Is one-sided differentiable inverse sufficient to conclude equal dimensions?

Question

Let $\Omega$ be an open subset of $\mathbb R^n$ and $\Upsilon$ be of $\mathbb R^m$. Let $f\colon \Omega\to \Upsilon$. Then this answer shows that if there exists a point $c\in\Omega$ where $f$ is differentiable, and if $f$ is invertible with the inverse being differentiable at $f(c)$, then $m = n$.

Note that differentiability at only one point is sufficient. The answer also comments that if there exists a homeomorphism between $\Omega$ and $\Upsilon$ (with these being nonempty), then too $m = n$.

But this left me wondering about the following questions:

1. If $f$ is not invertible, but has a right-inverse $g$ (that is differentiable at $f(c)$), is $m = n$ still? Or can we say nothing more than $m\le n$?

2. Same question about homeomorphism.

3. The answer said that the homeomorphism statement can be proved by Brouwer's fixed point theorem. Can you give me a reference for that?

Answer 1

1. It only means $m \le n$. As an example take $f : \mathbb R^n \to \mathbb R^{n+k}, f(x_1,\ldots, x_n) = (x_1,\ldots, x_n,0,\ldots, 0)$, and $g : \mathbb R^{n+k} \to \mathbb R^n, g(x_1,\ldots, x_{n+k}) = (x_1,\ldots, x_n)$.

2. Same example.

3. https://terrytao.wordpress.com/2011/06/13/brouwers-fixed-point-and-invariance-of-domain-theorems-and-hilberts-fifth-problem/