Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have problems to understand the Seifert-Van Kampen theorem when $U, V$ and $U\cap V$ aren't simply connected. I'm going to give an example:

Let's find the fundamental group of the double torus $X$ choosing as open sets $U$ and $V$: (see picture below)enter image description here

Then $U$ and $V$ are the punctured torus, so $\pi_1(U)=\pi_1(V)=\mathbb Z*\mathbb Z$, following the Van Kampen theorem, we have:

$\pi_1(X)=\pi_1(U)*\pi_1(V)/\overline C$, where $C=\{(i_*\gamma)(j_*\gamma)^{-1}:\gamma \in \pi_1(U\cap V)\}$, where $\overline C$ is the normal closure of C, and $i_*:\pi_1(U\cap V)\to \pi_1(U)$, $j_*:\pi_1(U\cap V)\to \pi_1(V)$ are the induced inclusion homomorphisms.

Then $\pi_1(X)=\mathbb Z*\mathbb Z*\mathbb Z*\mathbb Z/\overline C$

I have troubles to "see" this quotient group in order to get the fundamental group.

I need help, it's a long time I have this doubt I would appreciate if anyone help me.

Thanks

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

If $\pi_1(U)=\langle a,b\rangle$ and $\pi_1(V)=\langle c,d\rangle$ then (for the right choice of $a,b,c,d$ :) $i_*\gamma=aba^{-1}b^{-1}$ and $j_*\gamma=cdc^{-1}d^{-1}$, so you get a presentation $$\pi_1(X)=\langle a,b,c,d|aba^{-1}b^{-1}= cdc^{-1}d^{-1}\rangle.$$ In general, if you have presentations $\pi_1(U)=\langle A|S\rangle$ and $\pi_1(V)=\langle B|T\rangle$, you can get a presentation $\pi_1(X)=\langle A\sqcup B|S\sqcup T\sqcup R\rangle$, where the additional relations $R$ are of the form $i_*\gamma=j_*\gamma$, where $\gamma$ runs over generators of $\pi_1(U\cap V)$.

share|improve this answer
    
then $\gamma$ is the generator of $U \cap V$? –  user42912 Nov 30 '12 at 13:58
    
@RafaelChavez: yes (of $\pi_1(U\cap V)$) –  user8268 Nov 30 '12 at 14:59
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.