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Understanding theorem 15.12 in Kosniovski's A first course in algebraic topology

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Understanding theorem 15.12 in Kosniovski's A first course in algebraic topology

Proving two torus maps are homotopicExercise 1.1.18 in Hatcher's Algebraic TopologyNeed help understanding statement of Van Kampen's Theorem and using it to compute the fundamental group of Projective PlaneContractible spaces has trivial fundamental group.Exercise 2, chapter 4, Hatcher.Fundamental groups in path-connected spaceHow can I complete this proof of a algebraic topology theorem?Doubt in proof : homotopic maps induce same homomorphismLet $X$ be a connected CW complex and $G$ a group such that every $pi_1(X)to G$ is trivial. Show that every $Xto K(G, 1)$ is nullhomotopic.Proving directly that $S^2$ is simply connected: is a surjective loop homotopic to a non-surjective one?

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5

1

$begingroup$

This Theorem can be found in page 138 of the book A first Course in algebraic topology by Kosniowski.

15.12. Theorem Let $varphi, psi:X rightarrow Y$ be continuous maps between topological spaces and let $F$ be an homotopy between $varphi$ and $psi$. Let $f:Irightarrow Y$ be the path from $varphi(x_0)$ to $psi(x_0)$ given by $f(t)=F(x_0,t)$. Then, the homomorphisms

$$varphi_* : pi(X,x_0)rightarrowpi(Y,varphi(x_0))$$
and
$$psi_* : pi(X,x_0)rightarrowpi(Y,psi(x_0))$$

are related by $psi_* = u_f varphi_*$, where $u_f$ is the isomorphism

beginarraycccl
u_f: &pi(Y,varphi(x_0)) &longrightarrow &pi(Y,psi(x_0))
\ &[g] &longmapsto & [overlinef * g * f]
endarray

I was hoping for someone to throw some light on what does this theorem say and what can it be used for.

What I understand is that the fundamental groups of the images of a point $x_0$ under homotopic functions are isomorphic if there is a path connecting $varphi(x_0)$ and $psi(x_0)$, but it has already been proven that two path connected points have isomorphic fundamental groups, so this would be redundant.

algebraic-topology fundamental-groups

share|cite|improve this question

asked 8 hours ago

YaggerYagger

1,00911 gold badge66 silver badges1717 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I assume that $pi$ should be $pi_1$.
  $endgroup$
  – Arnaud Mortier
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ArnaudMortier Until now the author has only talked about $pi(X,x_0)$, the set of homotopy classes of loops with base point $x_0$ with the product $[f][g]:=[f*g]$
  $endgroup$
  – Yagger
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  That's exactly what any author I know would denote by $pi_1(X,x_0)$.
  $endgroup$
  – Arnaud Mortier
  8 hours ago
  

  
  
  
  

add a comment | 

5

1

$begingroup$

This Theorem can be found in page 138 of the book A first Course in algebraic topology by Kosniowski.

15.12. Theorem Let $varphi, psi:X rightarrow Y$ be continuous maps between topological spaces and let $F$ be an homotopy between $varphi$ and $psi$. Let $f:Irightarrow Y$ be the path from $varphi(x_0)$ to $psi(x_0)$ given by $f(t)=F(x_0,t)$. Then, the homomorphisms

$$varphi_* : pi(X,x_0)rightarrowpi(Y,varphi(x_0))$$
and
$$psi_* : pi(X,x_0)rightarrowpi(Y,psi(x_0))$$

are related by $psi_* = u_f varphi_*$, where $u_f$ is the isomorphism

beginarraycccl
u_f: &pi(Y,varphi(x_0)) &longrightarrow &pi(Y,psi(x_0))
\ &[g] &longmapsto & [overlinef * g * f]
endarray

I was hoping for someone to throw some light on what does this theorem say and what can it be used for.

What I understand is that the fundamental groups of the images of a point $x_0$ under homotopic functions are isomorphic if there is a path connecting $varphi(x_0)$ and $psi(x_0)$, but it has already been proven that two path connected points have isomorphic fundamental groups, so this would be redundant.

algebraic-topology fundamental-groups

share|cite|improve this question

asked 8 hours ago

YaggerYagger

1,00911 gold badge66 silver badges1717 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I assume that $pi$ should be $pi_1$.
  $endgroup$
  – Arnaud Mortier
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ArnaudMortier Until now the author has only talked about $pi(X,x_0)$, the set of homotopy classes of loops with base point $x_0$ with the product $[f][g]:=[f*g]$
  $endgroup$
  – Yagger
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  That's exactly what any author I know would denote by $pi_1(X,x_0)$.
  $endgroup$
  – Arnaud Mortier
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

This Theorem can be found in page 138 of the book A first Course in algebraic topology by Kosniowski.

15.12. Theorem Let $varphi, psi:X rightarrow Y$ be continuous maps between topological spaces and let $F$ be an homotopy between $varphi$ and $psi$. Let $f:Irightarrow Y$ be the path from $varphi(x_0)$ to $psi(x_0)$ given by $f(t)=F(x_0,t)$. Then, the homomorphisms

$$varphi_* : pi(X,x_0)rightarrowpi(Y,varphi(x_0))$$
and
$$psi_* : pi(X,x_0)rightarrowpi(Y,psi(x_0))$$

are related by $psi_* = u_f varphi_*$, where $u_f$ is the isomorphism

beginarraycccl
u_f: &pi(Y,varphi(x_0)) &longrightarrow &pi(Y,psi(x_0))
\ &[g] &longmapsto & [overlinef * g * f]
endarray

I was hoping for someone to throw some light on what does this theorem say and what can it be used for.

What I understand is that the fundamental groups of the images of a point $x_0$ under homotopic functions are isomorphic if there is a path connecting $varphi(x_0)$ and $psi(x_0)$, but it has already been proven that two path connected points have isomorphic fundamental groups, so this would be redundant.

algebraic-topology fundamental-groups

share|cite|improve this question

asked 8 hours ago

YaggerYagger

1,00911 gold badge66 silver badges1717 bronze badges

$endgroup$

share|cite|improve this question

asked 8 hours ago

YaggerYagger

1,00911 gold badge66 silver badges1717 bronze badges

share|cite|improve this question

asked 8 hours ago

YaggerYagger

1,00911 gold badge66 silver badges1717 bronze badges

asked 8 hours ago

YaggerYagger

1,00911 gold badge66 silver badges1717 bronze badges

asked 8 hours ago

YaggerYagger

1,00911 gold badge66 silver badges1717 bronze badges

1 Answer
1

active

oldest

votes

6

$begingroup$

This is essentially telling you that if two maps $Xto Y$ are homotopic, then they induce the "same" map at the level of homotopy groups, where the meaning of "same" here is only as broad as permitted by the fact that the two induced maps actually don't have the same codomain strictly speaking, because of the base point.

What could potentially be enlightening is to wonder what would happen if the two maps were not homotopic.

For instance consider these few examples:

• $X$ is connected, $Y$ is not, and $varphi$ and $psi$ land in two different components of $Y$. Obviously the induced maps in homotopy can't be considered equal in any reasonable way.




• $X=Bbb S^1$ with some base point that you call $1$, $Y=Bbb S^1times Bbb S^1$, $varphi(x)=(x,1)$ and $psi(x)=(1,x)$. Here, it is interesting to see that there is an isomorphism $u$ from the codomain to itself such that $psi_* = ucirc varphi_*$: indeed, $$varphi_star =Bbb ZtoBbb Z^2:nto (n,0)$$ and $$psi_star =Bbb ZtoBbb Z^2:nto (0,n)$$ so that the isomorphism $u$ is simply $$u=Bbb Z^2toBbb Z^2:(m,n)to (n,m)$$
  But to find this isomorphism you had to understand what was going on and build it yourself. The theorem is actually telling you that when the two maps are homotopic, you don't have to think, $u$ is God-given.

share|cite|improve this answer

answered 8 hours ago

Arnaud MortierArnaud Mortier

20.8k22 gold badges2323 silver badges6363 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That was enlightening!
  $endgroup$
  – Santana Afton
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SantanaAfton You're very welcome.
  $endgroup$
  – Arnaud Mortier
  2 hours ago
  

  
  
  
  

add a comment | 

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6

$begingroup$

This is essentially telling you that if two maps $Xto Y$ are homotopic, then they induce the "same" map at the level of homotopy groups, where the meaning of "same" here is only as broad as permitted by the fact that the two induced maps actually don't have the same codomain strictly speaking, because of the base point.

What could potentially be enlightening is to wonder what would happen if the two maps were not homotopic.

For instance consider these few examples:

• $X$ is connected, $Y$ is not, and $varphi$ and $psi$ land in two different components of $Y$. Obviously the induced maps in homotopy can't be considered equal in any reasonable way.




• $X=Bbb S^1$ with some base point that you call $1$, $Y=Bbb S^1times Bbb S^1$, $varphi(x)=(x,1)$ and $psi(x)=(1,x)$. Here, it is interesting to see that there is an isomorphism $u$ from the codomain to itself such that $psi_* = ucirc varphi_*$: indeed, $$varphi_star =Bbb ZtoBbb Z^2:nto (n,0)$$ and $$psi_star =Bbb ZtoBbb Z^2:nto (0,n)$$ so that the isomorphism $u$ is simply $$u=Bbb Z^2toBbb Z^2:(m,n)to (n,m)$$
  But to find this isomorphism you had to understand what was going on and build it yourself. The theorem is actually telling you that when the two maps are homotopic, you don't have to think, $u$ is God-given.

share|cite|improve this answer

answered 8 hours ago

Arnaud MortierArnaud Mortier

20.8k22 gold badges2323 silver badges6363 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That was enlightening!
  $endgroup$
  – Santana Afton
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SantanaAfton You're very welcome.
  $endgroup$
  – Arnaud Mortier
  2 hours ago
  

  
  
  
  

add a comment | 

6

$begingroup$

This is essentially telling you that if two maps $Xto Y$ are homotopic, then they induce the "same" map at the level of homotopy groups, where the meaning of "same" here is only as broad as permitted by the fact that the two induced maps actually don't have the same codomain strictly speaking, because of the base point.

What could potentially be enlightening is to wonder what would happen if the two maps were not homotopic.

For instance consider these few examples:

• $X$ is connected, $Y$ is not, and $varphi$ and $psi$ land in two different components of $Y$. Obviously the induced maps in homotopy can't be considered equal in any reasonable way.




• $X=Bbb S^1$ with some base point that you call $1$, $Y=Bbb S^1times Bbb S^1$, $varphi(x)=(x,1)$ and $psi(x)=(1,x)$. Here, it is interesting to see that there is an isomorphism $u$ from the codomain to itself such that $psi_* = ucirc varphi_*$: indeed, $$varphi_star =Bbb ZtoBbb Z^2:nto (n,0)$$ and $$psi_star =Bbb ZtoBbb Z^2:nto (0,n)$$ so that the isomorphism $u$ is simply $$u=Bbb Z^2toBbb Z^2:(m,n)to (n,m)$$
  But to find this isomorphism you had to understand what was going on and build it yourself. The theorem is actually telling you that when the two maps are homotopic, you don't have to think, $u$ is God-given.

share|cite|improve this answer

answered 8 hours ago

Arnaud MortierArnaud Mortier

20.8k22 gold badges2323 silver badges6363 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That was enlightening!
  $endgroup$
  – Santana Afton
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SantanaAfton You're very welcome.
  $endgroup$
  – Arnaud Mortier
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

This is essentially telling you that if two maps $Xto Y$ are homotopic, then they induce the "same" map at the level of homotopy groups, where the meaning of "same" here is only as broad as permitted by the fact that the two induced maps actually don't have the same codomain strictly speaking, because of the base point.

What could potentially be enlightening is to wonder what would happen if the two maps were not homotopic.

For instance consider these few examples:

• $X$ is connected, $Y$ is not, and $varphi$ and $psi$ land in two different components of $Y$. Obviously the induced maps in homotopy can't be considered equal in any reasonable way.




• $X=Bbb S^1$ with some base point that you call $1$, $Y=Bbb S^1times Bbb S^1$, $varphi(x)=(x,1)$ and $psi(x)=(1,x)$. Here, it is interesting to see that there is an isomorphism $u$ from the codomain to itself such that $psi_* = ucirc varphi_*$: indeed, $$varphi_star =Bbb ZtoBbb Z^2:nto (n,0)$$ and $$psi_star =Bbb ZtoBbb Z^2:nto (0,n)$$ so that the isomorphism $u$ is simply $$u=Bbb Z^2toBbb Z^2:(m,n)to (n,m)$$
  But to find this isomorphism you had to understand what was going on and build it yourself. The theorem is actually telling you that when the two maps are homotopic, you don't have to think, $u$ is God-given.

share|cite|improve this answer

answered 8 hours ago

Arnaud MortierArnaud Mortier

20.8k22 gold badges2323 silver badges6363 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Arnaud MortierArnaud Mortier

20.8k22 gold badges2323 silver badges6363 bronze badges

answered 8 hours ago

Arnaud MortierArnaud Mortier

20.8k22 gold badges2323 silver badges6363 bronze badges

answered 8 hours ago

Arnaud MortierArnaud Mortier

20.8k22 gold badges2323 silver badges6363 bronze badges