Elementary proof that $Gl_n(\mathbb R)$ and $Gl_m(R)$ are homeomorphic iff $n=m$

Question

Prove that $Gl_n(\mathbb R)$ and $Gl_m(R)$ are homeomorphic iff $n=m$

Since $Gl_n(\mathbb R)$ is homeomorphic to an open subset of $\mathbb R^{n^2}$, this boils down to proving that two open subsets of $\mathbb R^n$ and $\mathbb R^m$ are homeomorphic iff $n=m$.

This can be done via homology, of which I know absolutely nothing.

Do you know a proof that doesn't cover things an undergraduate isn't supposed to know ?

EDIT: I'll close the question, since it's a duplicate of Elementary proof that $\mathbb{R}^n$ is not homeomorphic to $\mathbb{R}^m$

You are basically asking for a proof that the dimension of a manifold is a topological invariant. There are proofs that do not use homology, and they are not short, but they may be accessible to a very energetic undergraduate. See for example the book by Hurewicz and Wallman, "Dimension Theory". — Lee Mosher, May 29 '14 at 19:06
related (essentially duplicate of?): Elementary proof that $\mathbb{R}^n$ is not homeomorphic to $\mathbb{R}^m$ — Grigory M, May 29 '14 at 19:07

score 1 · Accepted Answer · edited Apr 13 '17 at 12:20

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If $\mathrm{GL}_n(\mathbb{R})$ is homeomorphic to $\mathrm{GL}_m(\mathbb{R})$, then (using that these are $n^2$-dimensional resp. $m^2$-dimensional manifolds) it follows that there is a homeomorphism between $\mathbb{R}^{n^2}$ and $\mathbb{R}^{m^2}$. It is well-known that this implies $n=m$ and there a lot of proofs for this, also without homology. See SE/24873 and arXiv:1310.8090.

edited Apr 13 '17 at 12:20

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answered May 29 '14 at 19:09

Martin Brandenburg

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Elementary proof that $Gl_n(\mathbb R)$ and $Gl_m(R)$ are homeomorphic iff $n=m$

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