# Compute homology groups using Mayer-Vietoris sequence

We consider the following subspaces in $$\mathbb{R}^{3}$$: $$\begin{equation*} \begin{split} A &= \left\lbrace (x,y,z) | x^{2}+y^{2}=1, \ -1\le z\le 1 \right\rbrace, \\ B &= \left\lbrace (x,y,z) | x^{2}+y^{2}\le 1, \ z\in\lbrace -1,0,1\rbrace \right\rbrace, \\ C &= \left\lbrace (0,0,z) | -1\le z\le 1 \right\rbrace. \end{split} \end{equation*}$$ and let $$X=A\cup B\cup C$$. For compute $$H_{k}(X)$$ for all $$k\ge 0$$ using M-V, first I have that $$U\simeq\{*\}$$, $$V\simeq S^{2}$$ and $$U\cap V \simeq \{*\} \sqcup \{*\} \sqcup S^{1}$$. Since $$H_{k}(U\cap V) = H_{k}\left(\{*\} \sqcup \{*\} \sqcup S^{1}\right) = H_{k}(\{*\})\oplus H_{k}(\{*\}) \oplus H_{k}(S^{1}),$$ then $$H_{k}(U\cap V)=\begin{cases} \mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}, \ \ \text{if} \ k=0 \\ \qquad \mathbb{Z}, \ \ \ \ \ \ \ \ \text{if} \ k=1 \\ \qquad 0, \ \ \ \ \ \ \ \ \ \text{if} \ k\geq 2 \end{cases}$$ and since we know that $$H_{k}(U)=H_{k}(\{*\})=\begin{cases} \quad \mathbb{Z}, \ \ \text{if} \ k=0 \\ \quad 0, \ \ \ \text{if} \ k\neq 0 \end{cases} \ \ \ \ \ H_{k}(V)=H_{k}(S^{2})=\begin{cases} \quad \mathbb{Z}, \ \ \text{if} \ k=0,2 \\ \quad 0, \ \ \ \text{if} \ k\neq 0,2 \end{cases}$$

• For $$k=3$$, $$\cdots \xrightarrow{} H_{3}(U\cap V) \xrightarrow{i_{*}^{3}} H_{3}(U)\oplus H_{3}(V) \xrightarrow{\pi_{*}^{3}} H_{3}(X)\xrightarrow{\partial} H_{2}(U\cap V)\xrightarrow{i_{*}^{2}} H_{2}(U)\oplus H_{2}(V) \xrightarrow{} \cdots$$ which is $$0 \xrightarrow{i_{*}^{3}} 0\oplus0 \xrightarrow{\pi_{*}^{3}} H_{3}(X)\xrightarrow{\partial} 0 \xrightarrow{i_{*}^{2}} 0\oplus\mathbb{Z}$$ Then $$H_{3}\cong 0$$.
• For $$k=2$$, $$\cdots \xrightarrow{} H_{2}(U\cap V) \xrightarrow{i_{*}^{2}} H_{2}(U)\oplus H_{2}(V) \xrightarrow{\pi_{*}^{2}} H_{2}(X)\xrightarrow{\partial} H_{1}(U\cap V)\xrightarrow{i_{*}^{1}} H_{1}(U)\oplus H_{1}(V) \xrightarrow{} \cdots$$ which is $$0 \xrightarrow{i_{*}^{2}} 0\oplus\mathbb{Z} \xrightarrow{\pi_{*}^{2}} H_{2}(X)\xrightarrow{\partial} \mathbb{Z} \xrightarrow{i_{*}^{1}} 0\oplus0$$ Can you give me some hint to compute $$H_{2}(X)$$?
• For $$k=0,1$$, $$\cdots \xrightarrow{} H_{1}(U\cap V) \xrightarrow{i_{*}^{1}} H_{1}(U)\oplus H_{2}(V) \xrightarrow{\pi_{*}^{1}} H_{1}(X)\xrightarrow{\partial} H_{2}(U\cap V)\xrightarrow{i_{*}^{0}} H_{0}(U)\oplus H_{0}(V) \xrightarrow{} H_{0}(X) \xrightarrow{} 0$$ which is $$\cdots \xrightarrow{} \mathbb{Z} \xrightarrow{i_{*}^{1}} 0\oplus0 \xrightarrow{\pi_{*}^{1}} H_{1}(X)\xrightarrow{\partial} \mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z} \xrightarrow{i_{*}^{0}} \mathbb{Z}\oplus\mathbb{Z} \xrightarrow{\pi_{*}^{0}} H_{0}(X) \xrightarrow{\partial} 0$$ Can you give me some hint to compute $$H_{1}(X)$$ and $$H_{0}(X)$$?

I will appreciate any hint to solve and understand how to compute this homology groups.

I put the figures of $$X$$, $$U$$ and $$V$$.

• What are $U$ and $V$? Commented Jul 2, 2018 at 11:43
• I put a figure of $U$ and $V$. Commented Jul 2, 2018 at 12:22
• Thanks! I wish I could draw as well as you. I'll provide an answer below (which doesn't actually use your pictures - sorry!) Commented Jul 2, 2018 at 14:44
• As an alternative, you could observe that your space is homotopy equivalent to the wedge of 2 spheres and 2 circles. This would simplify a problem a lot but also could be considered as a cheating. Commented Dec 15, 2019 at 9:44

## 1 Answer

For $$k=2$$, after using the canonical isomorphisms $$0\oplus G\cong$$, you have an exact sequence $$0\rightarrow \mathbb{Z}\rightarrow H_2(X)\xrightarrow{\partial} \mathbb{Z}\rightarrow 0.$$

We can compute $$H_2(x)$$ purely algebraically from this (getting $$H_2(X) \cong \mathbb{Z}^2$$). In fact, we have the following general proposition:

Proposition: From a short exact sequence of abelian groups $$0\rightarrow H\xrightarrow{i} G\xrightarrow{\pi} \mathbb{Z}^n\rightarrow 0$$, it follows that $$G\cong H\oplus \mathbb{Z}^n$$.

Proof: Because the kernel of the last map $$\mathbb{Z}^n\rightarrow 0$$ is all of $$\mathbb{Z}^n$$, exactness shows $$\pi$$ must be surjective. So, there are elements $$g_i\in G$$ for $$i = 1...n$$ for which $$\pi(g_i) = e_i$$, where $$e_i = (0,..,0,1,0,...,0)$$ has a $$1$$ in the $$i$$th slot.

Now, define a homomorphism $$\psi:H\times \mathbb{Z}^n\rightarrow G$$ by $$\psi(h, \sum n_i e_i) = i(h) + \sum n_i g_i$$. We claim that $$\psi$$ is an isomorphism. Because $$i$$ is a homomorphism, $$\psi$$ is clearly a homomorphism. So we need only show its bijective.

To see that it is injective, assume $$i(h) + \sum n_i g_i = 0$$. Applying $$\pi$$ to this and using the fact that $$\pi\circ i = 0$$, we get $$0 = \sum n_i e_i$$, which implies all $$n_i$$ are $$0$$. Thus, we know that $$i(h) = 0$$. Since $$i$$ is injective (because $$\ker i$$ is equal to the image of the $$0$$ map), it follows that $$h = 0$$ as well.

To see that is is onto, let $$g\in G$$. We can write $$\pi(g) = \sum n_i e_i$$ for some integers $$n_i$$. Then notice that $$\pi(g - \sum n_i g_i) = 0$$, so by exactness, there is an $$h\in H$$ with $$i(h) = g - \sum n_i g_i$$. Then $$g = i(h) + \sum n_i g_i$$, so $$g = \psi(h, \sum n_i e_i)$$. $$\square$$

For $$k=0$$, we will ignore the exact sequence and instead compute directly.

Proposition: For any path space $$X$$, $$H_0(X) \cong \mathbb{Z}$$.

Proof: I'm thinking singular homology here, but a similar proof works for any of the usual homology theories. Choose any $$x\in X$$. Then, thinking of $$x$$ as as $$0$$-chain, $$\partial x = 0$$, so $$[x]\in H_0(X)$$. Note that $$[x]\neq 0$$: a $$1$$-chain is just an oriented curve $$\gamma$$ and $$\partial \gamma = 0$$ if its closed and is the difference of its end points, so $$\partial \gamma \neq x$$. Further, if $$y\in X$$ is any other point, then there is a curve $$\gamma$$ connecting them by hypothesis. Then $$x-y = \partial \gamma$$, so $$[x]=[y]$$. Thus, $$H_0(X)\cong \mathbb{Z}$$, generated by any of its points. $$\square$$

Now that we know the $$k = 0$$ case, we can figure out the $$k=1$$ case, again purely using algebra.

Proposition: Given an exact sequence $$0\rightarrow H_1(X)\rightarrow \mathbb{Z}^3\rightarrow \mathbb{Z}^2\rightarrow \mathbb{Z}\rightarrow 0$$, it follows that $$H_1(X)\cong \mathbb{Z}^2$$.

Proof: First note that $$H_1(X)\rightarrow \mathbb{Z}^3$$ is injective, so we may view $$H_1(X)$$ as a subgroup of $$\mathbb{Z}^3$$. Up to isomoprhism, the only subgroups of $$\mathbb{Z}^3$$ are $$0,\mathbb{Z},\mathbb{Z}^2,\mathbb{Z}^3$$.

Now, consider the last part of the squence $$\mathbb{Z}^2\rightarrow \mathbb{Z}\rightarrow 0$$. Exactness tells us the map $$\mathbb{Z}^2\rightarrow \mathbb{Z}$$ is surjective. The first isomorphism theorem for groups tells us $$\mathbb{Z}^2/\ker \cong \mathbb{Z}$$. Thus, clearly $$\ker$$ is non-trivial. The non-trivial proper subgroups of $$\mathbb{Z}^2$$ are all isomorphic to $$\mathbb{Z}$$, so $$\ker \cong \mathbb{Z}$$.

It follows that the first part of our exact sequence splits into $$0\rightarrow H_1(X)\rightarrow \mathbb{Z}^3\rightarrow \ker \rightarrow 0$$. From the first proposition, using the fact that $$\ker\cong\mathbb{Z}$$, it follows that $$\mathbb{Z}^3\cong \mathbb{Z}\oplus H_2(X)$$. It follows that $$H_2(X)\cong \mathbb{Z}^2$$.

• I just realized you asked for hints - and this is a complete answer. I can delete it if you prefer. Commented Jul 2, 2018 at 15:45