# Homology of doubly-pinched torus

I am trying to compute the homology of the doubly pinched torus, denoted $$X$$. I would like some input on if what I have done is correct. I have done this both using Mayer-Vietoris and a cell structure on $$X$$.

Mayer-Vietoris

Let $$U, V \subset X$$ be the two subspaces between the two pinched points of the torus. Then both $$U$$ and $$V$$ are homotopicly equivalent to $$S^2$$. Moreover, their intersection is the disjoint union of two points, i.e. $$U \cap V = *\bigsqcup*$$. Clearly the space is path connected so $$H_0(X) \cong \mathbb{Z}$$. The reduced M-V sequence yields: $$0 \rightarrow \widetilde{H}_2(U) \oplus \widetilde{H}_2(V) \rightarrow \widetilde{H}_2(X) \rightarrow \widetilde{H}_1(U \cap V) \rightarrow \widetilde{H}_1(U) \oplus \widetilde{H}_1(V) \rightarrow \widetilde{H}_1(X) \rightarrow \widetilde{H}_0(U \cap V) \rightarrow 0$$ We have $$\widetilde{H}_1(U \cap V) \cong 0$$ and $$\widetilde{H}_1(U) \oplus \widetilde{H}_1(V) \cong 0$$. Lastly we have $$\widetilde{H}_0(U \cap V) \oplus \mathbb{Z} \cong H_0(U \cap V) \cong \mathbb{Z} \oplus \mathbb{Z} \implies \widetilde{H}_0(U \cap V) \cong \mathbb{Z}$$

This gives the following isomorphisms due to exactness of M-V $$0 \rightarrow \widetilde{H}_2(U) \oplus \widetilde{H}_2(V) \cong \mathbb{Z} \oplus \mathbb{Z} \overset{\cong}{\rightarrow} \widetilde{H}_2(X) \rightarrow 0$$ $$0 \rightarrow \widetilde{H}_1(X) \overset{\cong}{\rightarrow} \widetilde{H}_0(U \cap V) \cong \mathbb{Z} \rightarrow 0$$ I obtain $$H_0(X) \cong \mathbb{Z}$$, $$H_1(X) \cong \mathbb{Z}$$ and $$H_2(X) \cong \mathbb{Z}\oplus \mathbb{Z}$$.

Cell Complex

One can build $$X$$ using $$2$$ cells in each degree $$0, 1$$ and $$2$$. This gives the following cellular chain complex $$0 \rightarrow \mathbb{Z} \oplus \mathbb{Z} \overset{\partial_1}{\rightarrow} \mathbb{Z} \oplus \mathbb{Z} \overset{\partial_0}{\rightarrow} \mathbb{Z} \oplus \mathbb{Z} \rightarrow 0$$

I believe that $$\partial_1 = 0$$ since attaching each 2 cell to its respective 1 cell would be to glue the boundary along the 1 cell in both directions, making the attaching map a degree 0 map for both 2 cells. For $$\partial_0$$ it is just the regular singular differential yielding $$\partial_0 = \begin{bmatrix} -1 & -1 \\ 1 & 1 \end{bmatrix} \overset{SNF}{\sim} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$ This gives the following homology groups $$H_0(X) \cong \mathbb{Z}^2 /\partial_0 \mathbb{Z}^2 \cong \mathbb{Z}^2 / SNF(\partial_0)\mathbb{Z}^2 \cong \mathbb{Z}$$ $$H_1(X) \cong \mathbb{Z}\langle\alpha - \beta \rangle \cong \mathbb{Z}$$ $$H_2(X) \cong \mathbb{Z} \oplus \mathbb{Z}$$ Here $$\alpha$$ and $$\beta$$ are the two 1-cells.

I believe that my Mayer-Vietoris arguement is correct, and I would like some input on the last argument, using the cellular chain complex. Any feedback is appreciated:)

Your cellular argument is correct. You make the $$\partial_2=0$$ computation more formal via the local degree formula, obtaining two preimages of any particular interior point of a $$1$$-cell inside the bounadries of the $$2$$-cells, and observing the map from a neighbourhood of one of the preimages is the same, up to a degree $$(-1)$$ reflection, as the map from a neighbourhood of the other preimage; altogether we get a sum $$1-1=0$$.
The Mayer-Vietoris argument is almost correct. Maybe you're already aware of this and implicitly doing it, but it's worth noting we do need some regularity assumptions on the subspace decomposition. You can take small open neighbourhoods of each $$U,V$$ which deformation retract onto $$U,V$$ - let's call these slightly 'thickened' subspaces $$U',V'$$ - and then apply Mayer-Vietoris to $$U'$$ and $$V'$$. Now and only now do we get to say that the interiors of $$U',V'$$ cover our whole space.
As discussed in comments, this space is also homotopy equivalent to $$S^1\vee S^2\vee S^2$$ which is a more powerful statement including the homology computation as a corollary.
• : ) I get it now. Yes, I believe the doubly pinched torus is htpy equivalent to $S^1\vee S^2\vee S^2$. Take your doubly pinched torus, take a semicircular arc along one of the "croissants" from the first pinched point to the second; this is contractible so we are free to collapse it, whereupon we find $S^2$ wedged with a singly pinched torus, and that's just $S^2\vee S^2\vee S^1$ up to homotopy Commented Nov 27, 2023 at 21:04
• @mNugget Actually we [AlgTop and I] have been quite sloppy about that. Intuitively it should be true but now that you've pointed it out, I realise we should have some additional hypotheses. As is so commonly useful in homotopy theory, taking products with the unit interval is a cocontinuous functor [basically it preserves quotients]. If $f:Y\simeq Z:g$ then we want to show $1\vee f:X\vee Y\simeq X\vee Z:1\vee g$. $(X\vee Y)\times I$ is a suitable quotient of $X\times I\sqcup Y\times I$. If the homotopies $Y\times I\to Y,Z\times I\to Z$ witnessing the equivalence are based, it can work. Commented Nov 28, 2023 at 9:24
• But now we need to check the homotopies involved in the equivalence [pinched torus] ~ [sphere wedge circle] are based. This really requires an inspection of the theorem: "If $(X,A)$ has the homotopy extension property and $A$ is contractible then $X\twoheadrightarrow X/A$ is a homotopy equivalence." If you examine the proof you'll realise we can make the homotopies relative to $A$, i.e. the map $X\times I\to X$ will map $A$ to itself at every timestep (and the map $(X/A)\times I\to X/A$ will fix the basepoint). This is enough to say pinched torus ~ sphere wedge circle by a based equivalence. Commented Nov 28, 2023 at 9:32