the $G$-action given by conjugation: $$sigma cdot f := X xrightarrowcdot sigma^-1 X xrightarrowf Y xrightarrowcdot sigma Y. $$
Note however that in a monoidally closed category, the external Hom can always be recoverd from the internal Hom as the set of its global elements:
$$ Hom(X,Y) cong Gamma(hom(X,Y)) := Hom( I, hom(X,Y)).$$
Here, $I$ is the unit object of the monoidal category.
In the above example, global elements of some $G$-set $Z$ are morphisms from the one-point $G$-set to $Z$. These correspond precisely to the fixed points of $Z$ under the $G$-action: $Gamma(Z) = Z^G$. In particular, the global elements of the internal Hom are mappings invariant under conjugation – i.e. morphisms of $G$-sets.
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1
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Thanks Jakob, this is a wonderful example!
$endgroup$
– mattecapu
8 hours ago
add a comment |
$begingroup$
Typically, internal homs of $newcommandCtextbfCC$ will look different from external homs just when “elements/points of $X$” (for objects $X in C$) are different from “maps $I to X$” (where $I$ is the monoidal unit); or slightly more precisely, when $C$ comes with a canonical forgetful functor $newcommandSettextbfSet U : C to Set$, which is different from the representable $C(I,-)$.
This is because (as noted in Jakob Werner’s answer) maps $I to [X,Y]$ correspond to (external) maps $X to Y$, and so if we want “points of the internal hom” to be different from external arrows, that implies that “points of the internal hom” must be different from “maps from $I$ to the internal hom”.
This idea suggests several examples:
$Gtext-Set$, for a group $G$ (as in Jakob’s answer). Here the monoidal unit is the terminal object $1$, and maps $1 to X$ correspond not to arbitrary points of $X$ but just to fixpoints of the $G$-action. So the external maps $X to Y$ (i.e. $G$-equivariant maps) correspond to fixpoints in the $G$-set $[X,Y]$; arbitrary points of $[X,Y]$ correspond to not-necessarily-equivariant functions $X to Y$.
The category of graded Abelian groups $newcommandAbmathrmAbnewcommandZmathbbZ Ab^Z$, with the graded tensor product $(X otimes Y)_n = coprod_i+j=n X_i otimes Y_j$. (Or graded modules over a ring, or $mathbfN$-graded, etc.) In this case, the monoidal unit is $Z$ in degree 0 and trivial in other degrees, and maps $I to X$ correspond to elements of $X_0$. So external maps $X to Y$ correspond to elements of $[X,Y]_0$, and elements of $[X,Y]$ in other degrees correspond to degree-shifting maps between $X$ and $Y$.
Categories of chain complexes, $newcommandChmathrmChCh(Ab)$. Mostly as in the previous case, but now maps $I to X$ correspond just to cycles in $X_0$, so external maps $X to Y$ correspond to degree-0 cycles in $[X,Y]$, while arbitrary elements of $[X,Y]_n$ correspond to maps $X to Y$ shifting degree by $n$ and not necessarily respecting the boundary operator.
Even if these are not as exotic as you were hoping for, hopefully the general principle “look for categories where maps out of the monoidal unit don’t correspond to ‘elements/points’ of objects” may help find more exotic examples.
$endgroup$
2
$begingroup$
The category of set and relation (monoidal for the products of sets) is also a good example. The exponential is given by the product of sets.
$endgroup$
– Simon Henry
7 hours ago
2
$begingroup$
@SimonHenry: why not make that an additional answer, since it has quite a distinct flavour from the examples here?
$endgroup$
– Peter LeFanu Lumsdaine
6 hours ago
add a comment |
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Here are a slightly different flavor of examples, of course as mentioned by Peter Lumsdain, the trick is always the same looking at categories where $Hom(I,X)$ looks quite different from the "underlying sets of $X$"
Consider the category whose objects are set, and whose morphisms are relations, i.e. subsets of $X times Y$ (with the usual composition of relation).
It has a monoidal structure given by the product of sets. The exponential for this monoidal structure is simply product, indeed,
A relation from $X$ to $Y times Z$ is the same as a relation from $X times Z$ to $Y$.
Other similar examples includes:
The 2-category of sets and span of sets (with the products of sets as monoidal structure).
The category of small categories and profunctor, with the products of categories as monoidal structure, here the exponential $[X,Y]$ is given by $Y times X^op$
The category of polynomial functor (which is actually cartesian closed) described in The cartesian closed bicategory of generalized species and structures by Fiore,Gambino,Hyland and Winskel. Where the exponential is a little more complicated, but still has the same flavor.
But one can somehow "explain the trick" behind these: One way to think about that these example is that these categories are equivalent to categories where the Hom objects corresponds more naturally to a natural structure one the set of morphisms, but the equivalence of categories somehow make the "underlying set" very different.
For example, the category of set and relation can be seen as the category of "Free suplattice":
A suplattice is an ordered set with all supremum, and suplattice morphisms are order preserving map, preserving the supremums. Given two suplattices $X$ and $Y$ the set of morphisms from $X$ and $Y$ is naturally a suplattice for the pointwise ordering (induce by the order of $Y$), and this corresponds to a monoidal symmetric closed structure on the category of suplattice. So this is really a "non-example" of what is asked: the hom objects are really exactly the set of morphisms with a natural structure induced on them.
Now the forgetfull functor from suplattice to sets has a left adjoint, sending any set to the suplattice $P(X)$ of subsets of $X$, sublattice morphisms from $P(X)$ to $P(Y)$ coincide with relation from $X$ and $Y$, and the monoidal structure described above is induced from the one on suplattice...
So our example become a non-example up an equivalence of categories. Similar description can be obtained for the others example I have mentioned.
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This is really a great example, thank you! Just to clarify: the external hom here is the power set of $X times Y$, while the internal hom is just $X times Y$ right? It is very dramatic, maybe the most striking here because of its simplicity.
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– mattecapu
1 hour ago
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Also the last part of very insightful, I love it. Thanks again!
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– mattecapu
1 hour ago
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Yes absolutely.
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– Simon Henry
1 hour ago
add a comment |
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To add to the other good answers here, there's a family of examples that could be seen as a bit trivial. But in a sense they give the simplest answer to your question.
Let $X$ be a partially ordered set, viewed as a category in the usual way: the objects of the category are the elements of $X$, and $textrmHom(x, y)$ has either $1$ or $0$ elements according to whether $x leq y$ or not. If $X$ is a meet-semilattice, i.e. any finite set of elements has a greatest lower bound, then the corresponding category has finite products. It's cartesian closed if for any two elements $y$ and $z$ there's an element $y to z$ with the property that for all $x in X$,
$$
x wedge y leq z iff x leq y to z.
$$
Here $wedge$ denotes greatest lower bound. So $y to z$ is the internal hom $z^y$. A poset with this property is more or less what's called a Heyting algebra.
I think this is an enlightening family of examples because in a poset, the external homs are pretty trivial (sets with at most one element), whereas the internal homs can be informative. For example, in a power set they give you the notion of complement (exercise!).
You can extend this family a bit. Again take an ordered set $X$, regarded as a category in the usual way. But now think about non-cartesian monoidal structures on it. For instance, take $X = mathbbR$ with its usual ordering. The cartesian monoidal structure is $mathrmmin$, but we could instead use $+$. Then the internal hom is given by
$$
y to z = z - y.
$$
Is that "unexpected"? That depends on your intuition. But in this example the external hom only tells you the sign of $z - y$, whereas the internal hom tells you its actual value. In other words, it produces the operation of subtraction, which historically has proved quite important.
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Yes! Heyting algebras are what I was looking at when I thought my question. So I was waiting for this answer (I would have written it myself tomorrow). Still, yours is surely better than what I would have written, in particular the last example I really cool.
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– mattecapu
14 mins ago
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Just to add to your last example: to see why the internal hom is $z-y$, it is a fact of life that in an Heyting algebra (aka postal Cartesian closed small category), $y to z = neg y lor y$, where the 'negation' of $y$ is $y to 0$ (this also solves the exercise about complements in power sets). Therefore if we unpack the universal property defining $to$ we get $x+y leq 0 iff x leq y to 0$, so $neg y = y to 0 = - y$ as expected.
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– mattecapu
4 mins ago
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$begingroup$
$newcommandHommathrmHomnewcommandSetmathrmSetnewcommandhommathrmhom$In this example one cannot really say that the internal Hom is “exotic”, but at least it is morally different from the external Hom.
Let $G$ be a group and let $Set(G)$ be the category of sets with a $G$-action. Morphisms of this category are mappings of sets which commute with the action. This category has finite products, given by the products of the underlying sets. In fact, it is cartesian closed: The internal Hom $hom(X,Y)$ is the set of all mappings $X to Y$, with the $G$-action given by conjugation: $$sigma cdot f := X xrightarrowcdot sigma^-1 X xrightarrowf Y xrightarrowcdot sigma Y. $$
Note however that in a monoidally closed category, the external Hom can always be recoverd from the internal Hom as the set of its global elements:
$$ Hom(X,Y) cong Gamma(hom(X,Y)) := Hom( I, hom(X,Y)).$$
Here, $I$ is the unit object of the monoidal category.
In the above example, global elements of some $G$-set $Z$ are morphisms from the one-point $G$-set to $Z$. These correspond precisely to the fixed points of $Z$ under the $G$-action: $Gamma(Z) = Z^G$. In particular, the global elements of the internal Hom are mappings invariant under conjugation – i.e. morphisms of $G$-sets.
$endgroup$
1
$begingroup$
Thanks Jakob, this is a wonderful example!
$endgroup$
– mattecapu
8 hours ago
add a comment |
$begingroup$
$newcommandHommathrmHomnewcommandSetmathrmSetnewcommandhommathrmhom$In this example one cannot really say that the internal Hom is “exotic”, but at least it is morally different from the external Hom.
Let $G$ be a group and let $Set(G)$ be the category of sets with a $G$-action. Morphisms of this category are mappings of sets which commute with the action. This category has finite products, given by the products of the underlying sets. In fact, it is cartesian closed: The internal Hom $hom(X,Y)$ is the set of all mappings $X to Y$, with the $G$-action given by conjugation: $$sigma cdot f := X xrightarrowcdot sigma^-1 X xrightarrowf Y xrightarrowcdot sigma Y. $$
Note however that in a monoidally closed category, the external Hom can always be recoverd from the internal Hom as the set of its global elements:
$$ Hom(X,Y) cong Gamma(hom(X,Y)) := Hom( I, hom(X,Y)).$$
Here, $I$ is the unit object of the monoidal category.
In the above example, global elements of some $G$-set $Z$ are morphisms from the one-point $G$-set to $Z$. These correspond precisely to the fixed points of $Z$ under the $G$-action: $Gamma(Z) = Z^G$. In particular, the global elements of the internal Hom are mappings invariant under conjugation – i.e. morphisms of $G$-sets.
$endgroup$
1
$begingroup$
Thanks Jakob, this is a wonderful example!
$endgroup$
– mattecapu
8 hours ago
add a comment |
$begingroup$
$newcommandHommathrmHomnewcommandSetmathrmSetnewcommandhommathrmhom$In this example one cannot really say that the internal Hom is “exotic”, but at least it is morally different from the external Hom.
Let $G$ be a group and let $Set(G)$ be the category of sets with a $G$-action. Morphisms of this category are mappings of sets which commute with the action. This category has finite products, given by the products of the underlying sets. In fact, it is cartesian closed: The internal Hom $hom(X,Y)$ is the set of all mappings $X to Y$, with the $G$-action given by conjugation: $$sigma cdot f := X xrightarrowcdot sigma^-1 X xrightarrowf Y xrightarrowcdot sigma Y. $$
Note however that in a monoidally closed category, the external Hom can always be recoverd from the internal Hom as the set of its global elements:
$$ Hom(X,Y) cong Gamma(hom(X,Y)) := Hom( I, hom(X,Y)).$$
Here, $I$ is the unit object of the monoidal category.
In the above example, global elements of some $G$-set $Z$ are morphisms from the one-point $G$-set to $Z$. These correspond precisely to the fixed points of $Z$ under the $G$-action: $Gamma(Z) = Z^G$. In particular, the global elements of the internal Hom are mappings invariant under conjugation – i.e. morphisms of $G$-sets.
$endgroup$
$newcommandHommathrmHomnewcommandSetmathrmSetnewcommandhommathrmhom$In this example one cannot really say that the internal Hom is “exotic”, but at least it is morally different from the external Hom.
Let $G$ be a group and let $Set(G)$ be the category of sets with a $G$-action. Morphisms of this category are mappings of sets which commute with the action. This category has finite products, given by the products of the underlying sets. In fact, it is cartesian closed: The internal Hom $hom(X,Y)$ is the set of all mappings $X to Y$, with the $G$-action given by conjugation: $$sigma cdot f := X xrightarrowcdot sigma^-1 X xrightarrowf Y xrightarrowcdot sigma Y. $$
Note however that in a monoidally closed category, the external Hom can always be recoverd from the internal Hom as the set of its global elements:
$$ Hom(X,Y) cong Gamma(hom(X,Y)) := Hom( I, hom(X,Y)).$$
Here, $I$ is the unit object of the monoidal category.
In the above example, global elements of some $G$-set $Z$ are morphisms from the one-point $G$-set to $Z$. These correspond precisely to the fixed points of $Z$ under the $G$-action: $Gamma(Z) = Z^G$. In particular, the global elements of the internal Hom are mappings invariant under conjugation – i.e. morphisms of $G$-sets.
answered 11 hours ago
Jakob WernerJakob Werner
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1
$begingroup$
Thanks Jakob, this is a wonderful example!
$endgroup$
– mattecapu
8 hours ago
add a comment |
1
$begingroup$
Thanks Jakob, this is a wonderful example!
$endgroup$
– mattecapu
8 hours ago
1
1
$begingroup$
Thanks Jakob, this is a wonderful example!
$endgroup$
– mattecapu
8 hours ago
$begingroup$
Thanks Jakob, this is a wonderful example!
$endgroup$
– mattecapu
8 hours ago
add a comment |
$begingroup$
Typically, internal homs of $newcommandCtextbfCC$ will look different from external homs just when “elements/points of $X$” (for objects $X in C$) are different from “maps $I to X$” (where $I$ is the monoidal unit); or slightly more precisely, when $C$ comes with a canonical forgetful functor $newcommandSettextbfSet U : C to Set$, which is different from the representable $C(I,-)$.
This is because (as noted in Jakob Werner’s answer) maps $I to [X,Y]$ correspond to (external) maps $X to Y$, and so if we want “points of the internal hom” to be different from external arrows, that implies that “points of the internal hom” must be different from “maps from $I$ to the internal hom”.
This idea suggests several examples:
$Gtext-Set$, for a group $G$ (as in Jakob’s answer). Here the monoidal unit is the terminal object $1$, and maps $1 to X$ correspond not to arbitrary points of $X$ but just to fixpoints of the $G$-action. So the external maps $X to Y$ (i.e. $G$-equivariant maps) correspond to fixpoints in the $G$-set $[X,Y]$; arbitrary points of $[X,Y]$ correspond to not-necessarily-equivariant functions $X to Y$.
The category of graded Abelian groups $newcommandAbmathrmAbnewcommandZmathbbZ Ab^Z$, with the graded tensor product $(X otimes Y)_n = coprod_i+j=n X_i otimes Y_j$. (Or graded modules over a ring, or $mathbfN$-graded, etc.) In this case, the monoidal unit is $Z$ in degree 0 and trivial in other degrees, and maps $I to X$ correspond to elements of $X_0$. So external maps $X to Y$ correspond to elements of $[X,Y]_0$, and elements of $[X,Y]$ in other degrees correspond to degree-shifting maps between $X$ and $Y$.
Categories of chain complexes, $newcommandChmathrmChCh(Ab)$. Mostly as in the previous case, but now maps $I to X$ correspond just to cycles in $X_0$, so external maps $X to Y$ correspond to degree-0 cycles in $[X,Y]$, while arbitrary elements of $[X,Y]_n$ correspond to maps $X to Y$ shifting degree by $n$ and not necessarily respecting the boundary operator.
Even if these are not as exotic as you were hoping for, hopefully the general principle “look for categories where maps out of the monoidal unit don’t correspond to ‘elements/points’ of objects” may help find more exotic examples.
$endgroup$
2
$begingroup$
The category of set and relation (monoidal for the products of sets) is also a good example. The exponential is given by the product of sets.
$endgroup$
– Simon Henry
7 hours ago
2
$begingroup$
@SimonHenry: why not make that an additional answer, since it has quite a distinct flavour from the examples here?
$endgroup$
– Peter LeFanu Lumsdaine
6 hours ago
add a comment |
$begingroup$
Typically, internal homs of $newcommandCtextbfCC$ will look different from external homs just when “elements/points of $X$” (for objects $X in C$) are different from “maps $I to X$” (where $I$ is the monoidal unit); or slightly more precisely, when $C$ comes with a canonical forgetful functor $newcommandSettextbfSet U : C to Set$, which is different from the representable $C(I,-)$.
This is because (as noted in Jakob Werner’s answer) maps $I to [X,Y]$ correspond to (external) maps $X to Y$, and so if we want “points of the internal hom” to be different from external arrows, that implies that “points of the internal hom” must be different from “maps from $I$ to the internal hom”.
This idea suggests several examples:
$Gtext-Set$, for a group $G$ (as in Jakob’s answer). Here the monoidal unit is the terminal object $1$, and maps $1 to X$ correspond not to arbitrary points of $X$ but just to fixpoints of the $G$-action. So the external maps $X to Y$ (i.e. $G$-equivariant maps) correspond to fixpoints in the $G$-set $[X,Y]$; arbitrary points of $[X,Y]$ correspond to not-necessarily-equivariant functions $X to Y$.
The category of graded Abelian groups $newcommandAbmathrmAbnewcommandZmathbbZ Ab^Z$, with the graded tensor product $(X otimes Y)_n = coprod_i+j=n X_i otimes Y_j$. (Or graded modules over a ring, or $mathbfN$-graded, etc.) In this case, the monoidal unit is $Z$ in degree 0 and trivial in other degrees, and maps $I to X$ correspond to elements of $X_0$. So external maps $X to Y$ correspond to elements of $[X,Y]_0$, and elements of $[X,Y]$ in other degrees correspond to degree-shifting maps between $X$ and $Y$.
Categories of chain complexes, $newcommandChmathrmChCh(Ab)$. Mostly as in the previous case, but now maps $I to X$ correspond just to cycles in $X_0$, so external maps $X to Y$ correspond to degree-0 cycles in $[X,Y]$, while arbitrary elements of $[X,Y]_n$ correspond to maps $X to Y$ shifting degree by $n$ and not necessarily respecting the boundary operator.
Even if these are not as exotic as you were hoping for, hopefully the general principle “look for categories where maps out of the monoidal unit don’t correspond to ‘elements/points’ of objects” may help find more exotic examples.
$endgroup$
2
$begingroup$
The category of set and relation (monoidal for the products of sets) is also a good example. The exponential is given by the product of sets.
$endgroup$
– Simon Henry
7 hours ago
2
$begingroup$
@SimonHenry: why not make that an additional answer, since it has quite a distinct flavour from the examples here?
$endgroup$
– Peter LeFanu Lumsdaine
6 hours ago
add a comment |
$begingroup$
Typically, internal homs of $newcommandCtextbfCC$ will look different from external homs just when “elements/points of $X$” (for objects $X in C$) are different from “maps $I to X$” (where $I$ is the monoidal unit); or slightly more precisely, when $C$ comes with a canonical forgetful functor $newcommandSettextbfSet U : C to Set$, which is different from the representable $C(I,-)$.
This is because (as noted in Jakob Werner’s answer) maps $I to [X,Y]$ correspond to (external) maps $X to Y$, and so if we want “points of the internal hom” to be different from external arrows, that implies that “points of the internal hom” must be different from “maps from $I$ to the internal hom”.
This idea suggests several examples:
$Gtext-Set$, for a group $G$ (as in Jakob’s answer). Here the monoidal unit is the terminal object $1$, and maps $1 to X$ correspond not to arbitrary points of $X$ but just to fixpoints of the $G$-action. So the external maps $X to Y$ (i.e. $G$-equivariant maps) correspond to fixpoints in the $G$-set $[X,Y]$; arbitrary points of $[X,Y]$ correspond to not-necessarily-equivariant functions $X to Y$.
The category of graded Abelian groups $newcommandAbmathrmAbnewcommandZmathbbZ Ab^Z$, with the graded tensor product $(X otimes Y)_n = coprod_i+j=n X_i otimes Y_j$. (Or graded modules over a ring, or $mathbfN$-graded, etc.) In this case, the monoidal unit is $Z$ in degree 0 and trivial in other degrees, and maps $I to X$ correspond to elements of $X_0$. So external maps $X to Y$ correspond to elements of $[X,Y]_0$, and elements of $[X,Y]$ in other degrees correspond to degree-shifting maps between $X$ and $Y$.
Categories of chain complexes, $newcommandChmathrmChCh(Ab)$. Mostly as in the previous case, but now maps $I to X$ correspond just to cycles in $X_0$, so external maps $X to Y$ correspond to degree-0 cycles in $[X,Y]$, while arbitrary elements of $[X,Y]_n$ correspond to maps $X to Y$ shifting degree by $n$ and not necessarily respecting the boundary operator.
Even if these are not as exotic as you were hoping for, hopefully the general principle “look for categories where maps out of the monoidal unit don’t correspond to ‘elements/points’ of objects” may help find more exotic examples.
$endgroup$
Typically, internal homs of $newcommandCtextbfCC$ will look different from external homs just when “elements/points of $X$” (for objects $X in C$) are different from “maps $I to X$” (where $I$ is the monoidal unit); or slightly more precisely, when $C$ comes with a canonical forgetful functor $newcommandSettextbfSet U : C to Set$, which is different from the representable $C(I,-)$.
This is because (as noted in Jakob Werner’s answer) maps $I to [X,Y]$ correspond to (external) maps $X to Y$, and so if we want “points of the internal hom” to be different from external arrows, that implies that “points of the internal hom” must be different from “maps from $I$ to the internal hom”.
This idea suggests several examples:
$Gtext-Set$, for a group $G$ (as in Jakob’s answer). Here the monoidal unit is the terminal object $1$, and maps $1 to X$ correspond not to arbitrary points of $X$ but just to fixpoints of the $G$-action. So the external maps $X to Y$ (i.e. $G$-equivariant maps) correspond to fixpoints in the $G$-set $[X,Y]$; arbitrary points of $[X,Y]$ correspond to not-necessarily-equivariant functions $X to Y$.
The category of graded Abelian groups $newcommandAbmathrmAbnewcommandZmathbbZ Ab^Z$, with the graded tensor product $(X otimes Y)_n = coprod_i+j=n X_i otimes Y_j$. (Or graded modules over a ring, or $mathbfN$-graded, etc.) In this case, the monoidal unit is $Z$ in degree 0 and trivial in other degrees, and maps $I to X$ correspond to elements of $X_0$. So external maps $X to Y$ correspond to elements of $[X,Y]_0$, and elements of $[X,Y]$ in other degrees correspond to degree-shifting maps between $X$ and $Y$.
Categories of chain complexes, $newcommandChmathrmChCh(Ab)$. Mostly as in the previous case, but now maps $I to X$ correspond just to cycles in $X_0$, so external maps $X to Y$ correspond to degree-0 cycles in $[X,Y]$, while arbitrary elements of $[X,Y]_n$ correspond to maps $X to Y$ shifting degree by $n$ and not necessarily respecting the boundary operator.
Even if these are not as exotic as you were hoping for, hopefully the general principle “look for categories where maps out of the monoidal unit don’t correspond to ‘elements/points’ of objects” may help find more exotic examples.
edited 7 hours ago
answered 9 hours ago
Peter LeFanu LumsdainePeter LeFanu Lumsdaine
9,3051 gold badge39 silver badges73 bronze badges
9,3051 gold badge39 silver badges73 bronze badges
2
$begingroup$
The category of set and relation (monoidal for the products of sets) is also a good example. The exponential is given by the product of sets.
$endgroup$
– Simon Henry
7 hours ago
2
$begingroup$
@SimonHenry: why not make that an additional answer, since it has quite a distinct flavour from the examples here?
$endgroup$
– Peter LeFanu Lumsdaine
6 hours ago
add a comment |
2
$begingroup$
The category of set and relation (monoidal for the products of sets) is also a good example. The exponential is given by the product of sets.
$endgroup$
– Simon Henry
7 hours ago
2
$begingroup$
@SimonHenry: why not make that an additional answer, since it has quite a distinct flavour from the examples here?
$endgroup$
– Peter LeFanu Lumsdaine
6 hours ago
2
2
$begingroup$
The category of set and relation (monoidal for the products of sets) is also a good example. The exponential is given by the product of sets.
$endgroup$
– Simon Henry
7 hours ago
$begingroup$
The category of set and relation (monoidal for the products of sets) is also a good example. The exponential is given by the product of sets.
$endgroup$
– Simon Henry
7 hours ago
2
2
$begingroup$
@SimonHenry: why not make that an additional answer, since it has quite a distinct flavour from the examples here?
$endgroup$
– Peter LeFanu Lumsdaine
6 hours ago
$begingroup$
@SimonHenry: why not make that an additional answer, since it has quite a distinct flavour from the examples here?
$endgroup$
– Peter LeFanu Lumsdaine
6 hours ago
add a comment |
$begingroup$
Here are a slightly different flavor of examples, of course as mentioned by Peter Lumsdain, the trick is always the same looking at categories where $Hom(I,X)$ looks quite different from the "underlying sets of $X$"
Consider the category whose objects are set, and whose morphisms are relations, i.e. subsets of $X times Y$ (with the usual composition of relation).
It has a monoidal structure given by the product of sets. The exponential for this monoidal structure is simply product, indeed,
A relation from $X$ to $Y times Z$ is the same as a relation from $X times Z$ to $Y$.
Other similar examples includes:
The 2-category of sets and span of sets (with the products of sets as monoidal structure).
The category of small categories and profunctor, with the products of categories as monoidal structure, here the exponential $[X,Y]$ is given by $Y times X^op$
The category of polynomial functor (which is actually cartesian closed) described in The cartesian closed bicategory of generalized species and structures by Fiore,Gambino,Hyland and Winskel. Where the exponential is a little more complicated, but still has the same flavor.
But one can somehow "explain the trick" behind these: One way to think about that these example is that these categories are equivalent to categories where the Hom objects corresponds more naturally to a natural structure one the set of morphisms, but the equivalence of categories somehow make the "underlying set" very different.
For example, the category of set and relation can be seen as the category of "Free suplattice":
A suplattice is an ordered set with all supremum, and suplattice morphisms are order preserving map, preserving the supremums. Given two suplattices $X$ and $Y$ the set of morphisms from $X$ and $Y$ is naturally a suplattice for the pointwise ordering (induce by the order of $Y$), and this corresponds to a monoidal symmetric closed structure on the category of suplattice. So this is really a "non-example" of what is asked: the hom objects are really exactly the set of morphisms with a natural structure induced on them.
Now the forgetfull functor from suplattice to sets has a left adjoint, sending any set to the suplattice $P(X)$ of subsets of $X$, sublattice morphisms from $P(X)$ to $P(Y)$ coincide with relation from $X$ and $Y$, and the monoidal structure described above is induced from the one on suplattice...
So our example become a non-example up an equivalence of categories. Similar description can be obtained for the others example I have mentioned.
$endgroup$
1
$begingroup$
This is really a great example, thank you! Just to clarify: the external hom here is the power set of $X times Y$, while the internal hom is just $X times Y$ right? It is very dramatic, maybe the most striking here because of its simplicity.
$endgroup$
– mattecapu
1 hour ago
$begingroup$
Also the last part of very insightful, I love it. Thanks again!
$endgroup$
– mattecapu
1 hour ago
$begingroup$
Yes absolutely.
$endgroup$
– Simon Henry
1 hour ago
add a comment |
$begingroup$
Here are a slightly different flavor of examples, of course as mentioned by Peter Lumsdain, the trick is always the same looking at categories where $Hom(I,X)$ looks quite different from the "underlying sets of $X$"
Consider the category whose objects are set, and whose morphisms are relations, i.e. subsets of $X times Y$ (with the usual composition of relation).
It has a monoidal structure given by the product of sets. The exponential for this monoidal structure is simply product, indeed,
A relation from $X$ to $Y times Z$ is the same as a relation from $X times Z$ to $Y$.
Other similar examples includes:
The 2-category of sets and span of sets (with the products of sets as monoidal structure).
The category of small categories and profunctor, with the products of categories as monoidal structure, here the exponential $[X,Y]$ is given by $Y times X^op$
The category of polynomial functor (which is actually cartesian closed) described in The cartesian closed bicategory of generalized species and structures by Fiore,Gambino,Hyland and Winskel. Where the exponential is a little more complicated, but still has the same flavor.
But one can somehow "explain the trick" behind these: One way to think about that these example is that these categories are equivalent to categories where the Hom objects corresponds more naturally to a natural structure one the set of morphisms, but the equivalence of categories somehow make the "underlying set" very different.
For example, the category of set and relation can be seen as the category of "Free suplattice":
A suplattice is an ordered set with all supremum, and suplattice morphisms are order preserving map, preserving the supremums. Given two suplattices $X$ and $Y$ the set of morphisms from $X$ and $Y$ is naturally a suplattice for the pointwise ordering (induce by the order of $Y$), and this corresponds to a monoidal symmetric closed structure on the category of suplattice. So this is really a "non-example" of what is asked: the hom objects are really exactly the set of morphisms with a natural structure induced on them.
Now the forgetfull functor from suplattice to sets has a left adjoint, sending any set to the suplattice $P(X)$ of subsets of $X$, sublattice morphisms from $P(X)$ to $P(Y)$ coincide with relation from $X$ and $Y$, and the monoidal structure described above is induced from the one on suplattice...
So our example become a non-example up an equivalence of categories. Similar description can be obtained for the others example I have mentioned.
$endgroup$
1
$begingroup$
This is really a great example, thank you! Just to clarify: the external hom here is the power set of $X times Y$, while the internal hom is just $X times Y$ right? It is very dramatic, maybe the most striking here because of its simplicity.
$endgroup$
– mattecapu
1 hour ago
$begingroup$
Also the last part of very insightful, I love it. Thanks again!
$endgroup$
– mattecapu
1 hour ago
$begingroup$
Yes absolutely.
$endgroup$
– Simon Henry
1 hour ago
add a comment |
$begingroup$
Here are a slightly different flavor of examples, of course as mentioned by Peter Lumsdain, the trick is always the same looking at categories where $Hom(I,X)$ looks quite different from the "underlying sets of $X$"
Consider the category whose objects are set, and whose morphisms are relations, i.e. subsets of $X times Y$ (with the usual composition of relation).
It has a monoidal structure given by the product of sets. The exponential for this monoidal structure is simply product, indeed,
A relation from $X$ to $Y times Z$ is the same as a relation from $X times Z$ to $Y$.
Other similar examples includes:
The 2-category of sets and span of sets (with the products of sets as monoidal structure).
The category of small categories and profunctor, with the products of categories as monoidal structure, here the exponential $[X,Y]$ is given by $Y times X^op$
The category of polynomial functor (which is actually cartesian closed) described in The cartesian closed bicategory of generalized species and structures by Fiore,Gambino,Hyland and Winskel. Where the exponential is a little more complicated, but still has the same flavor.
But one can somehow "explain the trick" behind these: One way to think about that these example is that these categories are equivalent to categories where the Hom objects corresponds more naturally to a natural structure one the set of morphisms, but the equivalence of categories somehow make the "underlying set" very different.
For example, the category of set and relation can be seen as the category of "Free suplattice":
A suplattice is an ordered set with all supremum, and suplattice morphisms are order preserving map, preserving the supremums. Given two suplattices $X$ and $Y$ the set of morphisms from $X$ and $Y$ is naturally a suplattice for the pointwise ordering (induce by the order of $Y$), and this corresponds to a monoidal symmetric closed structure on the category of suplattice. So this is really a "non-example" of what is asked: the hom objects are really exactly the set of morphisms with a natural structure induced on them.
Now the forgetfull functor from suplattice to sets has a left adjoint, sending any set to the suplattice $P(X)$ of subsets of $X$, sublattice morphisms from $P(X)$ to $P(Y)$ coincide with relation from $X$ and $Y$, and the monoidal structure described above is induced from the one on suplattice...
So our example become a non-example up an equivalence of categories. Similar description can be obtained for the others example I have mentioned.
$endgroup$
Here are a slightly different flavor of examples, of course as mentioned by Peter Lumsdain, the trick is always the same looking at categories where $Hom(I,X)$ looks quite different from the "underlying sets of $X$"
Consider the category whose objects are set, and whose morphisms are relations, i.e. subsets of $X times Y$ (with the usual composition of relation).
It has a monoidal structure given by the product of sets. The exponential for this monoidal structure is simply product, indeed,
A relation from $X$ to $Y times Z$ is the same as a relation from $X times Z$ to $Y$.
Other similar examples includes:
The 2-category of sets and span of sets (with the products of sets as monoidal structure).
The category of small categories and profunctor, with the products of categories as monoidal structure, here the exponential $[X,Y]$ is given by $Y times X^op$
The category of polynomial functor (which is actually cartesian closed) described in The cartesian closed bicategory of generalized species and structures by Fiore,Gambino,Hyland and Winskel. Where the exponential is a little more complicated, but still has the same flavor.
But one can somehow "explain the trick" behind these: One way to think about that these example is that these categories are equivalent to categories where the Hom objects corresponds more naturally to a natural structure one the set of morphisms, but the equivalence of categories somehow make the "underlying set" very different.
For example, the category of set and relation can be seen as the category of "Free suplattice":
A suplattice is an ordered set with all supremum, and suplattice morphisms are order preserving map, preserving the supremums. Given two suplattices $X$ and $Y$ the set of morphisms from $X$ and $Y$ is naturally a suplattice for the pointwise ordering (induce by the order of $Y$), and this corresponds to a monoidal symmetric closed structure on the category of suplattice. So this is really a "non-example" of what is asked: the hom objects are really exactly the set of morphisms with a natural structure induced on them.
Now the forgetfull functor from suplattice to sets has a left adjoint, sending any set to the suplattice $P(X)$ of subsets of $X$, sublattice morphisms from $P(X)$ to $P(Y)$ coincide with relation from $X$ and $Y$, and the monoidal structure described above is induced from the one on suplattice...
So our example become a non-example up an equivalence of categories. Similar description can be obtained for the others example I have mentioned.
edited 2 hours ago
answered 3 hours ago
Simon HenrySimon Henry
16.8k1 gold badge51 silver badges96 bronze badges
16.8k1 gold badge51 silver badges96 bronze badges
1
$begingroup$
This is really a great example, thank you! Just to clarify: the external hom here is the power set of $X times Y$, while the internal hom is just $X times Y$ right? It is very dramatic, maybe the most striking here because of its simplicity.
$endgroup$
– mattecapu
1 hour ago
$begingroup$
Also the last part of very insightful, I love it. Thanks again!
$endgroup$
– mattecapu
1 hour ago
$begingroup$
Yes absolutely.
$endgroup$
– Simon Henry
1 hour ago
add a comment |
1
$begingroup$
This is really a great example, thank you! Just to clarify: the external hom here is the power set of $X times Y$, while the internal hom is just $X times Y$ right? It is very dramatic, maybe the most striking here because of its simplicity.
$endgroup$
– mattecapu
1 hour ago
$begingroup$
Also the last part of very insightful, I love it. Thanks again!
$endgroup$
– mattecapu
1 hour ago
$begingroup$
Yes absolutely.
$endgroup$
– Simon Henry
1 hour ago
1
1
$begingroup$
This is really a great example, thank you! Just to clarify: the external hom here is the power set of $X times Y$, while the internal hom is just $X times Y$ right? It is very dramatic, maybe the most striking here because of its simplicity.
$endgroup$
– mattecapu
1 hour ago
$begingroup$
This is really a great example, thank you! Just to clarify: the external hom here is the power set of $X times Y$, while the internal hom is just $X times Y$ right? It is very dramatic, maybe the most striking here because of its simplicity.
$endgroup$
– mattecapu
1 hour ago
$begingroup$
Also the last part of very insightful, I love it. Thanks again!
$endgroup$
– mattecapu
1 hour ago
$begingroup$
Also the last part of very insightful, I love it. Thanks again!
$endgroup$
– mattecapu
1 hour ago
$begingroup$
Yes absolutely.
$endgroup$
– Simon Henry
1 hour ago
$begingroup$
Yes absolutely.
$endgroup$
– Simon Henry
1 hour ago
add a comment |
$begingroup$
To add to the other good answers here, there's a family of examples that could be seen as a bit trivial. But in a sense they give the simplest answer to your question.
Let $X$ be a partially ordered set, viewed as a category in the usual way: the objects of the category are the elements of $X$, and $textrmHom(x, y)$ has either $1$ or $0$ elements according to whether $x leq y$ or not. If $X$ is a meet-semilattice, i.e. any finite set of elements has a greatest lower bound, then the corresponding category has finite products. It's cartesian closed if for any two elements $y$ and $z$ there's an element $y to z$ with the property that for all $x in X$,
$$
x wedge y leq z iff x leq y to z.
$$
Here $wedge$ denotes greatest lower bound. So $y to z$ is the internal hom $z^y$. A poset with this property is more or less what's called a Heyting algebra.
I think this is an enlightening family of examples because in a poset, the external homs are pretty trivial (sets with at most one element), whereas the internal homs can be informative. For example, in a power set they give you the notion of complement (exercise!).
You can extend this family a bit. Again take an ordered set $X$, regarded as a category in the usual way. But now think about non-cartesian monoidal structures on it. For instance, take $X = mathbbR$ with its usual ordering. The cartesian monoidal structure is $mathrmmin$, but we could instead use $+$. Then the internal hom is given by
$$
y to z = z - y.
$$
Is that "unexpected"? That depends on your intuition. But in this example the external hom only tells you the sign of $z - y$, whereas the internal hom tells you its actual value. In other words, it produces the operation of subtraction, which historically has proved quite important.
$endgroup$
1
$begingroup$
Yes! Heyting algebras are what I was looking at when I thought my question. So I was waiting for this answer (I would have written it myself tomorrow). Still, yours is surely better than what I would have written, in particular the last example I really cool.
$endgroup$
– mattecapu
14 mins ago
$begingroup$
Just to add to your last example: to see why the internal hom is $z-y$, it is a fact of life that in an Heyting algebra (aka postal Cartesian closed small category), $y to z = neg y lor y$, where the 'negation' of $y$ is $y to 0$ (this also solves the exercise about complements in power sets). Therefore if we unpack the universal property defining $to$ we get $x+y leq 0 iff x leq y to 0$, so $neg y = y to 0 = - y$ as expected.
$endgroup$
– mattecapu
4 mins ago
add a comment |
$begingroup$
To add to the other good answers here, there's a family of examples that could be seen as a bit trivial. But in a sense they give the simplest answer to your question.
Let $X$ be a partially ordered set, viewed as a category in the usual way: the objects of the category are the elements of $X$, and $textrmHom(x, y)$ has either $1$ or $0$ elements according to whether $x leq y$ or not. If $X$ is a meet-semilattice, i.e. any finite set of elements has a greatest lower bound, then the corresponding category has finite products. It's cartesian closed if for any two elements $y$ and $z$ there's an element $y to z$ with the property that for all $x in X$,
$$
x wedge y leq z iff x leq y to z.
$$
Here $wedge$ denotes greatest lower bound. So $y to z$ is the internal hom $z^y$. A poset with this property is more or less what's called a Heyting algebra.
I think this is an enlightening family of examples because in a poset, the external homs are pretty trivial (sets with at most one element), whereas the internal homs can be informative. For example, in a power set they give you the notion of complement (exercise!).
You can extend this family a bit. Again take an ordered set $X$, regarded as a category in the usual way. But now think about non-cartesian monoidal structures on it. For instance, take $X = mathbbR$ with its usual ordering. The cartesian monoidal structure is $mathrmmin$, but we could instead use $+$. Then the internal hom is given by
$$
y to z = z - y.
$$
Is that "unexpected"? That depends on your intuition. But in this example the external hom only tells you the sign of $z - y$, whereas the internal hom tells you its actual value. In other words, it produces the operation of subtraction, which historically has proved quite important.
$endgroup$
1
$begingroup$
Yes! Heyting algebras are what I was looking at when I thought my question. So I was waiting for this answer (I would have written it myself tomorrow). Still, yours is surely better than what I would have written, in particular the last example I really cool.
$endgroup$
– mattecapu
14 mins ago
$begingroup$
Just to add to your last example: to see why the internal hom is $z-y$, it is a fact of life that in an Heyting algebra (aka postal Cartesian closed small category), $y to z = neg y lor y$, where the 'negation' of $y$ is $y to 0$ (this also solves the exercise about complements in power sets). Therefore if we unpack the universal property defining $to$ we get $x+y leq 0 iff x leq y to 0$, so $neg y = y to 0 = - y$ as expected.
$endgroup$
– mattecapu
4 mins ago
add a comment |
$begingroup$
To add to the other good answers here, there's a family of examples that could be seen as a bit trivial. But in a sense they give the simplest answer to your question.
Let $X$ be a partially ordered set, viewed as a category in the usual way: the objects of the category are the elements of $X$, and $textrmHom(x, y)$ has either $1$ or $0$ elements according to whether $x leq y$ or not. If $X$ is a meet-semilattice, i.e. any finite set of elements has a greatest lower bound, then the corresponding category has finite products. It's cartesian closed if for any two elements $y$ and $z$ there's an element $y to z$ with the property that for all $x in X$,
$$
x wedge y leq z iff x leq y to z.
$$
Here $wedge$ denotes greatest lower bound. So $y to z$ is the internal hom $z^y$. A poset with this property is more or less what's called a Heyting algebra.
I think this is an enlightening family of examples because in a poset, the external homs are pretty trivial (sets with at most one element), whereas the internal homs can be informative. For example, in a power set they give you the notion of complement (exercise!).
You can extend this family a bit. Again take an ordered set $X$, regarded as a category in the usual way. But now think about non-cartesian monoidal structures on it. For instance, take $X = mathbbR$ with its usual ordering. The cartesian monoidal structure is $mathrmmin$, but we could instead use $+$. Then the internal hom is given by
$$
y to z = z - y.
$$
Is that "unexpected"? That depends on your intuition. But in this example the external hom only tells you the sign of $z - y$, whereas the internal hom tells you its actual value. In other words, it produces the operation of subtraction, which historically has proved quite important.
$endgroup$
To add to the other good answers here, there's a family of examples that could be seen as a bit trivial. But in a sense they give the simplest answer to your question.
Let $X$ be a partially ordered set, viewed as a category in the usual way: the objects of the category are the elements of $X$, and $textrmHom(x, y)$ has either $1$ or $0$ elements according to whether $x leq y$ or not. If $X$ is a meet-semilattice, i.e. any finite set of elements has a greatest lower bound, then the corresponding category has finite products. It's cartesian closed if for any two elements $y$ and $z$ there's an element $y to z$ with the property that for all $x in X$,
$$
x wedge y leq z iff x leq y to z.
$$
Here $wedge$ denotes greatest lower bound. So $y to z$ is the internal hom $z^y$. A poset with this property is more or less what's called a Heyting algebra.
I think this is an enlightening family of examples because in a poset, the external homs are pretty trivial (sets with at most one element), whereas the internal homs can be informative. For example, in a power set they give you the notion of complement (exercise!).
You can extend this family a bit. Again take an ordered set $X$, regarded as a category in the usual way. But now think about non-cartesian monoidal structures on it. For instance, take $X = mathbbR$ with its usual ordering. The cartesian monoidal structure is $mathrmmin$, but we could instead use $+$. Then the internal hom is given by
$$
y to z = z - y.
$$
Is that "unexpected"? That depends on your intuition. But in this example the external hom only tells you the sign of $z - y$, whereas the internal hom tells you its actual value. In other words, it produces the operation of subtraction, which historically has proved quite important.
edited 11 mins ago
answered 27 mins ago
Tom LeinsterTom Leinster
19.7k4 gold badges75 silver badges130 bronze badges
19.7k4 gold badges75 silver badges130 bronze badges
1
$begingroup$
Yes! Heyting algebras are what I was looking at when I thought my question. So I was waiting for this answer (I would have written it myself tomorrow). Still, yours is surely better than what I would have written, in particular the last example I really cool.
$endgroup$
– mattecapu
14 mins ago
$begingroup$
Just to add to your last example: to see why the internal hom is $z-y$, it is a fact of life that in an Heyting algebra (aka postal Cartesian closed small category), $y to z = neg y lor y$, where the 'negation' of $y$ is $y to 0$ (this also solves the exercise about complements in power sets). Therefore if we unpack the universal property defining $to$ we get $x+y leq 0 iff x leq y to 0$, so $neg y = y to 0 = - y$ as expected.
$endgroup$
– mattecapu
4 mins ago
add a comment |
1
$begingroup$
Yes! Heyting algebras are what I was looking at when I thought my question. So I was waiting for this answer (I would have written it myself tomorrow). Still, yours is surely better than what I would have written, in particular the last example I really cool.
$endgroup$
– mattecapu
14 mins ago
$begingroup$
Just to add to your last example: to see why the internal hom is $z-y$, it is a fact of life that in an Heyting algebra (aka postal Cartesian closed small category), $y to z = neg y lor y$, where the 'negation' of $y$ is $y to 0$ (this also solves the exercise about complements in power sets). Therefore if we unpack the universal property defining $to$ we get $x+y leq 0 iff x leq y to 0$, so $neg y = y to 0 = - y$ as expected.
$endgroup$
– mattecapu
4 mins ago
1
1
$begingroup$
Yes! Heyting algebras are what I was looking at when I thought my question. So I was waiting for this answer (I would have written it myself tomorrow). Still, yours is surely better than what I would have written, in particular the last example I really cool.
$endgroup$
– mattecapu
14 mins ago
$begingroup$
Yes! Heyting algebras are what I was looking at when I thought my question. So I was waiting for this answer (I would have written it myself tomorrow). Still, yours is surely better than what I would have written, in particular the last example I really cool.
$endgroup$
– mattecapu
14 mins ago
$begingroup$
Just to add to your last example: to see why the internal hom is $z-y$, it is a fact of life that in an Heyting algebra (aka postal Cartesian closed small category), $y to z = neg y lor y$, where the 'negation' of $y$ is $y to 0$ (this also solves the exercise about complements in power sets). Therefore if we unpack the universal property defining $to$ we get $x+y leq 0 iff x leq y to 0$, so $neg y = y to 0 = - y$ as expected.
$endgroup$
– mattecapu
4 mins ago
$begingroup$
Just to add to your last example: to see why the internal hom is $z-y$, it is a fact of life that in an Heyting algebra (aka postal Cartesian closed small category), $y to z = neg y lor y$, where the 'negation' of $y$ is $y to 0$ (this also solves the exercise about complements in power sets). Therefore if we unpack the universal property defining $to$ we get $x+y leq 0 iff x leq y to 0$, so $neg y = y to 0 = - y$ as expected.
$endgroup$
– mattecapu
4 mins ago
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