Suppose $V$ is finite-dimensional, $U$ is a subspace of $V$, and $S \in \mathcal L(U,V)$. Prove there exists an invertible operator $T \in \mathcal L(V)$ such that $Tu = Su$ for every $u \in U$ if and only if $S$ is injective.

Suppose $S$ is injective. Let $u_1,\dots,u_n$ be a basis for $U$ then extend to a basis $u_1,\dots,u_n,v_1,\dots,v_m$ of $V$. since $S$ is injective $Su_1,\dots,Su_n$ is independent, meaning we can extend it to a basis $Su_1,\dots,Su_n,w_1,\dots,w_m$ of $V$.
Finally define $Tu_k = Su_k$ and $Tv_k = w_k$ for all $k$. $T$ is obviously surjective, implying $T$ is invertible by 3.69. (You can also show injectivity directly by using the independence of $Su_1,\dots,w_m$.)

Now suppose $S$ is not injective on $U$. This would imply $T$ is not injective on $U$ hence $T$ cannot be invertible.