The torus and the Klein's bottle:

We proceed along the same lines using the convenient form (25.5). Denoting by $X$ either the torus or the Klein's bottle and $p$ to be the origin, we see that $X - \{p\}$ deformation retracts to the figure eight loop and, in analogy with (26.6), the generators for the free group $\pi_1(X - \{p\})$ are given by

$$\Gamma_1(t) = \eta(\exp(i\pi t/2)),\quad \Gamma_2(t) = \eta(\exp(i\pi (t+1)/2)),\quad 0\leq t \leq 1$$

We take $U$ to be the open unit disc, $V$ to be $X - \{p\}$ and the class of (26.5) as the generator for $\pi_1(U\cap V, y_0)$ where the base point $y_0$ is $1/2$. Taking an auxiliary path $\beta$ joining $y_0$ and the point $x_0 = 1$ common to both $\Gamma_1(t)$ and $\Gamma_2(t)$, we get the generators

$$[\beta *\Gamma_1 *\beta^{-1}]\phantom{X}$$ and $$\phantom{X}[\beta *\Gamma_2 *\beta^{-1}] \eqno(26.9)$$

for $\pi_1(X-\{p\}, y_0)$. The deformation of the previous example (exercise (1)) can be employed here again and this time we get

$$i_*[\gamma] = [\beta *\Gamma_1 *\beta^{-1}][\beta *\Gamma_2 *\bet... ...ta *\Gamma_1^{-1} *\beta^{-1}][\beta *\Gamma_2^{-1} *\beta^{-1}] \eqno(26.10)$$

for the torus whereas for the Klein's bottle we get instead

$$i_*[\gamma] = [\beta *\Gamma_1 *\beta^{-1}][\beta *\Gamma_2 *\bet... ... [\beta *\Gamma_1 *\beta^{-1}][\beta *\Gamma_2^{-1} *\beta^{-1}] \eqno(26.11)$$

One could also work with the other models described in example (25.3) where the spaces are obtained by identifying the opposite edges of a square. The homotopy $\eta\circ F$ of the last example would have to be modified to $\eta\circ G\circ F$ where $G$ is a certain homeomorphism from the unit disc onto the square $[0, 1]\times [0, 1]$.

Denoting the generators (26.9) of $\pi_1(V, y_0)$ by $S$ and $T$ we are ready to apply corollary (26.7) since (26.10) gives us the image of the map $i_{*}$. The fundamental group of the torus is then

$$\langle S, T : ST = TS\rangle \cong\mathbb{Z}\times \mathbb{Z} \eqno(26.12)$$

and the fundamental group of the Klein's bottle is

$$\langle S, T : TST = S\rangle\cong \mathbb{Z}\ltimes \mathbb{Z}. \eqno(26.13)$$