# How can I prove that two sets are isomorphic, short of finding the isomorphism?

I have a set $E \subset X$ within a metric space ($X, d$). I want to prove that it is isomorphic to $\mathbb{R}^{n \times n}$, in the sense that there exists a continuous bijection between the two. Because $E$ is a fairly complicated set, it would be a huge pain to actually find an exact bijection, so instead I hope to identify a sufficient suite of conditions that I can test $E$ for that will suffice to show that the two are isomorphic.

Is there some sort of known method for doing this? Or will I have to find the exact function?

Thanks.

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What are $E$ and $X$? – Jonas Meyer Jan 21 '13 at 3:54
It need not be the case in general, i.e. $\Bbb Q$ is not homeomorphic/isomorphic to $\mathbb{R}$. – Clayton Jan 21 '13 at 3:55
@JonasMeyer $X$ is $\mathbb{R}^{n \times n} \times \mathbb{R}^n$ and $E$ is a strange set within that that would take a fair amount of effort to explain. I hope to find a test suite of conditions that applies to any $E$ we might choose. – GMB Jan 21 '13 at 3:59
@Clayton I don't follow: I don't believe there is a continuous bijection between $\mathbb{Q}$ and $\mathbb{R}$. – GMB Jan 21 '13 at 4:00
A continuous bijection of manifolds is an homeomorpism, as a consequence of the theorem of Invariance of Domain. It is important here that there be no boundaries (although this extends to manifolds with compact boundaries, iirc) – Mariano Suárez-Alvarez Jan 21 '13 at 4:44

The necessary and sufficient set of conditions for $E$ to be homeomorphic to $\mathbb R^{m}$ (in your situation $m=n^2$):