# Can we embed a closed manifold into a homotopy equivalent CW complex?

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.

• 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. Aug 18, 2021 at 9:47
• A closed manifold is a boundaryless, compact manifold. Aug 18, 2021 at 9:51
• 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.
– mme
Aug 18, 2021 at 11:32
• 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. Aug 18, 2021 at 19:00

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.

• Where is a good place to read the story described in your second paragraph?
– mme
Aug 18, 2021 at 18:50
• 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. Aug 18, 2021 at 18:54
• 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 Aug 29, 2021 at 6:05
• 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. Aug 29, 2021 at 6:16
• 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. Aug 29, 2021 at 9:44

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.