Can someone help me understand this proof of a modified version of Brouwer's theorem?

Question

Let $\Omega \subset\mathbb{R}^n$ and $\Gamma \subset\mathbb{R}^m$ be two open and non empty sets. If there exists a homeomorphism $h: \Omega \to \Gamma$, then $n=m$.

Proof of the case for $m =1$ (the part in italics is the extent to which I could follow the steps in the proof, the bold one is where I got lost):

Let $P \in \Omega$. Since $\Omega$ is open in $\mathbb{R}^n$, there exists $r> 0$, such that the ball of center $P$ and radius $r$, $B_r(P)$ is cointaned in $\Omega$. Since $B_r(P)$ is connected and $h$ is an homeomorphism, then $h(B_r(P))$ is a connected open set of $\mathbb{R}$, i.e, an open interval, say, $(a,b)$. Let $c \in (a, b)$, and let $Q \in B_r(P)$, such that $h(Q) = c$. Therefore $(a,b) - \{c\}$ is not a connected set of $\Gamma$, and again, because $h$ is an homeomorphism, we have that $B_R(P) - \{Q\}$ is not connected either, which is only possible when $n=1$.

The part in italics was fine, as I said before. But I don't get any of the other conclusions. I would be very grateful for any help here.

Update: I see why $(a,b)-\{c\} = h(B_r(P)) - \{h(Q)\}$ is not connected, and I think the conclusion follows because $B_R(P) - \{Q\}$ is the image of the inverse of $h$, and therefore not connected either. Is this correct? If not, why (so how does the conclusion follow?)?

Disclaimer: The book I'm using introduces this as Brower's theorem for some reason (I looked it up and don't see the relation, but anyway...), so that's the reason for the title.

Answer 1

Since $B_r(P)$ is convex and $h$ is an homeomorphism, then $h(B_r(P))$ is a convex open set of $\mathbb{R}$, i.e, an open interval, say, $(a,b)$.

This part is quite confusing: convexity of $h(B_r(P))$ has anything to do with convexity of $B_r(P)$ only by the virtue that convex sets are connected, image of connected set is connected, and connected open subsets of $\mathbb{R}$ are intervals. It is definitely not true that image of convex set is convex in general.

That's why the rest of the proof is completely wrong. Convexity is not preserved by homeomorphisms.

However, if you replace "convex" with "connected", the proof now works just fine.

And BTW, it's fine to call it Brouwer's theorem, but one might want to qualify it as Brouwer's invariance of domain theorem, to distinguish it from other famous Brouwer's (fixed point) theorem.

Update: After the update (replacing "convex" with "connected"), the proof works. The bold part is true, because if $h$ is a homeomorphism, it is also in particular a homeomorphism between $B_r(P)$ and $(a, b)$, and thus also between $B_r(P)$ with a point removed and $(a, b)$ with a point removed. However, these two spaces are not homeomorphic: removing any point from $(a, b)$ makes it disconnected, while there's no point in $B_r(P)$ such that removing it would make it disconnected.

Answer 2

This is a comment, but probably too long to be a comment. Have you copied the proof correctly? Is it possible that you made a translation error?

A ball minus a point in a Euclidean space is never convex, and convexity is not the relevant concept here. The relevant concept is that if you remove a point from in interval in $\mathbb R$, the resulting set is no longer connected, but if you remove a point from a ball in a higher dimensional Euclidean space, the result is still connected.