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I'm trying to solve a problem in Spivak's A comprehensive introduction to differential geometry. Here, the definition of a manifold is the next

A metric space $X$ is said to be a manifold if every point in $X$ has a neighbourhood homeomorphic to $\mathbb{R}^n$ for some $n$

Given $x\in X$, a neighborhood is just a set $U$ such that $p\in \mathrm{int}U$ (so $U$ is not necessarily open). However, after the enunciation of the Invariance of Domain theorem, he assures that the neighbourhood in the definition of a manifold must in fact be open and that this is an easy consequence of such theorem. This is what I've been trying to prove, although I couldn't succeed doing so.

My attempt so far was the next: Given a point $x$ in the manifold, a neighborhood $U$ of $x$ and a homeomorphism$\phi:U\to \mathbb{R}^n$, I tried to show that $U\subset \mathrm{int} U$, that is, for every $y\in U$ there's an open set $W$ in $X$ such that $y\in W\subset U$. For such a $y$ there exists a neighborhood $V$ of $y$ and a homeomorphism $\psi :V\to\mathbb{R}^m$ for some $m$. Set $A=U\cap \mathrm{int}V$. Since $A$ is open in $U$, $\phi(A)$ is open in $\mathbb{R}^n$, and we have a one-one continuous function $\psi\circ \phi^{-1}: \phi(A)\to \psi(A)$ from an open subset of $\mathbb {R}^n$ to some subset of $\mathbb {R}^m$. Here's where I'm stuck.

I can see that if $\psi(A)$ is open, then we're done. This would follow from ID theorem if $m=n$, but I can't see why this has to be true. From said theorem I was able to show that $n\leq m$ but I couldn't get further.

Could you please help me this? Also, I would appreciate more hints than complete answers. Thank you.

Brandon
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  • When you define a new homeomorphism $\psi$, why would the dimension of the manifold change? Shouldn't each homeomorphism on a neighborhood of a manifold be to $\mathbb{R}^n$? This would make more sense, since one often says "n-manifold" instead of just "manifold," and that would trivially fix your $m=n$ problem. – Glare Jan 18 '15 at 00:21
  • Isn't it the case that $\psi \circ \phi^{-1} : \phi(A) \to \psi(A)$ is a homeomorphism? In that case, invariance of dommain implies $m=n$, no? – Seub Jan 18 '15 at 00:25
  • @Glare not really, since you could take the disjoint union of one 1-manifold and a 2-manifold and get a manifold in the sense of the Spivak's definition (without regarding the metric). I think the term $n$-manifold were made just to avoid this kind of manifolds. – Brandon Jan 18 '15 at 00:32
  • @Seub yes, it is an homeomorphism, however, the theorem requires the target to be of the same dimension as the ambient space of the domain. For instance, you could consider the canonical embedding of $\mathbb{R}^n$ into $\mathbb{R}^{n+1}$: although is an homeomorphism with its image, the latter is not open in $\mathbb {R}^{n +1}$. – Brandon Jan 18 '15 at 00:35
  • I see. Okay I hope I got it right this time: it follows from Spivak's definition that $X$ is a manifold iff every point in $X$ has an open neighborhood homeomorphic to $\mathbb{R}^n$ for some $n$. Agreed? Then, in what you wrote, you can assume that $V$ is open. It follows that $\psi(A)$ is open. What do you think? (note that you were not using anywhere that $V$ is a neighborhood of $y$, I think) – Seub Jan 18 '15 at 00:55
  • If the manifold is connected every point has an open neighborhood homeomorphic to $\mathbb{R}^n$ for some fixed $n$. That's not specified in the definition, it is proved. Disconnected manifolds can have open neighborhoods of different dimensions at different points. – Matt Samuel Jan 18 '15 at 02:32

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