# Short Five Lemma for R-modules

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?

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.