Prove that $\operatorname{Kom}_0(\mathcal{A})$ is a equivalence of category of $D(\mathcal{A})$

Question

I'm working on Gelfand's Methods of Homological Algebra, Chapter 3:

If an abelian category $\mathcal{A} $ is semisimple then the functor $D(\mathcal{A}) \rightarrow \operatorname{Kom}_{0}(\mathcal{A})$ is an equivalence of categories.

To prove this Gelfand constructs two morphisms:

$f_{K} :\left(K^{n}, d^{n}\right) \longrightarrow\left(H^{n}\left(K^{\bullet}\right), 0\right), \quad g_{K} :\left(H^{n}\left(K^{\bullet}\right), 0\right) \longrightarrow\left(K^{n}, d^{n}\right)$

He claims that $K^{n}=B^{n} \oplus H^{n} \oplus B^{n+1}$ and $d^{n}\left(b^{n}, h^{n}, b^{n+1}\right)=\left(b^{n+1}, 0,0\right)$, and defines

$f_{K}^{n}\left(b^{n}, h^{n}, b^{n+1}\right)=h^{n}, \quad g_{K}^{n}\left(h^{n}\right)=\left(0, h^{n}, 0\right)$

So far so good, but then he claims that:

\begin{array}{l}{\text { First, it is clear that the composition } k \circ l \text { is isomorphic to the identity }} \\ {\text { functor in} \operatorname{Kom}_0 (\mathcal{A}) . \text { On the other hand, the functor } l \circ k : D(\mathcal{A}) \rightarrow D(\mathcal{A})} \\ {\text { maps a complex }\left(K^{n}, d^{n}\right) \in D(\mathcal{A}) \text { to }\left(H^{n}\left(K^{\bullet}\right), 0\right) \in D(\mathcal{A}) . \text { The above mor- }} \\ {\text { phisms of complexes } f_{K} : K^{\bullet} \rightarrow l \circ k\left(K^{\bullet}\right) \text { and } g_{K} : l \circ k\left(K^{\bullet}\right) \rightarrow K^{\bullet} \text { clearly }} \\ {\text { give mutually inverse isomorphisms in } D(\mathcal{A}), \text { so that }\left\{f_{K}\right\} \text { and }\left\{g_{K}\right\}, K^{\bullet} \in D} \\ {\text { are mutually inverse isomorphisms of functors id } D(\mathcal{A}) \text { and } l \circ k .}\end{array}

Here, $k:D(\mathcal{A}) \rightarrow \operatorname{Kom}_{0}(\mathcal{A})$ is given by the universal property of k, l is the composition of the embedding $\mathrm{Kom}_{0}(\mathcal{A}) \rightarrow \operatorname{Kom}(\mathcal{A}) $ with the localization $ Q : \operatorname{Kom}(\mathcal{A}) \rightarrow D(\mathcal{A})$.

My question is:

1) Why is $k\circ l$ isomorphic to identity? Given $(H^n(K^{\bullet}),0)$ in ${Kom}_{0}(\mathcal{A})$, it is first mapped to $(H^n(K^{\bullet}),0)$ as an element in $D(\mathcal{A})$, right? Then by the construction of $D(\mathcal{A})$, it should be mapped to $(H^n(H^n(K^{\bullet})),0)$, which I think should be $(0,0)$.

Edited: I now realize that $(H^n(H^n(K^{\bullet})),0) = (H^n(K^{\bullet}),0)$, hence the conclusion.

2) $g^n_K(f^n_K(b^n,h^h,b^{n+1}))=g^n_K(h^n)=(0,h^n,0)$, right? so why does $f_K$ and $g_K$ give mutually inverse isomorphisms in $D(\mathcal{A})$?

I'm quite confused, thanks for your help.

Answer 1

After some working I think I worked it out. Please correct me if I make any mistake:

1) Why is $k\circ l$ isomorphic to identity?

I now realize that $(H^n(H^n(K^{\bullet})),0) = (H^n(K^{\bullet}),0)$, hence the conclusion.

2) Why does $f_K$ and $g_K$ give mutually inverse isomorphisms in $D(\mathcal{A})$?

First, it is clear that $f_K\circ g_K$ is an isomorphism. Since $f_K$ and $g_K$ are isomorphisms in $D(\mathcal{A})$, so is their composition $g_K\circ f_K$. Hence the conclusion.

The trick here is to notice that an inverse is introduced in $D(\mathcal{A})$ for any quasi-isomorphism in $\operatorname{Kom}_0(\mathcal{A})$.