# On equivalent definitions of Ext

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?

• 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
Commented Sep 16, 2012 at 14:26
• 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.
Commented Sep 16, 2012 at 16:02
• 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.
Commented Sep 16, 2012 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
Commented Sep 16, 2012 at 16:37
• @Matt: sorry Matt, I'm not so comfortable with what you said.
– Mec
Commented Sep 17, 2012 at 0:22

$\DeclareMathOperator{Hom}{Hom}$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.

• 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
Commented Nov 11, 2012 at 16:57

$$\def\A{\mathcal{A}} \def\Ext{\operatorname{Ext}}$$Here's the abstract-nonsense approach: one could verify both definitions you gave for $$\Ext^i_\A(X,-)$$ satisfy the same universal property. Namely, they both constitute a $$\delta$$-functor that is universal (morphisms of $$\delta$$-functors are defined here).

There is a sufficient condition that a $$\delta$$-functor can satisfy for it being universal. Namely, we say that a $$\delta$$-functor $$(F^i)_{i\geq 0}$$ is erasable (or effaceable, after Grothendieck coined effaçable in French in his Tôhoku paper) if for each $$Y\in\operatorname{ob}\mathcal{A}$$ and $$i>0$$ there is an injection $$Y\to Z$$ such that $$F^iZ=0$$. In Proposition 2.2.1 of the Tôhoku paper, Grothendieck showed that erasability implies universality (in the SP this is 010T, and the result has been discussed several times in MSE, e.g. 1, 2).

Suppose then $$\mathcal{A}$$ has enough injectives (so that we have a definition of $$\Ext$$ via injective resolutions). It is trivial to show that the definition of the $$\Ext$$ functors via injective resolutions, as a $$\delta$$-functor, is erasable. On the other hand, to see that $$(\operatorname{Hom}_{D(\mathcal{A})}(X,(-)[i]))_{i\geq 0}$$ is erasable, consider an object $$Y$$ of $$\mathcal{A}$$ and take an injection $$Y\to I$$ into an injective object. For $$i>0$$ and by 05TG,

\begin{align*} \operatorname{Hom}_{D(\mathcal{A})}(X,I[i]) =\operatorname{Hom}_{K(\mathcal{A})}(X,I[i]) =0. \end{align*}

Dually, one shows that the contravariant $$\delta$$-functor $$(\operatorname{Hom}_{D(\mathcal{A})}(-,Y[i]))_{i\geq 0}$$ and the definition of $$\Ext$$ via projective resolutions are both coerasable.