# Model structure on the category of chain complexes in an abelian category gives rise to the derived category

Let $\mathcal A$ be an abelian category with enough projectives. Consider a morphism $f \colon X^\bullet \longrightarrow Y^\bullet$ in the category $C(\cal A)$ of chain complexes in $\mathcal A$.

Is it true that $f$ is a quasi-isomorphism if and only if $\mathcal A(P, f ) \colon \mathcal A (P,X^\bullet) \longrightarrow \mathcal A(P,Y^\bullet)$ is a quasi-isomorphism in $C (\mathsf{Ab})$ for every projective object $P$?

The reason I ask is because in an abelian category $\cal A$ with enough projectives, there is a projective class $(\mathcal P, \mathcal E)$ where $\mathcal P$ is the class of projectives and $\mathcal E$ is the class of epimorphisms, and under a few other assumptions the category of chain complexes $C( \mathcal A)$ becomes a model category whose weak equivalences are precisely those morphisms $f$ such that $\mathcal A(P,f)$ is a quasi-isomorphism for all $P \in \cal P$. See here for a reference.

Taking the localisation of $C(\mathcal A)$ at these weak equivalences should be the standard derived category $D(\mathcal A)$. However this is the localisation of $C(\mathcal A)$ at all quasi-isomorphisms, so ideally these two collections should be the same thing.

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The answer is yes, and the explanation is in the definition of 'enough projectives'. –  Fernando Muro May 28 '13 at 15:18
I can see that if $f$ is a quasi-isomorphism, then due to exactness of $\mathcal (P,- )$ this preserves quasi-isomorphisms. However the converse is still eluding me –  Paul May 29 '13 at 11:12
Never mind, I worked it out. It follows from $\mathcal A(P,-)$ collectively reflecting isomorphisms, and preserving homology. –  Paul May 29 '13 at 11:59