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.

  • $\begingroup$ I suppose you meant to say that each $U_i$ is an open subset of $M$? $\endgroup$
    – user14972
    Nov 24 '12 at 17:50
  • $\begingroup$ Where is this question from? $\endgroup$
    – levap
    Nov 24 '12 at 19:45
  • $\begingroup$ @Hurkyl, aren't they automatically open for being diffeomorphic to R^n? $\endgroup$
    – lee
    Nov 25 '12 at 5:38
  • $\begingroup$ @levap GTM33, page 21. $\endgroup$
    – lee
    Nov 25 '12 at 5:39
  • $\begingroup$ You can check out this 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. $\endgroup$
    – levap
    Nov 25 '12 at 6:08

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.)


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