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7

$begingroup$

I've been saying function $f$ "eclipses" $g$ if $f(g(x)) = f(x)$ for all $x$. For example, if $f(x) = lvert x rvert$ and $g(x) = -x$, then $f$ eclipses $g$.

Is there an established word for this property?

functions terminology

share|cite|improve this question

asked 8 hours ago

DoradusDoradus

1836

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Don't know that there is one. "Eclipse" is as good a candidate for a name as any....
  $endgroup$
  – fleablood
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is loosely related to the concept of en.wikipedia.org/wiki/Coequalizer, so maybe one could call $f$ "co-equalizing $g$ and the identity"
  $endgroup$
  – Hagen von Eitzen
  8 hours ago
  
  

  
  
  
  

add a comment | 

7

$begingroup$

I've been saying function $f$ "eclipses" $g$ if $f(g(x)) = f(x)$ for all $x$. For example, if $f(x) = lvert x rvert$ and $g(x) = -x$, then $f$ eclipses $g$.

Is there an established word for this property?

functions terminology

share|cite|improve this question

asked 8 hours ago

DoradusDoradus

1836

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Don't know that there is one. "Eclipse" is as good a candidate for a name as any....
  $endgroup$
  – fleablood
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is loosely related to the concept of en.wikipedia.org/wiki/Coequalizer, so maybe one could call $f$ "co-equalizing $g$ and the identity"
  $endgroup$
  – Hagen von Eitzen
  8 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

I've been saying function $f$ "eclipses" $g$ if $f(g(x)) = f(x)$ for all $x$. For example, if $f(x) = lvert x rvert$ and $g(x) = -x$, then $f$ eclipses $g$.

Is there an established word for this property?

functions terminology

share|cite|improve this question

asked 8 hours ago

DoradusDoradus

1836

$endgroup$

share|cite|improve this question

asked 8 hours ago

DoradusDoradus

1836

share|cite|improve this question

asked 8 hours ago

DoradusDoradus

1836

asked 8 hours ago

DoradusDoradus

1836

asked 8 hours ago

DoradusDoradus

1836

1 Answer
1

active

oldest

votes

7

$begingroup$

I would express this as "$f$ is invariant under $g$" or "$f$ is $g$-invariant", because of the analogy with group actions:

Compare this with: If $G$ is a group acting linearly on a vector space $V$, it induces an action on the dual $V^*$ by letting $g*f = f circ g$. If $g*f = f$ for all $g in G$, we would say $f$ is $G$-invariant.

Now, if $f : X to Y$ and $g : X to X$, then $g$ determines a monoid action of $mathbb N$ on $X$ by letting $n$ act by $g_n = g circ cdots circ g$. The monoid action induces a monoid action on the set of functions $X to Y$, by letting $n*f = f circ g_n$. A function $f$ is invariant under this monoid action iff $f circ g = f$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

punctured duskpunctured dusk

14.8k32887

$endgroup$

add a comment | 

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7

$begingroup$

I would express this as "$f$ is invariant under $g$" or "$f$ is $g$-invariant", because of the analogy with group actions:

Compare this with: If $G$ is a group acting linearly on a vector space $V$, it induces an action on the dual $V^*$ by letting $g*f = f circ g$. If $g*f = f$ for all $g in G$, we would say $f$ is $G$-invariant.

Now, if $f : X to Y$ and $g : X to X$, then $g$ determines a monoid action of $mathbb N$ on $X$ by letting $n$ act by $g_n = g circ cdots circ g$. The monoid action induces a monoid action on the set of functions $X to Y$, by letting $n*f = f circ g_n$. A function $f$ is invariant under this monoid action iff $f circ g = f$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

punctured duskpunctured dusk

14.8k32887

$endgroup$

add a comment | 

7

$begingroup$

I would express this as "$f$ is invariant under $g$" or "$f$ is $g$-invariant", because of the analogy with group actions:

Compare this with: If $G$ is a group acting linearly on a vector space $V$, it induces an action on the dual $V^*$ by letting $g*f = f circ g$. If $g*f = f$ for all $g in G$, we would say $f$ is $G$-invariant.

Now, if $f : X to Y$ and $g : X to X$, then $g$ determines a monoid action of $mathbb N$ on $X$ by letting $n$ act by $g_n = g circ cdots circ g$. The monoid action induces a monoid action on the set of functions $X to Y$, by letting $n*f = f circ g_n$. A function $f$ is invariant under this monoid action iff $f circ g = f$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

punctured duskpunctured dusk

14.8k32887

$endgroup$

add a comment | 

$begingroup$

I would express this as "$f$ is invariant under $g$" or "$f$ is $g$-invariant", because of the analogy with group actions:

Compare this with: If $G$ is a group acting linearly on a vector space $V$, it induces an action on the dual $V^*$ by letting $g*f = f circ g$. If $g*f = f$ for all $g in G$, we would say $f$ is $G$-invariant.

Now, if $f : X to Y$ and $g : X to X$, then $g$ determines a monoid action of $mathbb N$ on $X$ by letting $n$ act by $g_n = g circ cdots circ g$. The monoid action induces a monoid action on the set of functions $X to Y$, by letting $n*f = f circ g_n$. A function $f$ is invariant under this monoid action iff $f circ g = f$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

punctured duskpunctured dusk

14.8k32887

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

punctured duskpunctured dusk

14.8k32887

answered 8 hours ago

punctured duskpunctured dusk

14.8k32887

answered 8 hours ago

punctured duskpunctured dusk

14.8k32887