For an \(R\)-module \(X\), write \[ X^\vee=\Hom_R(X,R). \] Define the evaluation map \[ \theta_X:X\to X^{\vee\vee}, \qquad \theta_X(x)(\lambda)=\lambda(x). \] Show that \(\theta\) is natural: for every \(R\)-linear map \(f:A\to B\), \[ f^{\vee\vee}\circ\theta_A=\theta_B\circ f. \]
Solution
The dual map is \[ f^\vee:B^\vee\to A^\vee, \qquad f^\vee(g)=g\circ f, \] and the double dual map is \[ f^{\vee\vee}:A^{\vee\vee}\to B^{\vee\vee}, \qquad f^{\vee\vee}(\Phi)=\Phi\circ f^\vee. \]
Let \(a\in A\). To prove equality of the two elements of \(B^{\vee\vee}\), evaluate both at an arbitrary \(g\in B^\vee\).
On one hand, \[ \bigl(f^{\vee\vee}(\theta_A(a))\bigr)(g) =\theta_A(a)(f^\vee(g)) =(g\circ f)(a) =g(f(a)). \] On the other hand, \[ \theta_B(f(a))(g)=g(f(a)). \] Thus \[ f^{\vee\vee}(\theta_A(a))=\theta_B(f(a)) \] for every \(a\in A\), proving \[ f^{\vee\vee}\circ\theta_A=\theta_B\circ f. \]