Show that for any homomorphism \(f: A \to B\) of left \(R\)-modules the following diagram is commutative:
where \(\theta_A, \theta_B\) are as in Theorem 4.12 and \(f^*\) is the map induced on \(A^{**} \coloneqq \mathrm{Hom}_R(\mathrm{Hom}(A, R), R)\) by the map \[\overline f: \mathrm{Hom}(B, R) \to \mathrm{Hom}_R(A, R).\]
Solution
For a left \(R\)-module \(M\) write \(M^*=\operatorname{Hom}_R(M,R)\), and let \(\theta_M\colon M \to M^{**}\) be the canonical map, \(\theta_M(m)(g) = g(m)\) for \(m \in M\) and \(g \in M^*\). Let \(\overline f\colon B^* \to A^*\) be \(\overline f(g) = g \circ f\), and let \(f^{**}\colon A^{**} \to B^{**}\) be \(f^{**}(\varphi) = \varphi \circ \overline f\); this is the map the problem calls \(f^*\). The diagram asserts \(f^{**} \circ \theta_A = \theta_B \circ f\).
1. For \(a \in A\) and \(g \in B^*\), \((f^{**} \circ \theta_A)(a)(g) = g(f(a))\).
By the definitions of \(f^{**}\), \(\overline f\), and \(\theta_A\), \[(f^{**} \circ \theta_A)(a)(g) = \theta_A(a)(\overline f(g)) = \theta_A(a)(g \circ f) = g(f(a)).\]
2. For \(a \in A\) and \(g \in B^*\), \((\theta_B \circ f)(a)(g) = g(f(a))\).
This is the definition of \(\theta_B\) evaluated at \(f(a)\).
3. Q.E.D.
By steps 1 and 2, the maps \(f^{**} \circ \theta_A\) and \(\theta_B \circ f\) agree on every \(a\in A\) and every \(g\in B^*\), so \(f^{**} \circ \theta_A = \theta_B \circ f\).