Let us glue two Mobius Bands $A$, $B$ together along their boundaries. We want to apply Van-Kampen's theorem to their union. The intersection $L$ is homotopic to $S^1$. So we get $$\pi_1(X) = \mathbb{Z}*_\mathbb{Z}\mathbb{Z}$$ We look at what happens when we include the boundary into each space. Let $\pi_1(A)$ have generator $a$, $\pi_1(B)$ have generator $b$, $\pi_1(L)$ have generator $\sigma$. As the boundary goes around the generator twice, we get

$$i^*(\sigma)=a^2$$ $$j^*(\sigma) = b^2$$

Where $i,j$ the natural inclusions. Therefore $$\pi_1(X) = \langle a,b \mid a^2=b^2 \rangle $$.

But this doesn't seem right, as (apparently) $X$ is just the Klein bottle $K$ and $$\pi_1(K) = \langle a,b\mid abab^{-1} = 1 \rangle$$ But these groups don't look the same.


1 Answer 1


I'll use two different sets of letters to make the map clearer. You have, on one hand, $X=\langle a,b\;|\;a^2=b^2 \rangle $, and on the other hand, $K=\langle x,y\;|\;xyx=y \rangle$. Consider the map $f:X\to K$, $a\to xy$, $b\to y$. Now $$ f(a^2)=(xy)^2=xyxy=(xyx)y=y^2, $$ so that $a^2b^{-2}$ lies in the kernel, and the map is indeed well defined.

To get an inverse map, we simply send $x\to ab^{-1}$ and $y$ to $b$. Again, this sends $xyx$ to $(ab^{-1})b(ab^{-1})=a^2b^{-1}=b^2b^{-1}=b$, so the map is well defined.

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    $\begingroup$ I replaced $<>$ with the standard $\langle \rangle$, if you don't mind. $\endgroup$ Commented May 23, 2018 at 21:02
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    $\begingroup$ Well, I'll be damned, they are isomorphic! $\endgroup$ Commented May 23, 2018 at 21:11
  • $\begingroup$ Thanks @AlexProvost ! I was being lazy $\endgroup$
    – TomGrubb
    Commented May 23, 2018 at 21:13

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