From: rusin@math.niu.edu (Dave Rusin)
Newsgroups: sci.math.research
Subject: Re: Seifert/Van Kampen Theorem
Date: 26 Mar 1998 04:50:51 GMT

In article <sdkj-1903982023510001@cdale-dialup63.midwest.net>,
MS <sdkj@hgd.org> wrote:
>Let /\ mean intersection, and U be union.
>
>I have a question about the Seifert\Van Kampen theorem.
>Let A and B be 3-manifolds (with boundary). Let's say x is 
>a generator of A /\ B.  I cannot figure out what the image of
>x is under inclusion into A, but I can tell that x^n goes to
>a word a in Pi1(A).   If inclusion into B gives x = b, can I just 
>list b^n = a as one of the relators of Pi1(A U B), along with
>the others? 

Algebraically, you're asking the following. Given two groups  G  and  H
and maps from a cyclic group  g: <x> -> G and  h: <x> -> H, you wish
to compute the free product with amalgamation  X = G *_{<x>} H. You already
know that if  G = < gens | rels > and  H = <gens' | rels'>  are
presentations of these groups, then  
	X = < gens U gens' | rels U rels' U {g(x)=h(x)} >
You're hoping to replace this with the group presentation
	Y = < gens U gens' | rels U rels' U {g(x)^n=h(x)^n} >
for some  n.  Unfortunately  X  and  Y  are not in general isomorphic.
There is a natural surjection  Y -> X, but there's no reason to expect it
to be an isomorphism, and indeed it's easy to write down examples in
which  X  and  Y  are non-isomorphic, e.g.
	G = Z/2,   H = {1},  g=surjection,   n=2
gives  |X| = 1, |Y| = 2.

This example can be realized geometrically as well, even just with 
2-manifolds with empty boundary (cross with the interval if you want
3-manifolds): let  M  be a Moebius strip; identify opposite points of
its (circular) boundary to form  A = RP^2. Also let  B  be the unit sphere
in  R^3. Identify the center line of  M  (in  A)  with the equator of  B.
The resulting union of  A  and  B  is simply connected, but you'd be
presenting its fundamental group as  = pi1(A) = Z/2Z  since  a = b = identity.
for  n=2.

Perhaps I've misunderstood the geometric constraints under which
you're operating?

dave