Suppose $V$ is finite-dimensional and $T : V \to W$ is a surjective linear map of $V$ onto $W$. Prove that there is a subspace $U$ of $V$ such that $T|_U$ is an isomorphism of $U$ onto $W$. (Here $T|_U$ means the function $T$ restricted to $U$.)

Proof 1

From 2.34 we can find a subspace $U$ of $V$ such that $U\oplus\text{null T} = V$.
Clearly $\text{null }T|_U = \{0\}$ since being a direct sum means $U \cap \text{null }T = \{0\}$.
And obviously $\text{range }T|_U = \text{range }T = W$ since the nullspace doesn't contribute anything.
Making the last part rigorous, each $v \in V$ can be uniquely written $v = v_u + v_n$ for $v_u \in U$, $v_n \in \text{null }T$ meaning if $w = Tv$ then $w = Tv_u$ so the ranges are the same.

Proof 2

Let $v_1,\dots,v_n$ be a basis of $V$ and let $w_k = Tv_k$. we know $\text{span}(w_1,\dots,w_n) = W$ so we can remove vectors (by 2.31) to get a a basis $w_{k_1},\dots,w_{k_m}$ of $W$ then let $U = \text{span}(v_{k_1},\dots,v_{k_m})$.
Obviously $T|_U$ is surjective since $Tv_{k_j} = w_{k_j}$ and $w_{k_1},\dots,w_{k_m}$ is spanning. To see injectivity use the independence of $w_{k_1},\dots,w_{k_m}$

todo: come up with a proof that uses results from 3d