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define $X =\{(x,y,z)\in \mathbb{R^3} | x^2+y^2+z^2=1 \} \cup \{(0,0,t)\in \mathbb{R^3}|t\in[-1,1]\} $

(a) compute fundamental group and homology groups for $X$

(b) compute fundamental group and homology groups for $\mathbb{R^3}-X$

for (b) space is homotopic equivalent to $S^1 \cup S^2$, but for (a) I think fundamental group of X is $\mathbb{Z}$ but cannot show exact way to show it. any helps?

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    Hint: the segment meets the sphere in two points; you can move one of them around to make it coincide with the second, so that the segment becomes a loop. So, $X$ is homotopic to $S^2 \vee S^1$. – lisyarus Dec 02 '16 at 11:46
  • Note that $\Bbb R^3-X$ is not connected. – Arthur Dec 02 '16 at 12:16

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(a) For the fundamental group use Seifert-van Kampen. Note that you can divide the sphere plus the line into opens $U$, which is the line and the line of longitude plus a little bit, such that it is open, and $V=S^2$ plus a little bit, such that it is open. Then $U\cap V$ is homeo to a disc, hence $\pi_1(U\cap V)=0$.

By deformation retraction we get $\pi_1(X)=\pi_1(U)\ast\pi_1(V)=\pi_1(S^1)\ast\pi_1(S^2)=\mathbb{Z}\ast0=\mathbb{Z}$.

(b) In deed $H_n(\mathbb{R}^3\setminus X)=H_n(S^1\sqcup S^2)=H_n(S^1)\oplus H_n(S^2)$ by deformation retraction. For the second "=" see e.g. Homology of disjoint union is direct sum of homologies.

We conclude $H_n(\mathbb{R}^3\setminus X)=\begin{equation} \begin{cases} \mathbb{Z}\oplus\mathbb{Z} & \text{for } n=0 \\ \mathbb{Z} & \text{for } n=1 \\ \mathbb{Z} & \text{for } n=2 \\ 0 & \text{for } n\ge 3 \end{cases} \end{equation}$.

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Daniel Bernoulli
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