My question is regarding a step in a proof which is provided in Axler, "Linear Algebra Done Right" $3$rd. ed., page $81$:
Assuming $T \in \cal{L}$ $(V,W)$ is invertible. Then for $w \in W$, $w = T(T^{-1}w)$ which shows $w$ is in the range of $T$.
Specifically, it seems that the existence of $(T^{-1}w)$ for any $w \in W$ already assumes $T$ is surjective?
(Thinking, one could say $w = Iw = (TT^{-1})w = T(T^{-1}w)$. But I'm not sure that really helps.)
Thanks
Assuming $T:V\to W$ is invertible means that there exists a function $T^{-1}:W\to V$ such that $TT^{-1}$ is the identity on $W$ and $T^{-1}T$ is the identity on $V$. So $T^{-1}w$ exists for all $w\in W$. There is no assumption here that $T$ is surjective.
BTW, this result actually has nothing to do with linearity: any function, linear or not, is invertible if and only if it is bijective.
Why? One is assuming that there is a map $S : w \to v$ such that $T S$ is the identity on $W$ and $S T$ is the identity on $V$. Surjectivity of $T$ follows, but it is not a necessary step to claim the existence of such an $S$.