I'm unable to prove that if a mono $u:B\to A$ is less then mono $v:C\to A$ by $f:B\to C$ with $v\circ f=u$ and also $v\leq u$ by $u\circ g=v$ then $B\cong C$ by using some morphisms as above, but I believe that this holds. Is there a hint how this $\cong$ may look like together with a hint that it is $\cong$ ? This is certainly true in the category of Sets and functions but I suspect that it holds in general. Also, I think that this is further more equivalent to $u=v\circ \theta$ for some $\theta$ being an isomorphism. How can I see this equivalence ?
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hint: with the data you listed, do you have any morphisms between $B$ and $C$? Do they compose to the identity? – S.C. Jan 23 '23 at 20:46
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@S.C. I have also thought along these lines. I do have $f\circ g$ and $g\circ f$ but I cannot immediately see that they compose to $\mathbf{id}_A$. – user122424 Jan 23 '23 at 20:48
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@user122424 You’re almost done! Hint: $u$ and $v$ are monos. – azif00 Jan 23 '23 at 20:52
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Sorry I misunderstood your question and deleted my answer – Joe Jan 23 '23 at 20:54
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@azif00 I've got the first part of my OQ: $vfg=ug=v=v \mathbf{id}_C$. What about my last sentence in my OQ ? – user122424 Jan 23 '23 at 20:58
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1@user122424 If I understand correctly, just take $\theta=f$. – azif00 Jan 23 '23 at 20:59
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@azif00 Everything was trivial, after all. – user122424 Jan 23 '23 at 21:00
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@azif00 Is this line entirely correct including the $C$-index: $vfg=ug=v=v\circ \mathbf{id}_C$ ? – user122424 Jan 23 '23 at 21:01
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@user122424 That is correct. I just wrote an answer that summarizes everything in the comments. – azif00 Jan 23 '23 at 21:18
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Related : https://math.stackexchange.com/questions/4107996/why-a-hookrightarrow-b-hookrightarrow-c-simeq-a-implies-a-simeq-b – Arnaud D. Jan 24 '23 at 09:24
1 Answers
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Let $u \colon B \to A$ and $v \colon C \to A$ monomorphisms in a category. Then the following are equivalent:
- There are morphisms $f \colon B \to C$ and $g \colon C \to B$ such that $u=vf$ and $v=ug$.
- There is an isomorphism $\theta \colon B \to C$ such that $u=v\theta$.
Proof: If we assume (1), it suffices to prove that $f$ is an isomorphism (so, $\theta=f$ works for (2)). Indeed, since $$ u\text{id}_B=u=vf=(ug)f=u(gf) $$ and $u$ is mono, then $gf=\text{id}_B$. Also, since $$ v\text{id}_C=v=ug=(vf)g=v(fg) $$ and $v$ is mono, then $fg=\text{id}_C$. Hence, $f$ is an isomorphism.
On the other hand, if we assume (2), then the morphisms $f=\theta$ and $g=\theta^{-1}$ works for (1).
azif00
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