The identification topology

Recall that an equivalence relation ~ (a reflexive, symmetric, transitive relation) on a set X can be regarded as a method of partitioning X into disjoint subsets (called equivalence classes).
We shall denote the set of equivalence classes by X/~.
The identification or quotient topology gives a method of getting a topology on X/~ from a topology on X.

To see why we would want to do something like this, we will look at some examples.

Examples

1. Let X be the closed unit interval [0, 1] R.
   Define an equvalence relation on X by x ~ y if and only if x = y or {x, y} = {0, 1}.
   Then every point is its own equivalence class, except for {0, 1} which forms one class.
   So this relation has the effect of "identifying" 0 and 1 and leaving everything else alone.

   What would we like the topology to look like?

2. Let X = R and define ~ by x ~ y if and only if x - y Z.
   Then X/~ is the set of cosets R/Z of the additive group.
   What would we like the topology to be?
   A clue comes from group theory.
   Define f: R C - {0} by z exp(2pit). Then ker(f) = Z and the image of f is the unit circle in the C-plane.
   Thus, by the First Isomorphism Theorem of group theory, R/Z a circle and it would be a good idea to consider it with its subspace topology inherited from the plane.
   

If we have an equivalence relation ~ on X we get a natural projection map p: X X/~ got by mapping each point to its equivalence class.

Definition

The identification topology on X/~ is defined by:
A set A X/~ is open if and only if p ^-1(A) is open in X.

Remarks

1. This topology is the strongest ( most open sets) in which p is continuous.
   
2. Note that p ^-1(A) consists of all those points whose equivalence classes are in A.
   

Look again at the previous examples

We have X = [0, 1] and p : X X/~ identifying the end-points of the interval. If U is a set of X/~ containing the equivalence class {0, 1} then p ^-1 (U) contains both 0 and 1 and hence if it is open contains a set like  
So small open sets of X/~ (-neighbourhoods) are "the same" as those of X except "at the end-points" where they look like the sets    made by "gluing together two bits". It is easy to see that this is the same as the subspace topology on the circle as a subset off the plane.

Topologists like to consider spaces made by "shrinking a subset to a point".

Notation
If X is a topological space and A is a subset of X, we denote by X/A the equivalence classes X/~ under the relation x ~ y if x = y or x, y A.
So for x A, {x} is an equivalence class and A is a single class.

Examples

1. The above example is [0, 1]/{0, 1}
   
2. Let X be the closed unit disc and let A be the "boundary circle". Then X/A is homeomorphic to the unit sphere S² R³.
   
   Proof
   Look at the "teardrop" shown and "stereoscopic projection from the "point" P to a horizontal plane. The inverse image of an open set which contains the whole boundary is then an open set which contains P. Every other point is mapped in a one-one way. It is easy to verify that this map is a homeomorphism.