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I am doing an exercise where I am supposed to compute the fundamental group of $\mathbb{S}^1\times[0,1]$ using Van Kampen's theorem with the open cover $A=\mathbb{S}^1\times[0,3/4)$ and $B=\mathbb{S}^1\times(1/4,1]$. I know the answer is $\mathbb{Z}$. Now I know $\pi_1(A)=\pi_1(B)=\pi_1(A\cap B)=\mathbb{Z}$ so I get by the theorem

$$\pi_1(X)\cong\dfrac{\mathbb{Z}*\mathbb{Z}}{N},$$

where $N$ is the normal subgroup generated by $i_{AB}(w)i_{BA}(w)^{-1}$, for $w\in\mathbb{Z}$ (here $i_{XY}$ is the homomorphism $\pi_1(X\cap Y)\rightarrow\pi_1(X)$ induced by the inclusion $X\cap Y\rightarrow X$).

Now my goal is to give a description for the elements of $N$, in order to know know if a reduced word in $\mathbb{Z}*\mathbb{Z}$ lies or not in $N$ and I don't seem to know how to do that. Suppose I choose $a$, $b$ for the representatives of each copy of $\mathbb{Z}$. A word is in $N$ if $ab^{-1}=1$, that is, $a=b$? Don't know if this is the correct interpretation.

Many thanks!

1 Answers1

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Yes, that's correct. If $\{U, V\}$ is the cover of $S^1 \times [0, 1]$ you describe, then by van Kampen theorem

$$\pi_1(S^1 \times [0, 1]) \cong \pi_1(U) * \pi_1(V)/\langle i_U \cdot i_V^{-1}\rangle$$

Now, $\pi_1(U) \cong \pi_1(V) \cong \Bbb Z$. Note that $i_U : \pi_1(U \cap V) \to \pi_1(U)$ sends generator of $\pi_1(U\cap V)$, namely the center circle, to the generator of $\pi_1(U)$, the top circle. So $i_U$ is an isomorphism, as is $i_V$ by similar reasons.

If we denote the generator of $\pi_1(U) \cong \Bbb Z$ as $a$, and $\pi_1(V) \cong \Bbb Z$ as $b$, then

$$\pi_1(S^1 \times [0, 1]) \cong \langle a, b | ab^{-1} = 1 \rangle$$ from the above analysis. As $ab^{-1}$ implies $a = b$, one of the two generators is redundant and the group is simply infinite cyclic on one generator, which is $\Bbb Z$.

Balarka Sen
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  • How did you determine the relation $ab^{-1} =1$? How do you determine the relation that $N$ gives, we may not always have isomorphisms $i : \pi_1(U \cap V) \longrightarrow \pi_1(U)$. –  Oct 22 '16 at 00:08
  • @Elliot Think geometrically. Consider obvious deformation retracts $U\cap V \to S^1$ and $U \to S^1$, and look at the commutative square of of maps $\pi_1(U\cap V) \to \pi_1(U)$ and $\pi_1(S^1) \to \pi_1(S^1)$. You should be able to convince yourself that it's an isomorphism here (essentially because if you push a loop in $U \cap V$ to an"end of the cylinder" in $U$, it's still the generating loop because there's no further identifications throughout). – Balarka Sen Oct 22 '16 at 00:46
  • Of course it wouldn't always be an isomorphism, but most often pictorial arguments should be enough to see what $i_U$ and $i_V$ are. – Balarka Sen Oct 22 '16 at 00:47