notes6_product and quotient spaces

Chapter 6 Product and Quotient Spaces When we study topology, we do not dier from other areas of mathematics that much. We look at the dierent mathematical operations that are available to us and how they aect the mathematical structure we are studying. We also want to see how our mathematical structure aects these operations. The standard operations are addition, subtraction, multiplication and division. In topology addition corresponds to the union of spaces, A[B, and subtraction corre- sponds to the set dierence AnB. Note that addition is commutative, but dierence is not. How do multiplication and quotients carry to the area of topology. Actually, multiplication is quite simple to see, but quotients are not as easily introduced. 6.1 Finite Products LetXandYbe sets. The Cartesian product ofXandYis denoted by XYand is dened to be the set of all ordered pairs of points whose rst coordinate comes from Xand whose second coordinate comes from Y: XY=f(a;b)ja2Xandb2Yg: Now, ifXandYare topological spaces, then we would like to see what natural topology we should have on the product space, XY. The topology on XY should be built from the topologies on XandY, but how? If Uis open inXand Vis open inY, then we would want UVto be open in XY. However, the collection of all sets of this form is not closed under unions, and hence cannot be a topology. It is not dicult to see, in the plane, that the union of two of these rectangles (0 ;1)(0;1)[(2;3)(2;3) is not a set of the form UV, becauseUV would have four pieces, not two. LetB=fUVjUis open inXandVis open inYg. We have just seen thatBis not a topology. It is, however, a basis for a topology on XY. We need to show that it satises the conditions of Theorem 4.8. We must show that the union of the members of BisXYand that for each B1;B22Bandx2B1\B2, there is a member BxofBsuch thatx2BxB1\B2. 51 52 CHAPTER 6. PRODUCT AND QUOTIENT SPACES It should be clear that the union of the members of Bis all ofXY. For the second condition, let B1=U1V1andB2=U2V2whereUiis open inXandV1 is open inY. Now, B1\B2= (U1V1)\(U2V2) = (U1\U2)(V1\V2); so thatB1B22B. This is stronger than what is required in Theorem ??. Thus, Bis a basis for a topology on XY. The topology determined by Bis called the product topology onXY. We can generalize this denition: Denition 47 Let(X1;T1),(X2;T2), dots, (Xn;Tn)be a nite collection of nonempty topological spaces, and let Xdenote the Cartesian product X=nY i=1Xi=X1X2X3


Xn: LetBbe the family of all subsets of Xof the form O=nY i=1Oi=O1O2O3


On where each Oiis open inXi. Then Bis a basis for the product topology forX. The setXwith the product topology is a product space . The spaces X1,X2,:::, Xnare called the coordinate spaces orfactor spaces ofX. Each point x2Xis of the form x= (x1;x2;:::;xn); xi2Xi: Dene a function i:X!Xibyi(x) =xi. This is called the projection map on the ith coordinate space or the ith projection map . Theorem 57 The projection maps i:X!Xifrom the product space to the coor- dinate spaces are continuous. Proof: Leti:X!Xiand letUiXibe open. Then 1 i(Ui) =X1:::Xi1 UiXi+1:::Xn. This is a product of open sets, and is open. Thus, iis continuous. Theorem 58 Letf:Y!Xbe a function from a topological space to a product spaceX=Qn i=1Xi. Thenfis continuous if and only if ifis continuous for each projection map. What properties are preserved by products? c 1999, David Royster Introduction to Topology For Classroom Use Only 6.1. FINITE PRODUCTS 53 Theorem 59 The product of a nite number of Hausdor spaces is Hausdor. Proof: LetXi,i= 1;:::;n , be a Hausdor space for each n. Leta= (a1;:::;an) andb= (b1;:::;bn) be distinct points inQn i=1Xi. Since they are distinct, there is at least one coordinate space, Xj, for which aj6=bj. SinceXjis Hausdor, there are disjoint open sets, UjandVjinXjso thataj2Ujandbj2Vj. Then U=1 j(Uj) andV=1 j(Vj) are disjoint open sets of Xcontainingaandb, respectively. Theorem 60 The product of a nite number of connected spaces is connected. Proof: The proof is most easily done by induction. We will prove it for the case of n= 2 and indicate the manner of the rest of the proof. Assume that X1andX2are connected and let x12X1. Then,fx1gX2is a subspace of X1X2which is homeomorphic to X2using the function (x1;t)7!t; t2X2: SinceX2is connected,fx1gX2is connected. Now a similar argument shows that X1ftgis a connected set for each t2X2. Now,fx1gX2andX1ftghave the point (x1;t) in common. Thus, the set (fx1gX2)[[ t2X2(X1ftg) =X1X2 is connected. Now, by induction, suppose thatQn1 i=1Xiis connected and considerQn i=1Xi, where each Xiis connected. By this previous argument,Qn i=1Xiis connected since it is homeomorphic to the product of two connected spaces:Qn1 i=1XiandXn. Hence the nite product of connected spaces is connected. Theorem 61 The nite product of separable spaces is separable. Theorem 62 The nite product of rst countable spaces is rst countable. Theorem 63 The nite product of second countable spaces is second countable. Theorem 64 If(X1;d1),:::,(Xn;dn)are metric spaces, then the product topology on the product space is generated by the product metric: d((x1;:::;xn);(y1;:::;yn)) = nX i=1di(xi;yi)2!1=2 : c 1999, David Royster Introduction to Topology For Classroom Use Only 54 CHAPTER 6. PRODUCT AND QUOTIENT SPACES In order to prove an analogous result for compactness, we need the following lemma. Lemma 9 In order that a space Xbe compact it is sucient that there exists a basis BforXsuch that every open cover of Xby members of Bhas a nite subcover. Theorem 65 The nite product of compact spaces is compact. Proof: Using an inductive argument as in our last theorem, it suces to show that X1X2is compact if X1andX2are compact. LetB=fUVjUis open inX1andVis open inX2gbe the subbasis for the product topology. Let Obe an open cover of X1X2composed of members of B. By the above lemma, the compactness of X1X2will be proved if it can be shown that there is a nite subcover for X1X2derived from O. Forx2X1, the subsetfxgX2is compact and is therefore contained in the union of a nite number of members, say U1V1,U2V2,:::,UmVmofO, each of which meetsfxgX2. Then Ux=m\ i=1Ui is an open set in X1containingx. Note that UxX2=Ux m[ i=1Vi! =m[ i=1(UxVi)m[ i=1(UiVi): This last inclusion is due to the fact that UxUifor alli. Thus, for each x2X1 there is an open set Uxcontaining xfor which the set UxX2is contained in the union of a nite number of members of O. This family of open sets fUxjx2X1gis an open cover for X1. SinceX1is compact, this cover admits a nite subcover fUxigr i=1. Now, X1X2= r[ i=1Uxi! X2=r[ i=1(UxiX2) is the union of a nite number of the sets of the form UxiX2. Each of these sets is contained in the union of a nite number of members of O. This forms a nite subcover for X1X2of the members of O, makingX1X2compact. 6.2 Arbitrary Products We have previously dened a countable product of spaces as follows: 1Y i=1Xi=f(x1;x2;x3;:::)jxi2Xifor eachi= 1;2;:::g: c 1999, David Royster Introduction to Topology For Classroom Use Only 6.2. ARBITRARY PRODUCTS 55 We identify these with sequences of elements, each element coming from the appro- priate factor space. We had a dierent way to dene sequences when we studied calculus. If you will recall a sequence is a function f:Z+!X=n[ i=1Xi; f(i)2Xi: We will use this idea to extend the denition of products to arbitrary indexing sets. Denition 48 LetIbe an index set and fXj2Igbe a family of sets. The Cartesian product X=Q 2IXis the collection of all functions xwith domain I having the property that the value xofxatbelongs to the set X: X=Y 2IX=( x:I![ 2IXjx() =x2Xfor each2I) For2Ithe function :X!Xdened by(x) =xis called the projection map ofXon theth coordinate set. Note that this does not dier from our previous denition if the index set Iis nite. We get exactly the same space. We dene the product topology on the space X=Q Xusing the projection maps. Let Sconsist of all sets of the form 1 (U) whereUis open inX. This set forms a subbasis for the product topology. This means that the basis for the product topology for Xconsists of all nite intersections: n\ i=11 i(Ui); i2I;1in: Such a basic open set may be expressed in the product form as n\ i=11 i(Ui) =Y 2IV whereVi=Uifori= 1;:::;n andV=Xotherwise. Proofs of the following theorems are similar to there nite analogues. Theorem 66 Letf:Y!Xbe a function from a space Yinto a product space X=Q X. Thenfis continuous if and only if the composition foffwith each projection map is continuous. Theorem 67 The product of any family of Hausdor spaces is a Hausdor space. Theorem 68 The product of an arbitrary collection of connected spaces is connected. c 1999, David Royster Introduction to Topology For Classroom Use Only 56 CHAPTER 6. PRODUCT AND QUOTIENT SPACES This proof is based on the previous proof, but is not exactly the same. To prove the analogous theorem for compact spaces, we need some further results. The proofs are not trivial, but are not included here. The rst is needed for the second. Lemma 10 (The Alexander Subbasis Theorem) In order that a space Xbe com- pact, it is necessary and sucient that there exist a subbasis SforXsuch that every open cover of Xby members of Shas a nite subcover. Theorem 69 (The Tychono Theorem) The product of an arbitrary family of compact spaces is compact. 6.3 Quotient Spaces Quotient spaces are sometimes called identication spaces , because they result from agluing , or identication, process. Denition 49 LetXbe a space and let be an equivalence relation on X. LetX= denote the set of all equivalence classes [x] =fy2Xjxyg.X=is called the quotient space ofXby the equivalence relation . Dene a function q:X!X= byq(x) = [x]. This map qis called the quotient map . Dene a set AX=to be open ifq1(A)is open inX. This collection of open sets denes a topology on X= called the quotient topology . A particularly simple example of an equivalence relation giving rise to interesting quotient spaces is the following. Let AXand, for the sake of interest, let Ahave more than one member. Dene a relation, A, onXby:xAyifx;y2Aor if x=y62A. Thus, every point of Ais related to every other point of A, and if a point is not inAit is related only to itself. The quotient space is denoted by X=A and is called the identication space obtained by identifying the members of Ato a point. Example 44 The quotient space obtained by identifying the two endpoints 0 and 1 to a point is the circle, S1. Example 45 The quotient space of D2obtained by identifying the boundary circle, S1, to a point is homeomorphic to a sphere in R3. Example 46 LetI2= [0;1][0;1]. The quotient space of I2obtained by identifying the pairs of points (0 ;x2) and (1;1x2), 0x21, is homeomorphic to the M obius strip. c 1999, David Royster Introduction to Topology For Classroom Use Only 6.3. QUOTIENT SPACES 57 Example 47 LetX=I2with equivalence relation dened as follows: (1) ( x1;0) (x1;1) for each x12[0;1], and (2) (0 ;x2)(1;x2) for eachx22[0;1]. Then the quotient space X=is homeomorphic to a torus in R3. Example 48 Letbe an equivalence relation on Rdened byxyif and only if xy2Z. In other words, xx+nwheren2Z. We can actually think of these equivalence classes as [x] =fx+njx2[0;1] andn2Zg We have reduced our consideration from Rto [0;1]. It is clear that [0] = [1], so that R=is homeomorphic to our rst example, and R=is topologically equivalent to the circle. The quotient map here is equivalent to t7!(cos(t);sin(t)). Example 49 LetX=I2and deneonXby: 1. (x1;0)(x1;1) forx12[0;1], and 2. (0;x2)(1;1x2) forx22[0;1]. This is slightly dierent from 46 in that in the last stage the circles are identied with reversed orientations. The quotient space is the Klein bottle . Example 50 Dene an equivalence relation on the circle, S1, byzz, where we consider the circle as the set of all complex numbers of modulus 1 (or the set of vectors in the plane with norm 1). Then the quotient space S1=is topologically equivalent to S1. Example 51 LetX=I2and deneonXby: 1. (x1;0)(1x1;1) forx12[0;1], and 2. (0;x2)(1;1x2) forx22[0;1]. This is equivalent to taking the unit disk and applying the previous equivalence relation to the bounding circle. The resulting space is called the projective plane , and cannot be reasonably drawn in three dimensions. Example 52 LetXbe a topological space and let Abe the subset of XIgiven byA=Xf1g. The quotient space T(X) = (XI)=Ais called the cone overX. Note that our denition of the quotient topology guarantees that the quotient map is continuous. This is extremely useful in later results. c 1999, David Royster Introduction to Topology For Classroom Use Only 58 CHAPTER 6. PRODUCT AND QUOTIENT SPACES Theorem 70 LetX=be a quotient space with quotient map q:X!X=. Then a functionf:X=!Yis continuous if and only if the composite function fq:X! Yis continuous. Quotient topologies are not as uncommon as you might expect. A lot of topologies may be considered as quotient topologies, especially considering the characterization in the above theorem, Theorem 70 LetXbe a space and let f:X!Ybe a function onto the set Y. Dene a subset OinYto be open if its inverse image f1(O) is open in X. The family of open sets so dened on Yis called the quotient topology determined by f. Note that this actually is a topology. Clearly, f1(;) =;andf1(Y) =X, so that those conditions are satised. If fUj2Igare open in Y, then f1 [ 2IU! =[ 2If1(U) is open inX, so arbitrary unions of open sets are open in Y. IffUigare open in Y, then f1 n\ i=1Ui! =n\ i=1f1(Ui) is open inX, so the nite intersection of open sets is open. Theorem 71 LetXandYbe spaces and f:X!Ya continuous, onto function. Iffis either open or closed, then Yhas the quotient topology determined by f. Proof: LetTdenote the topology on Yand let Tfdenote the quotient topology induced by the function f. IfU2T, then, since fis continuous, f1(U) is open. This putsU2Tf. Thus, TTf. Suppose that fis an open function and let U2Tf. Thenf1(U) is open in X. Sincefis an open function with respect to T,f(f1(U) =Uis an open set in the topology T. Thus, TfT, and we are done. Example 53 [Adjunction Spaces] Let XandYbe topological spaces and let AX be non-empty. Assume that XandYare disjoint and that there is a continuous functionf:A!Y. Dene a function :X[Y!(XnA)[Yby (z) =8 >< >:f(z) ifz2A z ifz2XnA z ifz2Y The topology on X[Yis that a set is open if and only if its intersection with both XandYare open. Clearly, is onto. When we put the quotient topology induced byon the space ( XnA)[Y, we will denote this space by X[fYand call it the adjunction space via f. c 1999, David Royster Introduction to Topology For Classroom Use Only 6.3. QUOTIENT SPACES 59 a) IfYis a single point, then attaching Xto a single point by the function f:A! fagis the same as shrinking Ato a point. b) LetI2=IIbe the unit square in the plane. Let A=f(x;y)jx= 0 orx= 1 and 0y1g. LetY=I= [0;1] and dene f:I2!Ybyf(x;y) =y. ThenI2[fYis a cylinder formed by identifying the two vertical edges of I2. c) Let X be a topological space and let Y=fp0;p1gbe a two point discrete space not in X[1;1]. LetA=X(f1g[f 1g). Denef:A!Ybe dened by f(x;1) =p0andf(x;1) =p1. SinceYhas the discrete topology, fis continuous. The adjunction space ( X) =X[1;1][fYis called the suspension ofX. The equator is the image of Xf0gin (X). The image of X[0;1] in (X) is homeomorphic to the cone over X. Thus, the suspension ofXis two cones over Xidentied along the equator. As an example, ( S1) is homeomorphic to the 2-sphere, S2. c 1999, David Royster Introduction to Topology For Classroom Use Only