If I start with a, say, 3CW complex $X$ which can be embedded in $\mathbb{R}^5$, I can get a neighbourhood $U$ of $X$ which has the same homotopy type of $X$. Then $U$ is a $5$ dimensional open manifold. Can I get a close manifold (compact without a boundary) $M$, of dimension $6$ (or some higher dimension) such that $M$ and $X$ have the same homotopy type?

3$\begingroup$ As the answer indicates, this is impossible in general. For something a little weaker, you might be interested in Mike Davis' "reflection group trick". For any aspherical complex $X$, say, I think this enables you to produce a closed aspherical manifold $M$ such that $\pi_1(M)$ (virtually?) retracts to $\pi_1(X)$. I'm not certain if this grouptheoretic retraction can be improved to a topological retraction. This should be explained in Davis' book "The geometry and topology of Coxeter groups". $\endgroup$– HJRWAug 8, 2021 at 16:11

$\begingroup$ You are almost always going to have to attach some cells to get the homotopytype of a manifold. Perhaps a productive way to rephrase your question would be through the lens of starting with a compact (boundaryless) manifold. If you puncture that manifold, you get the homotopytype of a lowerdimensional CW complex. Which CW complexes do you get? $\endgroup$– Ryan BudneyAug 9, 2021 at 0:43
3 Answers
Take $X=S^3$. Then no closed manifold of dimension at least 6 has the same homotopy type.

6$\begingroup$ More generally, let X be an n dimensional complex and let m > n and let Y be a closed m manifold. Then by poincare duality Y has non vanishing cohomology in dimension m for aproppriate coefficients. But X has vanishing cohomology in dimension m for any coefficients. So X and Y are not homotopy equivalent. $\endgroup$– Tim Campion ♦Aug 8, 2021 at 14:46
More generally, suppose $n \le m$ are nonnegative integers, $X$ is a CW complex of dimension $\le n$, $M$ is a nonempty, closed $m$manifold, and $X$ and $M$ have the same homotopy type.
It is well known that a nonempty closed $m$manifold has nontrivial mod 2 homology in degree $m$, whereas a CW complex of dimension $\le n$ has no homology above dimension $n$.
Since $X$ and $M$ have the same homotopy type, they have isomorphic homology, so $H_m(X;\Bbb Z/2) \ne 0$. So the only possibility is that $m=n$.
The mod2 homology groups of any closed $n$manifold are `$n$palindromic' by Poincare duality, i.e., $H_m(M;\mathbb{Z}/2)\cong H_{nm}(M;\mathbb{Z}/2)$ for each $m$. If the mod2 homology of $X$ is not $n$palindromic for any $n$, there cannot be any closed manifold $M$ having the same mod2 homology as $X$.
The smallest such $X$ is the onepoint union of two circles.