From: chrisman@ucdmath.ucdavis.edu (Mark Chrisman)
Newsgroups: sci.math
Subject: Collapsing a simply-connected subspace
Date: 9 Nov 1994 12:48:31 GMT
Let X be a path-connected topological space, and let A be a simply
connected subset. Will the fundamental group of X/A be the same as
that of X? Or can someone give a counterexample?
[sig deleted here and below - djr]
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From: Benjamin.J.Tilly@dartmouth.edu (Benjamin J. Tilly)
Newsgroups: sci.math
Subject: Re: Collapsing a simply-connected subspace
Date: 9 Nov 1994 19:17:45 GMT
In article <39qgev$ko3@mark.ucdavis.edu>
chrisman@ucdmath.ucdavis.edu (Mark Chrisman) writes:
>
> Let X be a path-connected topological space, and let A be a simply
> connected subset. Will the fundamental group of X/A be the same as
> that of X? Or can someone give a counterexample?
>
Let X be an infinite wedge of circles, with the obvious metric. Let A
be the base point, plus on the n'th circle a closed interval that
starts from the base point and goes (n-1)/n around the circle. (This
makes sense since each circle is actually a 1-dim manifold everywhere
but the base point.) Now X/A is clearly a metric space, in fact it is
the infinite sea-shell, but its fundamental group is not the
fundamental group of X. In fact the two fundamental groups have
different cardinalities.
Since metric spaces are a special case of topological spaces, this does
for a counterexample.
Ben Tilly
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From: chrisman@ucdmath.ucdavis.edu (Mark Chrisman)
Newsgroups: sci.math
Subject: Re: Collapsing a simply-connected subspace
Date: 10 Nov 1994 07:30:58 GMT
In article <39r78p$hb1@dartvax.dartmouth.edu>
Benjamin.J.Tilly@dartmouth.edu (Benjamin J. Tilly) writes:
> In article <39qgev$ko3@mark.ucdavis.edu>
> chrisman@ucdmath.ucdavis.edu (Mark Chrisman) writes:
>
> >
> > Let X be a path-connected topological space, and let A be a simply
> > connected subset. Will the fundamental group of X/A be the same as
> > that of X? Or can someone give a counterexample?
> >
> Let X be an infinite wedge of circles, with the obvious metric. Let A
> be the base point, plus on the n'th circle a closed interval that
> starts from the base point and goes (n-1)/n around the circle. (This
> makes sense since each circle is actually a 1-dim manifold everywhere
> but the base point.) Now X/A is clearly a metric space, in fact it is
> the infinite sea-shell, but its fundamental group is not the
> fundamental group of X. In fact the two fundamental groups have
> different cardinalities.
>
> Since metric spaces are a special case of topological spaces, this does
> for a counterexample.
>
> Ben Tilly
Hmmm, I've never tried to collapse a metric space before. Tell me if
I'm getting this right; when you collapse a subspace A in a metric
space X with distance function d, you get a space whose set of points
is
(X-A) U {a}
and where distance is defined by
d'(x,y) = min {d(x,y), d(x,A)+d(y,A)}; (if x,y not in A)
d'(x,a) = d'(a,x) = d(x,A); (if x not in A)
d'(a,a) = 0.
If we simplify by assuming A is closed in X, then we don't have to
worry about distinct points in X/A having zero distance between them.
Anyway, I'll have to think about your example, because it's not
immediately apparent to me that X/A (collapsed metric space) is
homeomorphic to X/A (collapsed topological space). I don't think you
can just say "metric spaces are a special case of topological spaces".
Consider this similar example from algebra:
Groups are a special case of monoids. Consider the monoid M generated
by two elements a, a' with the relations aa' = a'a = 1. An element of
this monoid has one of the three forms
aaa...a
a'a'a'...a'
1.
This monoid also happens to be a group; it's the cyclic group generated
by a, where a' and a are inverses. Now, if we collapse the group by
setting a=1, we get M/a = {1} (the trivial group). But, if we collapse
M as a monoid, we get M/a = {1, a', a'a', ...} (a free monoid generated
by a').
-----
I've just thought about your example, and I think it works even if we
think of X/A as a collapse of a topological space rather than a metric
space.
==============================================================================
From: chrisman@ucdmath.ucdavis.edu (Mark Chrisman)
Newsgroups: sci.math
Subject: Re: Collapsing a simply-connected subspace
Date: 10 Nov 1994 07:34:05 GMT
> In article <39qgev$ko3@mark.ucdavis.edu>
> chrisman@ucdmath.ucdavis.edu (Mark Chrisman) writes:
>
> >
> > Let X be a path-connected topological space, and let A be a simply
> > connected subset. Will the fundamental group of X/A be the same as
> > that of X? Or can someone give a counterexample?
> >
A newsreader just sent me email asking about a similar question:
suppose now X is a manifold and A is simply connected. Under what
conditions will X/A be homeomorphic to X?
p.s. please post replies to the net rather than sending email to me.
==============================================================================
Newsgroups: sci.math
From: jdchrist@math.mit.edu (Dan Christensen)
Subject: Re: Collapsing a simply-connected subspace
Summary: you need the inclusion to be a cofibration
Date: Wed, 9 Nov 94 22:05:31 GMT
In article <39qgev$ko3@mark.ucdavis.edu>,
Mark Chrisman wrote:
>
>Let X be a path-connected topological space, and let A be a simply
>connected subset. Will the fundamental group of X/A be the same as
>that of X? Or can someone give a counterexample?
This won't be true in general. The extra condition that you need is that
the inclusion A --> X be a cofibration. This means that we have the
homotopy extension property:
Given any map f : X --> Y and any homotopy H : A x I --> Y that
agrees at time 0 with f restricted to A, we can extend H to a
homotopy K : X x I --> Y agreeing at time 0 with f.
Under this assumption, X/A and the mapping cone X \union_A CA are
homotopy equivalent (and hence have the same fundamental group). If A
is simply connected, the Seifert-Van Kampen theorem can be used to
show that X \union_A CA has the same fundamental group as X.
Here's a counterexample when the inclusion is not a cofibration.
(This is just off the top of my head; there are probably nicer
examples.) Let X be the circle with basepoint *, and let A be the
complement X - {*} of the basepoint, a contractible subspace. Then
X/A is the two point space with points A and *. The open sets are
{A, *}, {A} and {}. A map from the circle to X/A is simply a choice
of an open subset of the circle, so there are lots of them. But they
are all homotopic as the components of an open set can be deformed
through open sets to the empty set.
Exercise: Convince yourself directly that the inclusion A --> X is not
a cofibration.
Note: You may wonder why the mapping cone comes up here and in other
places. It should be thought of as the homotopy theoretic quotient of
X by A because it has nice properties whether or not the map A --> X
is a cofibration. It is also called the homotopy cofibre and is an
example of a more general construction called a homotopy colimit.
Hope this helps,
Dan Christensen