From: Philippe Gaucher <gaucher@irmasrv1.u-strasbg.fr>
To: categories@mta.ca
Subject: localization : more precise question
Date: Fri, 1 Dec 2000 22:13:06 +0100 (MET) [thread overview]
Message-ID: <200012012113.WAA10773@irmast2.u-strasbg.fr> (raw)
Re-bonjour,
Thank you for your answers. My question was very general. So here is
the example.
I am going to define the category C and the collection of morphisms S,
with respect to what I would like to localize.
The object of C are the oriented graph. Such an object X is a
topological space obtained by choosing a discrete set X^0 and by
attaching 1-dimensional cells *with orientations*. It is a
1-dimensional CW-complex with oriented arrows.
The morphisms of C are the continuous maps f from X to Y satisfying
this conditions :
1) f(X^0)\subset Y^0
2) f is orientation-preserving
3) f is non-contracting in the sense that a 1-cell is never contracted
to one point.
Remark I : in C, an arrow x--> is not isomorphic to a point.
Remark II : an arrow a-->b can be mapped on the loop a-->a with one
oriented arrow from a to a.
A morphism f of C is in S if and only if f induces an homeomorphism on
the underlying topological spaces. Now here is an example of f\in S
which is not invertible :
a--->b mapped on a-->x-->b
This morphism has no inverse in C because the image of x must be equal
to a or b by 1) and therefore one of the arrows would be contracted by
2), which contredicts 3).
I would like to know if C[S^{-1}] exists or no (in the same universe).
The irresistible conjecture is of course that C[S^{-1}] is equivalent
to the category whose objects are that of C and whose morphisms from A
to B are the subset of C^0(A,B) (the set of continuous maps from A to
B) containing all composites of the form
g_1.f_1^{-1}.g_2...g_n.f_n^{-1}.g_{n+1} where g_1,...,g_{n+1}, are
morphisms of C and f_1,...,f_n morphisms in S.
The Ore condition is not satisfied by S because of this example. The
Ore condition says that for any s:A-->B in S, and any f:X-->B, there
exists t:Y-->X in S and g:Y-->A such that s.g=f.t. Now the
counterexample : A is a--->b, B is a-->x-->b with s as above ; X is
a-->x with the inclusion f from X in B. Then necessarily Y=X and t=Id.
And s.g(x)=b and f.t(x)=x.
Thanks in advance. pg.
next reply other threads:[~2000-12-01 21:13 UTC|newest]
Thread overview: 5+ messages / expand[flat|nested] mbox.gz Atom feed top
2000-12-01 21:13 Philippe Gaucher [this message]
2000-12-02 18:05 ` Dan Christensen
2000-12-02 23:33 Philippe Gaucher
2000-12-03 22:39 ` Dan Christensen
2000-12-03 16:44 Philippe Gaucher
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