Clarifying the definition of the Dual Map I would appreciate help understanding the definition of the dual map $T'$ as presented in Axler's "Linear Algebra Done Right" 3rd ed. on page 103.


If $T \in \cal {L}$$(V,W)$ then the dual map of $T$ is the linear map $T' \in \cal {L}$$(W',V')$ defined by $T'(\phi)= \phi \circ T$ for $\phi \in W'$.


With $\phi \in W'$, on the LHS, $T'(\phi)$ maps to $V'$, where elements of $V'$ are linear functionals on $V$.
This is my question:
When looking at the RHS, $\phi \circ T$, $T$ maps to $W$ but since $\phi \in W'$, $\phi$ is a linear functional on $W$ and I would think should map to a scalar, not a linear functional on $V$, i.e., not an element of $V'$.
So things don't seen to line up. Thanks.
 A: It might be a bit easier to understand with different notation. If $T:V\to W$, then the dual map $T':W'\to V'$ satisfies $\phi[Tv]=(T'\phi)[v]$, for all $v\in V$ and $\phi\in W'$. That is, you can either first map $v$ to $Tv=w\in W$ and then apply $\phi$ to $w$, or first map $\phi$ to $T'\phi=\psi\in V'$ using $T'$ and then apply $\psi$ to $v$, getting the same scalar result either way.
A: $T'$ maps a linear functional $\phi\in W'$ to the linear functional $(\phi \circ T)\in V'$.
$$T': W'\to V' \\ \phi \stackrel{T'}\longrightarrow \phi\circ T$$
To see that $\phi \circ T$ is a functional consider its input and output.  The $T$ acts first and its input is an element of $V$.  Then the $\phi$ acts and its output is an element of $\Bbb F$.  So $\phi\circ T$ is a function $V\to \Bbb F$.  Graphically, these functions look like:
$$\require{AMScd}
\begin{CD}
    V @>T>> W @>\phi>> \Bbb F\\
    @. \large\searrow @. \large\nearrow \\
     @. \stackrel{\large\longrightarrow}{\phi\circ T}
\end{CD}$$
Because compositions of linear maps are linear, this shows that $\phi\circ T$ is in fact an element of $V':=\mathcal L(V,\Bbb F)$.
