Download LECTURE NOTES (WEEK 1), MATH 525 (SPRING

LECTURE NOTES (WEEK 1), MATH 525 (SPRING 2012)
CHARLES REZK
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I’ll start with a brief review of topological spaces.
Topological spaces. Definition of topological space. (X, TX ), where TX is a collection of subsets
of X (called open sets) such that
• ∅, X ∈ TX .
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• {Uα } ⊆ TX implies Uα ∈ TX .
• U, V ∈ TX implies U ∩ V ∈ TX .
Complements of open sets are closed sets. Sets can be both open and closed, or neither.
Example. Standard topology on Rn .
Example. Discrete topology on X.
Definition of continuous map. f : X → Y is continuous if V open in Y implies f −1 V open in X.
Example. Continuous maps Rm → Rn .
Example. Continuous maps X → Y with X discrete.
Constructions. Ways to build new spaces, characterized by universal properties.
Subspace. Given a space X and a subset A ⊆ X, the subspace topology on A is that given by
U ∈ TA if there exists V ∈ TX such that U = A ∩ V . Let j : A → X be the inclusion function.
Universal property of subspace. If T is a space, a function f : T → A is continuous if and only if
j ◦ f is continuous.
(A map k : Z → X which factors through a homeomorphism between Z and the subspace
k(Z) ⊆ X is an immersion.)
Quotient space. Let X be a space, let B = X/ ∼ be the set of equivalence classes for some
equivalence relation on X, and let q : X → B be the projection. The quotient topology on B is
that given by U ∈ TB if and only if q −1 U ∈ TX .
Universal property of quotient topology. If T is a space, a function g : B → T is continuous if and
only if f ◦ q is continuous.
Product.
Given a family {Xα }α∈I , the product is the space whose underlying set is the product
Q
X = Xα , and whose topology is the smallest one for which all sets p−1
α (Uα ) are open, where
Uα ⊆ Xα is open, and where pα : X → Xα is the projection map.
Universal property of product. If T is a space, a function f : T → X is continuous if and only if
pα ◦ f is continuous for all α.
Coproduct. Given
` a family {Xα }α∈I , the coproduct is the space whose underlying set is the
disjoint union X = Xα , and such that U ⊂ X is open if and only if U ∩ Xα is open in Xα for all
α. The inclusion maps iα : Xα → X are continuous.
Date: January 22, 2012.
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CHARLES REZK
Universal property of coproduct. If T is a space, a function g : X → T is continuous if and only if
g ◦ iα are continuous for all α.
Collapsing to a point. Given a space X and a subspace A ⊆ X, we define a space X/A as
the equivalence classes of an equivalence relation X q {∗}, where we identify a ∼ ∗ for all a ∈ A.
Topologize X/A as the quotient of X q {∗}, which is given the coproduct topology.
A continuous map f : X/A → T amounts to choosing a continuous map g : X → T and a point
t0 ∈ T such that g(a) = t0 for all a ∈ A.
Note that if A = ∅, then X/A = X q {∗}, and is not a quotient of X. We often write X+ for
this. One way to think of this, we always want X/A to have a basepoint.
Pushouts. The pushout of Y ← A → X. Note that X/A is an example of a pushout ∗ ← A → X.
F 20 Jan
Examples. Circle as interval with ends glued together. Wedge of two circles.
Möbius band.
Projective plane as Möbius band glued to disc.
Homeomorphism.
p with continuous inverse.
√ A homeomorphism is a continuous map
Example. x/ 1 − x2 : (−1, 1) → R, with inverse y/ 1 + y 2 : R → (−1, 1). (Not same as
continuous bijection, e.g., [0, 1) → S 1 .)
Connected and discrete. A space X is connected if the only subsets which are both open and
closed are ∅ and X.
Examples. R and I = [0, 1] are connected. R r {0} is not connected.
A space Y is discrete if every subset is open and closed. One way to characterize connectedness
is: X is connected if every continuous map f : X → Y to a discrete space Y is constant.
Compactness. A space X is compact if every open cover admits a finite subcover. The important
fact we need is the Bolzano-Weierstrass theorem: the compact subspaces of Rn are precisely
the closed and bounded subspaces.
Important fact. If f : X → Y is a continuous map and X is compact, then f (X), with the
subspace topology in Y , is a compact space.
This leads to the maximum principle.
Lesbegue number lemma. This is usually stated in terms of metric spaces. We need: if
X ⊆ Rn is compact, and {Uα } is an open cover of X, then there exists δ > 0 such that every subset
of diameter < δ is contained in the open cover. (Proof: consider all B (x; X) with x ∈ X which are
contained in some element of the cover, and use compactness of X.)
Example: maps out of intervals. Let I = [0, 1] be the unit interval. If Y is a space with open
cover {Uα }, and f : I → Y is continuous, then there exists N ≥ 1 such that f sends each subinterval
of the form [(a − 1)/N, a/N ], where a ∈ {1, . . . , N }, into one of the Uα .
Neighborhoods of compact sets of Rn .
Lemma 1.1. If K ⊆ U ⊆ Rn where K is a compact subspace
of Rn and U is open in Rn , then
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there exists an > 0 such that K ⊆ V ⊆ U , where V = x∈K B (x).
Proof. Consider all B2δ (x) such that x ∈ K and
S B2δ (x) ⊆ U . Since K is compact, we can choose
x1 , . . . , xn ∈ K and δ1 , . . . , δn such that K ⊆ Bδi (xi ) and B2δi (xi ) ⊆ U .
LECTURE NOTES (WEEK 1), MATH 525 (SPRING 2012)
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Let = min(δi ). If y ∈ V , then there exists x ∈ K and i ∈ {1, . . . , n} such that |y − x| < and
|x − xi | < , whence |y − xi | < 2 ≤ 2δi . Thus y ∈ B2δi (xi ) ⊆ U , V ⊆ U .
Spheres and disks. Let Dn = { x ∈ Rn | |x| ≤ 1 }, and let S n−1 = { x ∈ Rn | |x| = 1 }, given with
subspace topologies. Note that S n−1 is closed in Dn .
We want to construct a homeomorphism Dn /S n−1 ≈ S n . Since S n is defined as a subspace,
and Dn /S n−1 is defined as a quotient space, it makes most sense to start by building a map
f : Dn /S n−1 → S n .
Let X = Dn /S n−1 , and let q : Dn → X be the quotient map. As a set, X looks like (Dn −
S n−1 ) q {∗}. There is an evident continuous map f˜: Dn → S n ⊆ R × Rn = Rn+1 , given by
f˜(x) = (cos π|x|, (sin π|x|)/|x| · x)
and f˜(0) = (1, 0) (since (1/t) sin πt extends to a continuous function at t = 0). Since f˜(S n−1 ) =
{(−1, 0)}, it factors uniquely through a continuous bijection f : X → S n .
Claim. f is a homeomorphism.
We must show that f −1 is continuous; equivalently, that f takes open sets to open sets. It is
enough to have
Proposition 1.2. Let U ⊆ X = Dn /S n−1 , and let Ũ = q −1 U ⊆ Dn . Then U is open in X if and
only if (i) Ũ − S n−1 is open in Dn , and (ii) either Ũ ∩ S n−1 = ∅, or there exists > 0 such that
V ⊆ Ũ , where V = { x ∈ Dn | |x| > 1 − }.
Proof. It is clear that if Ũ satisfies (i) and (ii), then it is open, and therefore U is open. Conversely,
suppose U is open, and thus Ũ is open; clearly Ũ ∩ (Dn r S n−1 ) is open. If Ũ ∩ S n−1 6= ∅, then
S n−1 ⊆ Ũ . Thus, there exists suitable since S n−1 is compact.
It is clear that the restriction Dn r S n → S n r {(−1, 0)} of f is a homeomorphism, since yoeu
can describe its inverse explicitly. By the above proposition, it then suffices to show that f takes
q(V ) ⊂ Dn /S n−1 to an open set of S n , which is clear.
Department of Mathematics, University of Illinois at Urbana-Champaign, Urbana, IL
E-mail address: [email protected]