Download KOC¸ UNIVERSITY, Spring 2011, MATH 571 TOPOLOGY, FINAL

KOÇ UNIVERSITY, Spring 2011, MATH 571 TOPOLOGY, FINAL, JUNE 6,
INSTRUCTOR: BURAK OZBAGCI, 180 Minutes
Solutions by Sumeyra Sakallı.
PROBLEM 1 (10 points): Consider the set X = {a, b, c} with the topology T = {∅, X, {a}, {b}, {a, b}}.
(a) Is (X, T) a regular space? (b) Is (X, T) a normal space?
Solution:(a) The set {b, c} is T-closed since {a} ∈ T. Note that the only open set containing {b, c} is X
and it has nonempty intersection with any open set containing {a}. Thus we conclude that (X, T) is not
regular.
(b) The T-closed sets in X are precisely: ∅, X, {c}, {a, c}, {b, c}. But then, we observe that c belongs to
all of these sets except ∅. Thus (X, T) is normal, since the condition for being normal is satisfied trivially.
PROBLEM 2 (10 points): Prove or disprove: Every path-connected space is locally path-connected.
Solution: Let X ⊂ R2 be the union of the graph of sin(1/x), for x ∈ (0, 1/π] and a simple arc γ (disjoint
from this graph) connecting the points (1, 0) and a = (0, 1). Note that X is itself a path and thus it is
path connected. We claim that X is not locally path connected. Let V be a small neighborhood of a in
X (with relative topology induced from the usual topology in R2 ). Then each open neighborhood U of a
in V includes a piece of the arc γ and an arc β of the graph of sin(1/x). It is clear that the points in β
can not be connected to a by a path in V . Hence X is not locally path connected.
PROBLEM 3 (10 points): A topological space X is called completely regular if for any closed subset
F of X and any a ∈ X such that a ∈
/ F , there exists a continuous function f : X → [0, 1] such that
f (x) = 1 for every x ∈ F and f (a) = 0. Prove that a completely regular space is regular.
Solution: Suppose that X is a completely regular space. Let F be a closed set in X and a ∈
/ F . Then
there exists a continuous function f : X → [0, 1] such that f (x) = 1 for every x ∈ F and f (a) = 0. It
follows that f −1 ([0, 1/3)) and f −1 ((2/3, 1]) are disjoint open sets in X, containing a and F respectively.
We conclude that X is regular.
PROBLEM 4 (10 points): Prove or disprove: A second countable topological space is separable.
Solution: Let X be a second countable space and B = {Bn }n≥1 be a base of X. Now, ∀n ∈ N, choose
an arbitrary point sn from Bn and let S = {sn }n≥1 . Then S is countable. We claim that S is dense in
X, which proves that X is separable. Let x be a point in X and let V denote an open neighborhood of
x. Then V is the union of some Bn ’s in B, which implies that there exists Bm ∈ B such that Bm ⊆ V.
So in particular, sm ∈ Bm ⊆ V , showing that V ∩ S =
6 ∅. Hence x ∈ S and thus S = X.
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PROBLEM 5 (10 points): Show that P n is homeomorphic to the quotient space obtained from the
closed unit ball B n in Rn by identifying the antipodal points of its boundary S n−1 .
Solution:
f
Bn →
− P n = Sn/ ≈
&
↑?
Bn/
∼
Let π : S n → P n = S n / ≈ denote the map which identifies antipodal
points of S n ⊂ Rn+1 . Define a map
Pn
n
n
2
f : B → P by f (x1 , .., xn ) = π(x0 , .., xn ) where x0 ≥ 0,
i=0 xi = 1. We need the following three
facts to prove the result:
i) B n is compact since it is a closed and bounded subset of Rn and it is also Hausdorff. Now P n is
compact since P n = S n / ≈ = π(S n ) where π is continuous and S n is compact. In fact P n is also Hausdorff.
q
P
ii) f is continuous: Indeed, let g : B n → (S n )+ be given by g(x1 , .., xn ) = ( 1 − ni=1 x2i , x1 , ..xn ) =
Pn
2
(x0 , .., xn ) where x0 ≥ 0,
i=0 xi = 1. Then g is continuous and f = π ◦ g, so f is continuous.
iii) We claim that ∀ a, b ∈ B n , a ∼ b ⇔ f (a) = f (b).
n−1 , a = −b. Let a := (x , .., x ) ⇒
(=⇒) If a = b, then
1
n
Pfn (a) 2= f (b). So assume a 6= b, a ∼ b ⇒ a, b ∈ S
b = −(x1 , ..,P
xn ) and i=1 xi = 1. Now, take π(0, ±x1 , .., ±xn ) which in turn equal to f (±x1 , .., ±xn ), since
x0 = 0 and ni=0 (±xi )2 = 1. Furthermore, π identifies antipodals, so π(0, x1 , .., xn ) = π(0, −x1 , .., −xn ).
Therefore f (x1 , .., xn ) = f (−x1 , .., −xn ), so f (a) = f (b).
(⇐=) Conversely, take a = (x1 , .., xn ), b = (y1 , .., yn ) ∈ B n , f (a) = f (b). If a = b ⇒ a ∼ b.
n−1 and a = −b. Having f (a) = f (b) implies that π(x , .., x ) =
So take a 6= b. We want
0
n
Pna, b ∈ S2
Pn
2
π(y0 , .., yn ), x0 , y0 ≥ 0,
i=0 (±xi ) = 1,
i=0 (±yi ) = 1. Since π identifies antipodals and a 6= b, we
have (x0 , .., xn ) = −(y
0 = −y0 , hence x0 = 0 = y0 . Therefore, a = (x1 , .., xn ) = −(y1 , .., yn ) =
P 0 , .., yn ) ⇒ xP
n
2
n−1 ⇒ a ∼ b.
−b. What is more, ni=1 x2i = 1,
i=1 yi = 1 ⇒ a, b ∈ S
Hence by a theorem i) ii) iii) implies that B n / ∼ is homeomorphic to P n .
Problem 6 (10 points): Let X and Y be topological spaces and and let f : X → Y be an arbitrary
map. Prove the following:
f : X → Y is continuous ⇔ ∀A ⊆ X : f ( A ) ⊆ f (A)
Solution: (=⇒) Let A ⊆ X and let y ∈ f ( A ). Then there exists x ∈ A such that f (x) = y. We
want to show that y ∈ f (A). For a contradiction, assume that there exits an open neighborhood U of y
such that U ∩ f (A) = ∅ (∗). Then observe that x ∈ f −1 (U ) and f −1 (U ) is open. Since x ∈ A we have
f −1 (U ) ∩ A 6= ∅. Then ∃ x0 ∈ f −1 (U ) ∩ A ⇒ f (x0 ) ∈ U ∩ f (A) contradicting to (∗). Therefore y ∈ f (A).
(⇐=) Assume the right hand side holds and take a closed set U ⊆ Y . We want to show that f −1 (U ) is
closed in X. Let y ∈ f −1 (U ). Then f (y) ∈ f (f −1 (U )). By the assumption, f (f −1 (U )) ⊆ f (f −1 (U )) ⊆ U
which in turn equals to U , as U is closed. Thus f (y) ∈ U and so y ∈ f −1 (U ). Consequently, f −1 (U ) ⊆
f −1 (U ), which implies that f −1 (U ) is closed. As a result, f is continuous.
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PROBLEM 7 (10 points): If f : [0, 1] → [0, 1] is a continuous function, then show that f has a fixed
point.
Solution: If f (0) = 0 or f (1) = 1, then we are done. Otherwise, since f (0), f (1) ∈ [0, 1] we have f (0) > 0
and f (1) < 1. Let g(x) = f (x) − x. Note that g(0) > 0, g(1) < 0 and g is continuous. Then by the
Intermediate Value Theorem, there exists x0 ∈ (0, 1) such that g(x0 ) = 0 and hence f (x0 ) = x0 .
PROBLEM 8 (10 points): Show that for any n > 1, Rn is not homeomorphic to R.
Solution: Assume that there exists a homeomorphism f : R → Rn . Then for any x ∈ R, f (R − {x}) =
Rn − f (x). But for n > 1, Rn − f (x) is path connected, while R − {x} is not path connected. This
gives a contradiction because continuous image of a path connected space is path connected and f −1 is
continuous.
PROBLEM 9 (10 points): Prove or disprove: If there are simply connected open subsets U and V of
a topological space X such that X = U ∪ V and U ∩ V 6= ∅, then X is simply connected.
Solution: Consider X = {(x, y) ∈ R2 : 1 < x2 + y 2 < 4} with relative topology induced from the usual
topology on R2 . Let U = {(x, y) ∈ X : x > −1/2}, and V = {(x, y) ∈ X : x < 1/2}. Then U and V are
open simply connected subsets of X such that X = U ∪ V , and U ∩ V 6= ∅. Note that X is not simply
connected and in fact π1 (X) ∼
= Z.
PROBLEM 10 (10 points): Let A be a connected subset of a topological space X and let
A ⊆ B ⊆ A.
Show that B is connected.
Solution: Suppose B = U ∪ V for some disjoint relatively open sets in B. By definition U = W ∩ B for
some open set W in X. Then U ∩ A = W ∩ B ∩ A = W ∩ A is relatively open in A. Similarly V ∩ A
is relatively open in A. Hence A = (U ∩ A) ∪ (V ∩ A). But since A is connected either U ∩ A = ∅ or
V ∩ A = ∅. Without loss of generality, assume that U ∩ A = ∅, which implies that W ∩ A = ∅. Then no
point of W belongs to A, and thus A ⊆ X \ W . Thus we get
A ⊆ B ⊆ A ⊆ X \ W ⊆ X \ U.
Consequently U = ∅, otherwise B ⊆ X \ U and B = U ∪ V gives a contradiction. Hence B is connected.
PROBLEM 11 (10 points): A subspace A of a topological space X is a retract of X if there is a
continuous map f : X → A such that f (x) = x for all x ∈ A. True or false:
A = {z ∈ C : |z| = 1} is a retract of X = {z ∈ C : |z| ≤ 1}.
Solution: Suppose that f is a retraction of X onto A. Then the induced map f∗ : π1 (X) → π1 (A) is
surjective. This is impossible since π1 (A) ∼
= Z but π1 (X) is trivial.
e → X be a covering map, let x ∈ X, and let x
PROBLEM 12 (10 points): Let p : X
e1 , x
e2 ∈ p−1 ({x}).
e x
e x
Then show that p∗ π1 (X,
e1 ) and p∗ π1 (X,
e2 ) are conjugate in π1 (X, x). (Here you may assume that the
spaces are Hausdorff, path-connected and locally path-connected.)
e x
e x
e is path connected,
Solution: We want p∗ π1 (X,
e1 ) = [γ] p∗ π1 (X,
e2 ) [γ]−1 for some [γ] ∈ π1 (X, x). X
so there exists a path γ
e between x
e1 and x
e2 . We can move any loop based at x
e2 to a loop based at x
e1
−1
e
e
e
via isomorphism γ
e∗ : π1 (X, x
e2 ) → π1 (X, x
e1 ), γ
e∗ [e
α] = [e
γ ] [e
α] [e
γ ] where [e
α] ∈ π1 (X, x
e2 ). It means that
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e x
e x
e x
e x
∀ [α] ∈ π1 (X,
e1 ), ∃ [e
α] ∈ π1 (X,
e2 ) s.t. [α] = [e
γ ] [e
α] [e
γ ]−1 . Then π1 (X,
e1 ) = [e
γ ] π1 (X,
e2 ) [e
γ ]−1 .
e x
e x
e x
Projecting, we get: p∗ (π1 (X,
e1 )) = p∗ ([e
γ ] π1 (X,
e2 ) [e
γ ]−1 ) = p∗ ([e
γ ]) p∗ (π1 (X,
e2 )) p∗ ([e
γ −1 ]) = [p ◦
−1
−1
e
e
γ
e] p∗ (π1 (X, x
e2 )) [p ◦ γ
e ] = [p ◦ γ
e] p∗ (π1 (X, x
e2 )) [p ◦ γ
e] and [p ◦ γ
e] ∈ π1 (X, x) since p(e
x1 ) = p(e
x2 ) = x.
Therefore for [γ] = [p ◦ γ
e] result follows.