Exercise 11.5 (Duality is a Contravariant Functor)

Proof.

(a)

Denote our original linear maps by

$$\begin{array}{cc}\hfill B:V\to U& \hfill \\ \hfill A:U\to W& \hfill \\ \hfill A\circ B:V\to W,& (A\circ B)(v)=A(B(V))\hfill \end{array}$$

The dual ${(A\circ B)}^{\text{∗}}$ of the composition $(A\circ B)$ is then, by definition, given by:

$$\begin{array}{llll}\hfill {(A\circ B)}^{\text{∗}}:W^{\text{∗}}\to V^{\text{∗}}\text{such that }\underset{\in V^{\text{∗}}}{\underbrace{{(A\circ B)}^{\text{∗}}\stackrel{\in W\ast )}{\overbrace{(\omega )}}}}(v)& =\omega ((A\circ B)(v))& \hfill & \\ \hfill & =\omega (A(B(v))).& \hfill & \end{array}$$

On the other hand, our dual maps are given by

$$\begin{array}{ccc}\hfill B^{\text{∗}}:U^{\text{∗}}\to V^{\text{∗}},& \hfill \text{ such that }\hfill & B^{\text{∗}}(𝜗)(v)=𝜗(B(v))\hfill \\ \hfill A^{\text{∗}}:W^{\text{∗}}\to U^{\text{∗}},& \hfill \text{ such that }\hfill & A^{\text{∗}}(\omega )(u)=\omega (A(u))\hfill \\ \hfill B^{\text{∗}}\circ A^{\text{∗}}:W^{\text{∗}}\to V^{\text{∗}},& \hfill \hfill & (B^{\text{∗}}\circ A^{\text{∗}})(\omega )=B^{\text{∗}}(A^{\text{∗}}(\omega ))\hfill \end{array}$$

Having carefully uncovered of the definitions, we notice that the assignment condition for ${(A\circ B)}^{\text{∗}}$ coincides with the formula for $B^{\text{∗}}\circ A^{\text{∗}}$ since:

$$\begin{array}{llll}\hfill \forall <!--\; FUNCTION\; APPLICATION\; -->v\in V:B^{\text{∗}}(A^{\text{∗}}(\omega ))(v)& =A^{\text{∗}}(\omega )(B(v))=\omega (A(B(v)).& \hfill & \end{array}$$

(b)

Let $\omega \in V^{\text{∗}}$. We want to demonstrate that

$${\mathrm{Id}<!--\; FUNCTION\; APPLICATION\; -->}_V(\omega )=\omega .$$

Unraveling definitions, we see that

$$\forall <!--\; FUNCTION\; APPLICATION\; -->v\in V:({\mathrm{Id}<!--\; FUNCTION\; APPLICATION\; -->}_v)\ast (\omega )(v)=\omega ({\mathrm{Id}<!--\; FUNCTION\; APPLICATION\; -->}_V(v))=\omega (v),$$

as desired.