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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 $M$?

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    $\begingroup$ Dear harajm, just to make sure I got you right: you have an arbitrary bounded above complex $M$ in $A$ and you want to find a quasi-isomorphism $P\to M$ with $P$ bounded above with projective components? Could you also tell us your teacher's construction? $\endgroup$ – Hanno Jan 17 '15 at 8:06
  • $\begingroup$ Yes, this is precisely what I mean. I can't say to much about my teachers construction since I don't really understand it. But he proves the statement by induction on the components of the chain, assuming we have a quasi isomorphism on the $n:th$ place. Actually I am very unsure what his argument really is, sorry. But he is not using that each component $M$ has a "projective resolution" in the classical sence, because he later deduces this statement from the earlier. $\endgroup$ – harajm Jan 17 '15 at 16:25
  • $\begingroup$ But any way to prove this statement would help my enormously! $\endgroup$ – harajm Jan 17 '15 at 16:27

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