Re: Fundamental group question


W. Dale Hall napisal(a):
I've sorted out the text here in an attempt to
improve its logical flow. I also included a
corrected attribution line (to Z. Karno).

It appears that James needs to learn how to
use gmail/Google groups to format articles
correctly for sci.math. I'll point out that
I don't know how to do that either, but also
that I don't need to know how to do that,
since I don't use those clients.

James wrote:
Zbigniew Karno wrote:
James napisal(a):
Dear all, I am having trouble with the following
fundamental group question :
Let D be the closed unit disk in R^2. Now, let X
be the space obtained from D by first removing two disjoint open
subdisks in the interior of D, and then identifying all three
resulting boundary circles together via a homeomorphism preserving
clockwise orientation of these circles.
I wish to calculate the fundamental group of this
space (I am practicing questions to prepare for my qualifying
exam).
Well, I want to use Van Kampen. So I want to take
two open sets A and B, where A contains both subdisks (i.e. I draw
an oval that encircles both of the two subcircles, and then A is
the stuff inside, plus of course a tiny neighborhood to make it
open). B will be the stuff outside, plus a tiny neighborhood so
that A and B intersect in something homotopy equivalent to a
circle.

>> Incorrect! Assumptions in the Seifert-van Kampen theorem
are not satisfied. In the space X for these open sets A and B you
have chosen, the intersection of A and B is not path-connected
(even it is not connected) and, in fact, it has a homotopy type of
the disjoint union of two circles.


Before I even attempt to try this again, I need to understand what
you have said above, which to me is very incorrect. A /\ B is a
thick circle, which is path connected. Why not? I'd like to repeat
what A, B , and X are, just in case I wasn't clear. X is a closed
disk in R^2. Now, remove two disjoint subdisks from the interior of
X. Identifying all three resulting boundary circles together via a
homeomorphism preserving clockwise orientation of these circles.


One wonders why you felt that what Zbigniew wrote
was "very incorrect", instead of reviewing how you
had defined the space. At the very least, you could
have attempted to read and understand what it was
that he said. I'll try to spell it out.

Now to get A, draw a dotted oval around both subcircles (i.e. that
contains both) and call that dotted oval l_1. Now draw another
dotted oval around both subcircles but so that this new dotted oval
also has l_1 in its interior, and call this dotted oval l_2. Let A
be everything inside l_2, and let B be everything outside l_1. Then
A /\ B is a "thick open circle", i.e. it is S^1 x (0,1). So it is
path connected. Both A, B are open and path connected of course as
well.


Perhaps you forgot about the identifications that
you have made along the three circles. to be sure,
in the unit disc, the intersection of A with B is
precisely what you have said. However, *that* is
not the space you wish to study, now, is it?

Some part of the confusion stems from naming sets
in the unit disc, making some identifications to
produce the space X, and mis-reading what happens
when passing from the unit disc to X. In order to
distinguish the sets as they appear in the unit
disc from those in X, I'll append a tilde (~) to
the names of sets in the unit disc, and leave it
off when I talk about the corresponding sets in X.

Note that the intersection of A~ with B~ is a single
annulus. This is what you imagine the intersection
of A with B to be as well, and that is your error.

You have identified the two circles surrounding the
holes in A~ (in the unit disc) to a single circle; this
circle is then in A. The boundary of the unit disc,
within B~, is also identified with this same circle,
and so B contains this identification circle as well
and so the circle becomes a second component of the
intersection of A with B.

Note that if the sets A and B need to be open in X,
then you would have had to extend B~ into the holes
inside A~, and extend A~ out through those holes to
include a neighborhood of the boundary of the unit
disc in B~.

If you had done that, I imagine you might have avoided
your mistake.


I don't understand what you have said, and it must be a fundamental
concept I am missing. Please explain, and I thank you very much.
-James


Then I would think that, after labeling the outer
boundary circle by a, and the two inner boundary
circles by a, the fundamental group would be
<a | a = a^2>, i.e. the free group on a single
letter a, such that a = a^2, which of course is the trivial group.
The reason is that A intersect B is a circle,
which, in B, deformation retracts to the boundary circle a, and
which, in A, deformation retracts to a^2 I think. The reason for
the a^2 is that it seems that the two inner circles can be
homotoped to S^1 \/ S^1 (one-point union), and then since both
circles are labeled a, I get a^2.

I'll note that Zbigniew points your error out right here:

Incorrect reasoning! Observe that in X there is a circle which
was created as identification space of all three boundary circles
and this circle is contained in both A and B.

I don't think this is right. What am I doing wrong
here if anything?
Thank you for your help,

James

Rather, you have to use in your proof the following Corollary from
the Seifert-van Kampen theorem: Let X be the union of two subsets
U and V open in X. Assume U, V and the intersection of U and V are
path-connected and let x_0 be fixed point in the intersection of U
and V. If V is simply connected, then there is an s an isomorphism
k : pi_1(U , x_0) / N ---- > pi_1(X , x_0), where N is the least
normal subgroup of pi_1(U , (U , x_0) containing the image of the
homomorphism i* : pi_1(U cap V , x_0) ---- > pi_(U , x_0) induced
by the inclusion i of U cap V in U.

Using this fact a computation of the fundamental group of X is very
simple.

Best Regards, Z. Karno


That's it.

Dale.

Very nice explanation.

Thanks,
Z. Karno

.

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