Normality of Quotient Spaces

G.1 Normality of quotient spaces For a quotient space, the separation axioms--even the are difficult to verify. ausdorff property-- We give here three situations in which the quotient space is not only Hausdorff, but normal. Theorem G.1. is normal, then Proof. p: X-pY be a closed quotient map. X containing A such that p A *(A), p The proof is easy. (U) is a subset of is contained in The set C = X- N so that the set of If x no point of C, A. p so that it lies in be disjoint closed sets of p (B) N2 of and X containing them. B, sch that N1 and Definition. X; and let and Y N2 p Let f Now the map X and Zf Y p U, (p(C)) f (C) is an p U. of The set then (U) Y p(C) is is an open set p (x) contains is contained in Y. NOW let is continuous, p A (A) N, and U2 N1 beIopen sets of and U1 p' (U2 ) and Y X and and Choose disjoint open sets N1 and containing lies in N2' U2 0 be disjoint spaces; let be a continuous function. A be a closed subset We. define the to be the quotient space obtained from the union of (f(a)). ply Let a of p: XU Y A with the point Zf C and If f is closed in is closed in XY, C (C). Y C Y, Zf then is closed in A and hence closed in s that p(C) f(a) and with all into is a closed set of The set X. is closed in Y, We show p- (p(C)) so Therefore, Zf, by definition of the quotient topology. It now follows that is a homeomorphism of Y X be the quotient map. is a continuous injection of that it is also a closed map. set p are disjoint, so are Let equals the union of X. lies in by identifying each point the points of the thus Since U1 (U1 ) f: A-SY adjunction space N Since one-point sets are closed in are disjoint closed sets of Because of N; Y. Y. ad p(C) is a closed map, one-point sets are closed in B A U = Y- is a point of Now we verify normality of and X N. is closed. A, that contains Y, then there is an open set closed and disjoint from Y If is normal. First we show that if open set of containing Y Let p(Y) is a closed subspace of with p(Y). Zf, and that ply G.2 X Theorem G.2. Proof. If X and Y are notial, then so is Zf. A direct proof, using the definition of normality, is a bit elaborate. (See [D], p. 145.) An easier proof uses the Tietze theorem, as we now show. First, we note that belongs to p(Y), then Zf is T 1. z3} Let z be a point of Zf. If z is closed because one-point sets are closed in Y, and p)Y : Y-->Zf is a closed map. Otherwise, p-l (z) is a one-point set in X, and therefore closed; it.follows from the definition of a quotient map that z) is closed. Now let and B CX = p by setting and C be disjoint closed sets of (C)nX. Similarly, let h = g AX o - A, C By = p and h = 0 on (B) Y BX , and 7f. and . 4 O on By Y, choose a continuous function and 1 on Cy. defined when two of the sets overlap, Using normality of k : X. function from XC C X XUY h X ad h=continuous1 on C and h is unambiguously is continuous, by the pasting lemma. and the Tietze theorem, extend [-O1]. into Then g [0,1]; and g : Y -[,] Then define byfunsetting h = gof on A [0,1]h andto = Oinduces on Because each of these three sets is closed in function on / 2 Using normalilty of that equals BX = p(B) X Cy = p- (C) Y. h = 1 _ Y -& Let k h to a continuous together define a continuous it induces a continuous function G.3 F: Zf-;[O,1] on the quotient space that equals F ([O,½)) and F ((½,1]) 0 on B and 1 on C. TLe sets are then disjoint open sets about B and C, I respectively. Oe application of adjunction spaces occurs in point-set topology, when one is studying absolute retracts. (See Exercise 8, p.224.) Another application occurs in algebraic topology, when one constructs a CW complex; we will discuss this application shortly. Definition. of X Let whose union is sitbspaces X, if a set for each [. is open in X be a space and let X. A for each be a family of subspaces The topology of X is closed in whenever X (Or, equivalently,if a set X Xj U is said to be coherent with the A is open in XC X is closed in XX Un XO whenever .) There is a strong connection between coherent topologies and quotient spaces. It is described as follows: To begin, let us give topology and consider the product space of XX J X that is the union of the subspaces J. J the discrete Tlen we consider the subspace z J. X4X [41, for all This space is called the topological sum (or sometimes the disjoint union) of the spaces X. It is denoted JX . If we project Xx J onto X,1 we obtain a continuous map p: which maps each space p X X XpZh by the obvious homeomorphism onto is a quotient map if and only if the topology of subspaces X . subspaces X , then a map of the functions It follows that if fXK Theorem G.3. subspaces Let Xi , for these subspaces. i X f : X--Y X X . The map is coherent with the has the topology coherent with the is continuous if and only if each is continuous. X Z. If each be a space that is the union of countably many cosed Suppose the topology of Xi 1 is normal, then so is X X. is coherent with G.4 Proof. i, so p is a point of is closed in p} Let for If A arnd n>O, B X. is closed in ~Pn X X for each Xi is a T1 space. be closed disjoint sets in X. IDefine YO = AU B, and define AVBUXl1 ...UX f : - YO - >[C0,1] by letting it equal 0 Define a continuous function 1 thlen Therefore Yn and X, on B. on A In general, suppose one is given a continuous function ]. fn : Yn-[- , The space Xn+l is normal and Yn Xn+l iS closed in Xn+ If gn denotes the restriction of fn to the subspace YnCIXn+l, 1. we use the Tietze theorem to extend g: X+ 1 - [0,1] functions f n Because and g Yn to a continuous function g ar:d Xn+ 1 are closed subsets of Yn+l' the combine to define a continuous functionfn+l: Yn+l- [ °' C) l]1 that is an extension of f . Te; functions f n n f: X-->[0,1] that equals 0 orn A in turn combine to define a function ar;d has the topology coherent with the subspaces Given an element Sk of x all elements in Sa on the map Xn , B. f Because X is continuous. of S S let , x < such that be the subspace consisting of X . lTen is a closed interval X. so it is compact. The space S x S We show that SL, each of which X iS the union of the spaces is compact Hausdorff and thus normal. S - SD- , which is not normal, has the topology coherent with these subspaces. Let U be a subset of subspace, for each 4. such that S xS U (X x s is open in this T-,en the intersection U n (SX Sl ) is open in Sax Sfor each , and hence open in the union of the sets Un ( it is open in EJ The preceding theorem does not extend to uncountable coherent Example 1. unions. 1 S Sag, SQ ), as desired. S S Since U is G.5 It is an interesting question to ask under what conditions coherent topologies exist. One has the following two theorems: Theorem G.4. spaces X , Let for X be a set that is the union of the topological o(C-J. If there is a topological space as its underlying set, such that each is a topological space the topology of X X Proof. in X is a subspace of such that each X, D to be closed in It is immediate that . XT , X. X then there X and The topology of XT . We define a set for each having is a subspace of is coherent with the subspaces is finer than that of XC X~ XT 0 Xc and if Dn X is closed are closed. X The required properties about unions and intersections follow from the equations 4., E is closed in so that E of XC . Isn't this obvious? in XC , then A is closed in for some set X A that of XTI some closed in he set Theorem G.5. X; , for (J.. is closed in both A XO is closed in Thus the topology of X. Xc First, note that if Xi by definition. Xo4 is a subspace of XT. X4 XC XT , Thus . A is closed Conversely, suppose we have aBcause the topology of is also closed in is a subspace XC B = A Xi is finer than B = A AX4 for XC) as desired. 0 Let X be a set that is the union of the topological spaces If for each pair of indides X, aad from each of them, then Each Not quite. BEcause closed in X. E We must prove that each is closed in X. then XT. What else is there to prove? A , ) ), (ru X XT , is closed in is finer than the topology of B X)U *...V (Dn X is an arbitrary collection of closed sets. Note that if for each ID eb = (R LAX Djox, where o (D 1 DrI) nAXL = U. (D1V X Xa is a closed set in d,, the set XAn XA , and inherits the same topolcxy X has a topology coherent with the subspaces X in this topology. X. G.6 set to be closed in D is closed in A B Convarsely, let. AnXi then X, be a BnX To do this, we must show that is a subspace of X a closed subspace of ; B Then X XA B the set X, i. closed in B nX X1 by definition of is closed in X ; closed set of X is closed in Xn X X Since s. for each is closed in X. XC . is closed in B we show is closed in First, XC. XA is a closed subspace of We show that each space It is i. for each X X. immediate that this is a topology on if is closed in DhXd if XC by declaring a XC Once again, we define a topological space Proof. because the latter because X Xd is a This theorem does not hold if the word "closed" is omitted from the that is the union of three X There is an example of a set hypothesis. spaces such that the intersection of any two of the spaces is a subspace of each of them; but there is no topology on X (ee of the spaces are subspaces! Example 2. R subspace of R [Mu], p. 213.) in the product topology. consisting of all points inn. for Xi = 0 Consider at all of which all three Fn 1 such that x = (x],x 2 ,...) ... each of which is a closed subspace of the next. Their union is by Theorem G.5 has a topology coherent with the subspaces Theorem G.3 implies that IRl inherits as a subspace of IR6 other , n. topologies as well, ones that it in its various topologies. The subset inherits its usual topology from each of these topologies on subspaces R is normal in this topology. Now 1RO also has several Theorem denote the Then one has the sequence of subspaces , R C 2C which Let G.4 also applies to show that R Re. R Hence has a topology coherent with the IR ; this theorem also implies that the coherent topology is finer than each of these topologies on R. Since the one derived from the box topology is the finest of these, one has the following: Challenge question: subspaces IR Rn Is the topology on tlhe same as the topology that in the box topology? R that is coherent with the IRO inherits as a subspace of G.7 Here is a quick outline of how these notions are used in Final remark. algebraic topology. (See §38 of [Mu] for more details.) is the subset of n The unit ball in Rn consisting of all points whose euclidean distance from the origin is less than or equal to one; the unit sphere consists of those points for which this distance equals one. spaces are denoted h: Bn-->C, then Bn and Si , respectively. If there is a homeomorphism Bd C is called an n-cell, and we denote by C These the subspace h(Sn-). There is a class of spaces that is very important in algebraic topology called CW complexes; they were invented by J.H.C.Whitehead. In algebraic topology, one defines for a given space a number of groups, such as the homology H (X), the cohomology groups groups Hn(X), and the homotopy groups InX). Defining is one thing, but computing (or even getting useful information) The structure of CW complex gives one a tool for dealing with these is another. Ci complexes are quite versatile--many useful spaces, such as the groups. Grassman manifolds we shall mention shortly, have the complex. structure of a CW Or! the other hand, if one has some prescribed groups and wants to find a space for which the homology groups, say, are isomorphic to these groups, one can construct a CW complex that will do the job. A CW complex is constructed as follows: Then one takes a collection of disjoirt i-cells X . space, which we call f_ :Bd C -X0. and a family of continuous maps sum C , and uses the maps Now one takes a collection rD, and X1 g: , C f : Bd C - X0 by means of this map. it is called a 1-dimensional CW complex. X1; So far, so good. and a continuous map X0 o C Ole forms the topological to define a continuous map f, One forms the adjunction space obtained from This space is denoted One begins with a discrete 2 Bd D X1 - bI, means of this map. of disjoint 2-cells D , and forms an adjunction space from This space is denoted X and is a 2-dimensional CW complex. One has eventually an n-dimensional CW complex It is clear how to continue. X n, for each n. Are we finished? of the adjunction space Xn , n X-1 X. construction Recall that in the the projection map defines a homeomorphism of with a closed subspace of closed subspace of No. X. We normally identify n- X with this G.8 With this convention, we now have a sequence of spaces X1 C XO C X '-... ... each of which is a closed subspace of the next. Their union is given the topology coherent with these subspaces; it is called an (infinite-dimensional) CW complex. In order to work with the space we obtain, it is essential that it be a Hausdorff space. (Basically, so we know that compact sets are closed.) The theorems we have proved in this section do much more than that; they show that every CVT complex is normal. Final final remark. We have talked a lot about how quotient spaces are used in algebraic topology. Let us close by giving an example of how they are used in differential geometry. A very important space in differential geometry is the space n k-dimensional vector subspaces of R . of k-planes in n-space. The space through the origin in . R Gnk of It is called the Grassman manifold G is thus the space of all lines n,1 It is fairly intuitive what one means by saying that two k-planes are "close" tc one another, but how does one topologize this space rigorously? One topologizes it as a quotient space. To be specific, let V be the set of all k by n matrices, where k n, n,k whose rows are orthonormal vectors. Such a matrix satisfies the equation AA t = Ik There is an obvious topology on this set, for it can be considered as a subspace of R . as the equation In this topology, it is compact, for it is closed and bounded, AA = Ik shows. Let p: Vk Gn be k sends each matrix to the k-dimensional vector subspace of spanned by its rows. We topologize n k by requiring the map that JR p that is to, be a quotient map. Because p is continuous, it is immediate that next question is this: map p Is: it Hausdorff? is compact. The answer is "yes," The because the is in fact a closed map, so Theorem G.1 applies. To show p two matrices is closed, we examine first what the relationship is between A' and B whose rows span the same vector subspace of This occurs precisely when each row of the rows of A = CB, Gn,k B, and conversely. where C A Rn equals a linear combination of This statement can be expressed by the matrix equation is a nonsingular k by k matrix. It follows that C satisfies the G.9 equation CC = Ik, which means that C belongs computation with matrices verifies this fact: AA t = BB with the equations . ABt = C ard = Ik , to Vk,k' A quick The equation A.= CB, along implies that t Ik = C(BAt). t The first equation ives us, by transposing, the equation t BA C substituting this result into the second equation gives us the equation I k = CCt. Now we show that p then the set' p-l(p(S)) C belongs to image of Vk Vk kX S is compact ard S If S is a closed set in is the set of all matrices of the form and k is a closed map. A is an element of S. Thus p CA, (p(S)) under the map given by matrix multiplication. V where is the Now Vk1 is compact (being closed in Vn). Their cartesian n,kI product is compact, so the image under matrix multiplication (which is continuous) is also compct and therefore a closed bset of By definition of the quotient topology, it follows that in G V n,k p(S) is closed as desired. k There is of course a great deal more to say about Grassman manifolds. The space G is in fact a manifold (as the terminology implies); it is a manifold of dimension It we replace of n into n+l; k(n - k). IRn throughout by iR , then there is the obvious inclusion it gives rise to an incllsionmap of Vn, into Vn+l .k This in turn induces a continuous injective map on the quotient spaces i: G -G+, k Since all the spaces involved are compact Hausdorff, we can thus consider G to be a closed subspace of n,k the spaces Gk,k one has the space of all the coherent topology. k planes G n+lk If now we take thfe union of G+l,k ... in Arid Theorem G.3 IR3 . Gn,k . ) As you would expect, we give it implies that this space is normals