$T\mapsto[T]_{\mathcal B}^{\mathcal C}$ is a vector-space isomorphism $\mathcal L(V,W)\cong M_{m\times n}(F)$

For every basis vector $b_j$, coordinate uniqueness in [L2] gives $[(S+T)(b_j)]_{\mathcal C}=[S(b_j)]_{\mathcal C}+[T(b_j)]_{\mathcal C}$ and $[(\lambda T)(b_j)]_{\mathcal C}=\lambda[T(b_j)]_{\mathcal C}$, so $\Phi$ is linear column by column.

If $\Phi(S)=\Phi(T)$, then [L2] gives $S(b_j)=T(b_j)$ for every $j$; linearity and the unique expansion of every vector in $\mathcal B$ give $S=T$, so $\Phi$ is injective.

Given $A=(a_{ij})\in M_{m\times n}(F)$, prescribe $T(b_j):=\sum_{i<m}a_{ij}c_i$ and, for the unique expansion $v=\sum_{j<n}x_jb_j$ from [L2], define $T(v):=\sum_{j<n}x_jT(b_j)$. This is well defined and linear, and the $j$-th matrix column is the $j$-th column of $A$; hence $\Phi(T)=A$. Together with steps 1.1 and 2.1, $\Phi$ is a linear bijection. Its set-theoretic inverse is linear: if $A=\Phi(S)$ and $B=\Phi(T)$, then injectivity and linearity give $\Phi^{-1}(A+B)=S+T$ and $\Phi^{-1}(\lambda A)=\lambda S$. Thus $\Phi$ is a linear isomorphism by [L3].

If $n=0$, then $V$ is the zero space and both sides contain only their zero element; if $m=0$, then $W$ and $M_{m\times n}(F)$ are zero spaces and the only map is the zero map. Thus the construction also proves the isomorphism in every zero-dimensional case.