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

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 $\mathbb{R}^m$

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## marked as duplicate by studiosus, LeGrandDODOM, Asaf Karagila general-topology StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 29 '14 at 19:18

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
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.