Proofs of invariance of domain consequence

Question

So I know that (as a consequence of the invariance of domain) $\Bbb{R}^n\cong\Bbb{R}^m$ iff $m=n$. When I hadn’t seen any homology theory, I was asked to show this when $m=1,n=2$ (uses removing a single point then 1D is disconnected and 2D is still connected, that is 1D has 2 connected components and 2D has 1), and also $m=2,n=3$ we test for simply connectedness. This uses your classic remove a point and see what happens to connectedness which is a topological property. I always wondered why we never did arbitrary $m,n$ until I learned a bit of homology and de Rham cohomology theory which I chose as a project in an applied topology course. And I have seen a proof in Hatcher of seeing what happens to their relative homology groups when removing a point if they’re homeomorphic then their both $\Bbb{Z}$ when for the $i$th homology group we have say $i=m$ and trivial otherwise, so you deduce $m=n$. And similarly in Lee, I see that he uses de Rham cohomology and a similar type of reasoning. I want to know what are all the various proof methods one can use. Because in my right mind I cannot think of any elementary methods. And yes, I have seen this post about asking for an elementary proof which is not what I’m asking, per say. I would like to know how many independent proofs there are. And are the Lee proof and Hatcher proof basically using the same argument categorically speaking-wise? I mean they both use the fact that homeomorphic topological spaces have isomorphic homotopy, homology, de Rham cohomology group etc.

Added:

I began to think, when we do $m=1,n=2$ we remove a point and observe one is connected and the other is not. When $m=2,n=3$, we remove a points and use simply connectedness. It seems as though when dimension increases, the level of which the space is "connected" becomes stronger so you’re “more connected” and the connected components decrease when you remove a point. Just a thought.. so in a sense $\Bbb{R}^n$ is "more connected" than $\Bbb{R}^m$ if $m<n$ since you add axis so like for $\Bbb{R}^2$ and $\Bbb{R}^3$, if you remove the origin, then loops around origin in $2$ case (take $S^1$ for a loop which doesn’t contract to a point) are not homotopic to constant loop but if you add a dimension you can "hop" over to contract your loop. I.e., $\Bbb{R}^2\setminus \{\overline{0}\}$ is not simply connected but $\Bbb{R}^3\setminus\{\overline{0}\}$ still is.

Answer 1

Your are right when you say that $\mathbb{R}^m$ is more "connected" than $\mathbb{R}^n$ if $m > n$, and that can be expressed using homotopy groups of $\mathbb{R}^k\setminus \{0\}$, for $k\in \{n,m\}$. However one can see that in a more elementary way (but it is pretty much the same).

The most "elementary" proof I am aware of reduce the problem to Sard's Theorem and Sperner's Lemma. But I suppose here "elementary" is a matter of taste. I find it more elementary since I prefer to solve things using Analysis when it is possible. Sard's Theorem is a very useful basic tool in Differential Topology.

Step 1. Prove Brouwer's Fixed point Theorem using Sperner's Lemma.

Step 2. Show that the $n$-dimensional sphere $\mathbb{S}^n$ is not contractible (the identity is not homotopic to a constant function) using Brouwer's Fixed point Theorem.

I sumarize the argument by Dugundji. Fixed point theory: if there are $x_0\in \mathbb{S}^n$ and a continuous function $$h\colon \mathbb{S}^n \times [0,1]\mapsto \mathbb{S}^n$$

such that $h(x,0)=x_0$ and $h(x,1)=x$ for every $x\in \mathbb{S}^n$, define $r\colon B^{n+1}[0,1] \mapsto B^{n+1}[0,1]$, where $B^{n+1}[0,1]=\{x\in \mathbb{R}^{n+1}\colon |x|\leq 1\}$ as

$$ r(x) = \begin{cases} -x_0, & |x| \le \tfrac{1}{2}, \\ -h\!\left(\dfrac{x}{|x|},\, 2|x| - 1\right), & |x| \ge \tfrac{1}{2}. \end{cases} $$

Then $r$ is continuous and does not have a fixed point.

Step 3. Conclude using Step 2. that the inclusion $$f\colon \mathbb{S}^{n-1}\rightarrow \mathbb{R}^n\setminus \{0\}$$ given by $f(x)=x$ is not homotopic to a constant function in $C(\mathbb{S}^{n-1}, \mathbb{R}^n\setminus \{0\})$.

Step 4. Prove that if $n< m$ then every continuous function $g\colon \mathbb{S}^{n-1}\rightarrow \mathbb{R}^{m}\setminus \{0\}$ is homotopic to a constant in $C(\mathbb{S}^{n-1}, \mathbb{R}^m\setminus \{0\})$. To do this use Sard's Theorem. Since $g$ is just continuous first we find a $C^1$ function $\hat{g}$ homotopic to $g$ in $C(\mathbb{S}^{n-1}, \mathbb{R}^m\setminus \{0\})$ (use $C^1$ partitions of unit or something else).

Now consider the $C^1$ function

$$G\colon \mathbb{S}^{n-1}\times \mathbb{R}\mapsto \mathbb{R}^m$$

defined by $G(x,t)= t\hat{g}(x)$. Since

$$\dim \mathbb{S}^{n-1}\times \mathbb{R}=n < m$$

we conclude by Sard's theorem that the image of $G$ has zero measure. So pick $x_0\neq 0$ that is not in the image of $G$. Then

$$H\colon \mathbb{S}^{n-1}\times [0,1]\rightarrow \mathbb{R}^m$$

defined by $H(x,s) = (1-s)\hat{g}(x)+ sx_0$, is a homotopy between $\hat{g}$ and the function constant to $x_0$ in $C(\mathbb{S}^{n-1}, \mathbb{R}^m\setminus \{0\})$.

Step 5. Let $n < m$. Suppose there is a homeomorphism $q$ from $\mathbb{R}^n$ to $\mathbb{R}^m$. WLOG we assume that $q(0)=0$. Then $q$ is a homeomorphim between $\mathbb{R}^n\setminus \{0\}$ and $\mathbb{R}^{m}\setminus \{0\}$. Let $f\colon \mathbb{S}^{n-1}\rightarrow \mathbb{R}^n\setminus \{0\}$ be the inclusion $f(x)=x$. By Step 3. $f$ is not homotopic to a constant.

By Step 4. we have that $q\circ f$ is homotopic to a constant, that is, there is a continuous function $$H\colon \mathbb{S}^{n-1}\times [0,1]\rightarrow \mathbb{R}^m\setminus \{0\}$$ and $x_0\neq 0$ such that $H(x,0)=q\circ f(x)$ and $H(x,1)=x_0$ for every $x$.

But this implies that $$q^{-1}\circ H\colon \mathbb{S}^{n-1}\times [0,1]\rightarrow \mathbb{R}^n\setminus \{0\}$$ is a homotopy between $f$ and the constant function $q^{-1}(x_0).$ This contradicts Step 3.