I want to calculate the Klein bottle.
So I did it by Van Kampen Theorem. However, when I'm stuck at this bit. So I remove a point from the Klein bottle to get $\mathbb{Z}\langle a,b\rangle$ where $a$ and $b$ are two loops connected by a point.
Also you have the boundary map that goes $abab^{-1}=1$, so I did some calculations I got basically this
$\dfrac{\mathbb{Z}\langle ab,b\rangle}{\langle(ab)^2=b^2\rangle}$.
But, I don't think that is correct. But, yeah how you calculate the fundamental group of the Klein bottle?
Best Answer
The short answer is that your presentation is isomorphic to the (slightly more) "standard" presentation - so you computed a correct answer.
More detail -
According to Wolfram and using $K$ to denote the Klein bottle, we have $\pi_1(K) \cong \langle c,d \rangle/cdc^{-1}d$ whereas you have (according to Aaron Mazel-Gee's comment) $\pi_1(K)\cong \langle a,b\rangle /abab^{-1}$.
The question is, then, are these isomorphic?
Well, we have $cdc^{-1}d = e$ so, taking the inverse of both sides gives $d^{-1}cd^{-1}c^{-1} = e$. But this has the same form as the relation between $a$ and $b$, so now the isomorphism is clear: we map $a$ to $d^{-1}$ and $b$ to $c$.
That is, define $f:\langle a,b \rangle /abab^{-1}\rightarrow \langle c,d\rangle/cdc^{-1}d$ by $f(a) = d^{-1}$ and $f(b) = c$.
I claim this is well defined, for \begin{align*} f(abab^{-1}) &= f(a)f(b)f(a)f(b)^{-1} \\\ &= d^{-1}cd^{-1}c^{-1} \\\ &= (cdc^{-1}d)^{-1} \end{align*} as it should. (Technically, I'm defining $f$ on $\langle a,b\rangle$ and proving it descends to the quotient.)
Finally, rather than show this is 1-1 and onto, instead, show that $g:\langle c,d\rangle/ cdc^{-1}d\rightarrow \langle a,b\rangle/ abab^{-1}$ defined by $g(c) = b$ and $g(d) = a^{-1}$ is well defined, and the inverse to $f$, so $f$ is the desired isomorphism.