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



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> 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 $\mathbb{R}^{n}$?
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> 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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functions which covers(good covers) manifolds
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](http://en.wikipedia.org/wiki/Lusternik%E2%80%93Schnirelmann_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.


