Homology commutes with direct sum and product?

Question

I'm looking at exercise 1.2.1 from Weibel's An Introduction to Homological Algebra. (I need to show that homology commutes with direct sum and direct product.)

Is it possible to show that cokernels commute with direct sum and product?

Since $H_{n} (C)=\ker d_{n}/ \operatorname{im} d_{n+1}$, $H_{n} (C)=\operatorname{coker} d'_{n+1},$ where $d'_{n+1}:C_{n+1}\rightarrow \ker d_{n}\subset C_{n}$.

Or, if I can show that kernels, images, and quotients commute with direct sums and products, would that also prove the statement?

I can start with $\oplus(\ker d_{n})=\ker(\oplus d_{n})$, $\oplus(\operatorname{im} d_{n+1})= \operatorname{im}(\oplus d_{n+1})$, then $\oplus (\ker d_n /\operatorname{im} d_{n+1})=\ker \oplus d_n / \operatorname{im}\oplus d_{n+1}$ or $\oplus \ H_n (C)=H_n (\oplus \ C)$ where $ \oplus \ d$ is the differential for the chain complex $\oplus \ C. $ Similar statements for direct product. So my questions are, is it true that quotients, image, and kernel commute with direct sum and product? I think its easy to see that image and kernel commute with direct sum and product but I don't know how to prove it. I'm not so sure about quotients.

Answer 1

it is just a basic computation:

Answer 2

To treat exercise 1.2.1 in the most general setup, you can say that in a category in which colimits exists and the direct product functor is exact (AB4 type category), the homology functor of any degree does commute with direct sum. The proof is simple: take the following exact sequence $$0\to \text{im}\ d_{n+1}\to \ker d_{n}\to H_n(A_n^{i})\to 0$$ where $A^{i}$ is a complex for $i\in I$. Apply the exact functor direct sum. You immediately get what you want since by exactness direct sum commutes with $\ker$ (AB4) and $\text{im}$ (always).

For direct product everything dualises well (if the category verifies AB4*)