In topology, there’s the concept of gluing points or subspaces together. For example, take the closed interval *X* = [0, 1] and glue the endpoints 0 and 1 together. Pictorially, we get:

That looks like a circle, but to prove it’s homeomorphic to one, we need to define what’s meant by “gluing”. Let *X* be a topological space and ~ be an equivalence relation on *X*, which partitions *X* as a disjoint union of equivalence classes. Let *Y* := *X*/~ be the set of equivalence classes and we get a surjective map *p* : *X* → *Y*.

[ Conversely, given a surjective map *p* : *X* → *Y* of any two sets, we get a corresponding equivalence relation via ]

Defining the set *Y* is easy; the subtle issue is its underlying topology. For that, we need to explore the kind of properties we want for *Y*. Intuitively if *f* : [0, 1] → **R** is a continuous map such that *f*(0) = *f*(1), then gluing together 0 and 1 should induce a map [0, 1]/~ → **R**, which we *expect to be continuous*.

Universal Property of Quotient. For a surjective map we would like the topology on Y to satisfy:

- for any topological space Z and map , f is continuous if and only if is continuous.

This inspires the following definition.

Definition. Let be a surjective map from a topological space X to set Y. Thequotient topologyon Y is defined as follows.

- A subset is open if and only if is open in X.
It’s easy to show that this definition gives a topology on Y; the resulting topological space Y is called the

quotient space.

Theorem 1. The quotient topology satisfies the above universal property.

**Proof**.

(→) Under the quotient topology, *p* is continuous so if *f* is continuous, then so is . (←) Conversely, suppose is continuous. For each open subset *W* of *Z*, the set is open in *X*. By definition of the quotient topology, is open in *Y*. ♦

**Exercise**.

Prove that if *p* : *X* → *Y* is a surjective map from a topological space *X* to set *Y*, then any topology on *Y* satisfying the universal property must be defined as above. *Thus the universal property uniquely characterises the quotient topology*.

**Proof (Highlight to Read)**.

Suppose *Y* has two topologies *T* and *T’* satisfying the universal property with projection maps *p *: *X* → (*Y*, *T*) and *p’* : *X* → (*Y*, *T’*). Consider the identity map *f* : (*Y*, *T*) → (*Y*, *T’*) which satisfies *fp* = *p’*. Since *fp* is continuous, the universal property on *p* tells us *f* is continuous. Likewise, the inverse of *f* is continuous, so it is a homeomorphism. ♦

**Exercises**.

- Prove that if
*p*:*X*→*Y*and*q*:*Y*→*Z*are surjective maps such that*Y*has the quotient topology from*X*and*Z*has the quotient topology from*Y*, then*Z*has the quotient topology from*X*via*qp*:*X*→*Z*. - Prove that if
*p*:*X*→*Y*is bijective and*Y*has the quotient topology from*X*, then*p*is a homeomorphism. - Prove that if
*p*:*X*→*Y*is surjective, then the quotient topology can also be defined as follows: a subset*C*of*Y*is*closed*if and only if is*closed*in*X*. - Suppose
*i*:*X*→*Y*is injective and gives*X*the subspace topology from*Y*and*p*:*Y*→*Z*is surjective and gives*Z*the quotient topology from*Y*. If is surjective, does this give*Z*the quotient topology from*X*? [**Answer**: no, take the subspace [0, 1) of [0, 1], and the quotient topology on [0, 1] by gluing 0 and 1. Then the quotient topology is compact (see next problem), so [0, 1) → [0, 1]/~ is bijective and non-homeomorphic. By problem 2, the RHS is not a quotient topology from the LHS. ♦ ] - Prove that if
*p*:*X*→*Y*gives*Y*the quotient topology and*X*is compact (resp. connected), then so is*Y*. Again we see that compact and connected often go hand-in-hand in their properties.

## Proving Homeomorphisms

We saw that gluing 0 and 1 in [0, 1] gives a space that looks like a circle. Now that we’ve defined the quotient topology, let’s prove that in fact it is homeomorphic to a circle. Recall the useful result (see lemma here) that if *X* is compact and *Y* is Hausdorff, then a bijective continuous map *X**→Y* is a homeomorphism.

[ Question to ponder: homeomorphism is intuitively a “local” property. So can we replace “compact” with “locally compact” above? I.e. is it true that if *X* is locally compact and *Y* is Hausdorff, then a bijective continuous *X**→Y* is a homeomorphism? To get a hint of the answer, see the answer to exercise 4 above. ]

Let *Y* = [0, 1]/~, where ~ identifies 0 and 1, and *p* : [0, 1] → *Y* be the projection map. We wish to show , where Now:

is a continuous map satisfying *f*(0) = *f*(1) so it induces a map . By the universal property above, *g* is continuous since is continuous. Now, we invoke the lemma and conclude that *g* is a homeomorphism (together with the above exercise 5: since [0, 1] is compact, so is the quotient *Y*).

Now the acute reader may prove the following using the same technique.

**Example 1: Double Circle**

Let *Y* be the quotient topology from *X*=[0, 1] by identifying 0, 1/2 and 1. Prove that *Y* is homeomorphic to the union of two circles:

as a subspace of **R**^{2}.

[ Note: topologically, one calls this a *wedge sum* of two circles. Intuitively a wedge sum of a collection of topological spaces is obtained by fixing a point in each space and gluing all points together. This is a rather useful concept in algebraic topology, specifically homotopy, but that’s another story for another day. ]

**Example 2: Cylinder**

Let *X* = [0, 1] × [0, 1] and glue points (0, *y*) and (1, *y*) together, for all 0 ≤ *y* ≤ 1. Prove that we get the cylinder *S*^{1} × [0, 1].

Now one’s *very* tempted to believe: if gives *Y* the quotient topology from *X*, and gives *Y’* the quotient topology from *X’*, then gives *Y* × *Y’* the quotient topology from *X* × *X’*. But this is wrong, but a counterexample is fiendishly hard to find. For one, refer to *Munkres*, “*Topology*” (2nd ed.), section 22, page 145, exercise 6.

**Example 3: Torus**

Let *X* = [0, 1] × [0, 1] and glue points (0, *y*) and (1, *y*) together, as well as points (*x*, 0) and (*x*, 1) together. In particular, all the four corner points collapse to one. The result is a torus

**More Examples**

Let *X* = [0, 1] × [0, 1] and glue points (0, *y*) and (1, 1-*y*) together. We get a Mobius strip, obtained by taking a rectangular strip of paper, and gluing two opposite edges together after twisting the strip 180 degrees.

Let *X* = [0, 1] × [0, 1] and glue points (0, *y*) and (1, *y*) together, as well as (*x*, 0) and (1-*x*, 1) together. This gives a Klein bottle, which is an interesting example of a two-dimensional object which cannot be embedded in 3-space as a subspace. The proof for this is hard though.

Let be the unit disc on a plane. Collapse the outer circle comprising of points satisfying into a single point. The resulting topological space is homeomorphic to a 2-sphere:

Let *X* be the space **R** × {0, 1}, where {0, 1} is given the discrete topology. Identify the points (*x*, 0) with (*x*, 1) for all *x* ≠ 0. We get the “real line with double origin”:

Denoting the two origins by *p* and *p’*, we see that any open subsets *U*, *V* which contain *p*, *p’* respectively, must intersect since they must contain a positive real number. Thus, the space *X* is Hausdorff but its quotient isn’t.

In the next article, we’ll be looking at topology of group quotients. The situation is not as straightforward as one might imagine.

Hello! I was looking for quotient spaces and I found this nice blog. I have some questions:

1. Do you make the pictures? How?

2. Is it there a hint to prove that the unit disk with that equivalence relation is homeomorphic to the 2-sphere? I have studying with the book of Munkres, so I think that perhaps a theorem and a cororally in the section of quotient topology may help maybe without using Hausdorff spaces.

Hi. Really glad that the article is of help to you. Regarding your questions:

1. The pictures are created in Microsoft Word; then screen captured and pasted in Microsoft Paint.

2. The best proof I can think of is similar to that before example 1. To show that , first show that there is a continuous map . Then show that two points are equivalent if and only if . By the universal property above, this ensures that f factors through to give a bijective continuous map. Now, since our X is compact and Y is Hausdorff, we’re done.