Four lemma

Remarks

Part 1 is a diagram chase. Let the row maps be $r \colon A \to B$, $s \colon B \to C$, $t \colon C \to D$ and $r', s', t'$ below. If $\gamma(x) = 0$ for $x \in C$, then $\delta(t(x)) = t'(\gamma(x)) = 0$, so $t(x) = 0$ and $x = s(y)$ for some $y \in B$. Then $s'(\beta(y)) = 0$, so $\beta(y) = r'(u')$ with $u' = \alpha(u)$ for some $u \in A$; now $\beta(y - r(u)) = 0$, so $y = r(u)$ and $x = s(r(u)) = 0$. Part 2 is the dual statement, obtained by reversing all arrows. Example: take $A = A' = 0$, so that the rows are exact sequences $0 \to B \to C \to D$ and $0 \to B' \to C' \to D'$. If $\beta$ and $\delta$ are monomorphisms, part 1 shows that $\gamma$ is a monomorphism.

Going deeper

For a student

The two parts are proved by element chasing in modules, and hold in every abelian category by the Freyd–Mitchell embedding theorem or by pseudo-elements. Applying part 1 to the left four columns and part 2 to the right four columns of a five-column ladder gives the five lemma. The four lemma is useful when only one of injectivity or surjectivity is needed, so fewer hypotheses have to be checked.