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Suppose $X$ is a CW complex and $M$ is a closed manifold and suppose further that there exists a homotopy equivalence $X \simeq M$. Does there exists an embedding of $M$ into $X$ (i.e. an injective (potentially cellular) map)?

If this setting is to broad, I'm specifically interested in the case, where $M$ is a surface and $X$ is also $2$-dimensional (maybe even restrict it to aspherical surfaces).

Edit: mme provided a counterexample in dimension 3 (homotopy equivalent but not homeomorphic lens spaces), which can probably be generalized to higher dimensions. So only the two-dimensional case remains.

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    $\begingroup$ If $\mathbb{R}^n$ Or some open ball inside it count as closed manifolds then the answer is no as they are homotooy equivalent to a point. $\endgroup$
    – user51223
    Commented Aug 18, 2021 at 9:47
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    $\begingroup$ A closed manifold is a boundaryless, compact manifold. $\endgroup$
    – ThorbenK
    Commented Aug 18, 2021 at 9:51
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    $\begingroup$ No, in dimension 3. Choose two lens spaces M and X which are homotopy equivalent but not homeomorphic. Then a choice of homotopy equivalence f gives a counterexample: an injective map $M \to X$ between closed manifolds of the same dimension is a homeomorphism by invariance of domain. I don't dare make a guess in 2D. $\endgroup$
    – mme
    Commented Aug 18, 2021 at 11:32
  • $\begingroup$ No in dimension two. Take Bing's house of two rooms (eg, here: sketchesoftopology.wordpress.com/2010/03/25/bings-house) and attach a disk along the equator. This is homotopy equivalent to the two-sphere, but the homotopy equivalence is not realised by an embedding. $\endgroup$
    – Sam Nead
    Commented Aug 18, 2021 at 19:00

2 Answers 2

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Pick a torus, and add two discs along a meridian and a longitude. You get a 2-complex homotopic to a sphere that does not contain a sphere. This generalises easily to any genus by picking a genus-$g$ surface.

More generally, a finite 2-complex contains finitely many surfaces, and there are some moves (like the Matveev - Piergallini move) that preserve the (simple) homotopy type of the 2-complex, but can modify the surfaces it contains.

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  • $\begingroup$ Where is a good place to read the story described in your second paragraph? $\endgroup$
    – mme
    Commented Aug 18, 2021 at 18:50
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    $\begingroup$ I would maybe suggest the first two chapters of the book "Algorithmic Topology and Classification of 3-Manifolds" of Matveev. They are focused on spines of 3-manifolds. $\endgroup$
    – Bruno Martelli
    Commented Aug 18, 2021 at 18:54
  • $\begingroup$ that is a nice example, but how do you show that there is no embedding of the sphere into the 2-complex? Here by embedding I mean any continuous embedding. Cellular embedding seems to me a bad notion since it is rather restrictive and in general is also unclear what CW-structures for $M$ one is supposed to use $\endgroup$
    – Stefan Friedl
    Commented Aug 29, 2021 at 6:05
  • $\begingroup$ I would say that the image of the embedding is a surface without boundary, hence it must be a union of 2-cells. There is only one subsurface that is a union of 2-cells, and it is a torus. $\endgroup$
    – Bruno Martelli
    Commented Aug 29, 2021 at 6:16
  • $\begingroup$ Yes, good point. One can show, using the open mapping theorem, that the intersection of the image of a closed $n$-dimensional manifold and the interior of an $n$-dimensional cell is either empty or everything. $\endgroup$
    – Stefan Friedl
    Commented Aug 29, 2021 at 9:44
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While thinking about it with a friend, we came up with the following two dimensional counter example:

Take the standard knot diagram of the trefoil knot (as a self-intersecting curve in $\mathbb{R}^2$) and let $\bar{X}$ denote the "inner" of this curve i.e. the curve together with the 4 areas bounded by it. Let $X$ denote $\bar{X} \cup_\phi D^2$, where $\phi \colon S^1\to \bar{X}$ follows the trefoil knot. Since $\bar{X}$ is contractible, $X$ is homotopy equivalent to $S^2$. Using cellular homology, one can see that the "fundamental class" of $X$ hits the middle cell of $\bar{X}$ twice, hence there are no injective homotopy equivalences.

By taking "connected sums" of this counterexample with surfaces, one obtains counter examples for all closed surfaces.

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