Finite left exact functors are Hom functors with dual bimodule kernels

The $k$-dual $A^{\ast }$ with the actions $(a\cdot \lambda )(x)=\lambda (xa)$ and $(\lambda \cdot a)(x)=\lambda (ax)$ is a finite-dimensional $(A,A)$-bimodule (the actions commute by associativity of multiplication in $A$), and $\lambda \mapsto \lambda \cdot a$ is left $A$-linear for each $a$, since $((b\cdot \lambda )\cdot a)(x)=\lambda (axb)=(b\cdot (\lambda \cdot a))(x)$. Hence $F$ is defined on $A^{\ast }$ and, by [F3] applied to the bimodule $A^{\ast }$ in place of the regular module, $M=F(A^{\ast })$ is a finite-dimensional $(B,A)$-bimodule, the right action being $m\cdot a=F(\lambda \mapsto \lambda \cdot a)(m)$.

Define $F^{\mathrm{d}}(Y)=F(Y^{\ast }{)}^{\ast }$ for finite-dimensional left $A^{\mathrm{o}\mathrm{p}}$-modules $Y$ (equivalently right $A$-modules): $Y^{\ast }$ is a finite-dimensional left $A$-module by [F1], so $F(Y^{\ast })$ is in $B\text{-}\mathrm{m}\mathrm{o}\mathrm{d}$ and its dual is a left $B^{\mathrm{o}\mathrm{p}}$-module, so $F^{\mathrm{d}}$ is a functor $A^{\mathrm{o}\mathrm{p}}\text{-}\mathrm{m}\mathrm{o}\mathrm{d}\to B^{\mathrm{o}\mathrm{p}}\text{-}\mathrm{m}\mathrm{o}\mathrm{d}$, $k$-linear by [F1]. It is right exact: if $Y_1\to Y_2\to Y_3\to 0$ is exact, then dualizing gives the exact sequence $0\to Y_3^{\ast }\to Y_2^{\ast }\to Y_1^{\ast }$ by exactness of the duality [F1], left exactness of $F$ gives $0\to F(Y_3^{\ast })\to F(Y_2^{\ast })\to F(Y_1^{\ast })$, and dualizing again gives the exact sequence $F^{\mathrm{d}}(Y_1)\to F^{\mathrm{d}}(Y_2)\to F^{\mathrm{d}}(Y_3)\to 0$.

By [F2] applied to the finite-dimensional algebras $A^{\mathrm{o}\mathrm{p}}$ and $B^{\mathrm{o}\mathrm{p}}$, the right exact $k$-linear functor $F^{\mathrm{d}}$ is naturally isomorphic to $K{\otimes }_{A^{\mathrm{o}\mathrm{p}}}-$ with kernel $K=F^{\mathrm{d}}(A^{\mathrm{o}\mathrm{p}})$, a finite-dimensional $(B^{\mathrm{o}\mathrm{p}},A^{\mathrm{o}\mathrm{p}})$-bimodule. The dual of the regular left $A^{\mathrm{o}\mathrm{p}}$-module is $A^{\ast }$ with the left $A$-action $(a\cdot \lambda )(x)=\lambda (xa)$ of step 1.1, so $K=F(A^{\ast }{)}^{\ast }=M^{\ast }$ under the identifications of [F1]; thus $K$ is the $(A,B)$-bimodule dual of $M$.

For $X\in A\text{-}\mathrm{m}\mathrm{o}\mathrm{d}$ double duality of [F1] gives $F(X)\cong F(X^{\ast \ast })\cong F^{\mathrm{d}}(X^{\ast }{)}^{\ast }$, and step 2.1 gives $F^{\mathrm{d}}(X^{\ast })\cong K{\otimes }_{A^{\mathrm{o}\mathrm{p}}}{X}^{\ast }$; hence $F(X)\cong (K{\otimes }_{A^{\mathrm{o}\mathrm{p}}}{X}^{\ast }{)}^{\ast }$, naturally in $X$.

There is a natural left $B$-module isomorphism $(K{\otimes }_{A^{\mathrm{o}\mathrm{p}}}{X}^{\ast }{)}^{\ast }\cong {\mathrm{Hom}}_A(M^{\ast },X)$. Here $K=M^{\ast }$ is a left $A$-module and right $B$-module, and is regarded as a right $A^{\mathrm{o}\mathrm{p}}$-module by $u\cdot a^{\mathrm{o}\mathrm{p}}=au$; $X^{\ast }$ is a left $A^{\mathrm{o}\mathrm{p}}$-module by $a^{\mathrm{o}\mathrm{p}}\lambda =\lambda \cdot a$. For $A$-linear $\phi :M^{\ast }\to X$, the pairing $u\otimes \lambda \mapsto \lambda (\phi (u))$ is balanced since $\lambda (\phi (au))=\lambda (a\phi (u))=(\lambda \cdot a)(\phi (u))$. Conversely a functional $\omega$ defines $u\mapsto [\lambda \mapsto \omega (u\otimes \lambda )]$ into $X^{\ast \ast }\cong X$, and balancing makes this map $A$-linear. These constructions are inverse, $k$-linear and natural in $X$. The tensor product carries a right $B$-action $(u\otimes \lambda )b=(ub)\otimes \lambda$, so its dual has left action $(b\omega )(u\otimes \lambda )=\omega (ub\otimes \lambda )$. This corresponds to $(b\phi )(u)=\phi (ub)$, proving $B$-linearity.

Combining steps 3.1 and 3.2 gives a natural isomorphism $F(X)\cong {\mathrm{Hom}}_A(M^{\ast },X)$ of left $B$-modules for every finite-dimensional left $A$-module $X$, with $M=F(A^{\ast })$ a finite-dimensional $(B,A)$-bimodule by step 1.1.

(Equivalence on hom-categories.) The assignment lands in $k$-linear left exact functors by [F7]. For finite-dimensional $(B,A)$-bimodules $M,N$, a natural transformation $\eta :{\mathrm{Hom}}_A(M^{\ast },-)\Rightarrow {\mathrm{Hom}}_A(N^{\ast },-)$ with $B$-linear components corresponds by the Yoneda computation [F6] to the map $f={\eta }_{M^{\ast }}(1_{M^{\ast }}):N^{\ast }\to M^{\ast }$, which is $A$-linear by construction and right $B$-linear: naturality at $r_b:M^{\ast }\to M^{\ast }$ gives $r_b^{M^{\ast }}f={\eta }_{M^{\ast }}(r_b^{M^{\ast }})$, while $B$-linearity of ${\eta }_{M^{\ast }}$ gives ${\eta }_{M^{\ast }}(r_b^{M^{\ast }})=bf=f{r}_b^{N^{\ast }}$; conversely every $(A,B)$-bimodule map $N^{\ast }\to M^{\ast }$ gives such a natural transformation by precomposition. By the bimodule duality [F1] these correspond bijectively to $(B,A)$-bimodule maps $M\to N$; so the assignment $M\mapsto {\mathrm{Hom}}_A(M^{\ast },-)$ is full and faithful. It is essentially surjective by step 4.1: every $k$-linear left exact $F$ is naturally isomorphic to ${\mathrm{Hom}}_A(F(A^{\ast }{)}^{\ast },-)$ with $F(A^{\ast })$ a $(B,A)$-bimodule. Hence the assignment is an equivalence of categories, and it has quasi-inverse $F\mapsto F(A^{\ast })$: on objects this returns $F$ up to the isomorphism of step 4.1, and on morphisms it sends a natural transformation to its component at $A^{\ast }$, a $(B,A)$-bimodule map by naturality against the right-action maps of $A^{\ast }$. The other composite is naturally isomorphic to $M$: ${\mathrm{Hom}}_A(M^{\ast },A^{\ast })\cong M$ sends $\phi$ to the unique $m$ with $u(m)=\phi (u)(1)$ for every $u\in M^{\ast }$, using double duality. Its inverse sends $m$ to $u\mapsto [a\mapsto u(ma)]$; these formulas respect both actions and are natural in $M$.

Steps 1.1, 4.1 and 5.1 prove the statement: $M=F(A^{\ast })$ is a finite-dimensional $(B,A)$-bimodule, $F(X)\cong {\mathrm{Hom}}_A(M^{\ast },X)$ naturally in $X$, and $M\mapsto {\mathrm{Hom}}_A(M^{\ast },-)$ is an equivalence with quasi-inverse $F\mapsto F(A^{\ast })$. No commutativity of $A$ or $B$ was used, and all dualities, tensor products and presentations involved are finite-dimensional, so no choice is used.