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Let $X$ be the space obtained from two tori $S^1\times S^1$ by identifying a circle $S^1\times\{x_0\}$ in one torus with the corresponding circle $S^1\times\{x_0\}$ in the other. Calculate $\pi_1(X)$.

Well, my professor only explains what van-Kampen theorem is about (not in detail though) and handed a class an exercise and he's gonna put this in exam tomorrow.

So.. I'm bad at informal things.. How do I show this? Could someone please help..?

I think the problem is asking to calculate this enter image description here

Where should I cut?

Rubertos
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    Think about it as two tori stacked on top of each other. Take the top torus with half of the bottom torus, and the bottom torus with half of the top torus. The intersection will be half of the top torus and half of the bottom torus, which will deformation retract to $S_1$. – Robert Cardona Dec 17 '14 at 17:33
  • @RobertCardona is $X$ homeomorphic to genus 2 surface? – Rubertos Dec 17 '14 at 17:36
  • Not exactly, you only have one hole. The fundamental group doesn't come out to be very nice; it's going to be in terms of generators and commutators. Maybe something along the lines: $\langle a, b, c, d : aba^{-1}b^{-1}, cdc^{-1}d^{-1}, ac^{-1}\rangle$; the answer is kinda irrelevant in my opinion: this is primarily a nice exercise to learn how to use van Kampen. – Robert Cardona Dec 17 '14 at 17:38
  • @RobertCardona I haven't learned what deformation retract is, so could you please give me some more details how the fundamental group of the top part and intersection? – Rubertos Dec 17 '14 at 17:42
  • The fundamental groups of the top and bottom part are the same: they both deformation retract to the torus which has fundamental group $\mathbb Z^2$. By deformation retract, think: "deforms in a nice enough manner" for now. The intersection is half of the top torus glued together with half of the bottom torus. Each of the halfs will deformation retract to the area gluing both of them together which is $S^1$ which has fundamental group $\mathbb Z$. Then you have to do some calculations about the kernel to put it all together. – Robert Cardona Dec 17 '14 at 17:47
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    @Rubertos You are studying Van Kampen theorem but you don't know what a deformation retract is? – Balarka Sen Dec 17 '14 at 17:47
  • @BalarkaSen Yes. I even think I'm not studying this. My prof stated theorem and gave this exercise and that's all – Rubertos Dec 17 '14 at 17:49
  • Problem #3 here goes through all the gory details. – Robert Cardona Dec 17 '14 at 17:49
  • So.. I think he wants to do this completely informal. Like just saying mug cup and torus are homeomorphic and etc – Rubertos Dec 17 '14 at 17:50
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    Well, it's not that straight-forward, especially the calculation of the kernel, which is key here. Up until that calculation, you probably could be informal and get away with it. – Robert Cardona Dec 17 '14 at 17:53
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    Did you look at the link I added, problem #3? There's a little diagram on the top left corner; the intersection is in bold, you just revolve that around they $y$-axis to visualize. – Robert Cardona Dec 17 '14 at 17:56
  • @RobertCardona Great thanks. You saved my life. There is one more exercise he assigned and would you please check for this one too? I ask you a favor.. – Rubertos Dec 17 '14 at 18:04

3 Answers3

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Let $\alpha_1$ and $\beta_1$ be loops on $(T^2)_1$ which generate $\pi_1((T^2)_1)$ and $\alpha_2$ and $\beta_2$ be the corresponding loops in $(T^2)_2$. The space $X$ is what we get when we glue the tori together alone $\alpha_1$ and $\alpha_2$.

Let $U_1\subset X$ be a small open neighbourhood of $(T^2)_1$ in $X$ and similarly let $U_2\subset X$ be a small open neighbourhood of $(T^2)_2$ in $X$. The intersection $U_1\cap U_2$ is a small open neighbourhood of $\alpha_1=\alpha_2$ and so in particular has fundamental group which is generated by these elements. We can conclude from Van-Kampen's theorem that $$\pi_1(X)=((\mathbb{Z}^2\ast\mathbb{Z}^2)_{\langle \alpha_1,\beta_1,\alpha_2,\beta_2 \rangle})/\langle\alpha_1=\alpha_2\rangle$$

Honestly, the much quicker way to calculate this fundamental group, without worrying about Van-Kampen's theorem, is to see that $X$ is homeomorphic to the space $(S^1\vee S^1)\times S^1$, the product of a circle with a wedge of two circles (figure eight space). We can then use the fact that $\pi_1(A\vee B)\cong \pi_1(A)\ast \pi_1(B)$ and $\pi_1(A\times B)\cong \pi_1(A)\times\pi_1(B)$ to get $\pi_1(X)\cong (\mathbb{Z}\ast\mathbb{Z})\times\mathbb{Z}$ - here the three generaters from left to right are $\beta_1,\beta_2,\alpha_1(=\alpha_2)$.

Dan Rust
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Take $A$ to be one torus with title neighborhoods around the other torus and $B$ to be another torus with little neighborhoods around the another torus. $A \cup B$ then is the whole space $X$ and $A \cap B$ deformation retracts onto a circle.

So $\pi_1(X) \cong \pi_1(A) \star \pi_1(B)/\langle i_A^{-1}, i_B \rangle$ which is isomorphic to $\Bbb Z^2 \star \Bbb Z^2$ with the identification $(1, 0) \sim (1, 0)$. The fundamental group then is $\langle a, b, c, d |[a, b] = [c, d] = ac^{-1} = 1 \rangle$.

Balarka Sen
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  • Which part of torus represents (0,1)? – Rubertos Dec 17 '14 at 17:52
  • How did you get $[a, b] = [c, d] = ac^{-1} = 1$? and what is $\langle i_A^{-1}, i_B \rangle$? explicitly? – Robert Cardona Dec 17 '14 at 17:59
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    @RobertCardona $\pi_1(A)$ and $\pi_1(B)$ are the groups $\langle a,b | [a,b] = 1\rangle$ and $\langle c, d | [c, d] = 1 \rangle$ resp. and the identification $(1, 0) \sim (1, 0)$ gives the relation $a = c$. So the presentation follows. – Balarka Sen Dec 17 '14 at 18:05
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Let $U_1, U_2$ be the tori, and then $U_1\cap U_2$ is a circle. Recall that $\pi_1(U_i)=\mathbb Z^2$ and $\pi_1(S^1)=\mathbb Z$.

By VKT, the fundamental group is $\mathbb Z^2*\mathbb Z^2/\sim$, where $\sim$ identifies $(1, 0)$ in the first $\mathbb Z^2$ with $(1, 0)$ in the other. So we should get $\langle a, b, c, d\mid[a, b]=[c, d]=ac^{-1}=1\rangle=\langle a, b, d \mid [a, b]=[a, d]=1\rangle$.

Nishant
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    That doesn't work : $U_1$ and $U_2$ aren't open. You have to take small nbhds lying in the other torus. – Balarka Sen Dec 17 '14 at 17:36
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    Oops, you're right. I guess I forgot since in all the cases I used it, the obvious decomposition can be trivially extended to an open one... – Nishant Dec 17 '14 at 17:51