Problem is this: suppose a manifold $$M=\bigcup_{n\in\mathbb{N}} U_n,$$ where each $U_n$ is diffeomorphic to Euclidean space, and $U_n$ is contained in $U_{n+1}$. Then please show that $M$ is diffeomorphic to Euclidean space.
How to prove a manifold is diffeomorphic to Euclidean space?
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0I suppose you meant to say that each $U_i$ is an open subset of $M$? – 2012-11-24
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0Where is this question from? – 2012-11-24
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0@Hurkyl, aren't they automatically open for being diffeomorphic to R^n? – 2012-11-25
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0@levap GTM33, page 21. – 2012-11-25
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0You can check out [this](http://arxiv.org/pdf/math/0404372.pdf) article. It shows it under weaker hypothesis that doesn't involve smoothness, but it also seems that the original article which Hirsch refers to is also about topological spaces, with no mention of a smooth structure. It is interesting whether the smoothness assumption can possibly simply things further. – 2012-11-25
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0@lee: Mainly, I wanted confirmation that the choice of $U_n$'s is constrained by the hypothesis that $\bigcup U_n$ is actually a manifold. When I first read your question, my first thought was to set $U_n = \mathbb{R}^n$, and the union wouldn't have been a Euclidean space at all! – 2012-11-25
1 Answers
Here is a proof which works for $m=dim(M)\ne 4$, I am not sure about the remaining case $m=4$.
By shrinking each $U_n$ we obtain an exhaustion of $M$ by nested smooth closed $m$-dimensional balls: $$ B_1\subset B_2 \subset ... $$
By the h-cobordism theorem, for each $n$, the region $B_n- int(B_{n-1})$ is diffeomorphic to the annulus $S^{m-1}\times [0,1]$. By lining up these diffeomorphisms, we obtain a diffeomorphism $M- int(B_1)\to S^{m-1}\times [0,1)$.
What makes this "lining up" work is the fact that if $A$ is a manifold with boundary diffeomorphic to $S^k\times [0,1]$ and $h$ is a diffeomorphism from one boundary component of $A$ to $S^k\times \{0\}$, then $h$ extends to a diffeomorphism $A\to S^k\times [0,1]$.
Hence, $M$ is diffeomorphic to $R^m$.
In the topological category, this argument also shows that $M$ is homeomorphic to $R^m$ even if $m=4$. (Since the topological annulus conjecture is proven by Quinn in dimension 4.)