MATH535 lecture 3838.1

Last time : . defined degf for ft (s ', S' ) and used it to prove that any complex

polynomial pay of positive degree has a root.

. Observed that for any 3 sets X , Y , Z there is a bijectionHornsea (Z ,

Y"

)→ Hourset ( Z xx , Y ) , (h : z → Y

" ) t th : ZxXii ) where I text -- th th) ta)Here Hom

sa (A , B) = set of functions from A to Be BA.

Note we have the evaluation map evy : YXx X - Y,er, If, a) e- f- Cx)

So Hornsea ( 7,4×1 - Monger ( ZXX, Y) is That -- Wx ( HH ,a )

or, equivalently , I = ego ( hxidx)

Recalls the compaet-op.ee topology on co (XM) is generated by the sets

M (Kil) =L f :X- Y I f- (Kl EUI ,where KEX is compact, U E Y open .

We want to show : suppose AZ are locally compact Hausdorff , Y Hausdorff .

Then t he co ( Z , Hill) , I = Wx o'

(hxidx ) E C'

( Z xx, Y ) and-

: (z ,co ( X, Yl) - co ( Z xx , Y ) is a homeomorphism .

First step :

Lemma38 Suppose X is LCH ( hoc . compact Hausdorff ) . Then ev : CYXIY) xx - Y,If, a) he fix)

is continuous.

Proofs Let Me Y be open .We want to show : W

- ' (UI is open in (X, Y ) xx.Let Hin ) E W

-

YU) .Then fail -- ev (fault U.

Since f is continuous, f- ' (UI is

an open nbd of n in X.Since X is locally compact , A compact nbd K of a

so that K e f- '

(U) . But then f CK) EU. This implies that f e M (Kil) .

tf g e Mlk , U) t y e- K, gas) = err ( g , y) c- U. ⇒ er ( Mlk, U) x K ) E U.

⇒ Neck , h) x K E WY U) .•

'o ei

'

f UI is open .

Hence ev : co (X,Y ) x Y - Y is continuous . D

Grokary3 Suppose X is LCH. Then

the ft , O ( X, Y ) ) , Tn 't exo (hxidx ) e- Co ( Zx Y ,Y )

.

38.2

Proofs Since hxidx : Zx X - C ' IX.Y) x X and evx : CoA ,4) x X e Y are continuous,so is

T : = eve, o (hxidxl. D

Remand Suppose k : 2- xx - Y is continuous.Then b- z e Z

, Ftz) : X - Y , ICH la) : - k Hix)is continuous. This is because I #I = ko iz where iz :X- 2-xx is the inclusion

iz ca) z CE ,r ) txt X . We thus have a map

-

: ( 2- xx , Y ) - Homsee. ft , COCKY ) ) , k 1- I

.

The next lemma shows that the image of-

lands in co ( Z, co Hill) E House. (Z , COCKY) ) .

↳mma38c3_ Let 11,4, 't be three spaces . Then the (ZXX , Y) the map

K : Z - co ( X, Y ) , z te Te Lz) = k Cz , . ) is continuous , hence we have

a map-

: co (Z xx , Y ) - of -2 , CTX , Y) ) .

Proof Enough to show : t KEX compact ,they open , k

- '

(Mlk , U)) is open in -2.

Let t t k-

I MLK , U )) . Then kfhzsxk ) = ICH ( K) EU.⇒ aah x k e k

- '

(U) .Since his continuous

,k ' (UI is open, in 2-xx

. By Tube Lemma, 7 open nbdW of t sit Wx K E k

- ' ( U) . ⇒ t w c- IN U2 kdcwsxk) - ( Flw)) 1kt .⇒ a wt W KIWI c- Mlk , U ) .⇒ W E (Te)

"

( MLK , U )) . ⇒ I ' (Mlk ,U) ) is open in Z . D

Recaps If X is LCH,Y,Z are spaces we have maps

① -

i. Co ( Z ,

C-

(X, Y l) - co (z xx, Y ) , h teh,tile, x) = ditz ) ) ( x ) thx) EZXX

② - i ( 2- XX, Y ) - Colt , 01441) , kn I, KH - k tail ,htt EZ

.

It's easy to check that the two maps are inverses of each other.

He'd like to prove :

theorems Suppose X , Z are LCH, Y Hausdorff. Then the bijectionco ( Zxx, Y ) → Colt , (X, Y ) )

is a homeomorphism .

To prove 38.4 we need :

38.3

Lemma38 Let P be an LCH space , Q a space , B a sub basis for a topology on Q ,K -

- E KE P compact ) K is a nbd of some at PS .Then

I - l M IK ,U ) I KEK , U C- BS

is a sub basis for the compact - open topology on CYP,Q )

.

Proof Enough to show : tf k e P compact, t U '- Q open ,t f E M ( K ,U) , 7 Ui . . - Un EB

Ki , . - kn E K sit f E,§,

M (Ki , Ui) E M (K ,U) .

For each seek I Use c- B with text Uae EU and a compact nbd Ka of se with Ka e- f''

fun ) .( Note that f- E M (ka

,Ua) ta C- KI

.

Since K is compact 3 a , - - an EP sit . KE Ka,u - - o Kaen . ⇒ f t M (Kai , Uni ) .

Since age II,

M (Kai, Uni ) , g (K) e- GCU Kai ) -- U g ( Kai) EU Uai E U,

in,

M (Kai , Uno ) E Mlk, U) . D

proofofss.ly Consider

K = LK , x Kat 2- XX / K , is a compact- nbd of some # c- Z,Ke is a compact nbd of some at Xl

.

By 38.5, I M I Kixkz , U ) I Kixkz C-K,he 4in open4 is a subbasis of the compact - open

topology on co ( 2- xx, Y ) .

By 38.5 , l M ( K , , M (Ka, U)) I klxkz C-K,he th opens is a subbasis for the compact

open topology on co (Z , COCKY ) ) .Now he Mlk , xka , U)⇒ k CK, xkz) EU ⇐s CI ( ki )) (ka ) EU ⇐

ti (ki) E M (K2 ,U) ⇐ I E M (Ki , M ( Kz , U ) ) .

Hence the bijection- t.co/ZxX, Y ) - C ° (Z , co (X, Y ) ) takes a sub basis of the

topology to a sub basis.⇒-

is a homeomorphism . D

Compact -open topology and pullbacks ( restrictions .

Lemma3# Let 4 : X'- X be a continuous map and Y a space . t f e (X, Y)

,

44 Cfl : -- foul E (X', Y ) . The map44 : co (X,Y ) - (X' , Y ) is continuous (w.at . the compact - open topologies)

38.4

Proof f k EX' compact and HUEY open ,

④*5' ( Mlk ,U) ) - If e- COCKY) I 44ft e- Mlk , uh = If I Hot ) (K) e US

= If I f ( 41kt) E UG - M ( 4 (Kl , U) .

( Note that since 4 is continuous and K is compact , 4( K ) is compact ) .

⇒ Yt is continuous . D

Note that if X' EX is a subspace , then the map COCKY)→ cold, Y ), f 1-flyis continuous since the inclusion map i :X'as X is continuous and f 1×1 -

- fo i .