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Given an abelian category C, we can form the Yoneda extensions $YExt^i(X,Y)$ to the equivalent classes of $i$-extensions of X by Y. Given any abelian category C, we can always formulate the derived category D(C), and define $Ext^i_C(X,Y)$ to be $Hom_{D(C)}(X,Y[i])$.

Now we can naturally associate a Yoneda $i$-extension $$0\rightarrow Y\rightarrow Z_{i-1}\rightarrow \cdots\rightarrow Z_0\rightarrow X\rightarrow 0$$ of X by Y the element $Y[i]\leftarrow[Y\rightarrow Z_{i-1}\rightarrow \cdots\rightarrow Z_0]\rightarrow X$ of $Hom_{D(C)}(X, Y[i])$.

I am wondering when this map is inj (resp. surjective, bijective)?

We know this is an isomorphism under the assumption that C admits enough injective objects or projective objects.

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up vote 5 down vote accepted

It is true in general and Verdier did the computation in his thesis Des Catégories Dérivées des Catégories Abéliennes Chap III Sect 3 ... the thesis is on line!

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If we take $X$ and $Y$ in the abelian category $C$, and regard them as objects of the derived category of cochain compelexes (by placing them in degree 0), then $Hom_{D(C)}(X,Y[n]) = Ext^n(X,Y)$, where the left-hand side is Hom in the derived category from $X$ to a shift of $Y$, and the right-hand side is a Yoneda Ext group.

For this reason, when $A$ and $B$ are any two objects in the derived category, one write $Ext^n(A,B) = Hom(A,B[n]).$

Assuming that what I've written here is a faithful interpretation of your notation, then I can summarize things by saying that the answer to your question is yes.

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could you give a reference or a sketch of a proof of that result? – babubba Sep 25 '10 at 7:24
@John. For the category of modules, assuming you already know the bijection between elements of $Ext^n(A,B)$ and $n$-extensions of $A$ by $B$ (see Hilton-Stammbach, "A course in Homological Algebra" if not), you can find the rest of the bijection in the first chapter of Borel's "Algebraic D-modules" for instance. – a.r. Sep 26 '10 at 4:44
Dear Agusti, Thanks for giving the reference. I was also thinking of suggesting Borel's book, but couldn't remember whether this was treated there or not. – Emerton Sep 26 '10 at 5:24
Aren't you assuming that your category has enough projectives or injectives? – AFK Sep 26 '10 at 17:16
Dear YBL, I'm not sure if your question is directed at me or at Agusti, but if it is at me: I didn't think I was. I can't speak for Borel's book, because I haven't looked at it for years, but my impression is that the statement I made is true in general, via a direct consideration of complexes. – Emerton Sep 26 '10 at 21:43

You can look at F. Oort "Yoneda extensions in abelian categories" in Math Annalen.

He proves that the map is an isomorphism for artinian categories as the Yoneda Ext groups are the same in the category and in its category of pro-objects and the latter has enough projective objects.

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@YBL, I think the conclusion is that book is that: the Yoneda ext of X,Y in C are the same as the Yoneda extension of X and Y regarded as objects in the category of pro-objects. And the latter is the same as the EXT group defined via derived category, since the category of pro-objects admits enough projectives. So now the problems is that you still have the extension group of X by Y in the derived category of C itself. And I am also interested in the general situation. What Emerton said is quite surprising to me. I will check the reference given above. – Heer Sep 27 '10 at 15:03
Does the following help? Huber, Annette, Calculation of derived functors via Ind-categories. J. Pure Appl. Algebra 90 (1993), no. 1, 39-48. – mephisto May 11 '11 at 23:41
To mephisot: It's very helpful. Thanks a lot – Heer Jun 3 '11 at 14:37

For $A$ abelian artinian the higher Yoneda Ext $YExt^i(-,+)$ of objects $-,+\in A$ is the same as in $B = Pro (A)$. Now $A$ is thick in $B$ and yields $D^b(A)$ fully faithful in $D^b(B)$ given by those complexes with homology in $A$. Since $YExt^i_A(-,+) = Ext^i_B(-,+) = Hom_{D^b(B)}(-,+[i]) = Hom_{D^b(A)}(-,+[i])$ we get $YExt^i_A(-,+) = Hom_D^b(A)(-,+[i])$ for such an Artinian category $A$.

Similarly for $A$ abelian Noetherian using $Ind (A)$.

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