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I have a problem where I have to show that for two intersecting open subsets $U$ and $V$ of a topological manifold, if we have two homeomorphisms $\phi : U \to \mathbb R^n$ and $\psi: V \to \mathbb R ^ m$, and $\psi \circ \phi ^{-1} \mid_{\phi(U\cap V)}$ and $\phi \circ \psi ^{-1} \mid_{\psi(U\cap V)}$ are $C^1$, then $n = m$.

It's obvious that this must be true because we have homeomorphisms from open subsets of $\mathbb R^n$ to $\mathbb R^m$, but we don't have access to the topological machinery required to prove that. Also, we have the extra assumption that the maps are $C^1$, which suggests a more differential geometric approach. My problem is that I'm not sure what this approach is. I took vector calculus forever ago and there are so many ways to look at this.

I tried assuming that $n \ne m$ and looking at the Jacobian at a point and showing that it must either fail to be surjective or injective, but this isn't panning out (probably just because of my lack of technical skill), and I'm not sure if it will even work, or if it is the easiest way of going about this. I have a feeling that there is an easy <3 line proof.

This is homework, so suggestions things to try would be better than direct answers.

fhyve
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  • Hints: What is the Jacobian matrix of the inverse map? Is this matrix invertible? – Dimitris Sep 11 '14 at 22:52
  • The "differential" version of the invariance of dimension theorem is indeed easier to prove. You need to use the fact that $f$ induces a linear transformation on tangent spaces. Do you have access to such tools? – Ayman Hourieh Sep 11 '14 at 22:52
  • No, we just started introducing smooth manifolds, so just coordinate charts, atlases, etc., and whatever we have from vector calculus. – fhyve Sep 11 '14 at 22:56
  • It's essentially the same argument as the accepted answer below, but using fancier language. – Ayman Hourieh Sep 12 '14 at 20:49

2 Answers2

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I am elaborating on my comment. You have two maps $\phi\circ \psi^{-1} : \psi(U\cap V) \to \phi (U\cap V)$ and $\psi\circ \phi^{-1} : \phi (U\cap V)\to \psi(U\cap V)$, which are inverses of each other, and $C^1$ smooth. The Jacobian matrix of $\phi\circ \psi^{-1}$ is an $n\times m$ matrix. The Jacobian matrix of $\psi\circ \phi^{-1}$ is an $m\times n$ matrix and by the chain rule this matrix is the inverse of the Jacobian of $\phi\circ \psi^{-1}$. In particular, $n=m$ since the matrices have to be square.

Dimitris
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  • Really this is just using the chain rule applied to $f\circ g$ equal to the identity. The inverse function theorem tells you that maps with non-singular derivatives are local diffeomorphisms. – PVAL-inactive Sep 11 '14 at 23:05
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Without loss of generality assume $m\leq n$. Let $f=\phi\circ \psi^{-1},$ $g=\psi\circ\phi^{-1}$ on the sets you restricted to. Now $f\circ g$ is the identity on some open subset of $\Bbb R^n$. The chain rule says that $$J_x(f\circ g)=J_{g(x)}(f)\circ J_x(g)$$ As the derivative of the identity is $I_{\Bbb R^n}$, we have both $J_{g(x)}(f)$ and $J_x(g)$ must be rank $n$. So $n=m$.

PVAL-inactive
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