about free group and fundamental group.

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 ?

• Hint: Write up the presentation of the group. Jun 11 '20 at 20:21
• @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 $\frac{(\left\langle a, b | a b a^{-1} b^{-1}\right\rangle) *(\left\langle c, d | c d c^{-1} d^{-1}\right\rangle)}{\left\langle a c^{-1}\right\rangle} \cong$ ? Jun 11 '20 at 21:10
• You can make your expression more succinct by combining the presentations together to get $\langle a, b, c, d\mid aba^{-1}b^{-1}, cdc^{-1}d^{-1}, ac^{-1}\rangle$ Jun 11 '20 at 21:15
• @William .why ? Jun 11 '20 at 21:25
• That's how free product works. Jun 11 '20 at 21:42

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.
• @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 $T^2=S^1\times S^1$. Jun 13 '20 at 15:19