Homotopy type of smooth manifolds with boundary It seems very likely to me that every smooth connected  $n$-dimensional manifold with non-empty boundary has the homotopy type of a $(n-1)$-dimensional CW complex. Is that true and how to prove it? (or a counter-example?) Does this manifold need to be compact? What about $n$-dimensional open manifolds? Thank you.
 A: I know of two proofs in the compact case. Let $M$ be a compact smooth $m$-manifold with boundary $\partial M$. 
1) Morse theory (Sketch). For this I think we need $m \ge 4$.
 There is a Morse function $f: M\to \Bbb [0,\infty)$ with $f^{-1}(0) = \partial M$ and $f^{-1}(1) = \emptyset$. One can assume $f$ is self-indexing. This will give a handlebody structure on $M$ relative to $\partial M$, i.e., $M$ is obtained from attaching handles of index at most $m$. By the technique of cancelling $0$-handles, one can
alter the handle decomposition so that there are no $0$-handles. If we turn $f$ upside down, we obtain a handle decomposition of $M$ with no $m$-handles.  So $M$  is becomes a handlebody whose handles have index $< m$. This will imply (by an argument of Milnor) that $M$ has the homotopy type of a CW complex of dimension $\le m-1$.  
2) Wall's theory of finiteness ($m \ge 4$). 
First use that fact that $M$ is has the homotopy type of some CW complex (that is well-known and can be proved in different ways). By Poincaré duakity, Wall's condition $D_{m-1}$ is satisfied. Conseuqently, $M$ has the homotopy type of a CW complex of dimension $\le m-1$ by Wall's "Theorem E." Wall's paper can be found here:
C.T.C. Wall, Finiteness Conditions for CW-Complexes.
Annals of Mathematics
81 (1965), pp. 56–69
