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Let $f:A\to C$, $g:B\to C$ be morphisms in some category. I call $f,g$ "equivalent" iff there exists an isomorphism $h$ such that $f\circ h=g$ (and consequently $g\circ h^{-1}=f$).

Question: Is there an established term for this kind of equivalence?

Background: In a paper, I am defining this in a slightly more specific setting, and I would like to add a clarifying sentence such as "note that this is the same as the notion of ... for general categories".

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    $\begingroup$ Let $X$ be your category, set $X_C$ to be the category whose objects are morphisms of $X$ with target at $C$ and whose morphisms are the natural choice (ie give a commutative diagram in $X$). Then your notion of equivalent morphisms in $X$ is just the usual notion of isomorphic objects in $X_C$ $\endgroup$
    – Amr
    Commented Jul 13, 2021 at 12:42
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    $\begingroup$ Sometimes people consider the more general situation, and allow both pre- and post-composition with an isomorphism. In this case, the new morphism (your $g$) is sometimes called an "isomorph" of the original one (your $f$). This extra flexibility may be too much for your purposes, though, and you may with to signal that with a specific name, as in Francesco's answer, for instance. $\endgroup$
    – David Roberts ♦
    Commented Jul 13, 2021 at 13:13
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    $\begingroup$ @Amr's answer is correct and should be the accepted one. Note that this category is called a slice category. $\endgroup$
    – Mike Shulman
    Commented Jul 13, 2021 at 13:36
  • $\begingroup$ Combining Amr's and Mike's comments, I would come up with "note that this is, in general category-theoretic terms, the same as isomorphism in the slice category". I can live with that. But a less "composed" term for it would be nicer, if there is an established one. But of course, if it doesn't exist, it doesn't. :) $\endgroup$
    – Dominique Unruh
    Commented Jul 13, 2021 at 14:26

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Your condition implies that $f$ factors through $g$ and $g$ factors through $f$. Morphisms satisfying this condition are sometimes called right-equivalent, see for instance p. 139 of

Chen, Xiao-Wu; Le, Jue, A note on morphisms determined by objects, J. Algebra 428, 138-148 (2015). ZBL1321.18001.

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  • $\begingroup$ Right-equivalence is clearly implied by my "equivalence", but I think right-equivalence is strictly weaker. (I don't have an explicit counterexample at hand, though.) So it would be incorrect to state "note that this is the same as the notion of right-equivalence in Chen et al.". $\endgroup$
    – Dominique Unruh
    Commented Jul 13, 2021 at 14:21
  • $\begingroup$ @DominiqueUnruh: Actually, I never wrote "note that this is the same as the notion of right-equivalence in Chen et al." I just claimed that your condition implies right-equivalence, as you correctly remark. $\endgroup$
    – Francesco Polizzi
    Commented Jul 13, 2021 at 14:32
  • $\begingroup$ It is ok, for instance, to write something as "This is a particular case of right-equivalence in the sense of Chen-Le" (I do not have a counterexample, either, to show that the latter is actually weaker). $\endgroup$
    – Francesco Polizzi
    Commented Jul 13, 2021 at 14:34
  • $\begingroup$ Regarding a counterexample, suppose the common codomain of $f$ and $g$ is a terminal object. Then if the domains admit any morphisms at all between them, those morphisms will give a right equivalence between $f$ and $g.$ $\endgroup$
    – Kevin Carlson
    Commented Jul 13, 2021 at 17:10

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