7 QuotientsIf f : X → S is a surjective function from a space X to a set S, then we canuse f to define a topology on S.Definition 7.1. Let X be a space, S a set, and f : X → S a surjecti vefunction. The quotient topology on S (with respect to f : X → S) is thetopology wh er e the open sets of S are exactl y the sets U ⊂ S such that thepreimage f−1(U) is open.It is routine to verify that this is a topology. Note that if V ⊂ X is open,this definition does not imply that f(V ) is open.Example 7.2. Let X = [0, 1] with the standard topology and let S = [0, 1).Define f : X → S by f(x) = x for x ∈ [0, 1) and f(1) = 0. The quotienttopology on S has a basis of sets of the form (a, b) ⊂ [0, 1) and [0, b)∪(a, 1) ⊂[0, 1]. This t opology makes S homeomorphic to the unit circle in R2in thestandard topology.Since we are defi ni n g a topology by collapsin g together points of thespace, to define t he topology as generally as possible we want lose as littleinformation as possible when forming the quotient. Losing as little informa-tion as possible corresponds to taking as fine a topology as possible.Proposition 7.3. The quotient topology on S with respect to f : X → S isthe finest topology on S such that f : X → S is continu ou s .Proof. It is immediat e from the definition that f i s continuous with respectto the quotient topology. Now let S be another topology on S that makes fcontinuous. Suppose U ∈ S. Then f−1(U) is open i n X, and by definitionU is also open in the quoti ent topology. So S is coarser than the quotienttopology.For every quotient space S with r espect to f : X → S as above, thefunction f is a continuous surjection with respect to the quotient topol-ogy. However, not every continuous surjection f : X → Y of spaces can berealized this way.Definition 7.4. Let X and Y be spaces. A quoti ent map f : X → Y is acontinuous surjection such that the topology on Y is equal to the q u ot ie nttopology on Y with respect to the set map f : X → Y . Equivalently, acontinuous surjection f : X → Y is a quotient map if and only if the topologyon Y is the finest topology makin g f continuous.34Example 7.5. Let f : X → Y be a bijection and let X have the discret etopology and let Y have the indiscrete topology. Then f is continuous butnot a quotient map.There are a couple of easy criteria to recognize quotient maps.Definition 7.6. Let X and Y be spaces. A function f : X → Y is open iffor every open U ⊂ X, we h ave f (U ) open in Y . Similarly, f is closed if forevery closed A ⊂ X, we have f (A) closed i n Y .Note that a continuous bijection is a homeomorphism if and only if it isclosed if and only if it is open.Proposition 7.7. Every open continuous surjection is a quotient map, andevery closed continuous surjection is a quotient map.Proof. Suppose f : X → Y is an open continuous surjection. Suppose U ⊂Y . Because f is continuous, f−1(U) is open if U is. Because f is open,U = f (f−1(U)) is open if f−1(U) is. Then the topology on Y is the quotienttopology, and f is a quotient map.If f is closed and U ⊂ Y , then U = Y \f (X\f−1(U)) is open if U is. Sofor the same reasons as above, a closed continuous surjection is a quotientmap.The real utility of quotient spaces is that we can define sur j ec ti ve setmaps that identify any sets of points that we wish. The q u oti e nt space isthen the closest space to the original space in which these sets of points areidentified. Identifying sets of points corresponds to the intuitive notions ofgluing and collapsing.Definition 7.8. Let X be a space and let ∼ be an equivalence relation onS. The quotient of X with respect to ∼ is the quotient (in the sense offunctions) with respect to the surjective funct ion X 7→ X/∼, where X/∼ isthe set of ∼-equivalence c las se s. Often t he quotient space is also denotedX/∼.Example 7.9. Consider the interval [0, 1] in the subspace topology. We geta space homeomorphic to the circl e (from the previous example) by takingthe quotient with respect to the equivalence relation with x ∼ y if and onlyif x = y or {x, y} = {0, 1}.Perhaps l es s trivially, the quotient of R by the equivalence relation x ∼x + k for all k ∈ Z is also homeomorphic to the circle.35Example 7.10. Let X = [0, 1] × [0, 1] in the standar d topology. Definean equivalence relation by (a, b) ∼ (c, d) if and only if (a, b) = (c, d) or (a = c and {b, d} = {0, 1} ) or (b = d and {a, c} = {0, 1}) or (a, b), (c, d) ∈{(0, 0), (0, 1), (1, 0), (1, 1)}. Then X is a torus; X is homeomorphic to thesubspace of R3in the standard topology given by su rf ace of revolution of acircle not intersecting the axis of revolution (t h e sur fac e of a donut).Example 7.11. In general, the quotients of X with respect to the definingrelations for quasicompone nts, connected components, and path componentscan be interesting spaces. If X is locally connected, th en the connectedcomponent and quasicomponent quotients are discrete. If X is locally pathconnected, all these quotients are discrete.Example 7.12. The quotient of the topologist’s sine curve¯S by pathcomponents is homeomorphic to the Sierpinski space {0, 1} with open sets∅, {0}, {0, 1}.Example 7.13. The quotient of the space X of nested rectangles by quasi-components is homeomorphic to the subspac e {0} ∪ {1/(n + 1) |n ∈ N} ⊂ Rin the standard topology. The quasicomponent of the two vertical linesmaps to 0 in any such homeomorphism. The q u ot ie nt of X by connectedcomponents is the same space but with the point 0 doubled.The properties preserved under taking quotients (i n general) are notmuch better than the properties preserved under the images of general con-tinuous surjections. Quotients of connected spaces are connected. Se par a-tion properties are not necessarily preserved.Example 7.14. Let α be any irrational number. Define an equivalencerelation on R by d ec l ari n g x ∼ x + k + mα for all x ∈ R and k, m ∈ Z. LetX be R/ ∼ where R has the stand ard topology, and let p: R → X denotethe quotient map. Note t hat since each equivalence class is countable, Xis uncountable (otherwise ∼ would de compose R into a countable union ofcountable sets). Suppose x, y ∈ R …