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Shankar

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- Thread starter shankarvn
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Shankar

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mathwonk

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I have not used this concept too much, but I recall a little from reading kelley, general topology.

take any topological space X and any surjective map X-->Y to a set Y. then this map will be continuous with certain topologies on Y and not continuous with others.

i.e. the map is continuous iff every open set in Y ahs an open inverse image. so the more open sets Y has, the harder it is for the map to be continuous. E.g. if Y has the smallest possible topology, with only the empty set and Y open, the map is always continuous.

suppose we give Y the largest number of open sets, such that the map is still continuous. this i believe is called the quotient topology on Y.

i.e. given any surjective map X-->Y, we can view Y as made by taking a "quotient" of X by an equivalence relation.

just consider two points of X equivalent if they have the same image under the map.

Then given such a quotient space Y of X, the points of Y are actually subsets of X. Then the quotient topology on Y is expressed as follows: a set in Y is open iff the union in X of the subsets it consists of, is open in X.

This is a basic but simple notion. important, but nothing deep here except the idea of continuity, and the general idea of enhancing the structure of a set of equivalence classes.

the same thing goes on everywhere. given a group and a normal subgroup, one gets an equivalence relation on the elements of the group and makes a group out of these classes.

given a ring and an ideal one makes a ring out of the quotient object.

in general quotient constructions are techniques for constructing "maximal" targets for surjective maps subject to a certain condition.

I'm sorry. you asked for a physical picture. imagine the real line mapping to the circle by the map taking t to (cos(t),sin(t)). the "wrapping map" or polar coordinates map. then try to convince yourself that the quotient topology is exactly the usual topology on the circle.

take any topological space X and any surjective map X-->Y to a set Y. then this map will be continuous with certain topologies on Y and not continuous with others.

i.e. the map is continuous iff every open set in Y ahs an open inverse image. so the more open sets Y has, the harder it is for the map to be continuous. E.g. if Y has the smallest possible topology, with only the empty set and Y open, the map is always continuous.

suppose we give Y the largest number of open sets, such that the map is still continuous. this i believe is called the quotient topology on Y.

i.e. given any surjective map X-->Y, we can view Y as made by taking a "quotient" of X by an equivalence relation.

just consider two points of X equivalent if they have the same image under the map.

Then given such a quotient space Y of X, the points of Y are actually subsets of X. Then the quotient topology on Y is expressed as follows: a set in Y is open iff the union in X of the subsets it consists of, is open in X.

This is a basic but simple notion. important, but nothing deep here except the idea of continuity, and the general idea of enhancing the structure of a set of equivalence classes.

the same thing goes on everywhere. given a group and a normal subgroup, one gets an equivalence relation on the elements of the group and makes a group out of these classes.

given a ring and an ideal one makes a ring out of the quotient object.

in general quotient constructions are techniques for constructing "maximal" targets for surjective maps subject to a certain condition.

I'm sorry. you asked for a physical picture. imagine the real line mapping to the circle by the map taking t to (cos(t),sin(t)). the "wrapping map" or polar coordinates map. then try to convince yourself that the quotient topology is exactly the usual topology on the circle.

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mathwonk

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think of wrapping a straight thread around a circular spool.

in coordinates it is given by cosine and sine. i.e. we map the t- line onto the unit circle by sending the point with coordinate t on the t - line, to the point with coordinates (cos(t), sin(t)) on the unit circle.

somehow i feel i am not understanding your question. usually in this situation someone else does, and steps in and helps out. maybe they will.

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Thank you sir

Shankar

Shankar

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mathwonk

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mr shankar you are a gentleman and a scholar.

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Maybe a few years late to the party, but thats an excellent explanation. ThankYou.

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