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I have a question about the short five lemma. Since it is It is usually phrased something like:

Let the following be a commutative diagram of A-modules with exact rows

$$\require{AMScd} \begin{CD} 0 @>>> A & @>\psi>> B @>\phi>> C @>>> 0 \\ @. @VV\alpha_1V @VV\alpha_2V @VV\alpha_3V @. \\ 0 @>>> A' @>\psi'>> B' @>\phi'>> C' @>>> 0 \end{CD} $$ If $\alpha_1$ and $\alpha_3$ are isomorphisms, then so is $\alpha_2$

If I am given a diagram like the one above where $\alpha_2$ is not a homomorphism and just some map, does the conclusion still follow? In other words, if I have a diagram $$ \begin{CD} 0 @>>> A & @>\psi>> B @>\phi>> C @>>> 0 \\ @. @VV\alpha_1V @VV\alpha_2V @VV\alpha_3V @. \\ 0 @>>> A' @>\psi'>> B' @>\phi'>> C' @>>> 0 \end{CD} $$ where $\alpha_1$ and $\alpha_3$ are homomorphisms and $\alpha_2$ is any map (i.e. we do not assume it to be a homomorphism), does the lemma still hold? Or do we always assume $\alpha_1,\alpha_2$ and $\alpha_3$ to be homomorphisms from the start?

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No, it does not hold if $\alpha_2$ is not a homomorphism. Indeed, you can take for example the diagram of abelian groups

$$\require{AMScd} \begin{CD} 0 @>>> \mathbb{Z}/2\mathbb{Z}@>{i_1}>> \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z} @>{p_2}>> \mathbb{Z}/2\mathbb{Z} @>>>0 \\ & @V{}VV @V{f}VV @VV{\gamma}V \\ 0 @>>> \mathbb{Z}/2\mathbb{Z} @>>{j}> \mathbb{Z}/4\mathbb{Z} @>>{\pi}> \mathbb{Z}/2\mathbb{Z} @>>>0\end{CD} $$ where $i_1$ is just the inclusion of the first summand and $p_2$ the second projection (i.e. $i_1(x)=(x,0)$ and $p_2(x,y)=y$), $j$ is the morphism sending $\overline{1}$ to $\overline{2}$, $\pi$ is the quotient of the subgroup $\{\overline{0},\overline{2}\}$ of $\mathbb{Z}/4\mathbb{Z}$, with the quotient identified with $\mathbb{Z}/2\mathbb{Z}$ thanks to the second isomorphism theorem, and the outer vertical maps are just identities.

Then if you define $f$ by asking for example that $$f(\overline{0}, \overline{0})=\overline{0}$$ $$f(\overline{1}, \overline{0})=\overline{2}$$ $$f(\overline{0}, \overline{1})=\overline{1}=f(\overline{1}, \overline{1}),$$ the two squares commute, but $f$ is not bijective.

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  • $\begingroup$ Thank a lot. And what a coincidence - your counterexample was precisely the reason why I couldn't make sense of the theorem as it was stated in my textbook! $\endgroup$ – KurtKnödel Sep 13 at 9:46
  • $\begingroup$ Maybe not entirely a coincidence : the two short exact sequences I used are the smallest example of short exact sequences with the same extremes but non-isomorphic middle terms, so it's relatively natural to use them to built a counterexample. $\endgroup$ – Arnaud D. Sep 13 at 9:54

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