SOLUTIONS TO EXERCISES FOR. MATHEMATICS 205A Part 4. Function spaces

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1 SOLUTIONS TO EXERCISES FOR MATHEMATICS 205A Part 4 Fall 2014 IV. Functin spaces IV.1 : General prperties Additinal exercises 1. The mapping q is 1 1 because q(f) = q(g) implies that fr all x we have f(x) = p x q(f) = p x q(g) = g(x), which means that f = g. T prve cntinuity, we need t shw that the inverse images f subbasic pen sets in Y X are pen in C(X, Y ). The standard subbasic pen subsets have the frm W({x}, U) = p 1 x (U) where x X and U is pen in Y. In fact, there is a smaller subbasis cnsisting f all such sets W({x}, U) such that U = N ε (y) fr sme y Y and ε > 0. Suppse that f is a cntinuus functin such that q(f) lies in W({x}, U). By definitin the later cnditin means that f(x) U. The latter in turn implies that δ = ε d( f(x), y) > 0, and if d(f, g) < δ then the Triangle Inequality implies that d( g(x), y) < ε, which in turn means that g(x) U. Therefre q is cntinuus at f, and since f is arbitrary this shws q is a cntinuus mapping. 2. It suffices t shw that the map in questin is nt and distance-preserving. The map is nt because if u and v are cntinuus functins int Y and Z respectively, then we can retrieve f by the frmula f(x) = ( u(x), v(x) ). Suppse nw that f and g are cntinuus functins frm X t Y Z. Then the distance frm f t g is the maximum f d ( f(x), g(x) ). The latter is less than r equal t the greater f d ( p Y f(s), p Y g(s) ) and d ( p Y f(t), p Y g(t) ). Thus if Φ : C(X, Y Z) C(X, Y ) C(X, Z) then the distance frm Φ(f) t Φ(g) is greater than r equal t the distance frm f t g. This means that the map Φ 1 is unifrmly cntinuus. Cnversely, we claim that the distance frm f t g is greater than r equal t the distance between Φ(f) and Φ(g). The latter is equal t the larger f the maximum values f d ( p Y f(s), p Y g(s) ) and d ( p Z f(t), p Z g(t) ). If w X is where d(f, g) takes its maximum, it fllws that d ( f(w), g(t) ) = { max d ( p Y f(w), p Y g(w) ), d ( p Z f(w), p Z g(w) ) } which is less than r equal t the distance between Φ(f) and Φ(g) as described abve. 3. The distance between f and g is the maximum value f the distance between f(x) and g(x) as x runs thrugh the elements f x, which is the greater f the maximum distances between f(x) and g(x) as x runs thrugh the elements f C, where C runs thrugh the set {A, B}. But the secnd expressin is equal t the larger f the distances between d(f A, g A) and d(f B, g B). 1

2 Therefre the map described in the prblem is distance preserving. As in the previus exercise, t cmplete the prf it will suffice t verify that the map is nt. The surjectivity is equivalent t saying that a functin is cntinuus if its restrictins t the clsed subsets A and B are cntinuus. But we knw the latter is true. 4. Let ε > 0 be given. We claim that the distance between f g and f g is less than ε if the distance between f and f is less than ε and the distance between g and g is less than ε. Chse u 0 X and v 0 Z s that d ( f g (u 0, v 0 ), f g(u 0, v 0 ), ) is maximal and hence equal t the distance between f g and f g. The displayed quantity is equal t the greater f d ( f (u 0 ), f(u 0 ) ) and d ( g (v 0 ), g(v 0 ) ). These quantities in turn are less than r equal t d(f, f) and d(g, g) respectively. Therefre if bth f the latter are less than ε it fllws that the distance between f g and f g is less than ε. 5. (i) Fllw the hint. We then have V (h) V (f) = V (h f) = V (id), which is the identity. Likewise, we als have V (f) V (h) = V (f h) = V (id), which is the identity. (ii) Again fllw the hint. We then have U(f) U(h) = U(h f) = V (id), which is the identity. Likewise, we als have U(h) V (f) = V (f h) = V (id), which is the identity. (iii) Let f : A A and g : B B be the hmemrphisms. Let U(f 1 ) V (g) = V (g) U(f 1 ) equality hlds by assciativity f cmpsitin. By the first tw parts f the exercise this map is the required hmemrphism frm C(A, B) t C(A, B ). 6. It will be cnvenient t dente α(γ, γ ) generically by γ + γ. The cnstructin then implies that the distance between ξ + ξ and η + η is the larger d(ξ, η) and d(ξ, η ). Therefre the cncatenatin map is distance preserving. 7. Fllw the hint. If y y, then the values f k(y) at every pint is X, and hence it is nt equal t y, the value f k(y ) at every pint f X. Therefre k is 1 1. Next, we shall verify the set-theretic identities described abve. If y U then since k(y) is the functin whse value is y at every pint we clearly have k(y) W(K, U), and hence k(u) W(K, U) Image(k). Cnversely, any cnstant functin in the image f the latter is equal t k(y) fr sme y U. The identity k 1 ( W(K, U) ) = U fllws similarly. 8. As indicated in the hint, withut lss f generality we may assume that s < t. Given a functin f in Diff(A, B), the Mean Value Therem implies that f(t) f(s) = f (ξ) (t s) fr sme ξ (s, t). By hypthesis f (x) B fr all x, and therefre f(t) f(s) = f (ξ) t s B t s. Therefre, if ε > 0 and δ = ε/b, then s t < δ implies f(s) f(t) < ε. IV.2 : Adjint equivalences Additinal exercises 1. The map A : F(X Y, Z) F(X, F(Y, Z) ) sends h : X Y Z t the functin h such that [h (x)](y) = h(x, y). The argument prving the adjint frmula fr spaces f cntinuus 2

3 functins mdifies easily t cver these examples, and in fact in this case the prf is a bit easier because it is nt necessary t cnsider metrics r tplgies. 2. Let y 0 Y, and let L : Y [0, 1] Y be the map sending (y, t) t the pint (1 t)y + ty 0 n the line segment jining y t y 0. If we set H(x, t) = L( f(x), t), then H satisfies the cnditins in the hint and defines a cntinuus map in C(X, Y ) jining f t the cnstant functin whse values is always y 0. Thus fr each f we knw that f and the cnstant functin with value y 0 lie in the same arc cmpnent f C(X, Y ). Therefre there must be nly ne arc cmpnent. 3. By the adjint frmula there are hmemrphisms C ( X, C(Y, Z) ) = C(X Y, Z) = C(Y X, Z) = C ( Y, C(X, Z) ) and this yields the desired 1 1 crrespndence f sets. 3

4 V. Cnstructins n spaces V.1 : Qutient spaces Prblem frm Munkres, 22, pp (a) The hint describes a well-defined cntinuus map frm the qutient space W t the real numbers. The equivalence classes are simply the curves g(x, y) = C fr varius values f C, and they are parablas that pen t the left and whse axes f symmetry are the x-axis. It fllws that there is a 1 1 nt cntinuus map frm W t R. Hw d we shw it has a cntinuus inverse? The trick is t find a cntinuus map in the ther directin. Specifically, this can be dne by cmpsing the inclusin f R in R 2 as the x-axis with the qutient prjectin frm R 2 t W. This gives the set-theretic inverse t R 2 W and by cnstructin it is cntinuus. Therefre the qutient space is hmemrphic t R with the usual tplgy. (b) Here we define g(x, y) = x 2 + y 2 and the equivalence classes are the circles g(x, y) = C fr C > 0 alng with the rigin. In this case we have a cntinuus 1 1 nt map frm the qutient space V t the nnnegative real numbers, which we dente by [0, ) as usual. T verify that this map is a hmemrphism, cnsider the map frm [0, ) t V given by cmpsing the standard inclusin f the frmer as part f the x-axis with the qutient map R 2 V. This is a set-theretic inverse t the map frm V t [0, ) and by cnstructin it is cntinuus. Additinal exercises 0. We claim that every subset f X/R is bth pen and clsed. But a subset f the qutient is pen and clsed if and nly if the inverse image has these prperties, and every subset f a discrete space has these prperties. 1. We need t shw that U A is pen if and nly if r 1 [U] is pen in X. The ( = ) implicatin is immediate frm the cntinuity f r. T prve the ther directin, nte that r i = id A implies that U = i 1 [ r 1 [U] ] = A r 1 [U] and thus if the inverse image f U is pen in X then U must be pen in A. 2. Let p X and p A dente the qutient space prjectins fr X and A respectively. By cnstructin, j is the unique functin such that j p A = p X A and therefre j is cntinuus. We shall define an explicit cntinuus inverse k : X/R A/R 0. T define the latter, cnsider the cntinuus map p A r : X A/R 0. If yrz hlds then r(y)r 0 r(z), and therefre the images f y and z in A/R 0 are equal. Therefre there is a unique cntinuus map k f qutient spaces such that k p X = p A r. This map is a set-theretic inverse t j and therefre j is a hmemrphism. 3. (a) The relatin is reflexive because x = 1 x, and it is reflexive because y = αx fr sme α 0 implies x = α 1 y. The relatin is transitive because y = αx fr α 0 and z = βy fr y 0 implies z = βαx, and βα 0 because the prduct f nnzer real numbers is nnzer. (b) Use the hint t define r; we may apply the preceding exercise if we can shw that fr each a S 2 the set r 1 ({a}) is cntained in an R-equivalence class. By cnstructin r(v) = v 1 v, s r(x) = a if and nly if x is a psitive multiple f a (if x = ρ a then x = ρ and r(x) = a, while if 4

5 a = r(x) then by definitin a and x are psitive multiples f each ther). Therefre if xry then r(x) = ± r(y), s that r(x)r 0 r(y) and the map is a hmemrphism. S 2 /[x ±x] RP 2 4. Needless t say we shall fllw the hints in a step by step manner. Let h : D 2 S 2 be defined by h(x, y) = (x, y, 1 x 2 y 2 ). Verify that h preserves equivalence classes and therefre induces a cntinuus map h n qutient spaces. T shw that h is well-defined it is nly necessary t shw that its values n the R -equivalence classes with tw elements are the same fr bth representatives. If π : S 2 RP 2 is the qutient prjectin, this means that we need π h(u) = π h(v) if u = v = 1 and u = v. This is immediate frm the definitin f the equivalence relatin n S 2 and the fact that h(w) = w if w = 1. Why is h a 1 1 and nt mapping? By cnstructin h maps the equivalence classes f pints n the unit circle nt the pints f S 2 with z = 0 in a 1 1 nt fashin. On the ther hand, if u and v are distinct pints that are nt n the unit circle, then h(u) cannt be equal t ± h(v). The inequality h(u) h(v) fllws because the first pint has a psitive z-crdinate while the secnd has a negative z-crdinate. The ther inequality h(u) h(v) fllws because the prjectins f these pints nt the first tw crdinates are u and v respectively. This shws that h is 1 1. T see that it is nt, recall that we already knw this if the third crdinate is zer. But every pint n S 2 with nnzer third crdinate is equivalent t ne with psitive third crdinate, and if (x, y, z) S 2 with z > 0 then simple algebra shws that the pint is equal t h(x, y). Finally, prve that RP 2 is Hausdrff and h is a clsed mapping. If the first statement is true, then the secnd ne fllws because the dmain f h is a qutient space f a cmpact space and cntinuus maps frm cmpact spaces t Hausdrff spaces are always clsed. Since h is already knwn t be cntinuus, 1 1 and nt, this will prve that it is a hmemrphism. S hw d we prve that RP 2 is Hausdrff? Let v and w be pints f S 2 whse images in RP 2 are distinct, and let P v and P w be their rthgnal cmplements in R 3 (hence each is a 2- dimensinal vectr subspace and a clsed subset). Since Euclidean spaces are Hausdrff, we can find an ε > 0 such that N ε (v) P v =, N ε (w) P w =, N ε (v) N ε (w) =, and N ε ( v) N ε (w) =. If T dentes multiplicatin by 1 n R 3, then these cnditins imply that the fur pen sets N ε (v), N ε (w), N ε ( v) = T (N ε (v)), N ε ( w) = T (N ε (w)) are pairwise disjint. This implies that the images f the distinct pints π(v) and π(w) in RP 2 lie in the disjint subsets π [ N ε (v) ] and π [ N ε (w) ] respectively. These are pen subsets in RP 2 because their inverse images are given by the pen sets N ε (v) N ε ( v) and N ε (w) N ε ( w) respectively. 5. A set W belngs t (g f) T if and nly if (g f) 1 [W ] is pen in X. But (g f) 1 [W ] = f 1 [ g 1 [W ] ] 5

6 s the cnditin n W hlds if and nly if g 1 [W ] belngs t f T. The latter in turn hlds if and nly if w belngs t g (f T). 6. The bject n the left hand side is the family f all sets having the frm (f h) 1 [V ] where V belngs t T. As in the preceding exercise we have s the family in questin is just h (f T). (f h) 1 [V ] = h 1 [ f 1 [V ] ] 7. Let p : X Y X be prjectin nt the first crdinate. Then urv implies p(u) = p(v) and therefre there is a unique cntinuus map X Y/R X sending the equivalence class f (x, y) t x. Set-theretic cnsideratins imply this map is 1 1 and nt, and it is a hmemrphism because p is an pen mapping. 8. (a) If X/R is Hausdrff then the diagnal (X/R) is a clsed subset f (X/R) (X/R). But π π is cntinuus, and therefre the inverse image f (X/R) must be a clsed subset f X X. But this set is simply the graph f R. (b) If π is pen then s is pi π, fr the penness f π implies that π π takes basic pen subsets f X X int pen subsets f (X/R) (X/R). By hypthesis the cmplementary set X X Γ R is pen in X X, and therefre its image, which is (X/R) (X/R) (X/R) must be pen in (X/R) (X/R). But this means that the diagnal (X/R) must be a clsed subset f (X/R) (X/R) and therefre that X/R must satisfy the Hausdrff Separatin Prperty. (c) The cnditin n Γ R implies that each equivalence class is pen. But this means that each pint in X/R must be pen and hence the latter must be discrete. 9. (i) The binary relatin R is symmetric and transitive but nt symmetric. Therefre the equivalence relatin E generated by R cnsists f the unin R with the diagnal f D 2 ; in ther wrds, u E v if and nly if u = v r u R v, and if u v then u E v if and nly if u = αv, where u = v = 1 and α d = 1. (ii) In rder t prve the existence f the cntinuus mapping h we need t shw that h(u) = h(v) if u R v and u v; i.e., u = αv, where u = v = 1 and α d = 1. Under these cnditins we have h(z) = (0, z d ), s if u and v satisfy the given cnditins then h(u) = (0, u d ) = (0, α d v d ) because α d = 1. Therefre h is cnstant n equivalence classes, which implies the existence f h. Since D 2 is cmpact and C 2 is Hausdrff, by Therem III.1.9, the mapping h is a hmemrphism nt its image if and nly if it is 1 1. This is equivalent t shwing that h(u) = h(v) implies u R v. Suppse that h(u) = h(v); taking crdinates, we must have (1 u )u = (1 v )v and u d = v d. The latter implies that u d = v d, which in turn implies that u = v. There are nw tw cases. CASE 1: Suppse that u = v < 1. Then u = v implies 1 u = 1 v, and if we cmbine this with the equatin fr first crdinates we see that u = v. CASE 2: Suppse that u = v = 1. In this case h(u) = (0, u d ) and h(v) = (0, v d ), s if the tw image pints are equal then 1 = (u/v) d ; therefre if α = u/v, then α d = 1 and u = α v, s that u R v, which is what we wanted t shw. (iii) Fllw the hint. We knw that z z d maps D 2 t itself, and if z 0 then the equivalence class f z cnsists f all numbers f the frm α z, where α d = 1; in the exceptinal case where z = 0, 6

7 the equivalence class f z is merely {0}. In all cases we knw that u v implies u d = v d, s if h(z) = z d then u v implies h(u) = h(v), which means that h = h p, where p : D 2 D 2 /F is the qutient prjectin. The mapping h is nt because h is nt, s by Therem III.1.9 it is nly necessary t verify that h is 1 1, r equivalently that h(u) = h(v) implies u v. If u 0, then v 0 t and we have u p = v p ; as in (ii), this implies that u = α v fr sme α such that α d = 1. On the ther hand if u = 0 and h(v) = h(0) = 0, then v = 0 s that u v in this case t. V.2 : Sums and cutting and pasting Additinal exercises 1. ( = ) If X is lcally cnnected then s is every pen subset. But each A α is an pen subset, s each is lcally cnnected. ( = ) We need t shw that fr each x X and each pen set U cntaining x there is an pen subset V U such that x V and V is cnnected. There is a unique α such that x = i α (a) fr sme a A α. Let U 0 = i 1 α (U). Then by the lcal cnnectedness f A α and the penness f U 0 there is an pen cnnected set V 0 such that x V 0 U 0. If V = i α (V 0 ), then V has the required prperties. 2. X is cmpact if and nly if each A α is cmpact and there are nly finitely many (nnempty) subsets in the cllectin. The ( = ) implicatin fllws because each A α is an pen and clsed subspace f the cmpact space X and hence cmpact, and the nly way that the pen cvering { A α } f X, which cnsists f pairwise disjint subsets, can have a finite subcvering is if it cntains nly finitely many subsets. T prve the reverse implicatin, ne need nly use a previus exercise which shws that a finite unin f cmpact subspaces is cmpact. 3. Since the exercise asks fr details in a sketch t be filled in, we shall begin by reprinting this sketch: Let A S 2 be the set f all pints (x, y, z) S 2 such that z 1 2, and let B be the set f all pints where z 1 2. If T (x) = x, then T [A] = A and T [B] = B s that each f A and B (as well as their intersectin) can be viewed as a unin f equivalence classes fr the equivalence relatin that prduces RP 2. By cnstructin B is a disjint unin f tw pieces B ± cnsisting f all pints where sign(z) = ± 1, and thus it fllws that the image f B in the qutient space is hmemrphic t B + = D 2. Nw cnsider A. There is a hmemrphism h frm S 1 [ 1, 1] t A sending (x, y, t) t (α(t)x, α(t)y, 1 2t) where α(t) = 1 t2 4 and by cnstructin h( v) = h(v). The image f A in the qutient space is thus the qutient f S 1 [ 1, 1] mdul the equivalence relatin u v u = ±v. This qutient space is in turn hmemrphic t the qutient space f the upper semicircular arc S 1 + (all pints with nnnegative y-crdinate) mdul the equivalence relatin generated by setting ( 1, 0, t) equivalent t (1, 0, t), which yields the Möbius strip. The intersectin f this subset in the qutient with the image f B is just the image f the clsed curve n the edge f B +, which als represents the edge curve n the Möbius strip. We shall g thrugh the insertins needed at varius steps in this argument. 7

8 Let A S 2 be the set f all pints (x, y, z) S 2 such that z 1 2, and let B be the set f all pints where z 1 2. If T (x) = x, then T [A] = A and T (B) = B [etc.] This is true because if T (v) = w, then the third crdinates f bth pints have the same abslute values and f curse they satisfy the same inequality relatin with respect t 1 2. By cnstructin B is a disjint unin f tw pieces B ± cnsisting f all pints where sign(z) = ± 1, This is true the third crdinates f all pints in B are nnzer. There is a hmemrphism h frm S 1 [ 1, 1] t A sending (x, y, t) t (α(t)x, α(t)y, 1 2t) where α(t)s = 1 t2 4 One needs t verify that h is 1 1 nt; this is essentially an exercise in algebra. Since we are dealing with cmpact Hausdrff spaces, cntinuus mappings that are 1 1 nt are autmatically hmemrphisms. This qutient space [S 1 [ 1, 1] mdul the equivalence relatin u v u = ±v] is in turn hmemrphic t the qutient space f the upper semicircular arc S+ 1 (all pints with nnnegative y-crdinate) mdul the equivalence relatin generated by setting ( 1, 0, t) equivalent t (1, 0, t), which yields the Möbius strip. Let A and B be the respective equivalence relatins n S+ 1 [ 1, 1] and S 1 [ 1, 1], and let A and B be the respective qutient spaces. By cnstructin the inclusin S+ 1 [ 1, 1] S 1 [ 1, 1] passes t a cntinuus map f qutients, and it is necessary and sufficient t check that this map is 1 1 and nt. This is similar t a previus exercise. Pints in S 1 S+ 1 all have negative secnd crdinates and are equivalent t unique pints with psitive secnd crdinates. This implies that the mapping frm A t B is 1 1 and nt at all pints except perhaps thse whse secnd crdinates are zer. Fr such pints the equivalence relatins given by A and B are identical, and therefre the mapping frm A t B is als 1 1 and nt at all remaining pints. 4. We can and shall view X as A id B. Cnsider the map F 0 : A B A B defined by H 1 n A and the identity n B. We claim that this passes t a unique cntinuus map f qutients frm X t A h B; i.e., the map F 0 sends each nnatmic equivalence classes { (c, 1), (c, 2) } fr X = A id B t a nnatmic equivalence class f the frm { (u, 1), (h(u), 2) } fr A h B. Since F 0 sends (c, 1) t (h 1 (c), 1) and (c, 2) t itself, we can verify the cmpatibility f F 0 with the equivalence relatins by taking u = h 1 (c). Passage t the qutients then yields the desired map F : X A h B. T shw this map is a hmemrphism, it suffices t define Specifically, start with G 0 = F0 1, s that G 0 = H n A and the identity n B. In this case it is necessary t shw that a nnatmic equivalence class f the frm { (u, 1), (h(u), 2) } fr A h B gets sent t a nnatmic equivalence class f the frm { (c, 1), (c, 2) } fr X = A id B. Since G 0 maps the first set t { (h(u), 1), (h(u), 2) } this is indeed the case, and therefre G 0 als passes t a map f qutients which we shall call G. Finally we need t verify that F and G are inverses t each ther. By cnstructin the maps F 0 and G 0 satisfy F ([y]) = [F 0 (y)] and G([z]) = [G 0 (z)], where square brackets dente equivalence classes. Therefre we have G F ([y]) = G ([F 0 (y)]) = [G 0 (F 0 (y))] 8

9 which is equal t [y] because F 0 and G 0 are inverse t each ther. Therefre G F is the identity n X. A similar argument shws that F G is the identity n A h B. T cnstruct the example where X is nt hmemrphic t A h B, we fllw the hint and try t find a hmemrphism f the fur pint space {± 1} {1, 2} t itself such that X is nt hmemrphic t A h B is cnnected; this suffices because we knw that X is nt cnnected. Sketches n paper r physical experimentatin with wires r string are helpful in finding the right frmula. Specifically, the hmemrphism we want is given as fllws: ( 1, 1) A + (1, 2) A (1, 1) A + (1, 1) A (1, 2) A + ( 1, 1) A ( 1, 2) A + ( 1, 2) A The first f these implies that the images f S+ 1 {2} and S1 {1} lie in the same cmpnent f the qutient space, the secnd f these implies that the images f S 1 {1} and S1 + {1} bth lie in the same cmpnent, and the third f these implies that the images f S+ 1 {2} and S1 {2} als lie in the same cmpnent. Since the entire space is the unin f the images f the cnnected subsets S± 1 {1} and S1 ± {2} it fllws that A h B is cnnected. FOOTNOTE. The argument in the first part f the exercise remains valid if A and B are pen rather than clsed subsets. 5. (a) Fr each j let in j : X j k X k be the standard injectin int the disjint unin, and let P : X k (X k, x k ) k k be the qutient map defining the wedge. Define Y j t be P in j [ X j ]. By cnstructin the map P in j is cntinuus and 1 1; we claim it als sends clsed subsets f X j t clsed subsets f the wedge. Suppse that F X j is clsed; then P in j [F ] is clsed in the wedge if and nly if its inverse image under P is clsed. But this inverse image is the unin f the clsed subsets in j [F ] and k {x k} (which is a finite unin f ne pint subsets that are assumed t be clsed). It fllws that Y j is hmemrphic t X + j. The cnditin n Y k Y l fr k l is an immediate cnsequence f the cnstructin. The assertin that the wedge is Hausdrff if and nly if each summand is fllws because a subspace f a Hausdrff space is Hausdrff, and a finite unin f clsed Hausdrff subspaces is always Hausdrff (by a previus exercise). T verify the assertins abut cmpactness, nte first that fr each j there is a cntinuus cllapsing map q j frm k (X k, x k ) t (X j, x j ), defined by the identity n the image f (X j, x j ) and by sending everything t the base pint n every ther summand. If the whle wedge is cmpact, then its cntinuus under q j, which is the image f X j, must als be cmpact. Cnversely if the sets X j are cmpact fr all j, then the (finite!) unin f their images, which is the entire wedge, must be cmpact. T verify the assertins abut cnnectedness, nte first that fr each j there is a cntinuus cllapsing map q j frm k (X k, x k ) t (X j, x j ), defined by the identity n the image f (X j, x j ) and by sending everything t the base pint n every ther summand. If the whle wedge is cnnected, then its cntinuus under q j, which is the image f X j, must als be cnnected. Cnversely if the sets X j are cnnected fr all j, then the unin f their images, which is the entire wedge, must be cnnected because all these images cntain the base pint. Similar statements hld fr arcwise cnnectedness and fllw by inserting arcwise in frnt f cnnected at every step f the argument. 9

10 (b) T prve existence, first bserve that there is a unique cntinuus map F : k X k Y such that in j F = Fj fr all j. This passes t a unique cntinuus map F n the qutient space k (X k, x k ) because F is cnstant n the equivalence classes assciated t the qutient prjectin P. This cnstructs the map we want; uniqueness fllws because the cnditins prescribe the definitin at every pint f the wedge. (c) Strictly speaking, ne shuld verify that the s-called weak tplgy is indeed a tplgy n the wedge. We shall leave this t the reader. T prve [1], nte that ( = ) is trivial. Fr the reverse directin, we need t shw that if E is clsed in Y then h 1 [E] is clsed with respect t the s-called weak tplgy we have defined. The subset in questin is clsed with respect t this tplgy if and nly if h 1 [E] ϕ[x j ] is clsed in ϕ[x j ] fr all j, and since ϕ j maps its dmain hmemrphically nt its image, the latter is true if and nly if ϕ 1 h 1 [E] is clsed in X j fr all j. But these cnditins hld because each f the maps ϕ j h is cntinuus. T prve [2], nte first that there is a unique set-theretic map, and then use [1] t cnclude that it is cntinuus. (d) Fr each j let y j X j be a pint ther than x j, and cnsider the set E f all pints y j. This is a clsed subset f the wedge because its intersectin with each set ϕ[x j ] is a ne pint subset and hence clsed. In fact, every subset f E is als clsed by a similar argument (the intersectins with the summands are either empty r cntain nly ne pint), s E is a discrete clsed subset f the wedge. Cmpact spaces d nt have infinite discrete clsed subspaces, and therefre it fllws that the infinite wedge with the weak tplgy is nt cmpact. We shall cnclude this dcument by filling in sme details in the final remark in the exercises fr Sectin V.2. This remark is reprinted here fr the sake f cnvenience: Remark. If each f the summands in (d) is cmpact Hausdrff, then there is a natural candidate fr a strng tplgy n a cuntably infinite wedge which makes the latter int a cmpact Hausdrff space. In sme cases this tplgy can be viewed mre gemetrically; fr example, if each (X j, x j ) is equal t (S 1, 1) and there are cuntably infinitely many f them, then the space ne btains is the Hawaiian earring in R 2 given by the unin f the circles defined by the equatins ( x 1 ) 2 2 k + + y 2 = 1 2 2k. As usual, drawing a picture may be helpful. The k th circle has center (1/2 k, 0) and passes thrugh the rigin; the y-axis is the tangent line t each circle at the rigin. SKETCHES OF VERIFICATIONS OF ASSERTIONS IN THIS REMARK. If we are given an infinite sequence f cmpact Hausdrff pinted spaces { (X n, x n ) } we can put a cmpact Hausdrff tplgy n their wedge as fllws. Let W k be the wedge f the first k spaces; then fr each k there is a cntinuus map q k : n (X n, x n ) W k (with the s-called weak tplgy n the wedge) that is the identity n the first k summands and cllapses the remaining nes t the base pint. These maps are in turn define a cntinuus functin q : n (X n, x n ) k W k 10

11 whse prjectin nt W k is q k. This mapping is cntinuus and 1 1; if its image is clsed in the (cmpact!) prduct tplgy, then this defines a cmpact Hausdrff tplgy n the infinite wedge n (X n, x n ). Here is ne way f verifying that the image is clsed. Fr each k let c k : W k W k 1 be the map that is the identity n the first (k 1) summands and cllapses the last ne t a pint. Then we may define a cntinuus map C n k 1 W k by first prjecting nt the prduct k 2 W k (frget the first factr) and then frming the map k 2 W k. The image f q turns ut t be the set f all pints x in the prduct such that C(x) = x. Since the prduct is Hausdrff the image set is clsed in the prduct and thus cmpact. A cmment abut the cmpactness f the Hawaiian earring E might be useful. Let F k be the unin f the circles f radius 2 j that are cntained in E, where j k, tgether with the clsed disk bunded by the circle f radius 2 (k+1) in E. Then F k is certainly clsed and cmpact. Since E is the intersectin f all the sets F k it fllws that E is als clsed and cmpact. 11