index of morse functions and homotopical dimension Is it true that any manifold homotopy equivalent to a k-dimensional CW-complex admits a proper Morse function with critical points all of index <= k? I believe this is not true, so I would like to see a counterexample.
 A: Suppose that $M$ has a proper Morse function $f\ge 0$ with all critical points of index at most $k$. Then homotopically $M$ can be made of low-dimensional cells: it has homotopical dimension at most $k$. But also $M$, relative to its boundary $M^{\ge c}$, can be made by attaching high-dimensional cells. Thus the pair $(M,M^{\ge c})$ is $(dim(M)-k-1)$-connected. So for example if $k\le dim(M)-3$ and $M$ is simply connected then $M$ must also be "simply connected at infinity".
A: Take a contractible $3$-manifold which is not homeomorphic to $\mathbb R^3$ -- like the Whitehead manifold. 
If such a Morse function existed on the Whitehead manifold, it would be a Morse function with only one critical point, the minimum, and therefore the Whitehead manifold would be an open 3-ball.  The proof of this has two steps: (1) $f$ can have at most one critical point, WLOG a minimum by homotopy-type considerations, and it can't have less than one critical point by the "edit" below.  Step (2) if it has one critical point let it be $f(p)=0$, the minimum. By the Morse Lemma, $f^{-1}[0,\epsilon]$ is diffeomorphic to a closed 3-ball.  The flow the the gradient of $f$ gives a diffeomorphism between $f^{-1}[\epsilon,\infty)$ and $S^2 \times [\epsilon, \infty)$. Pasting these two diffeomorphisms together gives you a diffeomorphism between the manifold and $\mathbb R^3$. 
edit: Well, I guess there's the special case to consider that the Morse function could have no critical points but then you could deal with this by the argument that the Whitehead manifold isn't a product of a surface and $\mathbb R$, which by the classification of surfaces amounts to saying that the Whitehead manifold isn't homeo/diffeomorphic to $\mathbb R^3$. 
Welcome to MO, Victor!
