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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**

- 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? - 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(2*p**it*). 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.

In fact these two examples are the same. To see this, take as a representative of each equivalence class ( coset) an element in [0, 1). Then every equivalence class has a unique representative, but one should arrange things so that classes with representatives close to 1 should be near the one with representative 0.

So we fix the topology so that this happens.

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**

- This topology is the strongest (
*most*open sets) in which p is continuous.

- 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**

- The above example is [0, 1]/{0, 1}

- Let
*X*be the closed unit disc and let*A*be the "boundary circle". Then*X*/*A*is homeomorphic to the unit sphere*S*^{2}**R**^{3}.

**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.

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