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I have to compute the fundamental group of $\mathbb{R}^{3} \smallsetminus A$ where $A=\{(x,y,z): y=0,x^{2}+z^{2}=1\} \cup \{(x,y,z): y=z=0, x \ge 1\}$. In order to do this, I consider $B=\{(x,y,z): x^{2}+y^{2}+z^{2} > 1\}$: if $B$ and $A \cap B$ are simply connected, then the isomorphism $\beta_{*}: \pi(A \cap B,x_0) \mapsto \pi(B,x_0)$ induces an isomorphism $\pi(A,x_0) \mapsto \pi(A \cup B,x_0)$. So I can complete observing that $A \cup B$ is the complement of a circumference in the space $R^{3}$, then $\pi(A)=\pi(A \cup B) \equiv \mathbb{Z}$.

Can you help me, please?

TheWanderer
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1 Answers1

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You're right about $\pi(A\cup B)=\mathbb Z $ but that's truly unnecessary in finding the fundamental group of $\Bbb R - (A\cup B)$.

Since you've found $(A\cup B)\sim S^1$, you can see that $\Bbb R^3-(A\cup B)\sim \Bbb R^3 - S^1$. If you can see that it is homotopic to $\tau^1\cup D^2$ where $D^2$'s boundary is identified to the smallest horizontal loop in $\tau^1$. Then if you apply the Van Kampen Theorem with $U= \tau^1, V=D^2$ and $U\cap V=S^1$, we get that $\pi(\Bbb R^3-(A\cup B))=\frac{\Bbb Z x\Bbb Z}{i*(\alpha)\sim j*(\alpha)}$. We get that $i*(\alpha)=a$ and $j*(\alpha)=\{0\}$ so we find that $\pi(\Bbb R^3-(A\cup B))=\Bbb Z$.

ulucs
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