![[sol-5.pdf#page=27]] --- $\textbf{Exercise 13.}$ Show that the dual map of the identity operator on $V$ is the identity operator on $V^*$. $\textbf{Solution 13.}$ Let $I$ denote the identity operator on $V$. For $\varphi \in V^*$ we have $\begin{equation*} I^*(\varphi) = \varphi \circ I = \varphi. \end{equation*}$ Thus $I^*$ is the identity operator on $V^*$. --- $\textbf{Exercise 13.}$ Show that the dual map of the identity operator on $V$ is the identity operator on $V^*$. $\textbf{Solution 13.}$ Let $I$ denote the identity operator on $V$. For $\varphi \in V^*$ we have $\begin{equation*} I^*(\varphi) = \varphi \circ I = \varphi. \end{equation*}$ Thus $I^*$ is the identity operator on $V^*$. $\textit{Commentary:}$ This exercise demonstrates that the dual operation preserves the identity map. In other words, taking the dual of the identity map on a vector space $V$ yields the identity map on the dual space $V^*$. The proof is a one-liner using the definition of the dual map. This result is a special case of a more general property: the dual operation reverses the order of composition of linear maps. This property is encapsulated in the contravariance of the dual functor in category theory. $\textit{Example:}$ Let $V = \mathbb{R}^2$ and $I: \mathbb{R}^2 \to \mathbb{R}^2$ be the identity map. For any $\varphi = (a, b) \in (\mathbb{R}^2)^*$, we have $\begin{align*} I^*(\varphi)(x, y) &= \varphi(I(x, y)) \\ &= \varphi(x, y) \\ &= ax + by \\ &= \varphi(x, y). \end{align*}$ Thus $I^*(\varphi) = \varphi$ for all $\varphi \in (\mathbb{R}^2)^*$, so $I^*$ is the identity map on $(\mathbb{R}^2)^*$.