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 , MS 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: -> G and h: -> H, you wish to compute the free product with amalgamation X = G *_{} H. You already know that if G = < gens | rels > and H = 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