Download Math 535 - General Topology Fall 2012 Homework 9 Solutions

Math 535 - General Topology
Fall 2012
Homework 9 Solutions
Problem 1.
a. Let X be a topological space with finitely many connected components. Show that each
connected component is open in X.
Solution. Let X = X1 t . . . t Xn be the partition of X into connected components. Since
the connected components of X are always closed in X, the complement of any component
G
X \ Xk =
Xi
1≤i≤n
i6=k
is a finite union of closed subsets, hence closed in X. Therefore Xk is open in X.
b. Let X be a topological
of X such that X is
`
F space and {Ui }i∈I a collection of open subsets
the disjoint union X = i∈I Ui . Show that X is the coproduct X = i∈I Ui .
In particular, this conclusion applies to the situation in part (a).
Solution. We want to show that a subset U ⊆ X is open in X if and only if U ∩ Ui is open
in Ui for all i ∈ I.
(⇒) If U is open in X, then U ∩ Ui is open in Ui (subspace topology).
(⇐) Assume U ∩ Ui is open in Ui . Then U ∩ Ui is open in X since Ui is open in X. Therefore
G
U = (U ∩ Ui )
i∈I
is a union of open subsets of X, thus open in X.
c. Find an example of metrizable space X with a connected component C ⊂ X which is not
open in X.
Solution. (c.f. HW 8 #6b) Take X = {0} ∪ { n1 | n ∈ N} ⊂ R viewed as a subspace of R.
The singleton {0} is a connected component of X. Indeed, every other point n1 6= 0 is isolated
in X and thus is its own connected component. In particular, each n1 is not in the connected
component of 0.
However, the singleton {0} is not open in X. Indeed, any open ball around 0 contains infinitely
many points of the form n1 .
1
Problem 2. Show that the n-dimensional sphere S n is path-connected (for n ≥ 1).
Solution. (c.f. HW 4 #7) Write the standard unit sphere
S n = {x ∈ Rn+1 | kxk = 1}
as a union S n = Dun ∪ Dln of the upper and lower hemispheres
Dun = {x ∈ Rn+1 | kxk = 1 and xn+1 ≥ 0}
Dln = {x ∈ Rn+1 | kxk = 1 and xn+1 ≤ 0}.
Note that the intersection Dun ∩ Dln ∼
= S n−1 is non-empty (since n ≥ 1): it is the equator of S n .
Therefore it suffices to show that both hemispheres are path-connected.
Both hemispheres are homeomorphic to the standard unit disc
Dn := {x ∈ Rn | kxk ≤ 1}.
Writing Rn+1 ∼
= Rn × R, consider the map fU : Dn → Dun sending the disc to the upper
hemisphere:
p
fu (x) = (x, 1 − kxk2 ).
Then fu is a homeomorphism onto the upper hemisphere, with inverse the projection pRn : Dun →
Dn onto the first n coordinates. Likewise, the lower hemisphere Dln is homeomorphic to the
standard unit disc Dn . Therefore, it suffices to show that Dn is path-connected.
Dn ⊂ Rn is in fact a convex subset, hence path-connected via straight line segments. Indeed,
let x, y ∈ Dn and consider the straight line segment from x to y, given by
γ(t) = (1 − t)x + ty , t ∈ [0, 1].
For all t ∈ [0, 1], the norm of γ(t) is
kγ(t)k = k(1 − t)x + tyk
≤ k(1 − t)xk + ktyk
= |1 − t|kxk + |t|kyk
≤ |1 − t| + |t|
= (1 − t) + t
=1
which ensures γ(t) ∈ Dn .
2
Alternate solution. (c.f. HW 9 #5) Consider the map π : Rn+1 \ {0} S n defined by
π(x) =
x
.
kxk
Then π is continuous, since kxk is a continuous function of x which satisfies kxk > 0 for all
x ∈ Rn+1 \ {0}.
Moreover π is surjective, since we have π(x) = x for all x ∈ S n . Therefore it suffices to show
that Rn+1 \ {0} is path-connected to ensure that S n = π(Rn+1 \ {0}) is path-connected.
Let x, y ∈ Rn+1 \ {0} and consider the straight line segment γ in Rn+1 from x to y.
If γ does not go through the origin 0, then it is a path in Rn+1 \ {0} from x to y.
If γ does go through the origin, then pick a point z ∈ Rn+1 \{0} which is not on the line through
x, 0, and y. This is possible since the dimension is n + 1 ≥ 2. Then the unique line through x
and z does not contain 0, and likewise the unique line through z and y does not contain 0.
Let α be the straight line segment from x to z and β be the straight line segment from z to
y. Then α and β never go through 0, so that they are paths in Rn+1 \ {0}. Therefore their
concatenation α ∗ β is a path in Rn+1 \ {0} from x to y.
3
Problem 3. Consider the “infinite ladder” X ⊂ R2 consisting of two vertical “sides” {0} × R
and {1} × R along with horizontal “rungs” [0, 1] × { n1 } for all n ∈ N as well as [0, 1] × {0}. In
other words, X is the union
1
X = ({0, 1} × R) ∪ [0, 1] × { | n ∈ N} ∪ {0}
⊂ R2 .
n
Show that X is path-connected, but not locally path-connected.
Solution. X is path-connected. The left side {0} × R ∼
= R is path-connected, as is each
∼
“rung” [0, 1] × {t} = [0, 1]. Therefore the “comb”
1
Y = ({0} × R) ∪ [0, 1] × { | n ∈ N} ∪ {0}
n
is path-connected, since it is a union of path-connected subspaces that all intersect one of them,
namely the left side {0} × R. It follows that X is path-connected, since it is the non-disjoint
union
X = Y ∪ ({1} × R)
where both Y and the right side {1} × R are path-connected.
X is not locally path-connected. Let us show that X is not locally path-connected at the
point x = ( 21 , 0) ∈ X. Consider the neighborhood
1
U := (0, 1) × { | n ∈ N} ∪ {0} ⊂ X
n
of x. Note that the path components of U are the individual “rungs” (0, 1) × {t}.
Let V ⊆ U be a neighborhood of x, so that V contains an open ball Br (x) for some radius
r > 0. Pick distinct integers k1 , k2 ∈ N large enough so that k1i < r. Then the two points
ai := ( 12 , k1i ) satisfy
1 1
( , ) ∈ Br (x) ⊆ V ⊆ U.
2 ki
Because their vertical coordinates k11 6= k12 are distinct, the points a1 and a2 are in different
path components of U , i.e. there is no path in U from a1 to a2 . In particular, there is no path
in V from a1 to a2 , hence V is not path-connected.
4
Problem 4. Let (X, d) be a metric space. Given points x, y ∈ X and > 0, an -chain from
x to y in X is a finite sequence of points
x = z0 , z1 , . . . , zn−1 , zn = y
in X such that the distance from one to the next is less than , i.e. d(zi−1 , zi ) < for 1 ≤ i ≤ n.
Show that if X is connected, then for all > 0, any two points x, y ∈ X can be connected by
an -chain.
Solution. Let > 0 and let x ∈ X. Let U ⊆ X be the subset of all points that can be
reached by an -chain starting at x. Note x ∈ U , because of the trivial chain x = z0 (or the
constant chain x = z0 = z1 in case the trivial chain is not allowed).
U is open. Let y ∈ U , i.e. there is an -chain x = z0 , z1 , . . . , zn = y. For any y 0 ∈ B (y), the
list z0 , z1 , . . . , zn , zn+1 := y 0 is still an -chain in X, by the condition
d(zn , zn+1 ) = d(y, y 0 ) < .
This -chain from z0 = x to zn+1 = y 0 proves y 0 ∈ U , and thus B (y) ⊆ U .
U is closed. Let y ∈ U c , i.e. there is no -chain from x to y For any y 0 ∈ B (y), there
is no -chain from x to y 0 . If there were such an -chain z0 , z1 , . . . , zn = y 0 , then the list
z0 , z1 , . . . , zn , zn+1 . . . = y would still be an -chain in X, by the condition
d(zn , zn+1 ) = d(y 0 , y) < .
This proves y 0 ∈ U c , and thus B (y) ⊆ U c , so that U c is open.
Since X is connected and U is a non-empty clopen subset of X, we conclude U = X, i.e. all
points of X are connected to x by an -chain.
5
Problem 5. Let p ∈ Rn . Show that Rn \{p} is homotopy equivalent to the (n−1)-dimensional
sphere S n−1 .
Solution. Translation x 7→ x + p is a homeomorphism Rn \ {0} ∼
= Rn \ {p}, with inverse the
translation x 7→ x − p. Therefore it suffices to show the statement for p = 0, the origin in Rn .
Consider the map π : Rn \ {0} → S n−1 defined by
π(x) =
x
.
kxk
Then π is continuous, since kxk is a continuous function of x which satisfies kxk > 0 for all
x ∈ Rn \ {0}.
We claim that π is a homotopy equivalence, with homotopy inverse the inclusion i : S n−1 ,→ Rn .
For all x ∈ S n−1 , we have
π ◦ i(x) = π(x) =
x
x
= =x
kxk
1
which proves π ◦ i = idS n−1 .
For the other composite i ◦ π, consider the map
H : (Rn \ {0}) × [0, 1] → Rn \ {0}
defined by
H(x, t) = (1 − t)x + t
x
x
= ((1 − t)kxk + t)
kxk
kxk
which is well defined, i.e. never takes the value 0. Indeed, for all x ∈ Rn \ {0} and t ∈ [0, 1],
consider the equivalent conditions
((1 − t)kxk + t)
x
=0
kxk
⇔ (1 − t)kxk + t = 0 since x 6= 0
⇔ (1 − t)kxk = 0 and t = 0 since both terms are ≥ 0
⇔ t = 1 and t = 0
which never holds.
Moreover, H is continuous since sums and products of continuous functions are continuous, and
kxk is continuous and never zero on Rn \ {0}.
At t = 0, we have H(x, 0) = x, whereas at t = 1, we have
homotopy between idRn \{0} and i ◦ π.
6
x
kxk
= (i ◦ π)(x). Therefore H is a
Problem 6. Let X be a topological space and denote by π0 (X) the set of path components
of X.
a. Show that any continuous map f : X → Y induces a well-defined function
π0 (f ) : π0 (X) → π0 (Y ).
Solution. Let [x] ⊆ X denote the path component of a point x ∈ X. The induced function
π0 (f ) : π0 (X) → π0 (Y ) is given by
π0 (f )[x] = [f (x)].
To check that this is well defined, let x, x0 ∈ [x] be two representatives, so that there is a
path γ : [0, 1] → X from x to x0 . Then f ◦ γ : [0, 1] → Y is a path from f (γ(0)) = f (x) to
f (γ(1)) = f (x0 ). Thus f (x) and f (x0 ) are in the same path component of Y , i.e.
[f (x)] = [f (x0 )] ∈ π0 (Y )
and the formula for π0 (f ) is well defined.
b. Show that the induced function π0 (f ) only depends on the homotopy class of f . In other
words, if f ' f 0 are homotopic maps, then π0 (f ) = π0 (f 0 ).
Solution. Let H : X × [0, 1] → Y be a homotopy from f to f 0 . We want to show that for any
[x] ∈ π0 (X), the equality π0 (f )[x] = π0 (f 0 )[x] holds. This can be rewritten as [f (x)] = [f 0 (x)],
i.e. there is a path in Y from f (x) to f (x0 ).
Consider the path γ : [0, 1] → Y defined by γ(t) = H(x, t), which is indeed continuous since H
is. Then the endpoints of γ are γ(0) = H(x, 0) = f (x) and γ(1) = H(x, 1) = f 0 (x).
'
'
c. Show that a homotopy equivalence f : X −
→ Y induces a bijection π0 (f ) : π0 (X) −
→ π0 (Y ).
Solution. Let g : Y → X be a homotopy inverse of f , i.e. it satisfies g ◦ f ' idX and
f ◦ g ' idY . By part (b), we have
π0 (g ◦ f ) = π0 (idX )
π0 (f ◦ g) = π0 (idY ).
Moreover, π0 preserves composition:
π0 (g ◦ f )[x] = [g(f (x))] = π0 (g)[f (x)] = π0 (g) ◦ π0 (f )[x]
as well as identities:
π0 (idX )[x] = [idX (x)] = [x]
from which we obtain
π0 (g) ◦ π0 (f ) = π0 (g ◦ f ) = π0 (idX ) = idπ0 (X)
π0 (f ) ◦ π0 (g) = π0 (f ◦ g) = π0 (idY ) = idπ0 (Y ) .
Therefore π0 (f ) : π0 (X) → π0 (Y ) is an invertible function, or equivalently, a bijection.
7
Remark. This proves in particular that path-connectedness is a homotopy invariant. Given
homotopy equivalent spaces X and Y , then X is path-connected if and only if Y is pathconnected.
Remark. We have shown that π0 : Top → Set is a functor.
8