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I'm looking for a counterexample of the pullback lemma, i.e., a diagram

$$ \require{AMScd} \begin{CD} A @>{}>> B @>>> C\\ @VVV @VVV @VVV\\ D @>>> E @>>> F \end{CD} $$

such that left square and outer square are pullbacks but right square is not a pullback. I have tried hard but I can't find a counterexample.

Any hint would be appreciated.

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  • $\begingroup$ Reference for "pullback lemma": proofwiki.org/wiki/Pullback_Lemma $\endgroup$
    – HeinrichD
    Apr 17, 2017 at 7:52
  • $\begingroup$ You don't look for a counterexample of the pullback lemma (it is a proven lemma, it does not have counterexamples), but rahter of the variant that you have mentioned. $\endgroup$
    – HeinrichD
    Apr 17, 2017 at 7:53
  • $\begingroup$ Hint: Look for a diagram of vector spaces where most of the objects are zero. $\endgroup$ Apr 17, 2017 at 8:11

2 Answers 2

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Here's a counterexample that works in (almost) any category : let $p:X\to Y$ be a split epimorphism that is not an isomorphism, with section $s:Y\to X$. Then in the diagram $$\require{AMScd} \begin{CD} Y @>{id_Y}>> Y @>{id_Y}>> Y\\ @V{id_Y}VV @VV{s}V @VV{id_Y}V\\ Y @>>{s}> X @>>_p> Y, \end{CD}$$ the outer rectangle is a pullback since all its sides are identity maps, the left square is a pullback since $s$ is a monomorphism, but the right square cannot be a pullack if $s$ and $p$ are not isomorphisms.

Alternatively, you can take any monomorphism $m:A\to B$, and form the diagram $$\require{AMScd} \begin{CD} A @>{id_A}>> A @>{m}>> B\\ @V{id_A}VV @VV{m}V @VV{id_B}V\\ A @>>{m}> B @>>_{id_B}> B. \end{CD}$$ Here the left square is a pullback since $m$ is a mono, the rectangle is a pullback, but the right-hand square is not a pullback if $m$ is not an isomorphism.

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The paper The other pullback lemma by Michal Przybylek exactly deals with this kind of situation. The introduction mentions two counterexamples. See also this mathoverflow thread. There you can also find a very formal counterexample by Eric Wofsey.

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