Problem computing the fundamental group of the double torus.

To compute this I want to use the Seifert-van Kampen Theorem, then I choose $U=\text{Left Torus}/{point}$ and $V=\text{Right Torus}/{point}$ so then the intersection $U \cap V$ is a cylinder with out bottom covering and bottom covering, this is homotopic to the disc with out a point which is homotopic to the circle $S^{1}$ then $\pi_1({U})=<a,b>$, $\pi_1({V})=<c,d>$, and $\pi_1(S^{1})=<a>=\mathbb{Z}$.

My question is, How can I compute the fundamental group since I don't have any relations to work with here?

Thanks a lot in advance.

• I may be wrong but I think $U$ and $V$ are tori with an additional hole, so the fundamental group should be generated by three elements each. I feel the third generators get identified by Seifert-van Kampen, so you should end up with five generators. – blue May 2 '16 at 22:51
• How is that? It is suppose to be of to so that the double torus is of two – user162343 May 2 '16 at 22:53
• I was wrong. Typing out an answer now. – blue May 2 '16 at 23:23
• Good :) I will check it when you are done :) – user162343 May 2 '16 at 23:24

First consider the single torus, represented as the square with opposite edges identified, call the edges $a$ and $b$. $U$ is the torus with a single point removed, this retracts to just the borders of the square, so it is the union of two copies of $S^1$ and it has fundamental group $<a, b>$, the free group in two elements. Now consider a circle around the removed point, this represents a generator of the fundamental group of $U\cap V$, since $U\cap V$ is in this representation just the area between two circles around the removed point. This circle represents the element $aba^{-1}b^{-1}\in\pi_1(U)$. Similarly for $V$, the generator gets mapped to $cdc^{-1}d^{-1}$. Hence the fundamental group of the double torus is $<a, b, c, d \mid aba^{-1}b^{-1}(cdc^{-1}d^{-1})^{-1} >$.
Edit: Filling some details. First if you haven't seen it already, here's the picture that I am using for the argument. If you haven't seen this before, I strongly suggest that you look up this construction of the torus before moving on. Note that the ways denoted $A$ and $B$ in the picture are in fact loops.
Cut a point (or a small circle) out in the middle of that picture. Consider again a circle around the removed point in the torus as above. This circle generates the fundamental group of $U\cap V$. We wish to figure out where the generator gets mapped under the inclusion map $U\cap V \to U$ or more precisely under the induced map $\pi_1(U\cap V) \to \pi_1(U)$.
To do this, we take a representant, i.e. the circle and compose it with the inclusion map. Since the inclusion maps every point to itself, we still have the same circle. Now recall that the torus minus a point retracts to the border of the square. Under this retraction map, the circle gets mapped to the loop around the border of the square. This loop leads along all four edges, but remember that the edges are identified. In fact they are identified in such a way that when we move along an edge for the second time then we will pass it in the opposite direction. So our loop can be written as the composition of loops $ABA^{-1}B^{-1}$. This means that there is a similar equation for elements of the fundamental group. Technically, at this point you would have to go back from the square to the torus missing a point, where a similar equation holds, because the retraction induces an isomorphism of fundamental groups.