Sometimes I have trouble with the notation in the book "Linear Algebra Done Right" by Axler.
Currently, my issue is with the following theorem (Theorem 3.114 in the 3rd edition book)
Let $v_1, ..., v_n$ be a basis of linear space $V$, $\varphi_1,...,\varphi_n$ be a basis of $V'$, the dual space of $V$.
Let $w_1, ..., w_m$ be a basis of linear space $W$, and $\psi_1, ..., > \psi_m$ be a basis of $W'$.
Let $L(V,W)$ denote the linear space of linear maps from $V$ to $W$.
Suppose $T\in L(V,W)$ and $T'\in L(W', V')$ (dual map of $T$).
Let $M(T)$ denote the matrix of the linear map $T$ relative to the aforementioned bases of $V$ and $W$, and $M(T')$ denote the matrix of the linear map $T'$ relative to the aforementioned bases of $W'$ and $V'$.
Then $M(T')=(M(T))^t$, where the superscript $t$ denotes transpose.
The proof of this theorem starts as follows
Let $A=M(T)$ and $C=M(T')$.
Suppose $1\leq j\leq m$ and $1\leq k\leq n$.
From the definition of $M(T')$, we have
$$T'(\psi_j)=\sum_{r=1}^n C_{r,j}\varphi_r$$
The left side of the equation above equals $\psi\circ T$. Thus applying both sides of the equation above to $v_k$ gives
$$(\varphi\circ T)(v_k)=\sum_{r=1}^n C_{r,j}\varphi_r(v_k)$$
My question is about the statement in bold above.
As far as I can tell, $T'(\psi_j)$ means that we have a linear map $T':W'\to V'$, we give it a linear map in $W'$ as input (namely the linear map $\psi_j$), and we get a linear map in $V'$ as output (ie a linear map $V\to\mathbb{R}$).
$\psi\circ T$, on the other hand is a function composition, which I believe is defined for every $v\in V$ as
$$(\psi_j\circ T)(v)=\psi_j(Tv)\tag{1}$$
This is also a linear map $V\to\mathbb{R}$.
How do we know that $T'(\psi_j)=\psi_j\circ T$?
After reading the comments, the answer is obvious.
We know that $T'(\psi_k)=\psi_j\circ T$ by definition of the dual map of $T$, which in Axler is