(I am aware of the existence of this question).
Hello, there's a step of a proof that $H_n$ preserves coproducts in an abelian category that I do not understand, despite it looking fairly simple...
Let $(C_\bullet^i)_{i\in I}$ be chain complexes in an abelian category $\mathcal{C}$ and $n\in \mathbb{Z}$.
Let $C_\bullet = \coprod_{i\in I}C_\bullet^i$ and for $j\in I$, $q^j:C_\bullet^j\rightarrow C_\bullet$ the canonical injections.
For $j\in I$ wee have $H_n(q^j):H_n(C_\bullet^j)\rightarrow H_n(C_\bullet)$ which induces a unique map $\phi_n:\coprod_{i\in I}H_n(C_\bullet^i)\rightarrow H_n(C_\bullet)$.
Now the author writes
Since coproduct is exact, it preserves kernel and images, hence $\phi_n$ is an isomorphism.
And I can't seem to understand why this gives an isomorphism.
Since $\coprod Z_n(C^i_\bullet)\cong Z_n(C_\bullet)$ and $\coprod B_n(C^i_\bullet)\cong B_n(C_\bullet)$ I tried to use the exact sequence $0\rightarrow B_n \rightarrow Z_n \rightarrow H_n \rightarrow 0$ and somehow find that $\phi$ is an isomorphism but I couldn't find anything.
Let us denote the differential of $C_\bullet$, $\partial$ and the differential of $C^i_\bullet$, $\delta_i$.
$\partial$ is determined by the family of the $\partial_i : C^i_\bullet \to \coprod C^i_\bullet$ with $\partial_i = \imath^i \circ \delta_i$ with $\imath^i$ the injection of $C_\bullet$ in $\coprod C^i_\bullet$.
Then you have $$Z_\bullet (C_\bullet) = \ker \partial = \ker \coprod \partial_i = \coprod \ker \partial_i = \coprod \ker \delta_i = \coprod Z_\bullet(C^i)$$ $$B_\bullet (C_\bullet) = \text{im } \partial = \text{im } \coprod \partial_i = \coprod \text{im } \partial_i = \coprod B_\bullet(C^i)$$
Now note that each injection $\imath^i$ is a complex homomorphism an induce an homology morphism: $$\imath^i_*: H_\bullet(C^i) \to H_\bullet(C)$$ Then we define $$\phi = \coprod \imath^i_* : \coprod H_\bullet(C^i) \to H_\bullet(C)$$
It is an iso iff $\forall n, \coprod \imath^i_n$ is an iso. This is true and given by: $$H_n(C) = \coprod Z_n(C^i) \,/\,\coprod B_n(C^i) = \coprod \left(Z_n(C^i) \,/\,\coprod B_n(C^i) \right) \simeq \coprod \left(Z_n(C^i) / B_n(C^i)\right) = \coprod H_n(C^i)$$
You can easily see that since for fixed $i \neq j$, $B_n(C^j)$ does not live in the same space as $Z_n(C^i)$ hence its quotient has no effect.
Essentially $\imath^i_n$ sends $[c_i]_{B^i}$ to $[c_i]_{\coprod B^j}$, so it is very easy to see that $\coprod \imath^j_n$ is surjective and it is injective because of the reason explained above. $[c_i]_{\coprod B^j} = 0$ iff $c_i \in \coprod B^j$ iff $c_i \in B^i$ iff $[c_i ]_B^i = 0$.
I don't know if you need more clarification. Don't hesitate!