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Construction of Spaces Comments on Chap. 5 from Sieradski's book http://cis.k.hosei.ac.jp/~yukita/ The choice of topology influences the continuity. Prop. Given : An inclusion function i : X Y . A topology S on the set Y . Claim : i : ( X ,2 X ) (Y , S ) is continuous . Proof. U S ; i 1 (U ) 2 X Is there a smallest t opology T such that i : ( X , T ) (Y , S ) is continuous ? 2 1.1 Topological Subspace Construction Let i : X Y be an inclusion of a set X into the topologic al space (Y , S ). Then, (a) i * S {i 1 (V ) V X X | V S} is a topology on X . (b) i : ( X , T ) (Y , S ) is continuous if and only if T i * S . Therefore, i * S is the smallest t opology on X for which i is continuous . 3 Proof of (a) By definition U X ; U i * S V S ;U i 1 (V ) . Clearly, we have i 1 () and X i 1 (Y ), and thus , X i * S . Arbitrary union 1 1 * i ( V ) i ( V ) i S since Finite intersecti on 1 1 * i ( V ) i ( V ) i k k S since k k V S . V k S. k Remark. Taking arbitrary intersecti on does not make sense because V S is not guaranteed . 4 Proof of (b) i : ( X , T ) (Y , S ) is continuous V S ; i 1 (V ) T . This means i * S T . 5 Transitivity Def. Let i : X Y be an inclusion and Y have topology S . Then, ( X , i * S ) is said to have the subspace topology. Prop. If W is a subspace of X and X is a subspace of Y , then W is a subspace of Y . Proof. Let j : W X and i : X Y be inclusions . (i j ) 1 (V ) j 1 (i 1 (V )) for all open sets V Y . All open sets in X have the form i 1 (V ), and thus all open sets in W have the form j 1 (i 1 (V )). In other word s, all open sets in W have the form (i j ) 1 (V ) for some open set V Y . 6 Key feature Let i : X Y be a subspace inclusion. k : W X is continuous i k : W Y is continuous . Proof. is trivial. If i k : W Y is continuous , for each open set V Y , (i k ) 1 (V ) k 1 (i 1 (V )) k 1 (V X ) is open in W . Thus k 1 is continuous since all open subsets of X have the form V X for some open set V Y . 7 Restriction Theorem. Let i : X Y be an inclusion map and g : Y Z be a continuous map. Then, g | X : X Z is continuous . Proof . g | X g i. 8 Embedding Any continuous function h : W Y gives a continuous function h : W h(W ), provided that the image h(W ) Y is given the subspace topology. The continuous function h : W Y is called an embedding of W into Y provided that h : W h(W ) is a homeomorph ism. 9 Distinguishing Spaces embeddable non-embeddable Cylinder Moebius strip Sphere W1 and W2 is non - homeomorph ic if One has an embedding into a space Y while the other does not. 10 Topological Invariants non-embeddable Wire handcuff embeddable Wire eight the cross For a fixed space W , the existence of an embedding W Y is a topological invariant of Y . 11 Identification Topology (Dual to the subspace topology) Let p : X Y be a surjection and X have topology T . The indiscrete topology {, Y } is the smallest t opology on Y , and it makes the given function p : ( X , T ) (Y , {, Y }) continuous . Is there a largest to pology S on Y for which p : ( X , T ) (Y , S ) is continuous . 12 1.5 Identification Space Construction Let p : ( X , T ) Y be a surjection from the topologic al space ( X , T ) to the set Y . Then, (a) The collection p*T {V Y | p 1 (V ) T } is a topology on Y , and (b) p : ( X , T ) (Y , S ) is continuous if and only if S is contained in p*T . Therefore, p*T is the largest to pology on Y for which p is continuous . 13 Proof of (a) By definition , V p*T p 1 (V ) T . Since p 1 () T and p 1 (Y ) X T , we have {, Y } p*T . Let V p*T , namely p 1 (V ) T . Then, p 1 (V ) p 1 (V ) T , namely V p T . * Let Vi p*T , namely p 1 (Vi ) T . Then, p 1 (Vi ) p 1 (Vi ) T , namely i i V p T . i * i 14 Proof of (b) p : ( X , T ) (Y , S ) is continuous V S ; p 1 (V ) T . This means S p*T . 15 Identification Topology Surjection p : X Y identifies p 1 ({ y}) to the correspond ing point y Y . p*T on Y is called the identification topology determined by p and T . (Y , p*T ) is called an identification space of ( X , p ) and p is called an identification map. 16 Ex 4. X {1,2,3}, T {,{1},{1,2},{1,3}, X }, Y {a, b}. p : X Y is defined by p(1) p(3) a and p(2) b. Then, p*T {,{a}, Y }. 17 1.6 Theorem A continuous surjection p:XY that is either open or closed is an identification map. Proof. Let p is an open(close d) function. (i) Let V Y be a subset wit h p 1 (V ) is open(close d) in X . Then V p( p 1 (V )) since p is surjective , and V is open(close d) since p is open(close d). (ii) Let V Y be open(close d) in Y . Then p 1 (V ) is open(close d) since p is continuous . 18 Ex. 6, 7 Identification Maps e : 1 is defined by e(t ) (cos 2t , sin 2t ). p : 2 is defined by p( x, y) x. 19 1.7 Transitivity Let p : X Y and q : Y Z be indentific ation maps. Then q p : X Z is an identifica tion map. Proof. V Z is open if and only if q 1 (V ) Y is open, and q 1 (V ) Y is open if and only if p 1 (q 1 (V )) X is open. This proves that q p : X Z is an identifica tion map. 20 1.8 Key Feature Let p : X Y be an identifica tion map. k : Y Z is continuous if k p is continuous . Proof. For each subset W Z , 1 1 1 (k p ) (W ) p (k (W )) X is open. Hence, k 1 (W ) Y is open since p is an identifica tion. 21 1.9 Transgression Theorem Let p : X Y be an identifica tion map and let g : X Z be any continuous function t hat respects the identifica tion of p : x, x X ; p( x) p( x) implies g ( x) g ( x). Then there exists a unique continuous function h : Y Z such that h p g. Proof. Since y g ( p 1 ({ y})) is single valued, h is well - defined. From the key feature, h is continuous . 22 1.10 Corollary Let p : X Y and q : X Z be identifica tion maps that respect one another' s identifica tions. Then, X and Y are homeomorph ic. Proof. Applying Transgress ion Theorem twice, we have continuous functions h : Y Z and k : Z Y such that h p q and k q p. Then k h p k q p and h k q h p q; hence, k h 1Y and h k 1Z , by the surjectivi ty of p and q. 23 Quotient Modulo A Subspace U X U X/A A V V [A] qA : X X / A 24