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Let $M$ be a (not necessarily compact)) smooth manifold.

1.Is there a smooth map $f:M\to \mathbb{R}$ and an open covering $\mathbb{R}=\cup U_{\alpha}$ such that each $f^{-1}(U_{\alpha})$ is homeomorphic to $\mathbb{R}^{n}$?

2.Is there a smooth map $f:M \to \mathbb{R}^{k}$, for some $k \in \mathbb{N}$ and an open covering $\mathbb{R}^{k}=\cup U_{\alpha}$ such that $f^{-1}(U_{\alpha})$ is a good cover for $M$?

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  • $\begingroup$ Does your definition of "manifold" include "second countable"? $\endgroup$ Nov 2, 2014 at 23:47
  • $\begingroup$ @NateEldredge Yes I mean the standard definition of manifold.May be you are considering the long line? $\endgroup$ Nov 3, 2014 at 21:06
  • $\begingroup$ Right, some people consider the long line to be a manifold and others do not. $\endgroup$ Nov 4, 2014 at 2:08

2 Answers 2

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For (1), the answer is almost always no; the combinatorial properties of open covers of $\mathbb{R}$ are far too restricted. Suppose that $M$ is compact and connected and such an $f$ and $\{U_\alpha\}$ exist. Let the image of $f$ be $[a,b]$. Each $U_\alpha$ is a union of disjoint intervals; by connectedness, only one of those disjoint intervals can intersect $[a,b]$. Restricting the codomain of $f$ to $[a,b]$, we may thus assume each $U_\alpha$ is a single interval. Now take a minimal subcollection of $\{U_\alpha\}$ that still covers $[a,b]$: this will be a finite sequence of intervals $I_k=(x_k,y_k)$ going from left to right such that only consecutive intervals intersect. That is, we have $$x_1\leq x_2\leq y_1\leq x_3\leq y_2\leq x_4\leq y_3 \leq\dots$$

Let $U$ be the union of the $f^{-1}(I_k)$ for $k$ odd and $V$ be the union of the $f^{-1}(I_k)$ for $k$ even. Then both $U$ and $V$ are disjoint unions of copies of $\mathbb{R}^n$, and they cover $M$. This implies that the LS-category of $M$ must be at most 1. In particular, for instance, this implies that the cohomology of $M$ can have no nontrivial cup products. If $M$ is not compact, a similar (but a little bit more difficult) argument can be made to reach the same conclusion.

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  • $\begingroup$ what about the following weaker version :In first question we consider the weaker statment:' $f^{-1}(U_{\alpha})$ homeomorphic to an open subset of $\mathbb{R}^{n}$"? $\endgroup$ Nov 4, 2014 at 15:58
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For 2., this might be nuking a mosquito, but here is an argument for existence of such $f$ assuming that $M$ is paracompact. WLOG, assume $M$ is connected (and hence second-countable). Use the Whitney embedding theorem to find an embedding $f: M \to \mathbb{R}^k$ as a closed submanifold of a Euclidean space; use this embedding to think of $M$ as a subset of $\mathbb{R}^k$. The Euclidean metric on $\mathbb{R}^k$ restricts to a Riemannian metric on $M$. For each point $p \in M$, find a ball $B_{r(p)}(p)$ in $\mathbb{R}^k$ such that $f^{-1}(B_{r(p)}(p)) = B_{r(p)}(p) \cap M$ is geodesically convex in $M$ (see here for example). The $B_{r(p)}(p)$ plus the complement of $M$ then form an open cover $U_\alpha$ of $\mathbb{R}^k$ for which the inverse images $f^{-1}(U_\alpha)$ form a good open cover.

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    $\begingroup$ I could be wrong, but I suspect this particular mosquito is genetically engineered to resist non-nuclear weaponry. In other words, I think the result is no easier than the Whitney embedding theorem. $\endgroup$ Nov 3, 2014 at 4:28
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    $\begingroup$ @PaulSiegel I suspect you're right. I confess that when I wrote those words, I was in the midst of deploying greater nuclear power by first putting a Riemannian structure on $M$, and then using the Nash embedding theorem to isometrically embed in Euclidean space. Then it occurred to me that the above would involve much less weaponry. :-) $\endgroup$
    – Todd Trimble
    Nov 3, 2014 at 5:58
  • $\begingroup$ @ToddTrimble Thanks for your very interesting answer. What about the following possible definition: $\endgroup$ Nov 3, 2014 at 21:36
  • $\begingroup$ For a manifold $M$, we define the minimum $k$ such that a good covering of $M$ can be obtained via pull back of an open covering of $\mathbb{R}^{k}$ for some smooth function $f:M\to \mathbb{R}^{k}$. Is this invariant some how related to some thing as LS category as in the answer of @Eric? As you said this minimum is less than $2n+1$. Thanks again for your interesting answer. $\endgroup$ Nov 3, 2014 at 21:41

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