Let $\mathcal{A}$ be an abelian category and $\text{Ch}_*(\mathcal{A})$ the category of chain complexes $A_\bullet$ of objects in $\mathcal{A}$. We let $$H_i(A_\bullet):=\text{Coker}(\text{im}_{d_{i+1}}\to \text{ker}(d_i)).$$
Why is the functor $H_i:\text{Ch}_*(\mathcal{A})\to\mathcal{A}$ an additive functor? Where additive means that $\text{Hom}(A_\bullet,B_\bullet)\to \text{Hom}(H_i(A_\bullet),H_i(B_\bullet))$ is a group homomorphism.
I've drawn many diagrams, but can't get a hold of how to show this. I've seen that being left or right exact would imply this, but can't determine if either of those properties hold.
Best Answer
There are many way to prove this statements some more elegant than others but since you have tried to work out the proof by diagram chasing I assume you are looking for a solution by hand.
You have to prove that for every pair of morphisms $$f,g \in \text{Ch}_*(\mathcal A)[A_\bullet,B_\bullet]$$ the following equality holds $$H_i(f+g) = H_i(f)+H_i(g)\ .$$
By definition $H_i(f+g) \colon H_i(A_\bullet) \to H_i(B_\bullet)$ is the unique morphism that makes commute the following diagram: $$\require{AMScd}\begin{CD} \operatorname{im}d_{i+1}^A @>>> \ker d_i^A @>>> H_i(A) \\ @V{f+g}VV @VV{f+g}V @VV{H_i(f+g)}V \\ \operatorname{im}d_{i+1}^B @>>> \ker d_i^B @>>> H_i(B) \end{CD}$$ where the vertical arrows are the obvious ones.
So to prove the above mentioned equality you just need to prove the commutativity of the square $$\begin{CD} \ker d_i^A @>\pi_i^A>> H_i(A) \\ @V{f+g}VV @VV{H_i(f)+H_i(g)}V \\ \ker d_i^B @>>\pi_i^B> H_i(B) \end{CD}$$ i.e. that the equality $$\pi_i^B \circ (f+g) = (H_i(f)+H_i(g))\circ \pi_i^A$$ holds (where $\pi_i^X \colon \ker d_i^X \to H_i(X)$ is the cokernel of the inclusion $\operatorname{im} d_{i+1}^X \to \ker d_i^X$).
The said equality follows from the fact that $$\pi_i^B \circ f=H_i(f)\circ\pi_i^A,$$ $$\pi_i^B \circ g=H_i(g)\circ\pi_i^A,$$ and bilinearity of composition: $$\begin{align*} \pi_i^B \circ (f+g) &= (\pi_i^B \circ f)+(\pi_i^B \circ g) \\ &= (H_i(f)\circ\pi_i^A)+(H_i(g)\circ\pi_i^A) \\ &= (H_i(f)+H_i(g))\circ\pi_i^A \end{align*}$$
From this your claim follows.
Hope this helps.