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I am learning Homotopy category of chain complexes from wikipedia. There are two places I do not understand:

1.Let $K(A)$ (respectively, $D(A)$) be a homotopy category (respectively, derived category) of chain complexes. There is a canonical functor $K(A) \rightarrow D(A)$ if $A$ is abelian.

My idea is as follows: $D(A)$ is obtained by localizing $K(A)$ at the set of quasi-isomorphisms. The canonical functor $\pi: K(A) \rightarrow D(A)$ maps $f: M \to N$ to $1_{N} \circ f$. Is my understanding correct?

2.The homotopy category $\textrm{Ho}(C)$ of a differential graded category $C$ is defined to have the same objects as $C$, but morphisms are defined by $\textrm{Hom}_{\textrm{Ho}(C)}(X,Y)=H^{0}\textrm{Hom}_{C}(X,Y)$.

About 2, what is the meaning of $\textrm{Hom}_{\textrm{Ho}(C)}(X,Y)=H^{0}\textrm{Hom}_{C}(X,Y)$?

Edited to add: A differential graded category $C$ (dg category for short) is a category whose morphism sets are complexes: \begin{align*} \cdots \to \textrm{Hom}^{-1}(X,Y) \to \textrm{Hom}^{0}(X,Y) \to \textrm{Hom}^{1}(X,Y) \to \cdots \end{align*} $\textrm{Hom}_{C}(X,Y)=\oplus_{n \in \mathbb{Z}} \textrm{Hom}_{n}(X,Y)$.

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  • $\begingroup$ 1) Yes, $\pi$ is the localisation functor. 2) Since $C$ is differential graded category, $\mathrm{Hom}_C(X,Y)$ is a complex of abelian groups, and $\mathrm{Hom}_{\mathrm{Ho}(C)}(X,Y)$ is its zeroth cohomology, which is an abelian group. $\endgroup$ – user337830 Nov 30 '16 at 15:27
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    $\begingroup$ @bing What do you mean with $1_N$? I'm not sure I follow. $\endgroup$ – Denis Nardin Nov 30 '16 at 15:47
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    $\begingroup$ See the wiki article on dg categories for an explanation of how the category of chain complexes becomes a dg category. For chain complexes $X$ and $Y$, in the chain complex $Hom_C(X,Y)$ a $0$-chain is a map of graded abelian groups $X\to Y$, it is a $0$-cycle if it is a chain map, and two such maps differ by a boundary if and only if they are chain homotopic. $\endgroup$ – Tom Goodwillie Nov 30 '16 at 15:52
  • $\begingroup$ @ Denis Nardin I mean that $1_N$ is the identity morphism from $N$ to $N$. I view $1_N$ as a quasi-isomorphism from $N$ to $N$. $\endgroup$ – bing Dec 1 '16 at 1:45
  • $\begingroup$ @ user337830 I am a new. Could you explain more concretely about (2)? Thank you very much. $\endgroup$ – bing Dec 1 '16 at 3:35

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