Unclear steps in proof of matrix representation of $f\colon V \to W$ linear mapping We have the following theorem:

$f\colon V \to W$ is a linear mapping. And $\dim V = n$ and $\dim W = m$.
  Then there exists a basis $\mathcal{B}$ of $V$ and a basis $\mathcal{C}$ for $W$ such that
  $\mathcal{A_{f,\mathcal{B},\mathcal{C}}}$ = $\begin{bmatrix}I_k & | & 0_{k,n-k}\\0_{m-k,k} & | & 0_{m-k,n-k}\end{bmatrix}$ for a certain $k$ $\le{n,m}$.

Proof:
We write $V = V' \oplus \ker f$. Then we can say $V'$ is a complement of $\ker f$. Suppose that $\{b_1, \dots , b_k\}$ is a basis for $V'$ and $\{b_{k+1}, \dots , b_n\}$ is a basis for $\ker f$ then $\{b_1, \dots , b_n\}$ is a basis for $V$.
(Q: I understand the steps until here. The proof then goes as follows...)
Because $\ker f|_{V'} = \{0\}$ and $\operatorname{im} f|_{V'} = \operatorname{im} f$ therefore $\operatorname{im}f|_{V'} : V' \to \operatorname{im}f$ is an isomorphism.
(Q: I don't see the point of determining $\ker|_{V'}$ and $\operatorname{im}f|_{V'} = imf$. Why is this an isormophism? And why does this imply the next statements?)
This implies that $\{f(b_1), \dots , f(b_k)\}$ is a basis for $\operatorname{im}f$. We then extend this basis for $W$ to $\{f(b_1), \dots , f(b_k), c_{k+1}, \dots, c_{m}\}$. Compared to those basis's the matrix representation of $f$ is of the required form.
 A: The point of considering the map $f|_{V'}$ is that isomorphisms are very nice.  Here are the facts about isomorphisms that are relevant to the proof:


*

*If $f:X \to Y$ is an isomorphism and $\{v_1,\dots,v_n\}$ is a basis of $X$, then $\{f(v_1),\dots,f(v_n)\}$ is a basis of $Y$.

*If $\ker f = \{0\}$, then $f:X \to im(f)$ is an isomorphism (which is to say that $f$ is injective)


Now, because $\ker f|_{V'} = 0$, we know that $f|_{V'}:V' \to im(f|_{V'})$ is an isomorphism. By our first property of isomorphisms above, we'll be able to use $f$ to find a basis of $im(f|_{V'})$.
Moreover, because $im(f|_{V'}) = im(f)$, this basis we find will be a basis of $im(f)$.
All that remains is to find the matrix of $f$ relative to the bases $\{b_1,\dots,b_n\}$ and $\{f(b_1),\dots,f(b_k),c_{k+1},\dots,c_m\}$.
A: That is because, if $f\,\bigr\rvert_{V'}=0$, $\;f\,\bigr\rvert_{V'}$ is injective. Furthermore, $\;\DeclareMathOperator\img{Im}\img\bigl(f\,\bigr\rvert_{V'}\bigr)=\img f$. Finally, you can use the first isomorphism theorem.
