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Suppose we are in an abelian category $\mathscr{A}$. Given a fixed monomorphism $A \overset{i}{\hookrightarrow} B$, and an object $C$, I would like to express concisely the notion of the group of maps $A \to C$ modulo those which extend to $B$. I think it should have a concise expression in the $\operatorname{Ext}$ functor.

Background: For any chain complex $A_\bullet \in \operatorname{Ch}_\bullet(\mathscr{A})$, if we define $H^q(A_\bullet, M) \overset{\text{def}}{=}H^q(\operatorname{Hom}(C_\bullet, M))$, then the Kronecker pairing $$ H^q(A_\bullet, M) \times H_q A_\bullet \to M $$ gives us an adjoint map $$H^q(A_\bullet, M) \xrightarrow{\alpha} \operatorname{Hom}(H_q A_\bullet, M).$$ The Universal Coefficient theorem tells us that if $\mathscr{A}$ is a category of modules over a PID, then $\alpha$ is surjective, with kernel $\operatorname{Ext}^1(H_{q-1}A_\bullet, M)$.

If we just let $\mathscr{A}$ be any abelian category, it seems that $$ \operatorname{Ker}\alpha = \frac{\operatorname{Hom}(A_q/\operatorname{Ker}\partial^q, M)}{q\text{-coboundaries}} $$ which I can see view as maps $\operatorname{Hom}(\operatorname{Img}\partial^q, M)$ modulo those which extend to all of $A_{q-1}$. The maps that extend to all of $A_{q-1}$ are those maps $\operatorname{Img}\partial^q \xrightarrow{f} M$ such that the class of the short exact sequence $\operatorname{Img}\partial^q \hookrightarrow A_{q-1} \twoheadrightarrow A_{q-1}/\operatorname{Img}\partial^q$ goes to zero in the map $$ \operatorname{Ext}(A_{q-1}/\operatorname{Img}\partial^q,\operatorname{Img}\partial^q) \xrightarrow{\operatorname{Ext}(A_{q-1}/\operatorname{Img}\partial^q,f)} \operatorname{Ext}(A_{q-1}/\operatorname{Img}\partial^q, M). $$ So I'm looking for a neater way to express $\operatorname{Ker}\alpha$ in terms of the $\operatorname{Ext}$ functor.

Eric Auld
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1 Answers1

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Any map $i : A \to B$, not necessarily a monomorphism, induces a precomposition map $\text{Hom}(B, C) \to \text{Hom}(A, C)$, and you want to take the cokernel of this map. If $i$ is a monomorphism, it fits into a short exact sequence

$$0 \to A \to B \to B/A \to 0$$

which induces an Ext long exact sequence

$$0 \to \text{Hom}(B/A, C) \to \text{Hom}(B, C) \to \text{Hom}(A, C) \to \text{Ext}^1(B/A, C) \to \text{Ext}^1(B, C) \to \dots$$

from which it follows that the desired cokernel is the image of the map $\text{Hom}(A, C) \to \text{Ext}^1(B/A, C)$, or equivalently the kernel of the map $\text{Ext}^1(B/A, C) \to \text{Ext}^1(B, C)$. If $\text{Ext}^1(B, C) = 0$ (e.g. if $B$ is projective or $C$ is injective) then it will be exactly $\text{Ext}^1(B/A, C)$.

Qiaochu Yuan
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  • Thanks! I think a key realization for me is that the map $\operatorname{Hom}(A,C) \to \operatorname{Ext}^1(B/A,C)$ can be described as $f \mapsto \tilde{f}(\omega)$, if $\omega$ is the element represented by the short exact sequence in $\operatorname{Ext}^1(B/A,A)$, and $\tilde{f}$ is the map $\operatorname{Ext}^1(B/A,A) \to \operatorname{Ext}^1(B/A,C)$ induced by $f$. I'm not sure the best way to demonstrate that, though... – Eric Auld Jul 14 '16 at 23:51