There is one step the necessity of which I don’t understand in the proof of Inverse Function Theorem in Jerry Shurman’s $Calculus$ $and$ $Analysis$ $in$ $Euclidean$ $Space$ (available here https://www.professores.uff.br/diomarcesarlobao/wp-content/uploads/sites/85/2017/09/Multi_calculus_BOOK.pdf).
After having established injectivity of $f$ in some closed ball $\overline B$ around $a$ and introducing an open ball $W$ around $f(a)$ s.t. $|f(a) - y| < |f(x) - y|$ for every $y\in W$ and every $x\in \partial \overline B$, the author goes on to prove, using Critical Point Theorem and Chain Rule, that for every $y\in W$ there is one and only one $x\in int\overline B$ s.t. $f(x) = y$. Why do we need to do this, or, more concretely, why does the proof of this fact need to be so sophisticated? Why can’t we just take $V$ to be the pre-image of $W$, i.e. $V = f^{-1}(W)$ (every point of $W$ has a pre-image in $\overline B$, because $W$ is a subset of $f(\overline B)$; moreover, this pre-image should be unique due to injectivity of $f$ and not lie on the boundary of $\overline B$ by construction), and then $f$ becomes a bijective, and thus invertible map from V to W?
Okay, I was just being stupid. If we ignore injectivity for a second, then it is a straightforward manner to show that under a continuous mapping boundary points can become interior points and interior points can become boundary points. Thus, the point $a$ might have become a boundary point, which would then imply that $W$ is not really a subset of $f(\overline B)$. Maybe if injectivity is added back, this can be disproved (I fell like injectivity is still not enough, but a proof or a counterexample would be much appreciated). But nonetheless, it is much easier to show using the invertibility of the derivative matrix that $f(\overline B)$, even more so, $f(int\overline B)$ does indeed contain $W$