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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 $Ext^i=H^i(\mathbf{L}Hom)$? –  Matt Sep 16 '12 at 14:26
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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. –  t.b. Sep 16 '12 at 16:02
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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 (probably the preferred procedure for dealing with spectral sequences) or use this simple construction due to Keller (I describe the dual version). –  t.b. Sep 16 '12 at 16:10
    
@t.b. Yes. Of course. I was just assuming that situation to make the comment as short and easy to follow as possible. –  Matt Sep 16 '12 at 16:37
    
@Matt: sorry Matt, I'm not so comfortable with what you said. –  Mec Sep 17 '12 at 0:22
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1 Answer 1

$\DeclareMathOperator{Hom}{Hom}$The main point is that a projective resolution $\dots \to P^{-2} \to P^{-1} \to P^{0} \xrightarrow{\alpha^0} X$ gives you a quasi-isomorphism $\alpha \colon P^{\bullet} \to X[0]$. Indeed, the mapping cone of $\alpha$ is exact because it is the resolution (maybe up to an immaterial sign in the differentials).

A quasi-isomorphism becomes an isomorphism in the derived category (because that's what we invert), in particular precomposition with $\alpha^{-1}$ (or, if you prefer: composition with the roof $X[0] \xleftarrow{\alpha} P^\bullet \xrightarrow{1} P^\bullet$) gives an isomorphism $$\Hom\nolimits_{D(A)}(P^\bullet,Y) \xrightarrow{\cong} \Hom\nolimits_{D(A)}(X[0],Y) $$ for every complex $Y \in D(A)$.

From fact 2) of the question we have a composite isomorphism $$ \Hom\nolimits_{K(A)}(P^\bullet,Y) \xrightarrow{\cong} \Hom\nolimits_{D(A)}(P^\bullet,Y) \xrightarrow{\cong} \Hom\nolimits_{D(A)}(X[0],Y) $$ for all complexes $Y$ which is explicitly given by sending a (homotopy class of a) chain map $f\colon P^\bullet \to Y$ to the roof $X[0] \xleftarrow{\alpha} P^\bullet \xrightarrow{f} Y$, so it only remains to identify the complex $\Hom_{K(\mathscr{A})}(P^\bullet, Y)$.

I hope I got the signs and indices right in what is to follow: for any two complexes $A$ and $B$ over an additive category $\mathscr{A}$ one defines the total $\Hom$-complex $\Hom\nolimits^\bullet(A,B)$ of abelian groups by $$ \Hom\nolimits^k(A,B) = \prod_{n \in \mathbb{Z}}\Hom\nolimits_{\mathscr{A}}(A^n,B^{n+k}) $$ with differential $(f^n)_{n \in \mathbb{Z}} \mapsto \left(f^{n+1}d_A^n-(-1)^{k}d_B^{n+k}f^n \right)_{n\in\mathbb{Z}}$ and it is elementary to check (if indeed I got the signs right) that $$\boxed{ H^k\left(\Hom\nolimits^\bullet(A,B)\right) = \Hom\nolimits_{K(\mathscr{A})}(A,B[k]) }$$ because in order for $f = (f_n)_{n \in \mathbb{Z}}$ to be a cycle in $\Hom\nolimits^k(A,B)$ it is necessary and sufficient that $f \colon A \to B[k]$ defines a chain map and to be a boundary means that it is homotopic to zero. (In the sign conventions I'm used to shifting a complex by $1$ involves multiplying its differentials by $-1$)

If we take $A = P^\bullet$ and $B=Y[0]$ and $k\in\mathbb{Z}$ we see that the $\Hom$ complex collapses to $\Hom^k(P^{\bullet},Y[0]) = \Hom\nolimits_{\mathscr{A}}(P^{-k},Y)$ and computing its cohomology amounts to taking the cohomology of the complex $$ \dots \to 0 \to \Hom\nolimits_{\mathscr{A}}(P^0,Y) \xrightarrow{d^\ast} \Hom\nolimits_{\mathscr{A}}(P^{-1},Y) \xrightarrow{d^\ast} \Hom\nolimits_{\mathscr{A}}(P^{-2},Y) \xrightarrow{d^\ast} \cdots $$ which via $$ \operatorname{Ext}^k(X,Y) = \operatorname{Hom}_{D(A)}(X,Y[k]) \cong \operatorname{Hom}_{K(A)}(P^\bullet,Y[k]) = H^k(\operatorname{Hom}^\bullet(P^\bullet,Y[0])) $$ gives the identification you ask about.

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The next exercise would be to relate this to Matt's comment and for this I second the recommendation of consulting Hartshorne's residues and duality. The last chapter of Weibel's book also contains some useful things to know. –  none Nov 11 '12 at 16:57
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