Explanation for Proof of $[T(u)]_{\gamma}=[T]_{\beta}^{\gamma}[u]_{\beta}$ I am struggling with understanding the proof of the following theorem:

Theorem. Let $$ and $$ be finite-dimensional vector spaces over a field $$, having ordered basis $\beta$ and $\gamma$, respectively. Let $: \rightarrow $ be linear. Then for each $u∈$ we have: 
$$[T(u)]_{\gamma}=[T]_{\gamma}^{\beta}[u]_{\beta}$$

The proof for this theorem is given as:

Proof: Fix $u∈V$, and define the linear transformations $:F \rightarrow V$ by $(a)=au$ and $g:F \rightarrow W$ by $g(a)=aT(u)$ for all $a∈F$. Let $\alpha=\{1\}$ be the standard ordered basis for $F$. Notice that $g=T$. Identifying column vectors as matrices and using Theorem 2.11, we obtain 
$$[T(u)]_{\gamma}=[g(1)]_{\gamma}=[g]_{\alpha}^{\gamma}=[Tf]_{\alpha}^{\gamma}=[T]_{\beta}^{\gamma}[f]_{\alpha}^{\beta}=[T]_{\beta}^{\gamma}[f(1)]_{\beta}=[T]_{\beta}^{\gamma}[u]_{\beta}$$

I am having trouble understanding the logic behind this proof. It seems like it just plugs in 2 linear transformations ($f$ and $g$) that make this proof work. How does it show that there isn't an example of $f$ and $g$ that won't make this theorem work? How does this proof prove the theorem for all possible $T$ and $u$?
 A: The logic behind this proof is that it wants to utilize the already proved similar result, theorem 2.11.: for any conposable linear maps $F,G$ and fixed bases $\alpha,\beta,\gamma$ we have
$$[F\circ G]_\alpha^\gamma=[F]_\beta^\gamma \cdot [G]_\alpha^\beta\,.$$
For that end, we wish to represent the single vector $u$ as some linear map, so we define $f:F\to V$ which just 'picks out' $u$ (all such linear maps are determined by their value at $1$ since $\alpha=(1)$ is a basis for $F$), that is, $f(\lambda)=\lambda u$.
The main observation is that
$$[f]_\alpha^\beta=[f(1)]_\beta=[u]_\beta\,,$$
so when applying the theorem for $T$ and this specific $f$ (which depends on $u$, but $u$ was arbitrary), we get
$$[T\circ f]_\alpha^\gamma=[T]_\beta^\gamma\cdot [f]_\alpha^\beta=[T]_\beta^\gamma\cdot[u]_\beta$$
where the left hand side is $[T(f(1))]_\gamma=[T(u)]_\gamma$, and note that $g=T\circ f$.
A: It's not that all $f$ and $g$ must work. The logic here is that the authors pick some special $f$ and $g$ to make the notation easier. The claim needs to hold for all $V$, $W$, $F$, $T$, $\beta$, $\gamma$, and $u\in V$. The proof precisely demonstrates this holds. The claim does not claim it must hold for all $f$ and $g$.
