To calculate the fundamental group of the Klein bottle, $X$, we can find path connected open sets, $U$, $V$,and $U \cap V$, satisfying $U\cup V= X$. I know we can choose $U$ and $V$ as follows:[![$U$ with a fixed point $p$][1]][1][![$V$ with a fixed point $p$][2]][2] then $U\cap V$ is just [![$U\cap V$ with a fixed point $p$][3]][3] $U$ is homotopic equivalent to $S^1\vee S^1$ up to homeomorphism and $V$ is contractible. $U\cap V$ has a strong deformation retract to $S^1$. Then $\pi_1(U,p) \cong\pi_1(S^1, *)\cong \langle a,b \rangle$, $\pi_1(V,p)\cong \{1\}$ and $\pi_1(U\cap V, p)\cong \pi_1(S^1, *)\cong \langle x \rangle$.
By Seifert-Van-Kampen, $$\pi_1(X, p)\cong (\pi_1(U,p)*\pi_1(V,p))/N$$
where $$N=\langle \langle \iota_0(y) (\iota_1(y))^{-1}: y\in \pi_1(U\cap V,p) \rangle \rangle$$
the normal closure generated by elements in $\{ \iota_0(y) (\iota_1(y))^{-1}: y\in \pi_1(U\cap V, p) \}$
$\iota_0$ and $\iota_1$ are two maps as follows:
\begin{array}
& & \pi_1(U, p)& \\
& \nearrow{\iota_0} & & \\
\pi_1(U\cap V, p)& & & \\
& \searrow{\iota_1} & & \\
& \pi_1( V, p) & &
\end{array}
Intuitively, $abab^{-1}$ is a loop in $U$, but what does this mean, I need to make this expression more precise. And in some notes, some people say: in $\iota_0: \pi_1(U\cap V,p) \rightarrow \pi_1(U,p)\cong\langle a,b \rangle$, $x$ should be mapped to $abab^{-1}$. Why? Why not map $x$ to some generator of $\pi_1(U,p)$, say $a$ or $b$?
Some people say that I need to see what paths the circle $S^1$ traverses when it gets projected to the skeleton of that Klein bottle. Why? Is this idea precise? Can anybody elaborate on this?
[1]: https://i.stack.imgur.com/zSp1T.jpg
[2]: https://i.stack.imgur.com/TDSLg.jpg
[3]: https://i.stack.imgur.com/sQC4r.jpg
