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.

  • $\begingroup$ How do you decompose torus into $A\cup B$ such that $A,B$ and $A\cap B$ are homotopy equivalent to $S^1$? $\endgroup$
    – freakish
    Commented May 8, 2020 at 9:12
  • $\begingroup$ @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? $\endgroup$ Commented May 8, 2020 at 9:15
  • $\begingroup$ Ah, yes. For some reason I though $A,B$ have to be closed... $\endgroup$
    – freakish
    Commented May 8, 2020 at 9:16

1 Answer 1


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


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