# Fundamental group of $F_{3,0}\times S^1$

Let $$F_{3,0}$$ denote the $$2$$-punctured disk (or $$3$$-punctured sphere). I've seen in many textbooks that $$\pi_1(F_{3,0}\times S^1)=\langle a_1,a_2,a_3,h| a_ih=ha_i,\ 1\leq i\leq 3,\ a_1a_2a_3=1 \rangle$$ but I do not know how to prove it and I do not know from where to start.

• I'm not sure about this presentation, but the fundamental group is just $F(a,b) \times \mathbb{Z}$ where $F(a,b)$ is the free group on two generators. – Connor Malin Dec 10 '19 at 3:01

First of all $$\pi_1(X\times Y)\simeq\pi_1(X)\times\pi_1(Y)$$ as you can see here.
Now $$\pi_1(S^1)\simeq\mathbb{Z}$$ as we all know. On the other hand your $$F_{3,0}$$ is homotopy equivalent to the wedge sum of two circles $$S^1\vee S^1$$ and so $$\pi_1(F_{3,0})\simeq\mathbb{F}(a_1,a_2)$$ where on the right side we have the free group on two elements. And so
$$\pi_1(F_{3,0}\times S^1)\simeq \mathbb{F}(a_1,a_2)\times\mathbb{Z}$$
$$\langle a_1,a_2,h\ |\ a_ih=ha_i, i=1,2 \rangle$$
This is the same presentation that you have, simply because your last relation means $$a_3=(a_1a_2)^{-1}$$ and so $$a_3$$ is fully expressible via other generators. Meaning both the generator and the relation can be removed from the presentation.