Finite module duality is exact with commuting bimodule actions

For $\lambda \in X^{\ast }$ and $a\in A$ define $\lambda \cdot a$ by $(\lambda \cdot a)(x)=\lambda (ax)$; this is $k$-linear in $x$ because $x\mapsto ax$ is $k$-linear and $\lambda$ is, and the right-module axioms hold: $(\lambda +\mu )\cdot a=\lambda \cdot a+\mu \cdot a$ and $\lambda \cdot (a+a^{\mathrm{\prime }})=\lambda \cdot a+\lambda \cdot a^{\mathrm{\prime }}$ by linearity of $\lambda$ and additivity of the action of $A$, while $(\lambda \cdot a)\cdot a^{\mathrm{\prime }}=\lambda \cdot (a{a}^{\mathrm{\prime }})$ because both sides send $x$ to $\lambda (a(a^{\mathrm{\prime }}x))$, and $\lambda \cdot 1=\lambda$. Hence $X^{\ast }$ is a right $A$-module, equivalently by [F3] a left $A^{\mathrm{o}\mathrm{p}}$-module, and [F2] gives ${\dim }_k{X}^{\ast }={\dim }_{k}X<\mathrm{\infty }$, so $X^{\ast }$ is a finite-dimensional left $A^{\mathrm{o}\mathrm{p}}$-module.

For $A$-linear $u:X\to Y$ the map $u^{\ast }:Y^{\ast }\to X^{\ast }$, $u^{\ast }(\lambda )=\lambda \circ u$, is $k$-linear, since composition with the $k$-linear $u$ is additive and $k$-homogeneous by [L1]; moreover $(1_X{)}^{\ast }=1_{X^{\ast }}$ and $(v\circ u{)}^{\ast }=u^{\ast }\circ v^{\ast }$ for composable $A$-linear maps, while $u\mapsto u^{\ast }$ is additive and $k$-homogeneous, because $(u+v{)}^{\ast }(\lambda )=\lambda \circ (u+v)=\lambda \circ u+\lambda \circ v$ and $(cu{)}^{\ast }=c{u}^{\ast }$ for $c\in k$ by [L1].

Let $0\to X\stackrel{i}{\to }Y\stackrel{q}{\to }Z\to 0$ be a short exact sequence of finite-dimensional left $A$-modules. Then $i^{\ast }\circ q^{\ast }=0$, since $(i^{\ast }\circ q^{\ast })(\lambda )=i^{\ast }(\lambda \circ q)=\lambda \circ q\circ i$ and $q\circ i=0$ by [L3]; and $q^{\ast }$ is injective, since if $q^{\ast }(\lambda )=\lambda \circ q=0$ then $\lambda$ vanishes on $\mathrm{im}q=Z$ by surjectivity of $q$ from [L3], so $\lambda =0$. Thus $0\to Z^{\ast }\stackrel{q^{\ast }}{\to }{Y}^{\ast }\stackrel{i^{\ast }}{\to }{X}^{\ast }$ is exact at $Z^{\ast }$ and a complex at $Y^{\ast }$.

The map $i^{\ast }$ is surjective: let $(x_1,\dots ,x_m)$ be an ordered basis of the finite-dimensional space $X$; since $i$ is injective by [L3], the images $i(x_1),\dots ,i(x_m)$ are linearly independent and by [L4] are contained in an ordered basis $(y_1,\dots ,y_n)$ of $Y$ with $y_j=i(x_j)$ for $j\le m$; let $(y_1^{\ast },\dots ,y_n^{\ast })$ be its dual basis of $Y^{\ast }$ by [F2]. Given $\lambda ={\sum }_{j\le m}{c}_j{x}_j^{\ast }$ in $X^{\ast }$ with the dual basis $(x_j^{\ast })$ of [F2], put $\mu ={\sum }_{j\le m}{c}_j{y}_j^{\ast }\in Y^{\ast }$; then $i^{\ast }(\mu )(x_r)=\mu (i(x_r))=\mu (y_r)=c_r=\lambda (x_r)$ for every $r\le m$, so $i^{\ast }(\mu )=\lambda$ by [L5]. Hence $\mathrm{im}{i}^{\ast }=X^{\ast }$.

For an $(A,B)$-bimodule $X$ define $(b\cdot \lambda )(x)=\lambda (xb)$ and $(\lambda \cdot a)(x)=\lambda (ax)$ for $a\in A$, $b\in B$, $\lambda \in X^{\ast }$, $x\in X$; as in step 1.1 both are $k$-linear functionals and the module axioms hold for the left $B$- and right $A$-actions, so $X^{\ast }$ is at once a left $B$-module and a right $A$-module. The two actions commute: $((b\cdot \lambda )\cdot a)(x)=(b\cdot \lambda )(ax)=\lambda ((ax)b)=\lambda (a(xb))=(\lambda \cdot a)(xb)=(b\cdot (\lambda \cdot a))(x)$, using the bimodule identity $(ax)b=a(xb)$ of [L6]; hence $X^{\ast }$ is a $(B,A)$-bimodule.

In the situation of step 1.3, ${\dim }_{k}Y={\dim }_{k}X+{\dim }_{k}Z$: rank-nullity [L2] applied to the $k$-linear $q$ gives ${\dim }_{k}Y={\dim }_k(\ker q)+{\dim }_k(\mathrm{im}q)$, and $\ker q=\mathrm{im}i$ with $i$ injective and $q$ surjective by [L3], so ${\dim }_k(\ker q)={\dim }_{k}X$ and ${\dim }_k(\mathrm{im}q)={\dim }_{k}Z$.

The map $u^{\ast }$ is $A^{\mathrm{o}\mathrm{p}}$-linear: for $\lambda \in Y^{\ast }$ and $a\in A$, $u^{\ast }(\lambda \cdot a)(x)=(\lambda \cdot a)(u(x))=\lambda (au(x))=\lambda (u(ax))=(u^{\ast }\lambda )(ax)=(u^{\ast }(\lambda )\cdot a)(x)$ for all $x$, using the action of step 1.1 and $A$-linearity of $u$. Consequently, with step 1.1 for objects and step 1.2 for the morphism assignment, identities, composition and $k$-linearity on hom-spaces, $(-{)}^{\ast }$ is a contravariant $k$-linear functor from $A\text{-}\mathrm{m}\mathrm{o}\mathrm{d}$ to finite-dimensional left $A^{\mathrm{o}\mathrm{p}}$-modules.

Applying step 1.1 with the unital algebra $A^{\mathrm{o}\mathrm{p}}$ in place of $A$, the dual $X^{\ast \ast }$ of the finite-dimensional left $A^{\mathrm{o}\mathrm{p}}$-module $X^{\ast }$ is a finite-dimensional left $(A^{\mathrm{o}\mathrm{p}}{)}^{\mathrm{o}\mathrm{p}}=A$-module, and ${\mathrm{ev}}_X:X\to X^{\ast \ast }$, ${\mathrm{ev}}_X(x)(\lambda )=\lambda (x)$, is $k$-linear; it is $A$-linear because ${\mathrm{ev}}_X(ax)(\lambda )=\lambda (ax)=(\lambda \cdot a)(x)={\mathrm{ev}}_X(x)(\lambda \cdot a)=(a\cdot {\mathrm{ev}}_X(x))(\lambda )$ for the left action on $X^{\ast \ast }$ induced by step 1.1. Choosing an ordered basis $(x_1,\dots ,x_m)$ of $X$ with dual basis $(x_1^{\ast },\dots ,x_m^{\ast })$ of $X^{\ast }$, the dual family $({\epsilon }_1,\dots ,{\epsilon }_m)$ of $(x_1^{\ast },\dots ,x_m^{\ast })$ is a basis of $X^{\ast \ast }$ by [F2], and ${\mathrm{ev}}_X(x_i)(x_j^{\ast })={\delta }_{ij}={\epsilon }_i(x_j^{\ast })$ for all $i,j$; hence ${\mathrm{ev}}_X({\sum }_i{a}_i{x}_i)={\sum }_i{a}_i{\epsilon }_i$, which is zero only for the zero combination and realizes every element of $X^{\ast \ast }$, so ${\mathrm{ev}}_X$ is a $k$-linear isomorphism.

By rank-nullity [L2] applied to $i^{\ast }$ and $q^{\ast }$, and [F2] together with steps 1.3, 1.4 and 2.2: ${\dim }_k(\ker i^{\ast })={\dim }_k{Y}^{\ast }-{\dim }_k(\mathrm{im}{i}^{\ast })={\dim }_{k}Y-{\dim }_{k}X$, while ${\dim }_k(\mathrm{im}{q}^{\ast })={\dim }_k{Z}^{\ast }-{\dim }_k(\ker q^{\ast })={\dim }_{k}Z$; hence by step 2.2 both quantities equal ${\dim }_{k}Z$.

A map $u:X\to Y$ of $(A,B)$-bimodules, that is $u(axb)=au(x)b$ for all $a\in A$, $b\in B$, $x\in X$, has $u^{\ast }$ a map of $(B,A)$-bimodules: $u^{\ast }(b\cdot \lambda )(x)=(b\cdot \lambda )(u(x))=\lambda (u(x)b)=\lambda (u(xb))=(u^{\ast }\lambda )(xb)=(b\cdot (u^{\ast }\lambda ))(x)$ and $u^{\ast }(\lambda \cdot a)(x)=(\lambda \cdot a)(u(x))=\lambda (au(x))=\lambda (u(ax))=(u^{\ast }\lambda )(ax)=((u^{\ast }\lambda )\cdot a)(x)$ for all $x$, using the actions of step 2.1; with step 1.2 the assignment is functorial on the bimodule categories.

For $A$-linear $u:X\to Y$, $x\in X$ and $\lambda \in Y^{\ast }$ one has $(u^{\ast \ast }\circ {\mathrm{ev}}_X)(x)(\lambda )={\mathrm{ev}}_X(x)(u^{\ast }\lambda )=u^{\ast }(\lambda )(x)=\lambda (u(x))={\mathrm{ev}}_Y(u(x))(\lambda )$, so ${\mathrm{ev}}_Y\circ u=u^{\ast \ast }\circ {\mathrm{ev}}_X$, where $u^{\ast \ast }=(u^{\ast }{)}^{\ast }$ is the map of step 2.3; hence the isomorphisms ${\mathrm{ev}}_X$ of step 2.4 form a natural isomorphism $1\Rightarrow (-{)}^{\ast \ast }$. Therefore $(-{)}^{\ast }$ is a contravariant equivalence of categories with quasi-inverse $(-{)}^{\ast }$, since both composites are $(-{)}^{\ast \ast }$ and are naturally isomorphic to the identities, which proves (i).

In the situation of step 1.3, $\mathrm{im}{q}^{\ast }\subseteq \ker i^{\ast }$ because $i^{\ast }\circ q^{\ast }=0$, and by step 3.1 both are $k$-subspaces of $Y^{\ast }$ of dimension ${\dim }_{k}Z$; since a subspace of the same finite dimension equals the whole space, $\mathrm{im}{q}^{\ast }=\ker i^{\ast }$. With step 1.3 the dual sequence $0\to Z^{\ast }\stackrel{q^{\ast }}{\to }{Y}^{\ast }\stackrel{i^{\ast }}{\to }{X}^{\ast }\to 0$ is exact, which proves (ii).

For an $(A,B)$-bimodule $X$ the evaluation of step 2.4 is also $B$-linear on the right: ${\mathrm{ev}}_X(xb)(\lambda )=\lambda (xb)=(b\cdot \lambda )(x)={\mathrm{ev}}_X(x)(b\cdot \lambda )=({\mathrm{ev}}_X(x)\cdot b)(\lambda )$ for all $b\in B$, $\lambda \in X^{\ast }$, where the right $B$-action on $X^{\ast \ast }$ is the one induced by the left $B$-action on $X^{\ast }$ of step 2.1; so ${\mathrm{ev}}_X$ is a map of $(A,B)$-bimodules, and it is natural in the bimodule variable by the computation of step 3.3 applied to bimodule maps. By steps 3.2 and 3.3 the restriction of $(-{)}^{\ast }$ to finite-dimensional bimodules is a contravariant equivalence between finite-dimensional $(A,B)$-bimodules and finite-dimensional $(B,A)$-bimodules with quasi-inverse $(-{)}^{\ast }$, which proves (iii).

Steps 3.3, 4.1 and 4.2 prove (i), (ii) and (iii) respectively. The only choices made are finite bases, dual bases and basis extensions in finite-dimensional spaces, supplied without any choice principle by [F2] and [L4], so no choice is used.