Download Solutions to Homework 1

Solutions to Homework 1
1. (a) Let X and Y be sets and f : X → Y a surjection. Show that for
any other set Z, a map g : X → Z descends to Y —that is, there
is a map h : Y → Z with g = h ◦ f —if and only if it is constant
on the fibres of f .
Solution: If g = h ◦ f and x, x′ ∈ X lie in the same fibre of f —
that is, f (x) = f (x′ )—then g(x) = h(f (x)) = h(f (x′ )) = g(x′ ).
Conversely, suppose that g is constant on the fibres of f .
Given a point y ∈ Y , choose a point x ∈ X with f (x) = y, which
is possible since f is surjective, and define h(y) = g(x). Then h
is well-defined, for if x′ ∈ X is another point with f (x′ ) = y then
g(x′ ) = g(x) by hypothesis.
A slightly more sophisticated proof of the converse is the
following. Since f is surjective, there is a map of sets s : Y → X
with f ◦ s = 1Y . (Such a map is called a “section” of f . The
fact that every surjection has a section is special to the category
of sets: for example, if X and Y are topological spaces and f is
continuous, there may not be a continuous s with this property.)
Define h = g ◦ s, so h ◦ f = g ◦ s ◦ f = g ◦ 1Y = g.
(b) Now let X and Y be spaces and f : X → Y a quotient map.
Show that for any other space Z, a continuous map g : X → Z
descends to Y if and only if it is constant on the fibres of f .
Solution: We need only show that if g is constant on the fibres
of f then the map h constructed in part (a) is continuous. Let
U ⊂ Z be open. Then f −1 (h−1 (U )) = g−1 (U ) is open since g is
continuous, so h−1 (U ) is open since f is a quotient map.
2. (a) For each of the following, indicate whether you are familiar with
the fact, and familiar with the proof. (In case you weren’t familiar
with the proofs, here they are.)
i. If X is compact and F ⊂ X is closed then F is compact.
Proof: Let {Ui : i ∈ I} be an open cover of the space F .1
For each i ∈ I, choose a Vi ⊂ X with Ui = Vi ∩ F . Observe
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Do not assume that the index set I is countable.
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that {Vi : i ∈ I} ∪ {X \ F } is an open cover of X. Since X
is compact, there are i1 , . . . , in ∈ I with
X = Vi1 ∪ · · · ∪ Vin ∪ (X \ F ).
Intersect this with F to get
F = Ui1 ∪ · · · ∪ Uin ∪ ∅,
that is, {Ui1 , . . . , Uin } is a finite subcover.
ii. If X is compact and f : X → Y is continuous then the image
f (X) ⊂ Y is compact.
Proof: Let {Ui : i ∈ I} be an open cover of Y . Then
{f −1 (Ui ) : i ∈ I} is an open cover of X. Since X is compact,
there are i1 , . . . , ik such that
X = f −1 (Ui1 ) ∪ · · · ∪ f −1 (Uin ).
Apply f to this to get
f (X) = f (f −1 (Ui1 )) ∪ · · · ∪ f (f −1 (Uin ))
⊂ Ui 1 ∪ · · · ∪ Ui n .
iii. If Y is Hausdorff and K ⊂ Y is compact then K is closed.
Proof: First we show that for every y ∈
/ K there are open
sets U ⊃ K and V ∋ y with U ∩V = ∅. Since Y is Hausdorff,
for each k ∈ K there are open sets Uk ∋ k and Vk ∋ y with
Uk ∩ Vk = ∅. Then {Uk : k ∈ K} is an open cover of K, for
we have
[
[
K=
{k} ⊂
Uk .
k∈K
k∈K
Since K is compact, there are k1 , . . . , kn ∈ K such that K ⊂
Uk1 ∪· · ·∪Ukn . Now U = Uk1 ∪· · ·∪Ukn and V = Vk1 ∩· · · Vkn
satisfy U ⊃ K, V ∋ y, and U ∩ V = ∅, as claimed.
Now we show that Y \ K is open. We have seen that for
each y ∈
/ K there are Uy ⊃ K and Vy ∋ y with Uy ∩ Vy = ∅.
Then we have
[
[
Y \K =
{y} ⊂
Vy ⊂ Y \ K,
y ∈K
/
S
y ∈K
/
so Y \ K = y∈K
/ Vy , which is a union of open sets, hence is
open.
Note that the hypothesis that Y is Hausdorff is necessary: if we take Y = {0, 1} with the indiscrete topology and
K = {0} then K is compact, but not closed in Y .
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(b) Suppose that X is compact, Y is Hausdorff, and f : X → Y is a
continuous surjection. Show that f is a quotient map. You may
quote any of the facts from part (a).
Solution: We will show that F ⊂ Y is closed if and only if
f −1 (F ) is closed. If F ⊂ Y is closed then f −1 (F ) is closed since
f is continuous. Conversely, suppose that f −1 (F ) is closed. Since
X is compact, f −1 (F ) is compact, so f (f −1 (F )) is compact. Since
f is surjective, F = f (f −1 (F )). Thus F is a compact subspace
of a Hausdorff space, hence is closed.
3. I claim that (S 1 × I)/(S 1 × {0}) is homeomorphic to D 2 . (I should
have mentioned that I = [0, 1] ⊂ R.)
(a) Draw a picture.
(b) Argue that it is enough to write down a continuous surjection
S 1 ×I → D 2 which is constant on S 1 ×{0} and injective otherwise.
Solution: Let f : S 1 × I → D 2 be such a map, and g : S 1 × I →
(S 1 ×I)/(S 1 ×{0}) the natural map. The former is a quotient map
because S 1 × I is compact and D 2 is Hausdorff, and the latter is
a quotient map by definition. The fibres of f and g are the same:
S 1 × {0} and the single points of S 1 × (0, 1]. Thus we see that
the continuous map h : D 2 → (S 1 × I)/(S 1 × {0}) constructed
in problem 1(b) is a bijection. To show that h−1 is continuous,
we must show that if U ⊂ D 2 is open then h(U ) is open. If
U ⊂ D 2 is open then f −1 (U ) is open since f is continuous, so
h(U ) = g(f −1 (U )) is open since g is a quotient map.
(c) Write down such a map.
Solution: We regard S 1 and D 2 as subset of R2 in the usual way.
The map we want is
S 1 × I → D2
(x, t) 7→ tx.
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