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 $V$.

enter image description here

  • 1
    $\begingroup$ What are $U$ and $V$? $\endgroup$ Commented Jul 2, 2018 at 11:43
  • $\begingroup$ I put a figure of $U$ and $V$. $\endgroup$ Commented Jul 2, 2018 at 12:22
  • $\begingroup$ Thanks! I wish I could draw as well as you. I'll provide an answer below (which doesn't actually use your pictures - sorry!) $\endgroup$ Commented Jul 2, 2018 at 14:44
  • $\begingroup$ 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. $\endgroup$
    – Mihail
    Commented Dec 15, 2019 at 9:44

1 Answer 1


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 $H_2(X)\cong \mathbb{Z}^2$.

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

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