Rank of linear map and its dual: Understand the bijectivity in the proof. I have a question concerning the proof of the following statement about the rank of a linear map and its dual.
Theorem.
Let $f\colon V\to W$ be a linear map between finite dimensional vector spaces and $f\colon W^*\to V^*$ the corresponding dual map. Then $\textrm{rk}f=\textrm{rk}f^*$.
Proof. We split $f\colon V\to W$ into the the induced linear surjective map $f_1\colon V\to\textrm{im}f$ and the inclusion $f_2\colon \textrm{im}f\to W$. Then $f^*\colon W^*\to V^*$ is the composition of $f_2^*\colon W\to (\textrm{im}f)^*$ and $f_1^*\colon (\textrm{im}f)^*\to V^*$. Now, $f_2^*$ is surjective and $f_1^*$ is injective.


Therefore, $f_1^*$ maps $(\textrm{im}f)^*$ isomorphically to $\textrm{im}f^*$.


and thus $\textrm{rk}f=\textrm{dim}(\textrm{im}f)=\textrm{dim}(\textrm{im} f)^*=\textrm{dim}(\textrm{im}f^*)=\textrm{rk}f^*.$
My Question:
Could you please explain to me the marked sentence which I understand the following way:
Why is $f_1^*\colon (\textrm{im}f)^*\to\textrm{im}f^*$ an isomorphism, i.e. bijective?
I cannot find a reasonable justification.
 A: Let $g\in (Imf)^*, g:Imf\rightarrow K$, $f_1^*(g)=g\circ f_1=g'\circ f$ where $g':W\rightarrow K$ extends $g$. This implies that $f_1^*(g)\in Imf^*$, since $f_1^*$ is injective, we just need to show that it is surjective. 
Let $h\in Imf^*, h=f^*(g)=g\circ f, g\in W^*$, let $g_1=g_{\mid Imf}$, $g\circ f=g_1\circ f_1=f_1^*(g_1)$.
A: You know that $f_1: V \to \text{im}(f)$ is surjective by construction. Hence, the dual map $f_1^*: (\text{im}(f))^* \to V^*$ will be injective (I assume this is something you've proven).
Since $f_1^*$ is injective, it will be an isomorphism onto its image (if you restrict the target space of an injective map, it becomes bijective). But what is its image? Well, since $f = f_2 \circ f_1$, we have that $f^* = f_1^* \circ f_2^*$, and since $f_2^*$ is surjective (because $f_2$ is clearly injective), it follows that
\begin{align}
\text{image}(f^*) = \text{image}(f_1^*).
\end{align}
In other words, $f_1^*$ is an isomorphism onto $\text{image}(f^*)$.

(btw this is a very nice proof... I learnt something new and interesting from this $\ddot{\smile}$)

Edit:
The proof is a simple calculation:
\begin{align}
\text{image}(f^*) &:= f^*[W^*] \\
&= (f_1^* \circ f_2^*)[W^*] \\
&= f_1^*[f_2^*(W^*)] \\
&= f_1^*[(\text{im}(f) )^*] \tag{$f_2^*$ is surjective} \\
&:= \text{image}(f_1^*)
\end{align}
Notice that this equality is true simply because the "inner function" $f_2^*$ is surjective; it didn't matter that $f_1^*$ is injective to prove this equality. In other words, given any two functions $\alpha$ and $\beta$ (defined on any sets such that the composition is well defined) it is always true in general that if $\alpha$ is surjective then $\text{image}(\beta \circ \alpha) = \text{image}(\beta)$).
We need $f_1^*: (\text{im}(f))^* \to V^*$ being injective to establish the isomorphism onto its image (i.e the $\cong$ sign below):
\begin{align}
(\text{im}(f))^* \cong \text{image}(f_1^*) = \text{image}(f^*).
\end{align}
