# Why is homology functor additive?

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.

• Do you know that for abelian categories, additive is equivalent to "preserves finite biproducts" ? – Max Aug 19 '18 at 9:40

## 1 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.

• +1, great proof ! – Max Aug 19 '18 at 22:00