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?

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    $\begingroup$ Don't know that there is one. "Eclipse" is as good a candidate for a name as any.... $\endgroup$ – fleablood Jun 16 at 16:47
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    $\begingroup$ 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 Jun 16 at 17:08

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 $f \circ g = f$.

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    $\begingroup$ Cool, thanks for the answer. I understood the first paragraph; the rest are sure to be valuable for people smarter than I. $\endgroup$ – Doradus Jun 17 at 23:50

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