Topology: Quotient Topology and Gluing

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 pX → Y.

[ Conversely, given a surjective map pX → Y of any two sets, we get a corresponding equivalence relation via x_1 \sim x_2 \iff p(x_1) = p(x_2). ]

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 p:X\to Y, we would like the topology on Y to satisfy:

  • for any topological space Z and map f:Y\to Z, f is continuous if and only if f\circ p:X \to Z is continuous.

This inspires the following definition.

Definition. Let p:X\to Y be a surjective map from a topological space X to set Y. The quotient topology on Y is defined as follows.

  • A subset V\subseteq Y is open if and only if p^{-1}(V) 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.


(→) Under the quotient topology, p is continuous so if f is continuous, then so is f\circ p:X \to Z. (←) Conversely, suppose f\circ p is continuous. For each open subset W of Z, the set p^{-1}f^{-1}(W) = (f\circ p)^{-1}(W) is open in X. By definition of the quotient topology, f^{-1}(W) is open in Y. ♦


Prove that if pX → 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 X → (YT) and p’X → (YT’). Consider the identity map f : (YT) → (YT’) which satisfies fpp’. 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. ♦


  1. Prove that if pX → Y and qY → 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 qpX → Z.
  2. Prove that if pX → Y is bijective and Y has the quotient topology from X, then p is a homeomorphism.
  3. 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 p^{-1}(C) is closed in X.
  4. Suppose iX → Y is injective and gives X the subspace topology from Y and pY → Z is surjective and gives Z the quotient topology from Y. If p\circ i:X\to Z 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. ♦ ]
  5. Prove that if pX → 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 Y\cong S^1, where S^1=\{(x,y)\in\mathbf{R}^2 : x^2 + y^2 = 1\}. Now:

f:[0, 1] \to S^1, \quad t \mapsto (\cos(2\pi t), \sin(2\pi t))

is a continuous map satisfying f(0) = f(1) so it induces a map g:Y\to S^1. By the universal property above, g is continuous since g\circ p = f :[0, 1]\to S^1 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:

\{(x, y) : (x+1)^2 + y^2 = 1\} \cup \{(x,y) : (x-1)^2 + y^2 = 1\}

as a subspace of R2.


[ 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 S1 × [0, 1].



Now one’s very tempted to believe: if p:X\to Y gives Y the quotient topology from X, and q:X'\to Y' gives Y’ the quotient topology from X’, then r:=(p,q) : X\times X' \to Y\times Y' 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 S^1 \times S^1.


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 X = D^2 := \{(x, y)\in \mathbf{R}^2 : x^2 + y^2 \le 1\} be the unit disc on a plane. Collapse the outer circle comprising of points satisfying x^2 + y^2 = 1 into a single point. The resulting topological space is homeomorphic to a 2-sphere: S^2 := \{(x, y, z) \in \mathbf{R}^3 : x^2 + y^2 + z^2 = 1\}.

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”:

quotient_double_0Denoting the two origins by p and p’, we see that any open subsets UV which contain pp’ 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.

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2 Responses to Topology: Quotient Topology and Gluing

  1. Jenir says:

    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.

  2. limsup says:

    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 X/\sim \cong Y, first show that there is a continuous map f:X\to Y. Then show that two points x, x'\in X are equivalent if and only if f(x) = f(x'). By the universal property above, this ensures that f factors through g:X/\sim \to Y to give a bijective continuous map. Now, since our X is compact and Y is Hausdorff, we’re done.

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