Download Math F651: Homework 5 Solutions 1. (Solution by Jody Gaines

Math F651: Homework 5
Solutions
1. (Solution by Jody Gaines)
Show that (X1 × X2 ) × X3 is homeomorphic to X1 × X2 × X3 . Do not use the words “open”
or “closed” at any point in the proof.
Solution:
Consider the map f ∶ (X1 ×X2 )×X3 → X1 ×X2 ×X3 defined by ((x1 , x2 ), x3 ) ↦ (x1 , x2 , x3 ).
Note g ∶ X1 × X2 × X3 → (X1 × X2 )× X3 defined by (x1 , x2 , x3 ) ↦ ((x1 , x2 ), x3 ) is the inverse
function of f since
f ○ g(x1 , x2 , x3 ) = f ((x1 , x2 ), x3 ) = (x1 , x2 , x3 )
g ○ f ((x1 , x2 ), x3 ) = g(x1 , x2 , x3 ) = ((x1 , x2 ), x3 )
for each ((x1 , x2 ).x3 ) ∈ (X1 × X2 ) × X3 and (x1 , x2 , x3 ) ∈ X1 × X2 × X3 . Thus f is bijective.
Observe f3 = π3 ○ f and g3 = π3 ○ g are continuous since
f3 ((x1 , x2 ), x3 ) = π3 ○ f ((x1 , x2 ), x3 ) = π3 (x1 , x2 , x3 ) = x3 ,
g3 (x1 , x2 , x3 ) = π3 ○ g(x1 , x2 , x3 ) = π3 ((x1 , x2 ), x3 ) = x3 ,
which implies f3 and g3 are projection maps. Moreover the maps f1 = π1 ○ f and g1,2 =
(π1,2 ) ○ g, where π1,2 ∶ (X1 × X2 ) × X3 → X1 × X2 is defined by ((x1 , x2 ), x3 ) ↦ (x1 , x2 ), are
continuous since
f1 ((x1 , x2 ), x3 ) = π1 ○ f ((x1 , x2 ), x3 ) = π1 (x1 , x2 , x3 ) = x1
g1,2 (x1 , x2 , x3 ) = (π1,2 ) ○ g(x1 , x2 , x3 ) = (π1,2 )((x1 , x2 ), x3 ) = (x1 , x2 ),
which implies f1 and g1,2 are projection maps as well. It follows from the previous statement that f2 = π2 ○ f is continuous as well. By the Characteristic Property of the Product
Topology, f and g are continuous. Recall g is the inverse function of f . Hence f is a
homeomorphism.
2. (Solution by Aven Bross)
Prove the following.
a) A projection map from an arbitrary product space is an open map.
b) An arbitrary product of Hausdorff spaces is Hausdorff
c) A countable product of second countable spaces is second countable.
Solution, part a:
Let X = ∏α∈I X α be an arbitrary product space. Suppose U ⊆ X is open.
Let B be the collection of products of the form ∏α∈I U α with U α ⊆ X α open for each α ∈ I
and U α = X α for all but finitely many α ∈ I. Thus for B ∈ B and α ∈ I, π α (B) = U α where
U α ⊆ X α is open.
Math F651: Homework 5
Solutions
Recall B forms a basis for the infinite product topology on X, and for the standard product topology if X is a finite product space. Therefore U = ⋃β∈J B β , with J some set and
B β ∈ B for each β ∈ J. Therefore for a particular α ∈ I
⎛
⎞
π α (U) = π α ⋃ B β = ⋃ π α (B β ).
⎝β∈J ⎠ β∈J
As noted above, each π α (B β ) is open in X α . Thus π α (U) is a union of open subsets of X α
and therefore open. So π α is an open map.
Solution, part b:
Let X = ∏α∈I X α be an arbitrary product of Hausdorff spaces. For x ∈ X observe the α’th
component of x is π α (x).
Suppose x, y ∈ X such that x ≠ y. Then there exists β ∈ I such that π β (x) ≠ π β (y).
Therefore, since X β is Hausdorff, there exist disjoint open sets U β , Vβ ⊆ X β such that
π β (x) ∈ U β and π β (y) ∈ Vβ .
Define U α = X α and Vα = X α for all α ≠ β. Let U = ∏α∈I U α and V = ∏α∈I Vα . Observe U
and V are open by the definition of the product topology, and U ∩ V = ∅ since U β and
Vβ are disjoint. Thus X is Hausdorff.
Solution, part c:
′
Leet X = ∏∞
i=1 X i be a countable product of second countable spaces. Let B i be the countable basis for each X i . Then let us define Bi = Bi′ ∪ {X i }. Note each Bi is still a countable
basis for X i .
Observe the countable product ∏∞
i=1 B i is countable. Now let B be the set of products
∞
B
,
with
each
B
∈
B
,
and
where
B i = X i for all but fintely many i. Clearly B ⊆ ∏∞
∏i=1 i
i
i
i=1 B i
and thus B is countable. Moreover, it can be seen that B forms a basis for the infinite
product topology since each Bi is a basis for X i .
3. Problem 3-8 (Solution by Erika Burr)
Let X denote the product of countably infinitely many copies of R (which is just the set
of all infinite sequences of real numbers) endowed with the box topology. Define a map
f ∶ R → X by f (x) = (x, x, x, . . .). Show that f is not continuous even though each of its
component functions is continuous.
Solution:
Let X = ∏n∈N X n where X n = R for all n ∈ N and endow X with the box topology. Observe
that each component function f n ∶ R → X n is continuous since f n = π n ○ f = idR and
each X n = R has the usual topology. Note that for each n ∈ N, the set U n = (− n1 , n1 ) is
open in X n . Then U = ∏n∈N U n is open in X. But observe that
f −1 (U) = {x ∈ R ∣ f (x) ∈ U}
= {x ∈ R ∣ x ∈ U n for all n ∈ N}
= {0}.
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Math F651: Homework 5
Solutions
Then f −1 (U) is not open in R and f is not a continuous map from R to X with the box
topology.
4. [Lee 3-9] (Solution by Max Heldman)
Let X be as in the preceding problem. Let X + ⊆ X be the subset consisting of sequences
of strictly positive real numbers, and let z denote the zero sequence. Then z is in the
closure of X + , but there is no sequence of elements of X + converging to z. Thus X is not
first countable, and therefore not metrizable.
Solution:
Let U ⊆ X be an open set containing z. Then z ∈ ∏n∈N (a n , b n ) for some basis set
∏n∈N (a n , b n ) ⊆ U. Thus 0 ∈ (a n , b n ) for each n ∈ N. Then b n > 0 for each n ∈ N, so
there exists c n such that a n < 0 < c n < b n for each n ∈ N. Then (c n ) ∈ U ∩ X + , and z is a
limit point for X + .
Let ( f k ) be a sequence in X + . Let N = Π n∈N (−1,
f n (n)
2 ) for
(−1, f k (k)
2 ).
0 ∈ (−1,
f n (n)
2 ).
Observe first that z ∈ N since
all n ∈ N (recall f n (n) > 0). Now let k ∈ N. Since f k (k) >
f k (k)
2 ,
f k (k) ∉
Thus f k ∉ N. Then ( f k ) does not converge to z, since N is an open set
containing z that ( f k ) never enters. Therefore there is no sequence in X + converging to z.
Then by the sequence lemma X is not first countable, which implies X is not metrizable.
5. Problem 3-10 (Solution by Former Student Tyler Knowles)
Prove Theorem 3.41 (the characteristic property of disjoint union spaces): Suppose that
(X α )α∈A is an indexed family of topological spaces, and Y is any topological space. A map
f ∶ ∐α∈A X α → Y is continuous if and only if its restriction to each X α is continuous. The
disjoint union topology is the unique topology on ∐α∈A X α with this property.
Solution:
Let ι β ∶ X β ↪ ∐α∈A X α . Observe that the following diagram commutes:
ιβ
/∐
Xβ H
α∈A X α
HH
HH
HH
f
f˜β HHH $
Y
First note that ι β is continuous by the definition of the disjoint union topology. Suppose f
is continuous. Then f˜β = f ○ ι β is a composition of continuous functions and is therefore
continuous. Conversely, suppose f˜β is continuous for each β. Let U ⊆ Y be open in Y.
Then f˜β−1 (U) is open in X β . Note that
−1
f˜β−1 (U) = ι −1
β ( f (U)),
whence f −1 (U) is open in ∐α∈A X α . We conclude f is continuous.
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Math F651: Homework 5
Solutions
Now let τ1 and τ2 be topologies on ∐α∈A X α which satisfy the characteristic property of
disjoint union spaces. Observe that the following diagram commutes:

X β Ks K
ιβ
/ (∐
α∈A X α , τ 1 )
KKK
KKK
id
ι β KKK
% (∐α∈A X α , τ1 )
Since τ1 satisfies the characteristic property and since id is continuous, it follows that
each ι β is continuous. Similarly, the maps ι ′β ∶ X β → (∐α∈A X α , τ2 ) are continuous. Now
consider id12 ∶ (∐α∈A X α , τ1 ) → (∐α∈A X α , τ2 ).

X β Ks K
ιβ
/ (∐
α∈A X α , τ 1 )
KKK
KKK
id12
KKK
ι ′β
% (∐α∈A X α , τ2 )
By the characteristic property applied to τ2 , since each ι ′β is continuous so is id12 . Similarly,
id21 = id−1
12 is continuous. We conclude τ1 = τ 2 , meaning the disjoint union topology is the
unique topology on ∐α∈A X α with the characteristic property of disjoint union spaces.
6. Problem 3-13(a) (Solution by Jody Gaines)
Suppose X and Y are topological spaces and f ∶ X → Y is a continuous map. Prove that
if f admits a continuous left inverse, then it is a topological embedding.
Solution:
Let f ′ be a continuous left inverse of f . Then for all x1 , x2 ∈ X such that f (x1 ) = f (x2 ),
we have that
x1 = f ′ ○ f (x1 ) = f ′ ○ f (x2 ) = x2 .
Thus f is injective. Consider the map g ∶ X → f (X) defined by x ↦ f (x). Also let
g ′ = f ′ ∣ f (X) . Note for all x ∈ X, g ′ ○ g(x) = f ′ ∣ f (X) ( f (x)) = f ′ ( f (x)) = x. Likewise, for
all f (a) ∈ f (X), g ○ g ′ ( f (a)) = f ○ f ′ ( f (a)) = f (a). Thus g is bijective, and g ′ is the
inverse function of g. Since f is continuous, then by Corollary 3.10 (b), g is continuous.
Moreover, since f ′ is continuous, then by Corollary 3.10 (a), g ′ is continuous. Thus g is
a homeomorphism. Recall f is injective and g is a homeomorphism of f onto its image.
Hence f is a topological embedding.
7. Exercise 3.61 (Solution by Erika Burr)
Prove that a continuous surjective map q ∶ X → Y is a quotient map if and only if it takes
saturated open subsets to open subsets, or saturated closed subsets to closed subsets.
Solution:
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Math F651: Homework 5
Solutions
Lemma 7.1: Let q ∶ X → Y be a quotient map. A subset A ⊆ Y is closed if and only if q−1 (A)
is closed in X.
Proof of Lemma ??. Note that A ⊆ Y is closed if and only if AC is open. Since q is a quotient
C
map, AC is open in Y if and only if q−1 (AC ) = (q−1 (A)) is open in X. Finally, note that
C
(q−1 (A)) is open in X if and only if q−1 (A) is closed.
Proof of Claim. Let q ∶ X → Y be a continuous surjective map. We saw in class that q is a
quotient map if and only if it takes saturated open subsets to open subsets. We will show
that q is a quotient map if and only if q takes saturated closed subsets to closed subsets.
Suppose that q is a quotient map and that V ⊆ X is a saturated closed set. Since V is saturated,
there exists W ⊆ Y such that q−1 (W) = V . Then q−1 (W) is closed in X and by Lemma ??,
W is closed in Y . Hence q takes saturated closed subsets to closed subsets.
Conversely, suppose that q takes saturated closed subsets to closed subsets. To show that q
is a quotient map, we must show that U ⊆ Y is open if and only if q−1 (U) ⊆ X is open. By
the continuity of q, we know that if U ⊆ Y is open then q−1 (U) is open. Consider U ⊆ Y
C
such that q−1 (U) is open. Then (q−1 (U)) = q−1 (U C ) is a saturated closed set. By Lemma
??, then we know that U C is closed in Y . Hence U is open and q is a quotient map.
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