[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