While reading Axler's Linear Algebra Done Right, in Chapter 3 (Linear Maps) section D (Invertibility and Isomorphisms) I was briefly puzzled by his proof that invertibility implies and is implied by the combination of both injectivity and surjectivity. (3.63 in the 4th edition.) Specifically, it was the part where for some $T \in \mathcal{L}(\mathbf{V},\mathbf{W})$, and given $\mathbf{w} \in \mathbf{W}$, Axler states that $\mathbf{w} = T(T^{-1}\mathbf{w})$ implies that $\operatorname{range} T = \mathbf{W}$.
I was perplexed because it wasn't clear to me why the existence of some $S$ such that $S(T\mathbf{v}) = \mathbf{v}$ would necessarily also require that $T(S\mathbf{w}) = \mathbf{w}$.
I eventually realised that my concept for "inverse" was narrower than the definition the book gave. For example, I thought that if $T \in \mathcal{L}(\mathbb{R}^2,\mathbb{R}^3)$ is defined as $T(x,y) = (x,y,0)$, then the function $S(x,y,z) = (x,y)$ would serve as an inverse of $T$, because for any $\mathbf{x} \in \mathbb{R}^2$, $S(T\mathbf{x}) = \mathbf{x}$. The effect of $S$ effectively cancels out the effect $T$; we might say that $S$ has the inverse effect of $T$.
This $T$ is clearly not surjective, hence my surprise at the assertion that all invertible linear maps are injective and surjective.
I did eventually work out that this example doesn't qualify as an inverse because the definition of a linear map is symmetric: it requires not only that $ST = I$ but also that $TS = I$ which is not true for my example.
And it's that symmetry that in turn seems to guarantee surjectivity and injectivity of an inverse.
But I have a question: why the symmetry? Is there some way in which my initial misconception of an inverse (merely $ST = I$, with no requirement that also $TS = I$) is broken? Does this narrower concept even have a name? Or is it merely not interesting enough to be thought of as a thing in its own right?
