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Is there a technical name for a function $f$ such that $(g \circ f)(a_1, a_2,\cdots) \rightarrow g(a_1, a_2, \cdots)$? That is, is there a name for a function $f$ such that the result of composing $g$ with $f$ is $g$ invoked with $f$'s arguments?

EDIT: Accidentally reversed the order of operations and put $f \circ g$ when I meant $g \circ f$ (i.e. $g(f(\cdots)$).

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Identity operator? –  user13838 Nov 8 '11 at 13:46
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Wouldn't $f$ have to be the identity function $f(a_1,a_2,\ldots)=(a_1,a_2,\ldots)$? –  David Mitra Nov 8 '11 at 13:47
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$f$ is forced to be the identity only if we require $g\circ f=g$ for all $g$. If this identity is only required to hold for a single $g$ (which is not injective) it is possible for $f$ to differ from the identity. –  Bill Cook Nov 8 '11 at 14:06
    
Example: $g(x,y)=(1,1)$ then for any $f(x,y)$ we have $g\circ f(x,y)=g(f(x,y))=(1,1)=g(x,y)$. –  Bill Cook Nov 8 '11 at 14:11

2 Answers 2

To add to Bill's answer: in certain contexts we would say that "$f$ is a symmetry of the function $g$" or that "$g$ is invariant under $f$".

For example, if $g(x) = x^2$ and $f(x) = -x$, we say that $g$ is symmetric under reflection and $f$ is the reflection operation.

Another example, if $g(x) = \sin x$ and $f(x) = x+2\pi$, we say that $g$ is periodic with period $2\pi$, or that $g$ is invariant under translation by $2\pi$, or that $g$ is invariant under $f$.

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"Symmetry" or "invariant" is definitely better terminology than what I gave in my answer. Calling something an "identity of $g$" (referring to a single element instead of a whole set) is kind of weird. :) –  Bill Cook Nov 8 '11 at 14:26
    
A related concept (with lots of fun applications for dynamical systems) is that of an equivariant. We say that "$g$ is equivariant under $f$" if for all arguments $x$ we have $(g\circ f)(x) = (f\circ g)(x)$. For an example take $g(x)=x^3$ and $f(x)=-x$. –  Chris Taylor Nov 8 '11 at 15:01
    
@Chris: wouldn't that be '$g$ commutes with $f$'? –  wildildildlife Nov 8 '11 at 15:44
    
Yes; en.wikipedia.org/wiki/Equivariant_map –  Chris Taylor Nov 8 '11 at 15:58
    
@Bill: there are however times when left/right identity "sound" better. For example, let $g(x,y)$ be independent of $y$. And let $f(x,y) = (x,0)$ be the projection map. Sure you can say that $g$ has a symmetry. But the operation $f$ itself, being a projection map, I tend to think of as a "generalised" identity operator (since it is the identity operator for the subspace $X\times\{0\}$). This tend to be more relevant in (linear) algebraic contexts though. –  Willie Wong Nov 8 '11 at 16:00

In general for a binary operation $\circ$...

If $f \circ g=g$, then $f$ is a left identity for $g$ and if $g \circ f = g$, then $f$ is a right identity for $g$.

If $\circ$ is composition and $f$ is a right identity for all functions $g$, then $f$ must be the identity function: $f(x_1,\dots,x_n)=(x_1,\dots,x_n)$. The same is true if we change "right" to "left".

In fact, if $g$ has a left inverse (which is true iff it is one-to-one), then $\mathrm{id}=g^{-1} \circ g = g^{-1}\circ g\circ f= \mathrm{id}\circ f=f$.

If $g$ is not one-to-one, there are potentially many functions (other than the identity) such that $g\circ f=g$.

When "left" is replaced by "right", "one-to-one" should be replaced by "onto".

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