standard duality on modules over algebras
Let $k$ be a field and let $A$ be an associative unital algebra. Throughout we will assume that all $A$modules over $k$ are unital. If $M$ is a right $A$module, then the space of all linear mappings
$${\mathrm{Hom}}_{k}(M,k)$$ 
can be equipped with a left $A$module structure^{} as follows: for any $f\in {\mathrm{Hom}}_{k}(M,k)$ and $a\in A$ put
$$(af)(x)=f(xa).$$ 
Note that action direction need to be reversed, because
$$(abf)(x)=(bf)(xa)=f(xab).$$ 
Analogously ${\mathrm{Hom}}_{k}(,k)$ takes left $A$modules to right $A$modules. Also this action is compatible with functoriality of ${\mathrm{Hom}}_{k}(,k)$, which means that it takes $A$homomorphisms^{} to $A$homomorphisms. In particular we obtain a (contravariant) functor^{} from category^{} of left (right) $A$modules to category of right (left) $A$modules. Obviously $\mathrm{Hom}$ does not change the dimension of spaces, so we have well defined functors
$$D:\mathrm{mod}A\to A\mathrm{mod}$$ 
$$D:A\mathrm{mod}\to \mathrm{mod}A$$ 
which are restrictions^{} of $\mathrm{Hom}$ (here $\mathrm{mod}$ means finite dimensional modules left/right modules) and are known in literature as ,,standard dualities”.
Proposition^{}. Both $D$’s are quasi inverse^{} dualities of categories.
Proof. Let $M$ be a finite dimensional $A$module. We need to define a natural isomorphism between $M$ and $DD(M)$. Indeed, define
$${\tau}_{M}:M\to DD(M);$$ 
$${\tau}_{M}(m)(\alpha )=\alpha (m).$$ 
We will show that each $\tau $ is an isomorphism^{}.

1.
First we will show that $\tau $ is a monomorphism^{}. Assume that ${\tau}_{M}(m)=0$ for nonzero $m\in M$. This is if and only if $\alpha (m)=0$ for every linear mapping $\alpha :M\to k$. But $m$ is nonzero, so there is a basis of $M$ (as linear space) which contains $m$. In particular there is a linear mapping $f:M\to k$ such that $f(m)=1$. Contradiction^{}. Thus $m=0$, which completes^{} this part.

2.
$\tau $ is an epimorphism^{}. Indeed, let $F:D(M)\to k$ be a linear mapping. We need to show, that there is $m\in M$ such that
$$F(\alpha )=\alpha (m)$$ for any $\alpha \in D(M)$. Since $M$ is finite dimensional, then let $\{{e}_{1},\mathrm{\dots},{e}_{n}\}$ be a $k$basis of $M$. Of course $\{{e}_{1}^{*},\mathrm{\dots},{e}_{n}^{*}\}$ is a basis of $D(M)$, where ${e}_{i}^{*}$ is given by ${e}_{i}^{*}({e}_{j})=1$ if $i=j$ and ${e}_{i}^{*}({e}_{j})=0$ otherwise. Define
$${\lambda}_{i}=F({e}_{i}^{*})$$ and put
$$m=\sum _{i=1}^{n}{\lambda}_{i}\cdot {e}_{i}.$$ We leave it as a simple exercise, that $\tau (m)=F$.
What remains is to prove, that $\tau $ is natural. Consider an $A$homomorphism $f:X\to Y$. We need to show that the following diagram commutes:
$$\text{xymatrix}X\text{ar}{[r]}^{f}\text{ar}{[d]}_{{\tau}_{X}}\mathrm{\&}Y\text{ar}{[d]}^{{\tau}_{Y}}DD(X)\text{ar}{[r]}^{DD(f)}\mathrm{\&}DD(Y)$$ 
Indeed, if $x\in X$, then let $F={\tau}_{X}(x)$. We have that
$$DD(f)(F)=F\circ D(f)$$ 
and evaluating this at $\alpha \in D(M)$ we have
$$F(D(f)(\alpha ))=F(\alpha \circ f)=(\alpha \circ f)(x)=\alpha (f(x))={\tau}_{Y}(f(x))(\alpha ).$$ 
In particular we obtain that
$$DD(f)({\tau}_{X}(x))={\tau}_{Y}(f(x))$$ 
which means that
$$DD(f)\circ {\tau}_{X}={\tau}_{Y}\circ f$$ 
which completes the proof. $\mathrm{\square}$
Title  standard duality on modules over algebras 

Canonical name  StandardDualityOnModulesOverAlgebras 
Date of creation  20131211 15:25:39 
Last modified on  20131211 15:25:39 
Owner  joking (16130) 
Last modified by  joking (16130) 
Numerical id  5 
Author  joking (16130) 
Entry type  Theorem 
Classification  msc 16S99 
Classification  msc 20C99 
Classification  msc 13B99 