The Stacks project

Definition 13.27.1. Let $\mathcal{A}$ be an abelian category. Let $i \in \mathbf{Z}$. Let $X, Y$ be objects of $D(\mathcal{A})$. The $i$th extension group of $X$ by $Y$ is the group

\[ \mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {A}(X, Y) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(X, Y[i]) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(X[-i], Y). \]

If $A, B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ we set $\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {A}(A, B) = \text{Ext}^ i_\mathcal {A}(A[0], B[0])$.


Comments (3)

Comment #8785 by on

Is it proven somewhere in the Stacks Project that this definition for , indeed yields the -th derived functor of ? I cannot find it neither in 13.27 nor with the searcher.

Right after 13.27.1 it is commented that we have the expected long exact sequence, but why should this -functor be universal?

Comment #8829 by on

@#8785 Okay, I just realized: if has enough injectives, then the -functor is universal because it is erasable, i.e., it satisfies the hypothesis of 12.12.4: For an object , take an injection into an injective object. Then vanishes for all by 13.18.8. Dually, if has enough projectives, one can apply the analogous argument to to deduce erasibility (and thus universality).

Is the statement of this result to be found anywhere in the SP?

(Please, erase #8828, I posted it in the wrong tag.)

Comment #9257 by on

@#8785 Provided has enough injectives, this is immediate from Lemma 13.27.2 and the definition of a derived functor. In general the th derived functors may not be defined, but the Ext's are always defined.


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