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$\newcommand{\Id}{\operatorname{Id}}$Consider a morphism $f : A \rightarrow B$ which has a left inverse $g$, i.e. $g \circ f = \Id_A$. (That is, $f$ is a split monomorphism.) Of course, we don't necessarily have $f \circ g = \Id_B$, but it might be the case that $f \circ g \circ h = h$ for some $h : C \rightarrow B$. Does such a morphism $h$ have any particular name? (Aside from saying "it equalizes $g \circ f$ and $\Id_B$", I suppose.)

Intuition: In terms of functions, $f$ here is injective, and $g$ is surjective. If we have $g \circ f = \Id_A$ but $g \circ f \neq \Id_B$, it means there are elements in $B$ which are not in the range of $f$, and such elements can't be mapped to themselves by $g \circ f$. However, if we have a function $h : C \rightarrow B$ with $range(h) \subseteq range(f)$, then $f \circ g \circ h = h$, and $g$ also acts as a right inverse "on the range of $h$".

I have come across this situation several times while doing some research, and I was wondering if there is a name for it.

Sambo
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2 Answers2

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I've never really seen a name for such an $h$, as you ask.

but it might be the case that $f \circ g \circ h = h$ for some $h$

Indeed it always happens, for example a few interesting choices are $h\in \{f, fg\}$.

In ring/module theory, the fact that $f\circ g$ works is saying that it is an idempotent of $Hom(B,B)$.

If $A=B$ is a module, then one says $f$ is a von Neumann regular element of the ring $End(A)$ when $fgf=f$ for some $g$. Clearly one-sided invertible elements of a ring are all von Neumann regular.

For that matter you can see $f\circ q$ works for any $q$ at all, but I'm not sure if that has significance to anyone. One way to frame that would be "$f\circ g$ acts like the identity on the left of $f\circ Hom(B, A)$."

rschwieb
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Such $h$ are exactly those that factor through $f$ (in other words, $f$ is an equalizer of $f \circ g$ and $\mathrm{Id}_B$).

If $f \circ g \circ h=h$, then clearly, $h$ factors through $f$. Conversely, if $h=f \circ k$ for some $k:C \rightarrow A$, then $f \circ g \circ h=f \circ g \circ f \circ k=f \circ k=h$.

Dually, $g$ is a coequalizer of $f \circ g$ and $\mathrm{Id}_B$.

I cannot think of a better name than "morphism factoring through $f$", however.

Geoffrey Trang
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