Let us consider a category $\mathfrak{A}$ and the following diagram of morphisms:

$\require{AMScd}$ \begin{CD} A @>{f}>> B\\ @V{h}VV @V{k}VV\\ C @>{g}>> D \end{CD}

Assume that $k \circ f=g \circ h$ and that there exists a unique $w: B \longrightarrow C$ such that $w \circ f=h$. May I also deduce that $g \circ w=k$?

  • $\begingroup$ I added the labels to your vertical arrows. Check my edit to see how to do this. $\endgroup$ Sep 3, 2021 at 21:52

2 Answers 2


Another counterexample, using the category of abelian groups and group homomorphisms (just to keep things down to earth): $\require{AMScd}$ \begin{CD} 0 @>>> \mathbb{Z}/2\\ @VVV @VV{1}V\\ 0 @>>> \mathbb{Z}/2 \end{CD} This clearly commutes. There is a unique morphism $w: \mathbb{Z}/2 \to 0$ which happens to make the upper left triangle commute — that is, not only does the subset $\{w: w \circ f = h\} \subseteq \text{Hom}(B,C)$ have cardinality 1, but in fact $\text{Hom}(B,C)$ has cardinality 1. The bottom right triangle won't commute in this case.


No. Consider the category whose only objects are distinct objects $A, B, C, D$, where the only arrows $A \to A$, $B \to B$, $C \to C$, $D \to D$ are the identity arrows, and the following are all the other arrows:

  1. $f : A \to B$
  2. $h : A \to C$
  3. $w : B \to C$
  4. $e : A \to D$
  5. $k \neq k' : B \to D$
  6. $g : C \to D$

Note that with the exception of $B \to D$, there is at most one arrow $X \to Y$. So the only nontrivial composition is $g \circ w$, which we define to be $k'$.

This forms a category. And in this category, it is the case that $k \circ f = g \circ h$. But it is not the case that $g \circ w = k$.

  • $\begingroup$ I'm not 100% sure this gives a category. $\endgroup$
    – Randall
    Sep 3, 2021 at 20:34
  • $\begingroup$ @Randall Clearly, composition is well-defined. The identity law is satisfied automatically by definition, so the only thing to check is the associativity law. The only arrows with a common domain and codomain that could be unequal are $k, k' : B \to D$, so we need only check associativity for trios which compose to $B \to D$. But any such trio must contain an identity morphism, so associativity is trivial. $\endgroup$ Sep 3, 2021 at 20:35
  • $\begingroup$ I don't doubt that you're right. Is it clear that $k' \circ f = k \circ f$? You need that for the associativity $(gw)f = g(wf)$ to hold, right? $\endgroup$
    – Randall
    Sep 3, 2021 at 20:39
  • $\begingroup$ @Randall We see that both $k' \circ f$ and $k \circ f$ are from $A \to D$. The only arrow $A \to D$ is $e : A \to D$. $\endgroup$ Sep 3, 2021 at 20:40
  • 1
    $\begingroup$ Oh, I see. I didn't notice the edit. $\endgroup$
    – Randall
    Sep 3, 2021 at 20:41

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