I feel quite silly about this, and maybe I am missing something incredibly obvious. Let $V$ and $W$ be finite-dimensional $\mathbb{C}$-vector spaces. I want to show that if there is an injective homomorphism $\theta:V\to W$, then there is a surjective homomorphism $\theta':W\to V$.
I already have the intuition regarding the underlying sets. My trouble is figuring out how to define $\theta'$ such that it is in fact a homomorphism. Here is what I've tried so far:
Since $\theta$ is an injective homomorphism, we have that $V\cong \text{im}(\theta)$ by the first isomorphism theorem. Hence there is an inverse isomorphism $\chi:\text{im}(\theta)\to V$. Next we define a map $\theta':W\to V$ by $\theta'(w)=\begin{cases}\chi(w), \text{ if }w\in\text{im}(\theta)\\ 0\text{ otherwise}\end{cases}$.
My problem is that I don't think this is a homomorphism $W\to V$, despite being a surjection. For instance, if $w\in\text{im}(\theta), w'\in W\setminus \text{im}(\theta)$, then by this definition, $\theta'(w+w')=0$ since $w+w'\not\in\text{im}(\theta)$ (easy to see since $\text{im}(\theta)$ is a subspace of $W$), but $\theta'(w)+\theta'(w')=\theta'(w)$, which is not necessarily $0$. So, how should I define $\theta'$? I'm starting to question whether such a map even exists.
You can get beyond the obstacle by looking at bases. $\def\im{\operatorname{im}}$ Let $\{e_k\}$ be a basis of $\im\theta$. Extend it to a basis $\{e_j\}\cup\{f_j\}$ of $W$. Define $\pi:W\to\im\theta$ by $\pi(e_k)=e_k$ for all $k$, $\pi(f_j)=0$ for all $j$, and extend by linearity. Then $\chi\circ\pi:W\to V$ is a linear surjection.