functions which covers(good covers) manifolds 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  $M$? 

 A: 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. 
A: 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.
