Commutative diagram between modules and their double dual

Question

Hungerford Algebra, Exercise IV.$4.9$:

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For any homomorphism $\phi: A \rightarrow B$ of left $R$-modules the diagram is commutative, where $\theta_A(f)(a)=f(a)$ for $f \in A^*$ and $a \in A$ and $\theta_B(g)(b)=g(b)$ for $g\in B^*$ and $b \in B$. $\phi^*$ is the map induced on $A^{**}=\text{Hom}_R(\text{Hom}_R(A,R),R)$ by the map $\bar{\phi}:\text{Hom}_R(B,R) \rightarrow \text{Hom}_R(A,R) $ ($\bar{\phi}(g)=g\phi$).

\begin{matrix} & && & \theta_A &\\ & & &A &\longrightarrow & A^{**} \\ & & & \phi \big\downarrow & & \big\downarrow \phi^*\\ & & & B & \longrightarrow & B^{**}\\ & & & & \theta_B \end{matrix}

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In this problem, I can show $$\phi^*(\theta_A(a))=\theta_B(\phi(a))\ \ \forall a \in A.$$ However, I do not know how I should define $\phi^* (\xi)$ for $\xi \in A^{**}- \text{image}(\theta_A)$, or how I should extend $\phi^*$ from $\text{image}(\theta_A)$ to the whole $A^{**}$. If $A$ is free with a finite basis and $R$ has an identity, then $\theta_A$ is an isomorphism and there would be no problem. However, the exercise is stated for more general $R$-modules. Can anyone help me to extend $\phi^*$ for $A^{**}$?

Does such a $\phi^*$ always exist so that the diagram commutes? Or, if it exists then we know the diagram commutes?

Answer 1

The dual space mapping $A\to A^\ast$ is actually a (contravariant) functor, $(-)^\ast:\mathsf{RMod}\to\mathsf{RMod}^{\mathsf{op}}$. Our desired $\phi^{\ast\ast}:A^{\ast\ast}\to B^{\ast\ast}$ will be the image of $\phi$ under the composite functor $((-)^\ast)^{\mathsf{op}}\circ(-)^\ast:\mathsf{RMod}\to\mathsf{RMod}^{\mathsf{op}}\to\mathsf{RMod}$, more commonly written $(-)^{\ast\ast}$.

That just provides some motivation for the following definition:

$$\phi^{\ast\ast}:A^{\ast\ast}\to B^{\ast\ast}$$Is the mapping which takes $\alpha\in A^{\ast\ast}$ to the homomorphism $\alpha\circ\phi^\ast:B^\ast\to R$; $\phi^\ast$ being the homomorphism $B^\ast\to A^\ast$ that takes a homomorphism $f:B\to R$ to the homomorphism $f\circ\phi$.

You can continue the exercise by checking $\phi^{\ast\ast}$ is a genuine homomorphism and that it makes the diagram commute.