$H^r(Hom(A,I^\bullet))\cong H^r(Hom(P_\bullet,B))$ in the definition of Ext without the condition of having enough injectives or enough projectives?

Question

Let $\mathcal{C}$ be an abelian category.

$\bullet$ Let $A\in\mathcal{C}$. If $\mathcal{C}$ has enough injectives, the functor $\operatorname{Ext}^r(A,\cdot)$ is defined by choosing an injective resolution $B\to I^\bullet$ of $B$ and set $$ \operatorname{Ext}^r(A,B) := H^r(\operatorname{Hom}(A,I^\bullet)). $$

$\bullet$ Let $B\in\mathcal{C}$. If $\mathcal{C}$ has enough projectives, the functor $\operatorname{Ext}^r(\cdot,B)$ is defined by choosing a projective resolution $P_\bullet\to A$ of $A$ and set $$ \operatorname{Ext}^r(A,B) := H^r(\operatorname{Hom}(P_\bullet,B)). $$

The key point is that, if $\mathcal{C}$ has enough injectives and enough projectives, then the two definitions of $\operatorname{Ext}^r(A,B)$ coincide. The following is the proof given in J. S. Milne's note Class Field Theory, Appendix II.A, Proposition A.$13$:

However, it seems that this proof also works if $\mathcal{C}$ has only enough injectives or only enough projectives, but not both.

So I was wondering if this condition is even necessary, namely:

Let $\mathcal{C}$ be an abelian category in general, and $A,B\in\mathcal{C}$. Given an injective resolution $B\to I^\bullet$ of $B$ and a projective resolution $P_\bullet\to A$ of $A$, do we have $$ H^r(\operatorname{Hom}(A,I^\bullet))\cong H^r(\operatorname{Hom}(P_\bullet,B)) $$ in the category of abelian groups?

Thank you in advance, and please kindly forgive me if this turns out to be a silly question or a duplicate (I apologize if this is the case).

Answer 1

Without enough injectives or projectives, you need to be very careful with the definition of derived functor. With the right definition, you can prove that:

• $\textrm{Ext}^n (A, B)$ classifies Yoneda extensions of $A$ by $B$ of degree $n$.

• $\textrm{Ext}^*$ is (the cohomology of) the right derived functor of $\textrm{Hom}$, in both variables jointly as well as in each variable singly.

• If $P$ is a projective resolution of $A$ then $\textrm{Ext}^* (A, B)$ can be computed as $\mathrm{H}^* (\textrm{Hom} (P, B))$.

• If $I$ is an injective resolution of $B$ then $\textrm{Ext}^* (A, B)$ can be computed as $\mathrm{H}^* (\textrm{Hom} (A, I))$.

So what is such a definition? Unfortunately, it is quite involved.

Let $\mathcal{A}$ be an abelian category, let $\textbf{Ch} (\mathcal{A})$ be the category of unbounded chain complexes in $\mathcal{A}$, and let $\mathbf{D} (\mathcal{A})$ be the derived category, i.e. the localisation of $\textbf{Ch} (\mathcal{A})$ with respect to quasi-isomorphisms. This can be constructed somewhat more concretely as a category of fractions in the chain homotopy category $\mathbf{K} (\mathcal{A})$, i.e. $\textbf{Ch} (\mathcal{A})$ modulo chain homotopy, which is convenient for technical purposes.

Abusing notation, given an object $A$ in $\mathcal{A}$, we also write $A$ for the chain complex in $\mathcal{A}$ consisting of $A$ in degree $0$ and $0$ elsewhere. Given a chain complex $C$ in $\mathcal{A}$ and an integer $n$, we write $\Omega^n C$ for the shifted chain complex with $(\Omega^n C)_m = C_{n+m}$. (I use homological indexing.) Thus, given an object $B$ in $\mathcal{A}$, $\Omega^{-n} B$ is the chain complex with $B$ in degree $n$ and $0$ elsewhere.

Given objects $A$ and $B$ in $\mathcal{A}$, we now define: $$\textrm{Ext}^n_\mathcal{A} (A, B) = \textrm{Hom}_{\mathbf{D} (\mathcal{A})} (A, \Omega^{-n} B)$$ Equivalently: $$\textrm{Ext}^n_\mathcal{A} (A, B) = \textrm{Hom}_{\mathbf{D} (\mathcal{A})} (\Omega^n A, B)$$ The elements of $\textrm{Ext}^n_\mathcal{A} (A, B)$ are certain equivalence classes of diagrams of the form below, $$\require{AMScd} \begin{CD} \cdots @>>> 0 @>>> 0 @>>> \cdots @>>> A @>>> 0 @>>> \cdots \\ & @AAA @AAA & & @AAA @AAA \\ \cdots @>>> C_{n+1} @>>> C_n @>>> \cdots @>>> C_0 @>>> C_{-1} @>>> \cdots \\ & @VVV @VVV & & @VVV @VVV \\ \cdots @>>> 0 @>>> B @>>> \cdots @>>> 0 @>>> 0 @>>> \cdots \end{CD}$$ where the rows are chain complexes in $\mathcal{A}$, $C \to A$ is a quasi-isomorphism in $\textbf{Ch} (\mathcal{A})$, and $C \to \Omega^{-n} B$ is any morphism in $\textbf{Ch} (\mathcal{A})$. Given a projective resolution $P \to A$ and injective resolution $B \to I$, we have $$\textrm{Ext}^n (A, B) \cong \textrm{Hom}_{\mathbf{D} (\mathcal{A})} (P, \Omega^{-n} B) \cong \textrm{Hom}_{\mathbf{D} (\mathcal{A})} (P, \Omega^{-n} I) \cong \textrm{Hom}_{\mathbf{D} (\mathcal{A})} (\Omega^n P, I) \cong \textrm{Hom}_{\mathbf{D} (\mathcal{A})} (\Omega^n A, I)$$ because quasi-isomorphisms in $\textbf{Ch} (\mathcal{A})$ become isomorphisms in $\mathbf{D} (\mathcal{A})$. Furthermore:

• For any bounded below chain complex of projectives $P$ in $\mathcal{A}$, $\textrm{Hom}_{\mathbf{K} (\mathcal{A})} (P, -)$ sends quasi-isomorphisms to isomorphisms, so (abusing notation and omitting details) $\textrm{Hom}_{\mathbf{D} (\mathcal{A})} (P, -) \cong \textrm{Hom}_{\mathbf{K} (\mathcal{A})} (P, -)$; in particular, for any object $B$ in $\mathcal{A}$, $\textrm{Hom}_{\mathbf{D} (\mathcal{A})} (P, B) \cong \mathrm{H}^0 (\textrm{Hom}_\mathcal{A} (P, B))$.

• For any bounded above chain complex of injectives $I$ in $\mathcal{A}$, $\textrm{Hom}_{\mathbf{K} (\mathcal{A})} (-, I)$ sends quasi-isomorphisms to isomorphisms, so (abusing notation and omitting details) $\textrm{Hom}_{\mathbf{D} (\mathcal{A})} (-, I) \cong \textrm{Hom}_{\mathbf{K} (\mathcal{A})} (-, I)$; in particular, for any object $A$ in $\mathcal{A}$, $\textrm{Hom}_{\mathbf{D} (\mathcal{A})} (A, I) \cong \mathrm{H}_0 (\textrm{Hom}_\mathcal{A} (A, I))$.

See e.g. Stacks or any modern homological algebra textbook. The only point that is not well explained in the literature is the notion of a partial derived functor; you can refer here instead.