Exact Functors Commute with (Co)homology

Theorem.

Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories.

1. If $F:\mathcal{A}\to \mathcal{B}$ is an exact covariant functor and $C_{\bullet }$ is a chain complex in $\mathcal{A}$, then there is a natural isomorphism for each $i\in \mathbb{Z}$:

   $$F(H_i(C_{\bullet }))\cong H_i(F(C_{\bullet })).$$

2. If $G:\mathcal{A}\to \mathcal{B}$ is an exact contravariant functor and $C_{\bullet }$ is a chain complex in $\mathcal{A}$, then there is a natural isomorphism for each $i\in \mathbb{Z}$:

   $$G(H_i(C_{\bullet }))\cong H^i(G(C_{\bullet })).$$

Proof. We prove the covariant case; the contravariant case is analogous. An exact functor preserves kernels, images, and quotients. Let the chain complex be $C_{\bullet }$ with differentials $d_i$. The properties of an exact functor yield the following unbroken chain of natural isomorphisms.

$$\begin{array}{rl}F(H_i(C_{\bullet }))& =F(\ker (d_i)/\mathrm{im}(d_{i+1}))\\ & \cong F(\ker (d_i))/F(\mathrm{im}(d_{i+1}))\\ & \cong \ker (F(d_i))/\mathrm{im}(F(d_{i+1}))\\ & =H_i(F(C_{\bullet }))\end{array}$$

The first and last equalities hold by the definition of homology. The first isomorphism holds because $F$ preserves quotients, and the second holds because $F$ preserves kernels and images. $\square$