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

Comment #8785 by on

Is it proven somewhere in the Stacks Project that this definition for $\operatorname{Ext}^i(A,B)$, $A,B\in\mathcal{A}$ indeed yields the $i$-th derived functor of $\operatorname{Hom}$? 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 $\delta$-functor be universal?

Comment #8829 by on

@#8785 Okay, I just realized: if $\mathcal{A}$ has enough injectives, then the $\delta$-functor $(\operatorname{Ext}_\mathcal{A}^i(A,-))_{i\geq 0}$ is universal because it is erasable, i.e., it satisfies the hypothesis of 12.12.4: For an object $B\in\mathcal{A}$, take an injection $B\to I$ into an injective object. Then $\operatorname{Ext}_\mathcal{A}^i(A,I)$ vanishes for all $i>0$ by 13.18.8. Dually, if $\mathcal{A}$ has enough projectives, one can apply the analogous argument to $(\operatorname{Ext}_\mathcal{A}^i(-,B))_{i\geq 0}$ 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 $\mathcal{A}$ has enough injectives, this is immediate from Lemma 13.27.2 and the definition of a derived functor. In general the $i$th derived functors may not be defined, but the Ext's are always defined.

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