Download Problem set 1: Answers

Problem set 1: Answers
John Terilla
Fall 2014
Subspace and product topologies
Problem 1. Are the subspace and product topologies are consistent with each other?
Let {Xα }α∈A be a collection of topological spaces and let {Yα } be a collection of
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subsets; each Yα ⊂ Xα . There are two ways to put a topology on Y = α∈A Yα :
1. Take the subspace topology on each Yα , then form the product topology on Y .
2. Take the product topology on X, then equip Y ⊂ X with the subspace topology.
Is the outcome the same either way? If yes, prove it using only the universal properties.
Answer. Yes. To prove it, we’ll prove that the topology defined the first way satisfies
the universal property for the topology formed the second way. So, suppose that Z is a
topological space and f : Z → Y is a function. We will prove that f is continuous if
and only if i ◦ f : Z → X is continuous. Here’s a picture:
X
i◦f
Z
πα
i
f
Y
Xα
iα
ρα
Yα
So, suppose that i ◦ f : Z → X is continuous. Since for every α, the projection
πα : X → Xα is continuous, πα ◦ (i ◦ f ) : Z → Xα is continuous. The map
πα ◦ (i ◦ f ) = iα ◦ (ρα ◦ f ). By the universal property of the subspace topology on Yα ,
the map iα ◦ (ρα ◦ f ) being continuous implies ρα ◦ f : Z → Yα is continuous. So, we
have a collection of continuous maps: ρα ◦ f : Z → Yα so by the universal property of
the product topology on Y , the map f : Z → Y is continuous.
On the other hand, suppose that f : Z → Y is continuous. Because the projections
ρα : Y → Yα are all continuous, the compositions ρα ◦ f : Z → Yα are all continuous.
Since the inclusions iα : Yα → Xα are all continuous, the compotisions, iα ◦ ρα ◦ f :
Z → Xα are all continuous. Note the maps iα ◦ ρα ◦ f = πα ◦ (i ◦ f ), and the universal
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property of the product topology on X says that πα ◦ (i ◦ f ) being continuous for all α
implies the map i ◦ f : Z → X is continuous.
This proves that for every space Z and every function f : Z → Y , f is continuous
if and only if i ◦ f : Z → X is continuous. So, the two topologies agree.
Connectedness
The definitions of connected, path connected, locally connected and locally path connected are in Section 4 of May’s notes [1].
Definition 1. Define an arc in a space X to be an path I → X that is an embedding.
A space X is arc connected if every two points can be connected by an arc.
Prove (to yourself) that arc-connected ⇒ path-connected ⇒ connected. The following problem gives an example of a path connected space that is not arc connected.
Problem 2. Let X = {2, 3, 4, 5, . . .}. For any k ∈ X, let
Uk = {n ∈ X : n divides k}.
Check that the sets {Uk } form a basis for a topology on X and prove that it is path
connected and locally path connected but not arc-connected.
Answer. First, note that the {Uk } does define a base for a topology:
• The intersection of two basic open sets is a basic open set: Uj ∩ Uk = Ugcd(j,k)
• since k ∈ Uk , we have X = ∪k∈X Uk .
We’ll prove that X is path connected (hence connected) and locally path connected.
Let j, k ∈ X. The function p : [0, 1] → X defined by


if 0 ≤ t < 12

j
p(t) =
if t =
jk


k
if
1
2
1
2
<t≤1
is a path from j to k. To see that p is continuous, notice that if U is any open set
containing jk, then we have both j ∈ U and k ∈ U . So, p−1 (U ) = [0, 1], which
is open. Otherwise, U contains either j, k, both, or neither in which case p−1 (U ) =
1 −1
0, 2 , p (U ) = 21 , 1 , p−1 (U ) = 0, 12 ∪ 21 , 1 , or p−1 (U ) = ∅; open in every
case. Therefore, X is path connected.
To see that X is locally path connected, let Un be an open set and let j, k ∈ Un .
The function p : [0, 1] → X defined by

1


j if 0 ≤ t < 2
p(t) = n if t = 21


k if 1 < t ≤ 1
2
2
is a path from j to k entirely contained in Un . To see that p is continuous, notice that
if U is any open set containing n, then we have both j ∈ U and k ∈ U since j, k ∈ Un
and Un is the smallest open set containing n. So, p−1 (U ) = [0, 1], which is open.
Otherwise, U contains either j, k, both, or neither in which case p−1 (U ) = 0, 21 ,
p−1 (U ) = 12 , 1 , p−1 (U ) = 0, 21 ∪ 12 , 1 , or p−1 (U ) = ∅; open in every case.
Finally, note that X cannot be arc-connected since the cardinality of [0, 1] is strictly
larger than the cardinality of X.
Problem 3. Proposition 4.6 in May [1] states that the product of connected or path
connected spaces is connected or path connected. Pick one and prove it.
Answer. I’ll prove both. One is quite a bit easier.
Theorem 1. {Xα }α∈Λ is a collection of path connected topological spaces. The prodQ
uct X = α∈Λ Xα is path connected.
Proof. Suppose that Xα is path connected for every α ∈ Λ. Let a, b ∈ X. Since
each Xα is path connected, there exists a path pα : [0, 1] → Xα connecting aα to bα .
By the universal property of the product topology, there exists a continuous function
p : [0, 1] → X so that πα ◦ p = pα . This function p is a path from a to b.
Theorem 2. If {Xα }α∈Λ is a collection of connected topological spaces then the prodQ
uct X = α∈Λ Xα is connected.
Proof. Suppose that Xα is connected for every α ∈ Λ. Let a ∈ X be any point and
let F be the largest connected set containing a. We will prove that F = X. To do so,
we will show that every basic open set in X intersects F . This will prove that F = X.
Since F is a connected component, it is closed, so F = F and we will have proved
Q
that F = X. So, let U = α Uα be an arbitrary basic open set. We have Uα = Xα for
all α ∈ Λ except for α1 , . . . , αn , where Uαk may be a proper subset of Xαk . Choose
a point bαk ∈ Uαk for k = 1, . . . , n and define a sequence of sets E1 , E2 , . . . , En (see
Figure 1 below)
E1 : = {x ∈ X : xα1 ∈ Xα1 , xαk = aαk for k > 1}
E2 : = {x ∈ X : xα1 = bα1 , xα2 ∈ Xα2 , xαk = aαk for k > 2}
E3 : = {x ∈ X : xα1 = bα1 , xα2 = bα2 , xα3 ∈ Xα3 , xαk = aαk for k > 3}
..
.
En : = {x ∈ X : xα1 = bα1 , . . . , xαn−1 = bαn−1 , xαn ∈ Xαn }.
We make a few observations:
• For each k = 1, . . . , n, Ek ≃ Xαk , hence is connected.
• For each k = 1, . . . , n, Ek ∩ Ek+1 6= ∅.
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• a ∈ E1 .
• En ∩ U 6= ∅.
The first two observations imply that E := E1 ∪ E2 ∪ · · · ∪ En is connected. The third
implies that a ∈ E ⇒ E ⊂ F . The fourth implies that E ∩ U 6= ∅ ⇒ F ∩ U 6= ∅,
completing the proof.
Figure 1: A Picture of Ek . Here, Ek consists of all functions whose value at
α1 , . . . , αk−1 match bα1 , . . . , bαk−1 , whose value at αk is arbitrary, and whose value
at αk+1 , . . . , αn matches aαk+1 , . . . , aαn .
X α1
X αk
X α2
X αn
a
b α2
b α1
α1
α2
αk
αn Λ
Here, the four main ideas used in the proof are isolated.
Proposition 1. Let X be a topological space and let F be a set. If F ∩ U 6= ∅ for every
basic open set U , then F = X. Such sets whose closure equals the whole space are
sometimes called dense.
Proposition 2. Let X be a topological space and let a ∈ X be an arbitrary point.
Then there is a largest connected set F containing a. That is, if E is any connected set
containing a, then E ⊂ F . Furthermore, F is closed. Such largest connected sets are
called connected components and they form a partition X.
Proposition 3. Let {Xα }α∈Λ be an arbitrary collection of topological spaces. Let
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a = {aα } ∈ α∈Λ Xα be a fixed point in the product space and let β ∈ Λ be a fixed
index. Then the set
Y
Y := {x ∈
Xα : xα = aα for all α 6= β and xβ ∈ Xβ is arbitrary}
α∈Λ
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is homeomorphic to Xβ .
Proposition 4. Let X be a topological space and let {En }n∈N be a sequence of conS
nected subsets of X. If Ek ∩ Ek+1 6= ∅ for each k ∈ N, then E := n∈N En is
connected.
Problem 4. Let X be a space and f : X → S be surjective. Suppose that S is
connected in the quotient topology and that the all the fibers f −1 (s) are connected.
Prove that X is connected.
Answer. Suppose that X is not connected. That is, suppose there exist disjoint,
nonempty open sets U, V with X = U ∪V . For any s ∈ S, the set f −1 (s) is connected,
hence f −1 (s) ⊂ U or f −1 (s) ⊂ V . Let
U ′ = {s ∈ S : f −1 (s) ⊂ U } and V ′ = {s ∈ S : f −1 (s) ⊂ V }
Note that S ⊂ U ′ ∪ V ′ and that U ′ ∩ V ′ = ∅. Therefore, if we prove that U ′ and V ′
are open in S, we will have proved that if all the fibers f −1 (s) are connected and X is
disconnected, then S is disconnected. But, f −1 (U ′ ) = U and f −1 (V ′ ) = V so by the
definition of the quotient topology on S, U ′ and V ′ are open.
Compactness
See Section 5 of [1] for definitions.
Problem 5. Let τ be a topology on [0, 1] different from the usual one. Prove that if τ
is finer than the usual topology, then ([0, 1], τ ) can’t be compact and that if τ is coarser
than the usual topology, then ([0, 1], τ ) can’t be Hausdorff.
Answer. Suppose that τ is a topology on [0, 1] finer than the usual one. Then id :
(R, τ ) → (R, τusual ) is continuous. If id : (R, τ ) were compact, then id : (R, τ ) →
(R, τusual ) would be a coninuous a continuous bijection from a compact space onto
a Hausdorff space, so id would be a homeomorphism, contradicting the fact that τ is
finer than τusual .
On the other hand, if τ is coarser than the usual topology, id : (R, τusual ) → (R, τ )
is continuous. So, if (R, τ ) were Hausdorff, id : (R, τusual ) → (R, τ ) would be a
continuous bijection from a compact space onto a Hausdorff space, so id would be a
homeomorphism, contradicting the fact that τ is coarser than τusual .
Categories and functors
Definition. Let C and D be categories. A functor F : C → D consists of assignments:
• For each object X ∈ ob(C), there is an object F (X) ∈ ob(D),
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• For each pair of objects X, Y ∈ ob(C) and each morphism σ : X → Y in
mor(X, Y ), there is a morphism F (σ) : F (X) → F (Y ) in mor(F (X), F (Y )).
satisfying the following conditions:
• Compatibility with identites. For each object X ∈ ob(C), F (idX ) = idF (X)
• Compatibility with composition. For all objects X, Y, Z ∈ ob(C) and all morphisms f : X → Y and G : Y → Z, F (g ◦ f ) = F (g) ◦ F (f ).
Problem 6. For objects, X, Y ∈ ob(C), let X ≃ Y denote the object X is isomorphic
to the object Y . Let iso(X, Y ) be the set of isomorphisms X → Y and let aut(X) =
iso(X, X). Note that for any X, aut(X) is a group.
(a) Prove that if F : C → D is a functor and X ≃ Y in C, then F (X) ≃ F (Y ) in D.
Answer. Suppose X ≃ Y. Then there exists a morphism α : X → Y and β :
Y → X so that αβ = idY and βα = idX . Applying the functor F yields F (α) :
F (X) → F (Y ) and F (β) : F (Y ) → F (X). Applying F to αβ = idY yields
F (αβ) = F (id(Y )) ⇒ F (α)F (β) = idF (Y ) since F respects composition and
identities. Also, F (β)F (α) = idF (X) . This proves that F (X) ≃ F (Y ).
(b) Prove that if X ≃ Y as objects in a category C then aut(X) ≃ aut(Y ) as
groups.
Answer. Suppose X ≃ Y . Choose an isomorphism α : X → Y and β : Y →
X so that αβ = idY and βα = idX . The map
aut(X) → aut(Y )
f 7→ αf β.
defines a group isomorphism. It’s a homomorphism since
f g 7→ αf β = αf βαgβ
It’s an isomorphism since g 7→ βgα defines the inverse group homomorphism
from aut(Y ) → aut(X).
(c) Does X 7→ aut(X) define a functor from C → Groups?
Answer. This question isn’t well defined unless a map on morphisms is specified. If α : X → Y is an isomorphism, then the map aut(X) → aut(Y )
described in the previous part is a good map. However, there’s no obvious way
to define F (α) if α is a not an isomorphism.
Problem 7. For any topological space X, define a relation ≈ on X by x ≈ y if there
is a path from x to y.
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(a) Show that ≈ is an equivalence relation. The equivalence classes of ≈ are called
path components of X.
Answer. The constant function p : I → X defined by p(t) = x for all t ∈ I is a
path from x to x and shows that ≈ is reflexive.
If p : I → X is a path from x to y, then p−1 : I → X defined by p−1 (t) =
p(1 − t) is a path from y to x showing that ≈ is symmetric.
To see that ≈ is transitive, let p : I → X be a path from x to y and q : I → X
be a path from y to z. Then p · q : I → X defined by
(
p(2t)
if 0 ≤ t ≤ 21
p·q =
q(2t − 1) if 21 ≤ t ≤ 1.
is a path from x to z.
(b) Show that if f : X → Y is continuous, then f induces a well defined map from
the path components of X to the path components of Y .
Answer. The assignment
the path component of x 7→ the path component of f (x)
is well defined, for if x ≈ y, then there’s a path p : I → X connecting x
and y. Then f ◦ p : I → Y is a path connected f (x) and f (y). That is,
x ≈ y ⇒ f (x) ≈ f (y).
(c) Show that the assignment
X 7→ π0 (X) := the set of path components of X
(f : X → Y ) 7→ (π0 (f ) : π0 (X) → π0 (Y )) := the induced map on path components
defines a functor π0 : T op → Sets from the category of topological spaces to
the category of sets.
Answer. We need to check compatibility with identity morphisms and composition. Let X be a space and consider the identity function idX : X → X.
Then,
π0 (idX )(the path component of x) = the path component of idX (x)
= the path component of x.
So, π0 (idX ) is the identity function on path components; i.e., π0 (idX ) = idπ0 (X) .
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To check compatibility with composition, Let X, Y, Z be spaces and consider
continuous functions f : X → Y and g : Y → Z.
π0 (g) ◦ π0 (f )(the path component of x) = π0 (g)(the path component of f (x))
= the path component of g(f (x))
= π0 (g ◦ f )(the path component of x).
That is, π0 (g) ◦ π0 (f ) = π0 (g ◦ f ) as needed.
References
[1] J.P. May.
An outline summary of basic point
http://www.math.uchicago.edu/ may/MISC/Topology.pdf.
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set
topology.