Covariant and contravariant $\mathrm{Hom}$ are left exact

If $u_{\ast }f=0$, then $u(f(x))=0$ for every $x$, and injectivity of $u$ gives $f=0$; hence $u_{\ast }$ is injective.

Composability gives $v_{\ast }{u}_{\ast }=0$, so $\mathrm{im}{u}_{\ast }\le \ker v_{\ast }$.

If $g:X\to B$ satisfies $v_{\ast }g=0$, then $g(x)\in \ker v=\mathrm{im}u$ for every $x$. Injectivity of $u$ gives a unique $f(x)\in A$ with $u(f(x))=g(x)$; uniqueness makes $f$ linear, so $g=u_{\ast }f$.

If $p^{\ast }h=0$, surjectivity of $p$ gives $h=0$, so $p^{\ast }$ is injective; and $u^{\ast }{p}^{\ast }=0$ because $p\circ u=0$.

If $g:B\to X$ satisfies $u^{\ast }g=g\circ u=0$, then $g$ vanishes on $\mathrm{im}u=\ker p$. By [L1] it factors uniquely through $B\mathrm{/}\ker p$, and since $p$ is surjective the rule $\bar{g}(p(b))=g(b)$ gives the corresponding homomorphism $\bar{g}:C\to X$ with $g=\bar{g}\circ p=p^{\ast }(\bar{g})$.

Steps 1.1 to 1.3 prove the covariant sequence exact, and steps 1.4 and 1.5 prove the contravariant sequence exact.