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Compute the fundamental group of the space obtained from two tori $S^{1} \times S^{1}$ by identifying a circle $S^{1} \times\left\{x_{0}\right\}$ in one torus with the corresponding circle $S^{1} \times\left\{x_{0}\right\}$ in the other torus.

solution : Let $X$ be the surface, the identification of two tori $S^{1} \times S^{1}$ as described in the exercise. And we know the fundamental group of the torus is $\mathbb{Z} \times \mathbb{Z}$. Let's assume the two tori $T_{1}, T_{2}$ are identified by "stacking" one on the other. i.e. If $a, b$ and $c, d$ are the generators of the fundamental groups respectively. And $a$ and $c$ are their longitudes. The way we stack the two tori will make $a$ and $c$ identified. To use the Van Kampen's Theorem, let $A$ be the top torus $T_{1}$ together with a strip of open neighborhood of $a$ on itself and on the bottom torus $T_{2} .$ Similarly, let $B$ the bottom torus $T_{2}$ together with a strip of open neighborhood of $c$ on itself and the top one $T_{1}$ Then $A$ and $B$ are open subset of $X$ and $A \cap B$ is open and path connected. since $A$ and $B$ deformation retracts to $T_{1}$ and $T_{2}$ respectively, so $\pi_{1}(A)=\pi_{1}(B)=\mathbb{Z} \times \mathbb{Z}$. since $A \cap B$ deformation retracts to a circle, we have $\pi_{1}(A \cap B) \simeq \mathbb{Z}$, the generator has its image $a, c$ in $A, B$ respectively.

By Van Kampen, $\pi_{1}(X)$ is isomorphic to the quotient of $\pi_{1}(A) * \pi_{1}(B)$ by the normal subgroup generated by $\left\langle a c^{-1}\right\rangle$ , $ \pi_{1}(X) \cong \frac{(\mathbb{Z} \times \mathbb{Z}) *(\mathbb{Z} \times \mathbb{Z})}{\left\langle a c^{-1}\right\rangle} \cong(\mathbb{Z} * \mathbb{Z}) \times \mathbb{Z} $

i can't understand why $ \frac{(\mathbb{Z} \times \mathbb{Z}) *(\mathbb{Z} \times \mathbb{Z})}{\left\langle a c^{-1}\right\rangle} \cong(\mathbb{Z} * \mathbb{Z}) \times \mathbb{Z}$ ? is solution true ?

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  • $\begingroup$ Hint: Write up the presentation of the group. $\endgroup$
    – Berci
    Jun 11 '20 at 20:21
  • $\begingroup$ @Berci .for The torus we have : $\left\langle a, b | a b a^{-1} b^{-1}\right\rangle$ and another torus $\left\langle c, d | c d c^{-1} d^{-1}\right\rangle$ so $\frac{(\left\langle a, b | a b a^{-1} b^{-1}\right\rangle) *(\left\langle c, d | c d c^{-1} d^{-1}\right\rangle)}{\left\langle a c^{-1}\right\rangle} \cong $ ? $\endgroup$
    – 1200785626
    Jun 11 '20 at 21:10
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    $\begingroup$ You can make your expression more succinct by combining the presentations together to get $\langle a, b, c, d\mid aba^{-1}b^{-1}, cdc^{-1}d^{-1}, ac^{-1}\rangle$ $\endgroup$
    – William
    Jun 11 '20 at 21:15
  • $\begingroup$ @William .why ? $\endgroup$
    – 1200785626
    Jun 11 '20 at 21:25
  • $\begingroup$ That's how free product works. $\endgroup$
    – Berci
    Jun 11 '20 at 21:42
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Hint: Forget van Kampen's theorem, just observe that your space is homeomorphic to the product of the figure 8 graph and the circle. Now, use the formula for the fundamental group of product of two spaces. Now, use van Kampen's theorem to compute $\pi_1$ of the figure 8 graph.

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  • $\begingroup$ Why our space is homeomorphic to the product of the figure 8 graph and the circle ? Can you draw this? $\endgroup$
    – 1200785626
    Jun 13 '20 at 9:20
  • $\begingroup$ @amirbahadory: This is for you to figure out. Drawing picture of the product is not really helpful here, instead you should try to find two tori $T^2$ in the product of the figure 8 and $S^1$ and check what their intersection is. Keep in mind that $T^2=S^1\times S^1$. $\endgroup$ Jun 13 '20 at 15:19

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