# homology under exact functors

Let $\mathcal{A}$ be an Abelian category, $X$ be a complex, $F$ be a contravariant exact functor. I am wondering whether $F$ preserves the homology of $X$; that means whether $H^{i}(FX)=F(H^{-i}(X))$ for all $i$? (Obviously, this is true for covariant functors).

• A contravariant functor is a covariant functor into the opposite category. Commented Mar 18, 2017 at 2:35
• Yes, but what is the relationship between the homology in the opposite category and the original homology?
– luw
Commented Mar 18, 2017 at 3:10
• Well... What is the relation? Commented Mar 18, 2017 at 3:11
• In the opposite category, we have $H^{0}(FX^{op})=F(H^{0}(X^{op}))$, but what are their relation with $H^{0}(FX)$ and $FH^{0}(X)$?
– luw
Commented Mar 18, 2017 at 3:17

As suggested in the comments, a contravariant functor $F\colon \mathcal{A}\to \mathcal{B}$ is a functor $F\colon \mathcal{A}\to \mathcal{B}^\circ$ to the opposite category, which is also abelian. So there's no reason to consider the contravariant case separately.
But if you like, you may note that $F$ transforms a cohomological complex $(X^\bullet,d^\bullet)$ into the complex with objects $F (X^i)$ and differentials $F (d^{i-1})\colon F (X^i) \to F (X^{i-1})$, which is more homological: it's natural to declare that $F (X^i)$ sits in degree $i$. Note however that the differential on degree $i$ will be $F (d^{i-1})$.
$$H^i (X^\bullet) \cong \operatorname{coker} (\operatorname{im} d^{i-1} \rightarrowtail \ker d^i) \cong \ker (\operatorname{coker} d^{i-1} \twoheadrightarrow \operatorname{im} d^i).$$
As contravariant exact, $F$ maps kernels to cokernels and vice versa (and images to coimages, but these two are canonically isomorphic), hence
$$F H^i (X^\bullet) \cong \ker (\operatorname{coker} F(d^i) \twoheadrightarrow \operatorname{im} F (d^{i-1})) \cong \operatorname{coker} (\operatorname{im} F (d^i) \rightarrowtail \ker F (d^{i-1})),$$ which is $H_i (F(X^\bullet))$.