3
$\begingroup$

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 $g \circ w=k$?

$\endgroup$
1
  • $\begingroup$ 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

9
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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 $g \circ w = k$.

$\endgroup$
5
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.