If I start with a, say, 3-CW 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?
Algebraic Topology – How to Construct a Closed Manifold from a Finite CW Complex
at.algebraic-topologycw-complexesdifferential-topologygn.general-topologysmooth-manifolds
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I don't think that
torsion in the homology has been ruled out
Certainly, torsion in Cech cohomology has been ruled out for a compact subset. The "usual" universal coefficient formula, relating Cech cohomology to $\operatorname{Hom}$ and $\operatorname{Ext}$ of Steenrod homology, is not valid for arbitrary compact subsets of $\Bbb R^3$ (although it is valid for ANRs, possibly non-compact). The "reversed" universal coefficient formula, relating Steenrod homology to $\operatorname{Hom}$ and $\operatorname{Ext}$ of Cech cohomology is valid for compact metric spaces, but it does not help, because $\operatorname{Ext}(\Bbb Z[\frac1p],\Bbb Z)\simeq\Bbb Z_p/\Bbb Z\supset\Bbb Z_{(p)}/\Bbb Z$, which contains $q$-torsion for all primes $q\ne p$. (Here $\Bbb Z_{(p)}$ denotes the localization at the prime $p$, and $\Bbb Z_p$ denotes the $p$-adic integers. The two UCFs can be found in Bredon's Sheaf Theory, 2nd edition, equation (9) on p.292 in Section V.3 and Theorem V.12.8.)
The remark on $\operatorname{Ext}$ can be made into an actual example. The $p$-adic solenoid $\Sigma$ is a subset of $\Bbb R^3$. The zeroth Steenrod homology $H_0(\Sigma)$ is isomorphic by the Alexander duality to $H^2(\Bbb R^3\setminus\Sigma)$. This is a cohomology group of an open $3$-manifold contained in $\Bbb R^3$, yet it is isomorphic to $\Bbb Z\oplus(\Bbb Z_p/\Bbb Z)$ (using the UCF, or the Milnor short exact sequence with $\lim^1$), which contains torsion. Of course, every cocycle representing torsion is "vanishing", i.e. its restriction to each compact submanifold is null-cohomologous within that submanifold.
By similar arguments, $H_i(X)$ (Steenrod homology) contains no torsion for $i>0$ for every compact subset $X$ of $\Bbb R^3$.
It is obvious that "Cech homology" contains no torsion (even for a noncompact subset $X$ of $\Bbb R^3$), because it is the inverse limit of the homology groups of polyhedral neighborhoods of $X$ in $\Bbb R^3$. But I don't think this is to be taken seriously, because "Cech homology" is not a homology theory (it does not satisfy the exact sequence of pair). The homology theory corresponding to Cech cohomology is Steenrod homology (which consists of "Cech homology" plus a $\lim^1$-correction term). Some references for Steenrod homology are Steenrod's original paper in Ann. Math. (1940), Milnor's 1961 preprint (published in http://www.maths.ed.ac.uk/~aar/books/novikov1.pdf), Massey's book Homology and Cohomology Theory. An Approach Based on Alexander-Spanier Cochains, Bredon's book Sheaf Theory (as long as the sheaf is constant and has finitely generated stalks) and the paper
- Sergey A. Melikhov, Steenrod homotopy, Russ. Math. Surveys 64 (2009) 469-551; translated into Russian in: Uspekhi Mat. Nauk 64:3 (2009) 73-166, doi:10.1070/RM2009v064n03ABEH004620, arXiv:0812.1407
As for torsion in singular $4$-homology of the Barratt-Milnor example, this is really a question about framed surface links in $S^4$ (see the proof of theorem 1.1 in the linked paper).
Here is an example where ${\rm Diff}(M)$ with the compact-open topology is not homotopy equivalent to a CW complex. Take $M$ to be a surface of infinite genus, say the simplest one with just one noncompact end. I will describe an infinite sequence of diffeomorphisms $f_n:M\to M$ converging to the identity in the compact-open topology and all lying in different path-components of ${\rm Diff}(M)$. Assuming this, suppose $\phi:{\rm Diff}(M) \to X$ is a homotopy equivalence with $X$ a CW complex. The infinite sequence $f_n$ together with its limit forms a compact set in ${\rm Diff}(M)$, so its image under $\phi$ would be compact and hence would lie in a finite subcomplex of $X$, meeting only finitely many components of $X$. Thus $\phi$ would not induce a bijection on path-components, a contradiction.
To construct $f_n$, start with an infinite sequence of disjoint simple closed curves $c_n$ in $M$ marching out to infinity, and let $f_n$ be a Dehn twist along $c_n$. The $f_n$'s converge to the identity in the compact-open topology since the $c_n$'s approach infinity. We can choose the $c_n$'s so that they represent distinct elements in a basis for $H_1(M)$ and then the $f_n$'s will induce distinct automorphisms of $H_1(M)$. If two different $f_n$'s were in the same path-component of ${\rm Diff}(M)$ they would have to induce the same automorphism of $H_1(M)$ since any path joining them would restrict to an isotopy of any simple closed curve in $M$ (see the next paragraph below) and a basis for $H_1(M)$ is represented by simple closed curves.
If $g_t$ is a path in ${\rm Diff}(M)$ then the images $g_t(c)$ of any simple closed curve $c$ vary by isotopy since this is true as $t$ varies over a small neighborhood of a given $t_0$, so since the $t$-interval $[0,1]$ is compact, a finite number of these neighborhoods cover $I$ and the claim follows.
Remark: The $f_n$'s were chosen to be Dehn twists just for convenience. Many other choices of diffeomorphisms would work just as well. One can easily see how to generalize to higher dimensions.
Best Answer
Take $X=S^3$. Then no closed manifold of dimension at least 6 has the same homotopy type.