Does every finite $CW$ complex have the homotopy type of a smooth manifold? How about infinite $CW$ complexes? 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?
 A: 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.)
A: This is a partial answer.
If $X$ is a finite CW-complex, then in general there does not exist a compact manifold which is homotopy equivalent to $X$.
It clearly suffices to consider connected CW-complexes because we can argue componentwise.
So let $X$ be a finite connected CW-complex of dimension $n$ such that its $n$-th homology group with coefficients in $\mathbb Z_2$ has the property $H_n(X;\mathbb Z_2) \ne \mathbb Z_2$. An example is $S^n \vee S^n$. Assume that $X$ is homotopy equivalent to a compact manifold $M$ of dimension $m$. It must clearly be connected.
A well-known fact (see e.g. Hatcher Theorem 3.26) is that $H_k(M;\mathbb Z_2) = 0$ for $k > m$ and $H_m(M;\mathbb Z_2) = \mathbb Z_2$. Hence $n \ge m$ is impossible. But also $n < m$ is impossible because in that case $H_m(X) = 0$.
This leaves open whether there always exists a non-compact manifold $M$ which is homotopy equivalent to $X$. In the example $X  = S^1 \vee S^1$ this is the case.
Update:
In Moishe Kohan's answer you see that the answer is "yes" for finite (connected) CW-complexes and non-compact (connected) manifolds. Note that if $X$ is homotopy equivalent to such $M$, then necessarily $\text{dim} M  >  \text{dim} X$ provided $H_{\text{dim} X}(X) \ne 0$. See Hatcher Proposition 3.29.
