The injective and surjective Four Lemmas

Assume $a$ is surjective and $b,d$ are injective, and let $z\in C$ satisfy $c(z)=0$. Then [C3] gives $d(w(z))=w^{\mathrm{\prime }}(c(z))=0$, so injectivity of $d$ gives $w(z)=0$.

Assume $b,d$ are surjective and $e$ is injective, and let $z^{\mathrm{\prime }}\in C^{\mathrm{\prime }}$. By surjectivity of $d$, choose $q\in D$ with $d(q)=w^{\mathrm{\prime }}(z^{\mathrm{\prime }})$.

Exactness gives $y\in B$ with $v(y)=z$. By [C2], $v^{\mathrm{\prime }}(b(y))=c(v(y))=0$, so exactness gives $y^{\mathrm{\prime }}\in A^{\mathrm{\prime }}$ with $u^{\mathrm{\prime }}(y^{\mathrm{\prime }})=b(y)$.

By [C4], $e(x(q))=x^{\mathrm{\prime }}(d(q))=x^{\mathrm{\prime }}{w}^{\mathrm{\prime }}(z^{\mathrm{\prime }})=0$; injectivity of $e$ gives $x(q)=0$. Exactness gives $z\in C$ with $w(z)=q$.

Surjectivity of $a$ gives $y^{\mathrm{\prime }}=a(t)$ for some $t\in A$. Then [C1] gives $b(y-u(t))=0$; injectivity of $b$ gives $y=u(t)$, and exactness gives $z=v(y)=v(u(t))=0$. Thus $c$ is injective.

Now $w^{\mathrm{\prime }}(c(z)-z^{\mathrm{\prime }})=d(w(z))-w^{\mathrm{\prime }}(z^{\mathrm{\prime }})=0$ by [C3], so exactness gives $y^{\mathrm{\prime }}\in B^{\mathrm{\prime }}$ with $v^{\mathrm{\prime }}(y^{\mathrm{\prime }})=c(z)-z^{\mathrm{\prime }}$. Surjectivity of $b$ gives $y^{\mathrm{\prime }}=b(y)$; then [C2] yields $c(z-v(y))=z^{\mathrm{\prime }}$. Thus $c$ is surjective.

Steps 1.1, 2.1, and 3.1 prove the injective Four Lemma, while steps 1.2, 2.2, and 3.2 prove the surjective Four Lemma.