Prove this version of the four-lemma writing out all of the details. Let $R$ be a commutative ring. Given the diagram of $R-$modules with exact rows $\require{AMScd}$ \begin{CD} A @>{f}>> B @>{g}>> C @>{h}>> D\\ @V{\alpha}VV @VV{\beta}V @VV{\gamma}V @VV{\delta}V\\ A' @>{r}>> B' @>{s}>> C' @>{t}>> D'\end{CD} Also prove that $\alpha$ is an epimorphism, $\beta$ and $\delta$ are monomorphisms, then the morphism $\gamma$ is a monomorphism.
While studying complexes/homology, consequences of Snake's lemma, I came across this problem. I would be much thankful to anyone who can help me in solving this.
It's a standard diagram chase. It's a really good exercise to do these on your own, and I advise you to try and do it before reading my solution. If you must, read my solution, and then try to prove the other version of the four lemma (where two maps are epimorphisms and one is a monomorphism).
Anyway. Suppose that $c \in C$ is such that $\gamma(c) = 0$, we aim to prove that $c = 0$.
We have $\delta(h(c)) = t(\gamma(c)) = t(0) = 0$, and since $\delta$ is monic, $h(c) = 0$. The top row is exact, hence $c = g(b)$ for some $b$.
But then $s(\beta(b)) = \gamma(g(b)) = \gamma(c) = 0$. It follows that $\beta(b) = r(a')$ for some $a' \in A'$, by exactness of the bottom row.
Now $\alpha$ is epic, hence $a' = \alpha(a)$ for some $a \in A$. We then get $\beta(f(a)) = r(\alpha(a)) = r(a') = \beta(b)$. Since $\beta$ is monic, this implies $f(a) = b$.
But now remember that $c = g(b)$, which as we just saw is equal to $g(f(a))$. Since the top row is exact, this vanishes and $c = 0$ as expected.