Suppose $V$ is finite-dimensional and $T \in \mathcal L(V,W)$. Prove that $T$ is surjective if and only if there exists $S \in \mathcal L(W,V)$ such that $TS$ is the identity map on $W$.

Suppose $T$ is surjective, then for every $w \in W$ there exists a $v \in V$ with $Tv = w$. Define $Sw = v$ so $TSw = w$ as desired.

Like in 3b/20 $S$ inherits linearity from $T$

• $S(w_1+w_2)$
• $S(\lambda w)$

TODO: finish (making $S$ preserve linearity is a bit trickier. I want to do it without introducing $U\oplus N(T)$ but I'm not sure how)

For the backward direction suppose $TS$ is the identity map on $W$, then for any $w \in W$ pick $v = Sw$ so that $Tv = w$, thus $T$ is surjective.

Old bad proof

First suppose $T$ is surjective, meaning $\text{range }T = W$. Notice how $\text{range }T \subseteq V$ implies $\text{range }T = W$ is finite-dimensional.
Let $w_1,\dots,w_m$ be a basis for $W$ and let $v_1,\dots,v_m$ be a list in $V$ such that $w_j = Tv_j$ for all $j$. The list is clearly independent (see 3b/20), so we may extend it (2.33) to a basis $v_1,\dots,v_n$ of $V$.
Define $Sw_j = v_j$ for all $j$ (possible as $m \le n$), then for any $w \in W$ we have

After using linearity and the definition. This completes the forward direction

For the backward direction suppose $TS$ is the identity map on $W$, then for any $w \in W$ pick $v = Sw$ so that $Tv = w$, thus $T$ is surjective.

This is almost exactly the same as 3b/20, I could extract some lemmas to avoid duplication, but I feel like the book should provide all the theorems needed to make short arguments. Perhaps I'm proving this wrong? Or perhaps answers are expected to be this long and contain duplication.

todo: shorten proof