I just started to learn homological algebra and I find it quite hard, so I am sorry if the question is unclear and confused. In fact I am confused.
Let $K^-(A)$, where $A$ is an abelian category with enough projectives, denote the category of complexes in $A$ with morphisms given by complex morphisms modulo homotopy. How to show that there is a quasi isomorphism from a complex of projectives to any complex $M$?
My teacher proved this in class and defined projective resolution in these terms, but I can not follow my notes.
However, looking into texts that do this the "classical way". I know that there exists a projective resolution of $M$ (in the classical sense), i.e., an exact sequence of projective elements in $K^-(A)$ with $P_0$ mapping surjectively to $M.$ However I can't seem to deduce my teacher's definition from this one. I know also that the projective objects of $K^-(A)$ are split exact complexes of projectives but how do I turn all of these exact sequences on the n:th place into one projective element of $A$ having isomorphic n:th homology with $M$?