Problem understanding how to compute fundamental group of connected sum of torus

I have attempted trying to compute the fundamental group of a 2 torus, however I don't know how to proceed to "simplify" the result after applying van Kampen's Theorem.

I calculated the fundamental group of the torus $T$ to be $\pi_1 (S^1\times S^1)\cong\mathbb{Z}\times\mathbb{Z}$.

Then, let $U=T\setminus f(Int(D^2))$, and $V=T\setminus g(Int(D^2))$, where $f$ and $g$ are embeddings.

$U\cap V=S^1$ is path-connected.

Hence we may use Seifert-van Kampen Theorem,

$\pi_1 (T\#T)=\pi_1(T)\coprod_{\pi_1 (S^1)} \pi_1 (T)$ the free product of $\pi_1(T)$ and $\pi_1(T)$ with amalgamation through group homomorphisms $j_1*:\pi_1(S^1)\to\pi_1(T)$ and $j_2*:\pi_1(S^1)\to\pi_1(T)$.

However, I am unsure how to simplify the result to a simplified form, e.g. $\langle a,b,c,d\mid \text{relations}\rangle$.

Thanks for any help! (My background on free groups is quite weak)

Note: I have tried reading the following answers but can't understand them fully due to lack of background. Hopefully, someone can explain in a simple way how to proceed for the last step (simplification of fundamental group)!

• If I'm understanding your set-up correctly, $U$ and $V$ are homotopy equivalent to once-punctured tori (which has fundamental group that is free on 2 generators), so $\pi_1 ( T \# T )$ is $\pi_1 ( U) \ast_{\pi_1 ( S^1)} \pi_1 (V)$ rather than the amalgamated free product of $\pi_1 T$ with itself. Now you'll want to be careful about describing the image of $\pi_1 U \cap V$ under the inclusion induced homomorphisms to get the amalgamating subgroup right. – Charlie Oct 7 '15 at 5:18

If I understand your problem in the right way, you can calculate the fundamental group of the connected sum of two tori in the following way: You divide the connected sum of the two tori into $U$, $V$ and $U\cap V$ like you see in the picture above in the sense of Seifert-van Kampen.

The picture for $U$ and $V$ is very similar. By deformation retraction you get: where the red circle is the hole, you get. This deformation retracts to a bouquet of two copys of an $S^1$, i.e. $S^1\vee S^1$.

$U\cap V$ deformation retracts to an $S^1$.

Now we have to look at the inclusions $i_1:U\cap V\to U$ and $i_2:U\cap V\to V$. The generator (loop) in $U\cap V$, let's call it $c$, is sent by deformation retraction in $U$ or in $V$ to the boundary.

Let us say $a$ and $b$ are the generators in $U$ and $\alpha$ and $\beta$ are the generators in $V$. Then $i_{1\ast}(c)=aba^{-1}b^{-1}$ and $i_{2\ast}(c)=\alpha\beta\alpha^{-1}\beta^{-1}$ (cf. picture above and the boundary!).

So we have $$\pi_1(U\cup V) = \pi_1(S^1\vee S^1)\ast_{\pi_1(S^1)}\pi_1(S^1\vee S^1) = (\pi_1(S^1)\ast \pi_1(S^1))\ast_{\pi_1(S^1)}(\pi_1(S^1)\ast \pi_1(S^1))=\langle a,b,\alpha,\beta|aba^{-1}b^{-1}\alpha^{-1}\beta^{-1}\alpha\beta\rangle$$ (If you don't understand the last step, see "Algebraic Topology" (Hatcher, p. 43).)

Of course it makes sense to use the presentation as a polygon with vertices identified, if you consider the general case, $T^{\#n}$ ($n\in \mathbb{N}$). But for $n=2$, I think it is more intuitive to draw something like the first picture.

MfG Daniel

• No problem @yoyostein :-) – Daniel Bernoulli Oct 18 '15 at 16:16