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:)
Best Answer
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.