# First homology torsion element of a space satisfying certain conditions

My question is relating to topological spaces $$X$$ that can be expressed as $$X = A \cup B$$, where $$A$$, $$B$$, and $$A \cap B$$ are all homotopy equivalent to $$S^1$$. In particular, I am interested in the torsion element their first homology.

There does exist such a space $$X$$ with first homology torsion free: The torus $$T$$ has: $$H_1(T) = \mathbb{Z} \oplus \mathbb{Z}$$.

There also exists a space satisfying the above with torsion element of order $$2$$: the Klein bottle $$K$$ has $$H_1(K) = \mathbb{Z} \oplus \mathbb{Z}_2$$. - The Klein bottle can be given as the union of two Möbius bands, which are homotopy equivalent to $$S^1$$.

But can we construct a space $$X$$, satisfying the above, with torsion element of order $$3$$? Say: $$H_1(X) = \mathbb{Z} \oplus \mathbb{Z}_3$$.

Or, more generally, can we construct a space $$X$$, satisfying the above, with torsion element of order $$n$$ for some $$n \in \mathbb{N}$$? Say: $$H_1(X) = \mathbb{Z} \oplus \mathbb{Z}_n$$.

My first step has been to try to find any surface with first homology torsion element of order $$3$$, and then subsequently seeing if I can express it as a union of two spaces $$A$$ and $$B$$, as above.

I've tried adding cross caps to the sphere, but the torsion element of these spaces always seems to remain $$2$$. The same applies if you take any 2-manifold with any number of "holes" and cross caps.

All help would be highly appreciated.

• How do you decompose torus into $A\cup B$ such that $A,B$ and $A\cap B$ are homotopy equivalent to $S^1$? Commented May 8, 2020 at 9:12
• @freakish Couldn't we say two open cylinders overlapping in one end and "kissing" in the other end, one of them having a boundary and the other not? Commented May 8, 2020 at 9:15
• Ah, yes. For some reason I though $A,B$ have to be closed... Commented May 8, 2020 at 9:16

Here's one possible construction for arbitrary $$n$$.

Let $$f: S^1\to S^1$$ denote a map of degree $$n$$, and $$c: S^1\to S^1$$ a null map.

Then you can construct the double mapping cylinder $$S^1\overset{c}\leftarrow S^1 \overset{f}\rightarrow S^1$$.

Concretely, you start from $$S^1\coprod (S^1\times [-1,1]) \coprod S^1$$, and then you identify $$S^1\times \{-1\}$$ to the leftmost $$S^1$$ via $$c$$, and $$S^1\times \{1\}$$ to the rightmost $$S^1$$ via $$f$$.

Let's call that $$X$$. Then you'll want to take $$A$$ to be the image of $$S^1\coprod (S^1\times [-1, \epsilon))$$ for some small $$\epsilon >0$$ (I'm doing this to have it be open to simplify the argument later, but it doesn't change much), and $$B$$ to be the image of $$(S^1\times (-\epsilon, 1])\coprod S^1$$.

Then $$A$$ is just the mapping cylinder of $$c$$, and $$B$$ the mapping cylinder of $$f$$, in particular, $$A\simeq S^1$$ and $$B\simeq S^1$$ (via the projections onto the leftmost $$S^1$$, and rightmost $$S^1$$ respectively)

Moreover, $$A\cap B$$ is homeomorphic to $$S^1\times (-\epsilon,\epsilon)$$, so it's also homotopy equivalent to $$S^1$$.

Now I specifically chose $$A,B$$ to be open to be able to apply the Mayer-Vietoris long exact sequence : we get $$H_1(A\cap B)\to H_1(A)\oplus H_1(B)\to H_1(X)\to 0$$ (there's a $$0$$ there because $$H_0(A\cap B)\to H_0(A)\oplus H_0(B)$$ is injective)

Now $$H_1(A\cap B)\to H_1(A)$$ is $$0$$ because the inclusion is (by construction) nullhomotopic, and $$A\cap B\to B$$ is $$f$$ when you do the identifications $$A\cap B\simeq S^1$$ and $$B\simeq S^1$$, so that $$H_1(A\cap B)\to H_1(B)$$ is just multiplication by $$n$$ when you identify them both with $$\mathbb Z$$.

So it follows that $$H_1(X) \cong \mathbb (Z\oplus \mathbb Z)/(0\oplus n\mathbb Z) \cong \mathbb Z\oplus \mathbb Z/n$$

Note that if you take $$n=0$$, you get a construction which is different from the torus : it's just a sphere with a circle attached at its north pole and another circle attached at its south pole (and $$A=$$ the northern circle plus the northern hemisphere, which of course retracts onto the circle, and $$B$$ similarly with the southern hemisphere - the intersection then retracts onto the equator)

To find this example I just pretended $$A,B$$ were open and applied the Mayer-Vietoris sequence to see what sort of examples there could be This example is never a surface though, and actually there can't be a compact surface example.

Indeed, a compact surface is either orientable, in which case it's a connected sum of tori and has no torsion in its homology; or it's a nonorientable surface, and those only have $$2$$-torsion.

I'm not sure about non compact surfaces, but I would assume it can't work either