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The proofs of this post are perfectly fine, but I thought there was another proof using the "invariance of domain" theorem. I tried to write it explicitly but it does not seem to work.

By absurd, let $\ \varphi:H^n \longrightarrow \mathbb{R}^n$ be a homeomorphism and let's call $\psi: \mathbb{R}^n \longrightarrow H^n$ its inverse which does go from an open subset of $\mathbb{R}^n$ to $\mathbb{R}^n$ and is injective as required by the invariance of domain theorem.

The problem now is that $\psi$ is not continuous as a map $\mathbb{R}^n \longrightarrow \mathbb{R}^n $ but only as a map $\mathbb{R}^n \longrightarrow H^n$ (recall: $\psi^{-1}(open)$ should be open, but $U$ open in $H^n$ means $\exists\ V\subset\mathbb{R}^n$ open such that $U = H^n\cap V$, i.e. $U$ is not necessarily open in $\mathbb{R}^n$).

If $\psi$ were continuous then one would obtain a contradiction: by the theorem $\psi(\mathbb{R}^n)=H^n$ should be open in $\mathbb{R}^n$ but it is not, so there cannot be an homeomorphism $H^n\cong \mathbb{R}^n$.

Question: can one use the invariance of domain to proove that there is no homeomorphism between $H^n$ and $\mathbb{R}^n$?

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Noix07
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    I think it is ok and it works as an alternative proof. If $\psi$ is continuous as a map to $H^n$ and the inclusion $H^n\subseteq \Bbb R^n$ is obviously continuous, then the composition is continuous too. – Peter Franek Nov 01 '16 at 11:13
  • Yeah you're right!! if we use the $\forall \epsilon$... characterization of continuity it will work.... Oh and your edit is even better!!! I validate your answer!! – Noix07 Nov 01 '16 at 11:19
  • The continuity issue is a good remark but since you have injectivity I don't see what could stop you from using this method... – Edu Nov 01 '16 at 11:19

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If $Z$ is a subspace of the topological space $X\subset Y$, then $f: X \rightarrow Z$ is continuous if and only if the extended domain $f:X \rightarrow Y$ is continuous. The reason is that, if $U \subset Z$ is open in $Z$, then $U=V\cap Z$ for some $V$ that is open in $Y$, meaning $f^{-1}(U)=f^{-1}(V \cap Z)=f^{-1}(V)$ (because $f(X)\subset Z$), so continuity of the "extension" implies continuity of the original function, and if $V$ is open in $Y$, $f^{-1}(V)=f^{-1}(V\cap Z)$ with $V\cap Z$ being open in $Z$, hence continuity of the original function implies continuity of the extension.

So yes, your argument is perfectly valid.

JKRT
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