Consider the following theorem (Axler, Linear Algebra Done Right 3rd Ed., Theorem 3.108)
$T$ surjective is equivalent to $T'$ injective.
Suppose $V$ and $W$ are finite-dimensional and $T\in L(V,W)$. Then $T$ is surjective if and only if $T'$ is injective.
Note that $T'$ is the dual map of $T$.
I am interested in an interpretation of this result, not just a proof.
But to start, let me provide my attempt at a proof of the statement that $T'$ injective implies $T$ surjective (at the end of this post I will consider the reverse statement.)
Suppose $T'$ is injective.
Then dim null $T'$=0 and since it can be shown that
$$\text{dim null}\ T'=\text{dim null}\ T + \text{dim}\ W-\text{dim}\ V $$
then we have
$$\text{dim}\ V=\text{dim null}\ T+\text{dim}\ W$$
$$\text{dim}\ W=\text{dim}\ V - \text{dim null}\ T =\text{dim range}\ T$$
$$\implies T\ \text{surjective}$$
Next, here is my interpretation of this result.
$T'$ is a specific linear map in $L(W',V')$, namely the linear map that maps each linear functional $\varphi\in W'$ to the linear functional $\varphi\circ T\in V'$.
When we say that $T'$ is injective, we are saying that for each pair of distinct $\varphi_1$ and $\varphi_2$ in $W'=L(W,\mathbb{F})$, the corresponding $\varphi_i\circ T$'s are distinct.
Now, suppose that $T$ is not surjective. Pictorially, we have something like
If range $T\neq W$, ie if $T$ is not surjective, then we can have two different $\varphi_1,\varphi_2\in W'$ that map range $T$ the same way to $\mathbb{F}$, thus making $T'$ not injective.
Another way to see this is that $\varphi_1-\varphi_2 \in W'$ maps range $T$ to $0\in \mathbb{F}$, and thus the nullspace of $T'$ is not empty and so is not injective, contradicting our assumption that it is injective.
Pictorially, for $T'$ to be injective we have to have something like
ie, $T$ must be surjective (also means that dim $W\leq$ dim $V$).
Is this interpretation correct?
Finally, let me consider the statement that $T$ surjective implies $T'$ injective.
I am not sure if the following is an adequate proof.
By definition of $W'$, for each pair of distinct $\varphi_1,\varphi_2\in W'$ we have
$$\varphi_1\circ T\neq \varphi_2\circ T$$
and thus $T'$ is injective.