Fundamental group of sphere with 2 handles with 2 mobius bands.

Question

Is it real to calculate?? We have a sphere with with 2 handles and with 2 glued mobius bands (red on picture).

So, i think we need to use Van Kampen Theorem 2 times.

1) $ X = X_1 \cup X_2 $ , where $X_1$ and $X_2$ are toruses with glued 1 mobius band. Then $X_1 \cap X_2$ is just a $S^1$.

2) $X_2 = X_1 = Y_1 \cup Y_2 $,where $Y_1 = T \setminus {D} \setminus {D}$ , where D is disk. $Y_2 = M$ , where $ M $ is mobius band. $Y_1 \cap Y_2 = S^1$

3) So, we knows that:

$$\pi_1 (Y_1) = \pi_1 (T \setminus {D} \setminus {D}) = \mathbb{Z} \times \mathbb{Z}$$ $$\pi_1 (Y_2) = \pi_1 (M) = \pi_1 (S^1) = \mathbb{Z}$$

How can we find $\pi_1(X_1)$ and $\pi_1(X)$

Answer 1

Since there is no one, who answered your question, I will give you an answer which is not complete, but answers your question, if "it is real to calculate" and gives an idea of the calculation. Maybe this is enough for you and already answers your question.

Here you should use Seifert-van Kampen twice. For a) and b) look in the comment above.

First: $U$ and $V$ are just the first and the second $X$ (with a "hole", because of the connected sum) plus something, s.t. they are open. Because of this "something" you get $U\cap V\simeq S^1$.

Second: W.L.O.G. we consider $U$. Let $W_1$ be the torus $T^2$ (with its hole) and a neighbourhood of the möbius strip $M$ with an attached intersection. $W_2$ is the möbius strip with a neighbourhood of $T^2$ including an attached intersection. So $$W_1\cap W_2\simeq S^1,\quad W_1\simeq T^2\setminus\mbox{ hole}\simeq S^1\vee S^1,\quad W_2\simeq M\simeq S^1$$ by deformation retractions. This should be easy to calculate. The only difficulty is, to consider how the inclusions $i_1:W_1\cap W_2\hookrightarrow W_1$ and $i_2:W_1\cap W_2\hookrightarrow W_2$ look like.

a) The $i_1$ should send the generating loop of $W_1\cap W_2$ to the generating loop of the torus. But the second inclusion $i_2$ maps a generator of $W_1\cap W_2$ to 2 times a generator of $W_2$.

b) The $i_2$ is the same, but the first inclusion maps the generating loop to the surface of the torus, where we can shrink it (graphicly: the loop gets mapped to "the boundary of a disc", which is part of the torus suface and is contractible). So it maps the generator to the trivial element.

Let $a,b$ be the generators of $\pi_1(W_1)\cong\pi_1(S^1\vee S^1)\cong \mathbb{Z}\ast\mathbb{Z}$, $c$ the generator of $\pi_1(W_2)\cong\pi_1(S^1)\cong \mathbb{Z}$ and $d$ the generator of $\pi_1(W_1\cap W_2)\cong\pi_1(S^1)\cong \mathbb{Z}$.

By van Kampen: $\pi_1(W_1\cup W_2)\cong (\mathbb{Z}\ast\mathbb{Z})\ast\mathbb{Z}/N$, where $N$ is the normal subgroup generated by $i_1(w)i_2^{-1}(w)$ with $w\in\mathbb{Z}\cong\pi_1(W_1\cap W_2)$.

a) $\pi_1(X\setminus \mbox{ hole})\cong\pi_1(W_1\cup W_2)\cong \langle a,b,c|a=c^2\rangle\cong \mathbb{Z}\ast\mathbb{Z}$.

b) $\pi_1(X\setminus \mbox{ hole})\cong\pi_1(W_1\cup W_2)\cong \langle a,b,c|c^2=e_0\rangle\cong \mathbb{Z}\ast\mathbb{Z}\ast\mathbb{Z}_2$.

Let's get back to the first calculation and use van Kampen:

We know $U,V\simeq X\setminus \mbox{ hole}$ and therefore $\pi_1(U)\cong \pi_1(X\setminus \mbox{ hole})\cong \pi_1(V)$. So we get by Seifert-van Kampen:

$\pi_1(U\cup V)\cong \pi_1(U)\ast\pi_1(V)/N,$

where $N$ is of course the normal subgroup of Seifert-van Kampen. You now have to calculate $j_1:U\cap V\hookrightarrow U$ and $j_2:U\cap V\hookrightarrow V$ for case a) or b). Maybe you can get an idea of it by drawing a picture. But this is a little bit tricky.

To answer your question: I would say, because of the latter calculation, that it is not nice to calculate the fundamental group, but I would not claim that it is completely impossible.