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Say that we have $f: \mathbb{R}^m \rightarrow \mathbb{R}^n$, that is $C^1$. Also, we know that $f^{-1}: \mathbb{R}^n \rightarrow \mathbb{R}^m$ exists and is $C^1$ too. Intuitively, I think this means that the Jacobian determinant is non-zero and that the Jacobian $\mathbf{J}(f)$ must be square $k \times k$, where $k = m = n$ --- so it means that $m = n$ --- but the Jacobian determinant is only defined for square matrices, so that line of thinking seems tautological.

Is my intuition correct? How would such an argument actually be built up? Please provide hints instead of laying out the answer (I really need to learn this stuff), and I will ask further questions if needed in the comments!

bzm3r
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    Your hunch that $m=n$ is correct, though it's a non-trivial result. See https://en.wikipedia.org/wiki/Invariance_of_domain – 211792 Sep 13 '16 at 02:23
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    @AustinC This question does come from a topology exercise I am attempting to tackle, so the connection is not surprising. But, my instructor assured me I only need the most basic facts to come to the conclusion -- so I think what I have to do shouldn't be as tricky as having to recreate Brouwer's result. – bzm3r Sep 13 '16 at 02:31
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    @AustinC Are you sure invariance of domain is relevant here? That is much stronger, and describes homeomorphisms between Euclidean spaces. Here we have differentiable maps, which have much more structure. – Neal Sep 13 '16 at 02:55
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    To OP - what about $f:\Bbb{R}^2\to \Bbb{R}$ defined by $f(x,y) = x$, and $f^{-1}$ defined by $f^{-1}(t) = (t,0)$? These are both $C^1$ maps. Are you also making assumptions about surjectivity and injectivity? – Neal Sep 13 '16 at 02:57
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    @Neal Yes, I am also making assumptions about surjectivity and injectivity, because I built $f$ by chaining together homeomorphisms, which are bijective. So, $f$ is bijective too, unlike in your example. That should have been something I thought about...$m = n$ also means that I can have bijective functions, otherwise you can't have bijectivity...(information from some dimension would be squeezed out)? – bzm3r Sep 13 '16 at 03:03
  • @Neal Now I think the bijectivity is key to showing that $m = n$. Would some sort of counting argument work? "There are a lot more points in $\mathbb{R}^n$ than in $\mathbb{R}^m$ (my real analysis is weak -- compare cardinality of $\mathbb{R}^n$ vs $\mathbb{R}^m$), if $n > m$...so how could a bijective function exist, similarly if $n < m$...so $n = m$" – bzm3r Sep 13 '16 at 03:08
  • @Neal Oh, the continuity and the bijectivity go hand in hand: http://math.stackexchange.com/questions/197735/injective-function-from-mathbbr2-to-mathbbr ! – bzm3r Sep 13 '16 at 03:14
  • @Neal Sorry, I didn't mean to imply that it was necessary to prove invariance of domain, only that the conclusion $m=n$ is correct (and indeed, holds even with weaker conditions on $f$). – 211792 Sep 13 '16 at 03:20

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