TITLE: Does every finite $CW$ complex have the homotopy type of a smooth manifold? How about infinite $CW$ complexes?
QUESTION [5 upvotes]: Let M be a smooth manifold. I was wondering if every finite or infinite CW complex has the homotopy type of a smooth manifold because it seems way easier to compute De Rham cohomology groups than homology groups.
If the answer to the questions are yes, is it hard to find a smooth approximation of a CW complex?

REPLY [9 votes]: For finite complexes - yes, for infinite complexes in general - no.

Why yes: First, each finite CW  complex $W$ is homotopy-equivalent to a finite simplicial complex $C$. (This should be in Hatcher's "Algebraic Topology.") Next, each finite simplicial complex embeds as a subcomplex in a simplex which then embeds in some ${\mathbb R}^n$ as a part of a triangulation of the latter. Now, you have a subcomplex $X$ of a triangulated ${\mathbb R}^n$. Take the regular neighborhood $N_X$ of $X$ in ${\mathbb R}^n$: It will be homotopy-equivalent to $X$ and a compact (PL) submanifold with boundary. (Both should be in Rourke and Sanderson's "PL topology".)  Lastly, take the interior of $N_X$. This is your manifold.

With a bit more work, this proof extends to the case of countable CW complexes of finite dimension. (The main difference is that you first prove the existence of a finite-dimensional locally finite simplicial complex $C$ homotopy-equivalent to $W$, this is in Hatcher's book. Then the simplicial version of Whitney's theorem gives you a proper embedding as a subcomplex in ${\mathbb R}^n$. The rest is the same.)

Why no (even for countable complexes). Consider the wedge of spheres of all dimensions. The resulting CW complex $X$ has the property that
$$
H_k(X)\ne 0, \forall k\in {\mathbb N}. 
$$
However, if $M$ is an $m$-dimensional manifold then $H_k(M)=0$ for all $k>m$. (This is a consequence of the Poincare duality, again is in Hatcher's book.)

As another, finite-dimensional, example, take $X$ to be an uncountable set with discrete topology, or (if you prefer a connected example), an uncountable wedge of circles. Then $H_1(X)$ has uncountable rank which is impossible for a manifold. (I am assuming that manifolds are required to be 2nd countable, which is a standard assumption, although some people disagree.)