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Let $A$ be an abelian category and $X$, $Y$ two objects of $A$. Let's define Ext in this way:

Ext$^i_A(X,Y)$=Hom$_{D(A)}(X[0],Y[i])$

Where $X[0]$ is the complex with all zeros except in degree 0 where it has $X$, and $Y[i]$ is the complex with all zeros except in degree $-i$ where it has $Y$.

Now I would like to prove that this definition is equivalent to the usual definition of Ext, i.e.:

take a projective resolution of $X$: $\cdots\rightarrow P^{-1}\rightarrow P^0\rightarrow X\rightarrow 0$ then Ext$^n_A(X,Y)$ is the $n$th cohomology group of the complex $0\rightarrow\mathrm{Hom}(P^0,Y)\rightarrow\mathrm{Hom}(P^{-1},Y)\rightarrow\mathrm{Hom}(P^{-2},Y)\rightarrow\cdots$.

I have an hint: denote by $K(A)$ the homotopy category. It seems to be useful to prove that we have an isomorphism

$\mathrm{Hom}_{K(A)}(X^\bullet,Y^\bullet)\rightarrow\mathrm{Hom}_{D(A)}(X^\bullet,Y^\bullet)$ in the following cases:

1) $Y^\bullet\in\mathrm{Ob}\;Kom^+(I)$, i.e. $Y^\bullet$ is a bounded complex on the left of injectives objects;

2) $X^\bullet\in\mathrm{Ob}\;Kom^-(P)$, i.e. $X^\bullet$ is a bounded complex on the right of projective objects.

I'm pretty sure that we need only one between 1 and 2, and the other is useful if we want to prove the caracterization of Ext with injective resolutions, but I can do that if you could show me how to do it with projective resolutions.

By the way the map $\mathrm{Hom}_{K(A)}(X^\bullet,Y^\bullet)\rightarrow\mathrm{Hom}_{D(A)}(X^\bullet,Y^\bullet)$ is $f\mapsto$ the equivalence class of the roof $X\leftarrow X\rightarrow Y$, where $id:X\rightarrow X$ and $f:X\rightarrow Y$.

And if you need the definition of roof just look to one of my previous questions: why is this composition well defined?

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    Are you comfortable with the following? If $\mathcal{A}$ has enough projectives, then there is an equivalence $Q: Kom^b(P(\mathcal{A}))\to D^b(\mathcal{A})$ just by including the bounded complex of projectives into the bounded homotopy category and then the localization (inverting quasi-isos). Let $F:\mathcal{A}\to \mathcal{B}$ be a right exact functor. One way to *define* the total derived functor $\mathbf{L}F$ is via the composition $Q\circ \overline{F} \circ Q^{-1}$. Before writing an answer, do you believe this is equivalent to the way you've defined $Ext^i=H^i(\mathbf{L}Hom)$?2012-09-16
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    This is done in detail (assuming enough injectives) in I.§6 of Hartshorne's *Residues and duality*. @Matt: don't you need a finiteness condition on $\mathcal{A}$ such as "every object has a finite projective resolution" to reduce to *bounded* complexes? Otherwise $Q$ will fail to be essentially surjective, even onto $\mathcal{A}$. I agree with the rest of what you say, of course.2012-09-16
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    To find for every right bounded complex a quasi-isomorphism from a right bounded complex of projectives you can either use [Cartan-Eilenberg resolutions](http://en.wikipedia.org/wiki/Cartan–Eilenberg_resolution) (probably the preferred procedure for dealing with spectral sequences) or use [this simple construction](http://math.stackexchange.com/questions/80513/) due to Keller (I describe the dual version).2012-09-16
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    @t.b. Yes. Of course. I was just assuming that situation to make the comment as short and easy to follow as possible.2012-09-16
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    @Matt: sorry Matt, I'm not so comfortable with what you said.2012-09-17
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    I just tried to write a proof for you and I realized something. You should define $Ext^i(X^\bullet, Y^\bullet)=Hom_{D(A)}(X^\bullet [0], Y^\bullet [i])$ for complexes and then prove for just single objects in $A$ otherwise $Ext$ won't be defined for enough things to write a proof.2012-09-18

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