Use Mayer-Vietoris sequence to compute homology groups of 3-torus

Question

In Example 2.39 in Hatcher, he used cellular homology to compute the homology groups of the 3-torus. I am studying for my exam and we did not cover the cellular homology. So I am thinking of using Mayer-Vietoris sequence. So we are considering the standard representation of the 3-torus X as a quotient space of the cube.

I am going take A=small ball inside the cube. $B=X\setminus A'$ (A' small neighborhood of A) so that $A \cap B $ deformation retracts onto the sphere $S^2$. I know the homology groups of $A$ and of $A \cap B$. I also know that $B$ deformation retracts to the quotient space of the union of all square faces of the cube.

My problem is this: How can I determine the homology groups of B? And once I do that how can I see the map from $H_2(S^2)$ to $H_2(B)$?

PS: One of the answer suggested a really nice other decomposition. However, I might want to need to compute the homology of B first as the problem recommended!

Answer 1

First, I think Matteo Tesla proposed a great decomposition that simplifies the problem. Since OP requested to keep the original MV argument, I decided to complete it.

Let $A=D^3,B$ be as what OP stated in the question.

Determine $H_*(B)$.

$B$ deformation retracts onto the surface of the cube, which consists of six squares with opposite edges identified, i.e., it consists of six $T^2$, whose homology groups are known. Thus, $H_2(B)=\bigoplus_{i=1}^3{H_2(T^2)}=\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}$ because opposite faces are identified on the edges, which are also generators of the $2$nd homology group of each $T^2$. Similarly, $H_1(B)=\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}$. You can work out these expressions by drawing a flat diagram of the surface of the cube and labelling all equivalence classes. (I can also edit the post to include my drawing if you want...)

Although all of the six faces are tori, their generators of $H_1,H_2$ are identified. A brief way to determine the homology group is just observing this graph, but you can also regard them as different tori and apply MV sequence multiple times, then mod out those identified images, which is more convincing but also more complicated.

Compute $H_*(T^3)$:

We compute $H_3(T^3)$ by a part of MV sequence:

$$0\to H_3(T^3)\overset{\phi_3}{\to}\mathbb{Z}\overset{\psi_3}{\to}\bigoplus_{i=1}^3\mathbb{Z}\to...$$

Your question specifically asks for how to determine $\psi$, so let's focus on that. Consider the following commutative diagram similar to that of Seifer-Van Kampen Thm $$ \require{AMScd} \begin{CD} H_2(S^2)@>i>>H_2(A)\\ @Vj=\psi VV @VlVV\\ H_2(B)@>k>>H_2(T^3) \end{CD} $$ We can ignore $H(A)$ because $A\simeq\{*\}$. And, Let $\alpha,\beta,\gamma$ be the three generators of $H_2(B)$ that are oriented counterclockwise and $\delta$ the generator of $H_2(S^2)$.

Then, $\psi(\delta)=\alpha+\beta+\gamma-\alpha-\beta-\gamma=0$ (use the diagram of the flat surface to help you). Geometrically, the diagram is induced by the chain complex, so $\psi$ actually sends cycles to cycles. $\delta$, as a generator of $H_2(S^2)$ is mapped into $B$ (observing $\delta$ in $B$) it deformation retracts onto the surface. The surface consists of three pairs of faces with opposite orientation when it is identified (you can try to make one, even though they're all oriented counterclockwise in the diagram), so we get the expression as desired because all groups are abelian. Thus $\text{im}(\psi)=0,\text{ker}(\psi)=\Bbb{Z}$, which implies $H_3(T^3)\cong\mathbb{Z}$.

For $H_2(T^3)$, we already know that the map $\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}\overset{}{\to} H_2(T^3)$ is surjective because we have $H_2(T^3)\to H_1(S^2)=0$. Now because $\text{im}(\psi)=0$, the map $\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}\overset{}{\to} H_2(T^3)$ is also injective. Hence, $H_2(T^3)\cong\bigoplus_{i=1}^3\mathbb{Z}$.

I guess I can stop here to make this post focus on the main problem on that map.

Answer 2

I assume that by 3-torus you mean $S^1 \times S^1 \times S^1$. You can decompose the first component, $S^1=A \cup B$ $A\times S^1\times S^1$ is homotopic to a 2-torus, also the other part. The intersection is homotopic to 2 disconnected 2-tori, so you have to know the homology of $S^1 \times S^1$ first. To do computation you have to consider also the maps involved.

For the 2-torus you obtain $$ 0\to H_2(T) \to H_1(S^1\times (S^1\setminus\{-1,1\})) \to H_1(S^1\times (S^1\setminus \{-1\}))\oplus H_1(S^1\times (S^1\setminus \{1\})) \to H_1(T) \to \dots $$

To study the map $d:H_1(S^1\times (S^1\setminus\{-1,1\})) \to H_1(S^1\times S^1\setminus \{-1\})\oplus H_1(S^1\times S^1\setminus \{1\})$, you consider the generator of the domain which are, $[\gamma,P],[\gamma,Q]$ ($P, Q$ in different connected component of $S^1 \setminus \{-1,1\}$). This generator are mapped by $d$ to $([\gamma,P],-[\gamma,P])$ and $([\gamma,Q],-[\gamma,Q])$ respectively (this are the same because $S^1 \times (S^1 \setminus P)$ is connected). So $d$ has non trivial kernel $[\gamma,P]-[\gamma,Q]$, so $H_2(T)\cong \mathbb{Z}$.

Let's do the hard part and compute $H_1(S^1 \times S^1)\cong \mathbb{Z}^2$.

We can splite the sequence at the level of $H_1(S^1\times S^1)$:

$$0\to\text{Coker} (\phi)\to H_1(T)\to \text{Im}(\delta)\to 0$$ is exaxt. Where $\phi: H_1(S^1\times (S^1\setminus\{-1,1\})) \to H_1(S^1\times (S^1\setminus \{-1\}))\oplus H_1(S^1\times (S^1\setminus \{1\}))$, $\delta:H_1(T) \to H_0(S^1\times (S^1\setminus\{-1,1\}))$. It remain to prove that $\text{Im}(\delta)\cong \mathbb{Z}$, so that the sequence split. Also $\text{Coker}(\phi)\cong \mathbb{Z}$, so $H_1(T)\cong \mathbb{Z}\oplus \mathbb{Z}$.

For the 3-torus you can proceed in the same way.

The decomposition you are taking I don't think it is useful but I might be wrong. In dimension two your $B$ is $S^1 \times S^1\setminus D$ where $D$ is a small disc which is homotopic to a bucket of two circumference. You have to use again MV.