•
Gluing. Topologists talk of gluing points together. If X is a topological space, gluing the points x and y in X means considering the quotient space obtained from the equivalence relation a \sim b if and only if a = b or a = x, b = y (or a = y, b = x). • Consider the unit square I^2 = [0, 1] \times [0, 1] and the equivalence relation \sim generated by the requirement that all boundary points be equivalent, thus identifying all boundary points to a single equivalence class. Then I^2 / \sim is
homeomorphic to the
sphere S^2. File:Collapsing a subspace.svg|thumb|For example, [0,1]/\{0,1\} is homeomorphic to the circle S^1. •
Adjunction space. More generally, suppose X is a space and A is a
subspace of X. One can identify all points in A to a single equivalence class and leave points outside of A equivalent only to themselves. The resulting quotient space is denoted X/A. The 2-sphere is then homeomorphic to a
closed disc with its boundary identified to a single point: D^2 / \partial{D^2}. • Consider the set \R of
real numbers with the ordinary topology, and write x \sim y
if and only if x - y is an
integer. Then the quotient space X / {\sim} is
homeomorphic to the
unit circle S^1 via the homeomorphism which sends the equivalence class of x to \exp(2 \pi i x). • A generalization of the previous example is the following: Suppose a
topological group G
acts continuously on a space X. One can form an equivalence relation on X by saying points are equivalent if and only if they lie in the same
orbit. The quotient space under this relation is called the
orbit space, denoted X / G. In the previous example G = \Z acts on \R by translation. The orbit space \R / \Z is homeomorphic to S^1. •
Note: The notation \R / \Z is somewhat ambiguous. If \Z is understood to be a group acting on \R via addition, then the
quotient is the circle. However, if \Z is thought of as a topological subspace of \R (that is identified as a single point) then the quotient \{ \Z \} \cup \{ \,\{r\} : r \in \R \setminus \Z \} (which is
identifiable with the set \{ \Z \} \cup (\R \setminus \Z)) is a countably infinite
bouquet of circles joined at a single point \Z. • This next example shows that it is in general true that if q : X \to Y is a quotient map then every
convergent sequence (respectively, every
convergent net) in Y has a
lift (by q) to a convergent sequence (or
convergent net) in X. Let X = [0, 1] and \,\sim ~=~ \{ \,\{ 0, 1 \} \, \} ~\cup~ \left\{ \{ x \} : x \in (0, 1) \, \right\}. Let Y := X / {\sim} and let q : X \to X / {\sim} be the quotient map q(x) := [x], so that q(0) = q(1) = \{ 0, 1 \} and q(x) = \{ x \} for every x \in (0, 1). The map h : X / {\sim}\to S^1 \subseteq \Complex defined by h([x]) := e^{2 \pi i x} is well-defined (because e^{2 \pi i (0)} = 1 = e^{2 \pi i (1)}) and a
homeomorphism. Let I = \N and let a_{\bull} := \left(a_i\right)_{i \in I} \text{ and } b_{\bull} := \left(b_i\right)_{i \in I} be any sequences (or more generally, any nets) valued in (0, 1) such that a_{\bull} \to 0 \text{ and } b_{\bull} \to 1 in X = [0, 1]. Then the sequence y_1 := q\left(a_1\right), y_2 := q\left(b_1\right), y_3 := q\left(a_2\right), y_4 := q\left(b_2\right), \ldots converges to [0] = [1] in X / {\sim} but there does not exist any convergent lift of this sequence by the quotient map q (that is, there is no sequence s_{\bull} = \left(s_i\right)_{i \in I} in X that both converges to some x \in X and satisfies y_i = q\left(s_i\right) for every i \in I). This counterexample can be generalized to
nets by letting (A, \leq) be any
directed set, and making I := A \times \{ 1, 2 \} into a net by declaring that for any (a, m), (b, n) \in I, (m, a) \; \leq \; (n, b) holds if and only if both (1) a \leq b, and (2) if a = b \text{ then } m \leq n; then the A-indexed net defined by letting y_{(a, m)} equal a_i \text{ if } m = 1 and equal to b_i \text{ if } m = 2 has no lift (by q) to a convergent A-indexed net in X = [0, 1]. ==Properties==