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[Topology I, Final Exam — Solutions] The exam consists of 6 questions. Questions 1 to 4 are take-home. The other two questions will be given on December 16 in class. The solutions to the take-home questions are due on Wednesday, December 16, 10:30 AM. Please write your solutions on the provided paper. Use this sheet as a cover sheet and write your name on top of every page, including this one. Problem 1 Let f : X −→ Y be a continuous map, Y Hausdorff. Prove that f = g ◦ h, where h : X −→ Z is a closed continuous map and g : Z −→ Y is an open continuous map. [Hint: Take Z = X × Y .] Solution: by Let Z = X × Y and define h : X −→ Z and g : Z −→ Y h(x) = (x, f (x)), g(x, y) = y. Clearly, h and g are continuous and f = g ◦h. As we’ve seen in problem 16.4, g is an open map. Let C ⊂ X be a closed set and let Γ = h(X) be the graph of f . Then h(C) = Γ ∩ (C × Y ). Thus, since C × Y is closed in X × Y , to show that h is a closed map, it will suffice to show that Γ is closed in X × Y . So let (x, y) 6∈ Γ, i.e. f (x) 6= y. Since Y is Hausdorff, there are disjoint open sets W, V ⊂ Y such that f (x) ∈ W and y ∈ V . Let U = f −1 (W ). Then (x, y) ∈ U × V . Also, if (x0 , y 0 ) ∈ U × V , we have f (x0 ) 6= y 0 , i.e. Γ ∩ (U × V ) = ∅. This shows that Γ is open. F Problem 2 Let U ⊂ R2 be a non-empty bounded open set. Prove that U has uncountably many boundary points. Solution: Let V = R2 − Ū and let B = Ū ∩ V̄ be the boundary of U . Suppose B is countable. Then the space X = R2 −B is a connected space (problem 24.9). On the other hand, U and V are disjoint open subsets of X and X = U ∪ V . Finally, U is non-empty (given) and so is V (since U is bounded). Thus {U, V } is a separation of a connected space. The contradiction proves that B is uncountable. F Problem 3 Let (X, d) be a metric space. Prove that is every continuous function f : X −→ R is bounded, then X is compact. Solution: We will show, by contradiction, that X limit-point compact. So let A ⊂ X be an infinite set and suppose A has no limit point. Then A is closed (since it contains all its limit points) and the subspace topology on A is discrete: if a ∈ A is any point, there is a neighborhood U of a in X such that U ∩ (A − {a}) = ∅. Let g : A −→ R be any unbounded map (pick any injective sequence (an ) in A, let g(an ) = n and g(x) = 0 for all x ∈ A − {an | n ≥ 1}). Since A is discrete, g is continuous. Then, since A is closed, Urysohn’s extension theorem allows to extend g to a continuous function f : X −→ R such that f |A = g. Then f is unbounded, contrary to the assumption. The contradiction shows that X is limit-point compact. F Problem 4 Let X, Y, Z be topological spaces and f : Y −→ Z a continuous function. Prove that if C(X, Y ) and C(X, Z) are given the compact-open topology, then the map f? : C(X, Y ) −→ C(X, Z), given by f? (g) = f ◦ g, is continuous. Solution: Let C ⊂ X be compact and U ⊂ Z open. Then f −1 (U ) is open and g ∈ f?−1 (S(C, U )) ⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒ f? (g) ∈ S(C, U ) f ◦ g ∈ S(C, U ) f (g(C)) ⊂ U g(C) ⊂ f −1 (U ) g ∈ S(C, f −1 (U )). Thus f?−1 (S(C, U )) = S(C, f −1 (U )) and f? is continuous. F Problem 5 Prove the following generalization of the pasting lemma: Let X = ∪α∈J Aα , where (Aα )α∈J is a locally finite family of closed sets, and for each α ∈ J, let fα : Aα −→ Y be a continuous function. If for every α, β ∈ J we have (1) fα |Aα ∩Aβ = fβ |Aα ∩Aβ , then there is a unique continuous function f : X −→ Y such that f |Aα = fα for all α ∈ J. Solution: Letting f (x) = fα (x) for x ∈ Aα defines a function f : X −→ Y whose restriction to Aα agrees with fα . The compatibility condition (1) guaranties that the definition is correct. Also, f is uniquely determined by its restrictions to the Aα . It remains to show that f is continuous. Let C ⊂ X be a closed set. For each α ∈ J, f −1 (C) ∩ Aα = fα−1 (C) is closed in Aα and thus in X, since Aα is closed in X. The family {f −1 (C) ∩ Aα } is a refinement of the locally finite family {Aα }. Therefore it is locally finite itself. It follows that [ f −1 (C) = (f −1 (C) ∩ Aα ) α∈J is closed. Thus f is continuous. F Problem 6 Let A0 , A1 , . . . , An be pairwise disjoint closed sets in a normal space X. Prove that there is a continuous function f : X −→ [0, n] such that f (Ak ) = {k} for all k = 0, 1, . . . , n. Solution: Here are two different solutions, one using the Tietze Extension Theorem, the other using Urysohn’s lemma. First solution: Let A = ∪ni=0 Ai . This is clearly a closed subspace of X. Let g : A −→ [0, n] be the function defined by g(x) = k if x ∈ Ak . If C ⊂ [0, n] is any set, we have [ g −1 (C) = Ak , where K = Z ∩ C. k∈K −1 Thus g (C) is closed, being a finite union of closed sets, and g is continuous. By the Tietze Extension theorem, g can be extended to a continuous function f : X −→ [0, n]. Second solution: using Urysohn’s lemma, for each m = 1, 2, . . . , n find a continuous function fm : X → [0, 1] such that ( 0 if x ∈ A0 ∪ A1 ∪ . . . ∪ Am−1 fm (x) = 1 if x ∈ Am ∪ Am+1 ∪ . . . ∪ An and let f = f1 + f2 + · · · + fn . Clearly f is a continuous function f : X −→ [0, n]. Also, if x ∈ Ak , then fm (x) = 1 for m ≤ k and fm (x) = 0 for m > k. Thus f (x) = k. F