Download Applied Topology, Fall 2016 1 Topological Spaces

Applied Topology, Fall 2016
The Basics of Point Set Topology
Slightly rearranged, but mostly copy-pasted from
• Hatcher’s Notes on Introductory Point-Set Topology,
• Renzo’s Math 490 Introduction to Topology,
• Gunnar Carlsson’s Topological Pattern Recognition for Point Cloud Data.
One way to describe the subject of topology is to say that it is qualitative geometry.
The idea is that if one geometric object can be continuously transformed into another,
then the two objects are to be viewed as being topologically the same. For example, a
circle and a square are topologically equivalent. Physically, a rubber band can be stretched
into the form of either a circle or a square, as well as many other shapes which are also
viewed as being topologically equivalent. On the other hand, a figure eight curve formed
by two circles touching at a point is to be regarded as topologically distinct from a circle
or square. A qualitative property that distinguishes the circle from the figure eight is the
number of connected pieces that remain when a single point is removed: when a point
is removed from a circle what remains is a single arc, whereas for a figure eight if one
removes the point of contact of its two circles, what remains is two separate arcs, two
separate pieces. The term used to describe two geometric objects that are topologically
equivalent is homeomorphic. One of the basic problems of topology is to determine when
two given geometric objects are homeomorphic. This can be quite difficult in general.
Our first goal will be to define exactly what the ‘geometric objects’ are that one studies
in topology. These are called topological spaces. The definition turns out to be extremely
general, so that many objects that are topological spaces are not very geometric at all.
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Topological Spaces
The definitions of ‘metric space’ and ‘topological space’ were developed in the early 1900’s,
largely through the work of Maurice Frechet (for metric spaces) and Felix Hausdorff (for
topological spaces). The main impetus for this work was to provide a framework in
which to discuss continuous functions, with the goal of examining their attributes more
thoroughly and extending the concept beyond the realm of calculus.
Definition 1.1. A topological space is a pair (X, T ), where X is a set and T a collection
of subsets of X, such that:
(T1) Both ∅ and X are in T .
(T2) The union of any collection of sets in T is in T .
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(T3) The intersection of any finite collection of sets in T is in T .
The collection T is called a topology on X and the elements of T are called open sets
with respect to T .
It is always possible to construct at least two topologies on every set X by choosing
the collection T of open sets to be as large as possible or as small as possible:
• The collection Tdiscrete of all subsets of X defines a topology on X called the discrete
topology;
• If we let Ttrivial consist of just X itself and ∅, this defines the trivial topology.
These examples illustrate how one can have two topologies T and T 0 on a set X, with
every set that is open in the T topology is also open in the T 0 topology, so T ⊂ T 0 . In
this situation we say that the topology T 0 is finer than T and that T is coarser than T 0 .
Thus the discrete topology on X is finer than any other topology and the trivial topology
is coarser than any other topology. Of course, given two topologies on a set X, it need
not be true that either is finer or coarser than the other.
The spaces we work with often have very complex open sets. So once we define a
structure on a set, we often try to understand what the ‘manageable’ data is needed to
specify the structure. The notion of a topological basis will help us to describe different
topologies more systematically.
1.1
Basis of a Topology
Definition 1.2. Let X be a set. A topological basis on X is a collection B of subsets of
X such that
(B1) For every x ∈ X, there is an element B ∈ B such that x ∈ B.
(B2) If x ∈ B1 ∩B2 where B1 , B2 are in B, then there is B3 in B such that x ∈ B3 ⊂ B1 ∩ B2 .
Lemma 1.3. (Generating a topology). Let B be a topological basis on X. Define TB to
be the collection of subsets U ⊂ X satistying
(G1) For every x ∈ U , there is B ∈ B such that x ∈ B ⊂ U .
Then TB defines a topology on X. The ∅ trivially satisfies the condition, so that ∅ ∈ TB .
Remark. TB is called the topology generated by a basis B. On the other hand, if (X, T )
is a topological space and B is a topological basis such that TB = T , then we say B is a
basis of T . Note that T itself is a basis of the topology T . So there is always a basis for
a given topology.
Proof. We need to check the three axioms:
(T1) ∅ ∈ TB as we assumed X ∈ TB by (B1).
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(T2) Consider a collection of subsets Ui ∈ TB , i ∈ I. We need to show
[
U :=
Ui ∈ TB .
i∈I
By the definition of the union, for each x ∈ U , there is Ui such that x ∈ Ui . Since
Ui ∈ TB , there is B ∈ B such that x ∈ B ⊂ Ui . For this B ∈ B, x ∈ B ⊂ U . Thus
U ∈ TB .
(T3) Consider U1 , U2 ∈ TB . We need to show that
U := U1 ∩ U2 ∈ TB .
If the intersection is empty, there is nothing to show. So let x ∈ U . By the
definition of the intersection, x ∈ Ui for i = 1, 2. For i = 1, 2, Bi ∈ TB exists such
that x ∈ Bi ⊂ Ui . According to (B2) B12 exists such that x ∈ B12 ⊂ B1 ∩ B2 .
Example 1.4. Let R be the set of all real numbers. Let B be the collection of all open
intervals:
(a, b) := {x ∈ R | a < x < b}
Then B is a basis of a topology and the topology generated by B is called the standard
topology of R (usual, Euclidean).
So far we have three different topologies on R, the standard topology, the discrete
topology and the trivial topology. Here are two more, the first with fewer open sets than
the usual topology, the second with more open sets:
• Let T consist of the empty set together with all subsets of R whose complement
is finite. The axioms (1) − (3) are easily verified (homework). Every set in T is
open in the usual topology, but not vice versa. This topology is called the Cofinite
topology on R.
• Let T consist of all sets T such that for each x ∈ T there is an interval [a, b) with
x ∈ [a, b) ⊂ T . Intervals [a, b) are certainly in T so this topology is different from
the usual topology on R. Every interval (a, b) is in T since it can be expressed as a
union of an increasing sequence of intervals [an , b) in T . It follows that T contains
all sets that are open in the usual topology since these can be expressed as unions
of intervals (a, b). This topology is called the Sorgenfrey topology on R.
In the introductory lecture I said that data sets are often given in terms of point clouds,
ie. finite metric space. We now take a look at our primary example of a topological space,
a metric space.
Example 1.5. Metric Spaces
Definition 1.6. A metric on a set X is a function d : X × X → R such that
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M1 d(x, y) ≥ 0 for all x, y ∈ X and d(x, y) = 0 iff x = y.
M2 d(x, y) = d(y, x) for all x, y ∈ X.
M3 d(x, y) ≤ d(x, z) + d(z, y) for all x, y, z ∈ X.
A pair (X, d) is called a metric space.
This last condition is called the triangle inequality because for the usual distance
function in the plane it says that length of one side of a triangle is always less than or
equal to the sum of the lengths of the other two sides.
Given a metric on X one defines the open ball of radius r centered at x to be the set
Br (x) = {y ∈ X | d(x, y) < r}.
The closed ball of radius r centered at x is the set
Br (x) = {y ∈ X | d(x, y) ≤ r}.
Proposition 1.7. The collection of all balls Br (x) for r > 0 and x ∈ X forms a basis for
a topology on X.
A space whose topology can be obtained in this way via a basis of open balls with
respect to a metric is called a metric space.
Proof. First a preliminary observation: For a point y ∈ Br (x) the ball Bs (y) is contained in Br (x) if s ≤ r − d(x, y), since for z ∈ Bs (y) we have d(z, y) < s and hence
d(z, x) ≤ d(z, y) + d(y, x) < s + d(x, y) ≤ r.
Now to show the condition to have a basis is satisfied, suppose we are given a point
y ∈ Br1 (x1 ) ∩ Br2 (x2 ). Then the observation in the preceding paragraph implies that
Bs (y) ⊂ Br1 (x1 ) ∩ Br2 (x2 ) for any s ≤ min{r1 − d(x1 , y), r2 − d(x2 , y)}.
A simple example of a metric space is R2 = {(x, y) | x, y ∈ R} equipped with the
standard distance function
p
d2 ((x1 , y1 ), (x2 , y2 )) = (x2 − x1 )2 + (y2 − y1 )2
measuring the distance between the two points (x1 , y1 ) and (x2 , y2 ) in the plane. The ball
B (x1 , y1 ) coincides with the set of all points inside (but not on) the circle of radius centered at (x1 , y1 ).
Just as there are many different topologies on R, so there are many different types of
distance functions on R2 .
Different metrics on the same set X can give rise to different bases for the same topology. For example, if we instead take d∞ (x, y) = max{|x1 −y1 |, |x2 −y2 |} then the ‘balls’ are
squares. Here x = (x1 , y1 ) and y = (y1 , y2 ). Another metric is d1 (x, y) = |x1 − y1 | + |x2 − y2 |,
which has balls that are also squares, but rotated 45 degrees from the squares in the previous metric.
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2
Closed Set, Closure, Interior, and limit points
Definition 2.1. A subset A of a topological space X is closed if its complement X \ A is
open.
Example 2.2. In R with the usual topology, a closed interval [a, b] is a closed subset.
Similarly, in R2 with its usual topology a closed disk, the union of an open disk with its
boundary circle, is a closed subset.
Given a subset A of a topological space X , then for each point x ∈ X exactly one of
the following three possibilities holds:
1. There exists an open set U in X with x ∈ U ⊂ A.
2. There exists an open set U in X with x ∈ U ⊂ X \ A.
3. Every open set U with x ∈ U meets both A and X \ A.
Points x such that (1) holds form a subset of A called the interior of A, written int(A).
The points where (2) holds then form int(X \ A). Points x where (3) holds form a set
called the boundary or frontier of A, written ∂A. The points x where either (1) or (3)
hold are the points x such that every open set U containing x meets A. Such points are
called limit points of A, and the set of these limit points is called the closure of A,
written A. Note that A ⊂ A, so we have int(A) ⊂ A = int(A) ∪ ∂A, this last union being
a disjoint union. We will use the symbol t to denote union of disjoint subsets when we
want to emphasize the disjointness, so A = int(A) t ∂A and X = int(A) t ∂A t int(x \ A).
Example 2.3. In R with the usual topology the intervals (a, b), [a, b], [a, b), and (a, b]
all have interior (a, b), closure [a, b] and boundary {a, b}. Similarly, in R2 with the usual
topology, if A is the union of an open disk B1 (0, 0) with any subset of its boundary circle
S 1 then int(A) = B1 (0, 0), A = B1 (0, 0) ∪ S 1 , and ∂A = S 1 . For a somewhat different
type of example, let A = Q in X = R with the usual topology on R. Then int(A) = ∅
and A = ∂A = R.
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3
Subspaces
A subset of a topological space has a naturally induced topology, called the subspace
topology. In geometry, the subspace topology is the source of all funky topologies.
Definition 3.1. Let (X, T ) be a topological space. Let Y be a subset of X. The collection
TY := {Y ∩ U | U ∈ T } is a topology on Y , called the subspace topology.
If we take X to be R2 with its usual topology, then every subset of R2 becomes a
topological space. In particular, geometric figures such as circles and polygons can now
be viewed as topological spaces. Likewise, geometric figures in R3 such as spheres and
polyhedra become topological spaces, with the subspace topology from the usual topology
on R3 .
Example 3.2. Let Y = [0, 1). The set [0, 21 ) is open in the subspace topology, since
[0, 12 ) = [0, 1) ∩ (− 12 , 12 ) and (− 21 , 12 ) is open.
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Continuity and Homeomorphisms
Definition 4.1. A function f : X → Y between topological spaces is continuous if f −1 (U )
is open in X for each open set U in Y .
For brevity, continuous functions are sometimes called maps or mappings. (A map in
the everyday sense of the word is in fact a function from the points on the map to the
points in whatever region is being represented by the map.)
Proposition 4.2. Given a function f : X → Y and a basis B for Y , then f is continuous
if and only if f −1 (B) is open in X for each B ∈ B.
Proof. One direction is obvious since the sets in B are open. In the other direction,
S
suppose f −1 (B) is open for each B S
∈ B. ThenSany open set U in Y is a union α Bα of
basis sets Bα , hence f −1 (U ) = f −1 ( α Bα) = α f −1 (Bα ) is open in X, being a union of
the open sets f −1 (Bα ).
Proposition 4.3. If f : X → Y and g : Y → Z are continuous, then their composition
g ◦ f : X → Z is also continuous.
Proof. This uses the easy set-theory fact that (g ◦ f )−1 (A) = f −1 (g −1 (A)) for any A ⊂ Z.
Thus if f and g are continuous and A is open in Z then g −1 (A) is open in Y so f −1 (g −1 (A))
is open in X. This means g ◦ f is continuous.
Proposition 4.4. If f : X → Y is continuous and A is a subspace of X, then the restriction f |A of f to A is continuous as a function A → Y .
Proof. For an open set O ⊂ Y we have (f |A )−1 (U ) = f −1 (U ) ∩ A, which is an open set in
A since f −1 (U ) is open in X.
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Definition 4.5. A continuous map f : X → Y is a homeomorphism if it is oneto-one and onto, and its inverse function f −1 : Y → X is also continuous.
Example 4.6. Stereographic Projection
Stereographic projection is an important homeomorphism between the plane R2 and
the sphere S 2 minus a point. The 2-sphere S 2 is the set of points (x, y, z) ∈ R3 such that
x2 + y 2 + z 2 = 1. Let S 2 \ {N } denote the 2-sphere minus its north pole, i.e. the point
(0, 0, 1).
There exists a homeomorphism f : S 2 \ {N } → R2 , which can be described as follows.
First, identify the set P = {(x, y, z) ∈ R3 | z = 0}with R2 ; the map P → R2 given by
(x, y, 0) → (x, y) is a homeomorphism.
For a point p, let f (p) denote the unique point in P such that the intersection of the
segment N f (p) and S 2 is p. In coordinates, this map is precisely
f (x, y, z) = (
x
y
,
)
1−z 1−z
We can also compute that
f −1 (u, v) = (
2u
2v
u2 + v 2 − 1
,
,
).
u2 + v 2 + 1 u2 + v 2 + 1 u2 + v 2 + 1
Via similarly defined maps, one can show that that the n-sphere minus a point is
homeomorphic to Rn .
https : //www.youtube.com/watch?v = V X − 0Laeczgk
Figure 1: Homeomorphism between a doughnut and a coffee cup.
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https : //www.youtube.com/watch?v = qV 7iw635iJ8
https : //www.youtube.com/watch?v = dwrhCSORERA
A basic problem in topology is to decide whether two given topological spaces X and
Y are homeomorphic or not. To show that X and Y are homeomorphic, we have to find
a continuous bijection between them whose inverse is also continuous; this comes down to
being able to construct continuous functions. On the other hand, to show that X and Y
are not homeomorphic, we have to prove that there does not exist any homeomorphism
between them. This can be difficult or even impossible. In practice, one tries to do
this with the help of certain ‘invariants’: for example, topological properties such as
connectedness or compactness that (as it will turn out) are the same for homeomorphic
spaces.
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Connectedness
Some spaces are in a sense ‘disconnected’, being the union of two or more completely
separate subspaces. For example the space X ⊂ R consisting of the two intervals A = [0, 1]
and B = [2, 3] should certainly be disconnected, and so should a subspace X of R2 which
is the union of two disjoint circles A and B. As these examples show, it is reasonable to
interpret the idea of A and B being ‘completely separate’ as saying not only that they are
disjoint, but no point of A is a limit point of B and no point of B is a limit point of A.
Since we are assuming that X is the union of A and B, this is equivalent to saying that A
and B each contain all their limit points. In other words, A and B are both closed subsets
of X . Since each of A and B is the complement of the other, it would be equivalent to say
that both A and B are open sets. Thus we have arrived at the following basic definition:
Definition 5.1.
version 1 A topological space (X, T ) connected if the only subsets of X which are both
open and closed are the empty set and the whole space X.
version 2 A topological space (X, T ) connected if X = U ∪ V where U and V are open
sets which are disjoint, then either U or V is the empty set.
If non-empty U and V from version 2 exist, we say they form a separation of X.
[Convince yourself of the equivalence of these definitions.]
Example 5.2. The space R \ {0} is not connected, because (∞, 0) and (0, ∞) form a
separation.
Example 5.3. The trivial topology Ttrivial on a set X is connected since in this topology
the only open sets are ∅ and X. If X is a set with at least two distinct elements then the
discrete topology Tdiscrete on X is not connected: because if x0 ∈ X then the singleton set
{x0 } is both open and closed (in the discrete topology, every subset of X is both open
and closed) but {x0 } =
6 ∅ and {x0 } =
6 X.
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Theorem 5.4. An interval [a, b] in R is connected.
Proof. We may assume a < b since it is obvious that [a, a] is connected. Suppose [a, b] is
decomposed as the disjoint union of sets U and V that are open in [a, b] with the subspace
topology, so they are closed in [a, b]. After possibly changing notation we may assume
that a is in U . Since U is open in [a, b] there is an interval [a, a+) contained in U for some
> 0, and hence there is an interval [a, c] ⊂ U , with a < c. The set C = {c | [a, c] ⊂ U }
is bounded above by b, so it has a least upper bound L ≤ b. (A fundamental property
of R is that any set that is bounded above has a least upper bound.) We know that
L > a by the earlier observation that there is an interval [a, c] ⊂ U with c > a. Since
no number smaller than L is an upper bound for C, there exist intervals [a, c] ⊂ A with
c ≤ L and c arbitrarily close to L. These numbers c are in U , hence L must also be in U
since U is closed. Thus we have [a, L] ⊂ U . Now if we assume that L < b we can derive
a contradiction in the following way. Since U is open and contains [a, L], it follows that
A contains [a, L + ] for some > 0. But this means that C contains numbers bigger
than L, contradicting the fact that L was an upper bound for C. Thus the assumption
L < b leads to a contradiction, so we must conclude that L = b since we know L ≤ b.
We already saw that [a, L] ⊂ U , so now we have [a, b] ⊂ U . This means that V must
be empty, and we have shown that it is impossible to decompose [a, b] into two disjoint
nonempty open sets. Hence [a, b] is connected.
For homework, you will prove
Proposition 5.5. If (X, TX ) is a connected topological space and f : (X, TX ) → (Y, TY )
continuous, then f (X) ⊆ Y is connected in the induced topology.
Example 5.6. Let f : [0, 2π] → S 1 be given by ϕ 7→ (sin ϕ, cos ϕ). This is a continuous
function. Since images of continuous functions are continuous, so is S 1 .
There is a more refined way to apply this fact that connectedness is preserved by
homeomorphisms, using the following idea. In a connected space X, a point x ∈ X is
called a cut point if removing x from X produces a disconnected space X \ {x}. Cut
points can also be used to show that R is not homeomorphic to Rn for n > 1, since the
latter spaces have no cut points whereas in R every point is a cut point.
Example 5.7. The Cartesian plane R2 is not homeomorphic to the real line R. This can
be proven by contradiction; suppose there is a homeomorphism f : R2 → R. Choose p,
and consider f (p) ∈ R. If f is a homeomorphism, then f restricts to a homeomorphism
between R2 \ {p} and R \ {f (p)}. But R2 \ {p} is connected, while R \ {f (p)} is not!
Thus, f cannot be a homeomorphism.
If a space X is not connected, then it decomposes as the union of a collection of disjoint
connected subspaces that are maximal in the sense that none of them is contained in any
larger connected subspace. These maximal connected subspaces are called the connected
components of X.
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Given a point x ∈ X, we define
[
C(x) = {C ⊂ X | C is connected and x ∈ C}
as the union of all connected subspaces of X that contain the point x. The set C(x)
is called the connected component of x; we will show that it is the maximal connected
subspace of X containing the point x. Any two connected components of X are either
equal or disjoint, so we get a partition of the space X into maximal connected subsets.
A couple of lemmas are needed to prove this.
Lemma 5.8. Let X = C ∪ D be a separation of a topological space. If Y ⊂ X is a
connected subspace, then Y ⊂ C or Y ⊂ D.
Proof. We have Y = (Y ∩ C) ∪ (Y ∩ D), and both sets are open in the subspace topology
and disjoint. Since Y is connected, one of them must be empty; but then we either have
Y = Y ∩ C, which means that Y ⊂ C; or Y = Y ∩ D, which means that Y ⊂ D.
Lemma 5.9. Let Y be a connected subspace of a topological space X. If Y ⊂ Z ⊂ Y ,
then Z is again connected.
Proof. Suppose that there was a separation Z = C ∪ D. By Lemma 5.8, the connected
subspace Y has to lie entirely in one of the two sets; without loss of generality, we may
assume that Y ⊂ C. Now C = Z ∩ A for some closed set A⊂ X, because C is closed
in the subspace topology of Z. Since Y ⊂ A, we also have Z ⊂ Y ⊂ A, and therefore
Z = C. This contradicts the fact that D is nonempty.
If we join together any number of connected subspaces at a common point, the result
is again connected.
Proposition 5.10. Let {Yi }i∈I be a collection of connected subspaces of a topological space
X. If the intersection ∩i Yi is nonempty, then the union Y = ∪i Yi is again connected.
Proof. We argue by contradiction. Suppose that Y = C ∪ D was a separation. Choose a
point x ∈ ∩i Yi ; without loss of generality x ∈ C. Since each Yi is connected, and since Yi
and C have the point x in common, the previous lemma tells us that Yi ⊂ C. This being
true for every i ∈ I, we get Y ⊂ C, which contradicts the fact that D is nonempty.
Proposition 5.11. The sets C(x) have the following properties:
• Each subspace C(x) is connected.
• If x, y ∈ X, then C(x) and C(y) are either equal or disjoint.
• Every nonempty connected subspace of X is contained in a unique C(x).
Proof.
• C(x) is a union of connected subspaces that all contain the point x, and therefore
connected by Proposition 5.10.
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• Suppose that C(x) and C(y) are not disjoint. Take any z ∈ C(x) ∩ C(y). We shall
argue that C(x) = C(z); by symmetry, this will imply that C(x) = C(z) = C(y).
We have z ∈ C(x), and since C(x) is connected, we get C(x) ⊂ C(z). Hence
x ∈ C(z), and for the same reason, C(z) ⊂ C(x).
• Let Y ⊂ X be nonempty connected subspace. Choose a point x ∈ Y ; then
Y ⊂ C(x) by construction. Uniqueness follows from (b).
Definition 5.12. The space X is said to be path-connected if for any two points x, y ∈ X
there exists a continuous function f from the unit interval [0, 1] to X with f (0) = x and
f (1) = y. (This function is called a path from x to y.)
Example 5.13. The rational numbers Q are not path connected.
There is a similar definition for path connectedness: given a point x ∈ X, we set
S
P (x) =
{P ⊂ XP is path connected and x ∈ P }
= {y ∈ X | x and y can be joined by a path in X}
and call it the path component of x. The following result can be proved in the same way
as Proposition 5.11.
Proposition 5.14. The path components of X have the following properties:
• Each subspace P (x) is path connected.
• If x, y ∈ X, then P (x) and P (y) are either equal or disjoint.
• Every nonempty path connected subspace of X is contained in a unique P (x).
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Homotopy Groups
The fundamental idea of algebraic topology is that one should develop methods for counting the occurrences of geometric patterns in a topological space in order to distinguish
it from other spaces, or to suggest similarities between different space. Trying to figure
out how many ‘pieces’ a topological space is made of is only the first step. Other natural questions include: ‘Are there any loops or higher dimensional voids in a topological
space?’
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In examining these two spaces, we see that the left hand space contains a single loop,
while the second one contains two distinct loops. Thus, a count of loops is an interesting
quantity to consider, from an intuitive point of view. However, it appears difficult to make
precise mathematics out of this intuition. It is reasonably easy to make sense of what
one means by a loop in a space X, i.e. a continuous map f : S 1 → X. So in this case,
the pattern associated to a loop is the circle itself, and an occurrence of the pattern is a
continuous map from the circle S 1 to X. However, there are almost always infinitely many
loops in a space. For example, any loop can always be reparametrized by precomposing
with any self homeomorphism of the circle. Another difficulty, though, is the situation
illustrated by the following subset of the plane.
The interesting feature is the hole in the center, and both the loops (as well as an
infinity of others) capture that feature, in the sense that they ‘go around’ the hole. This
makes for an even larger set of loops, and the idea here is to create a kind of count which
captures the feature using the presence of loops around it, rather than producing an
infinity of loops. The key insight to be had here is that the idea of counting occurrences
of patterns directly is unworkable, but that counting equivalence classes of occurrences of
patterns under an equivalence relation is workable.
Definition 6.1. Two paths f, g : [0, 1] → Y are homotopic, we write f ' g, if there is
a continuous map H : [0, 1] × [0, 1] → Y so that H(x, 0) = f (x) and H(x, 1) = g(x) for
all x ∈ [0, 1], and H(0, t) = x0 , H(1, t) = x1 for all t ∈ [0, 1]. The relationship of being
path homotopic is an equivalence relation. We denote the equivalence class of paths path
homotopic to f by [f ].
Remark. The fact that one most choose equivalence classes of occurrences of a pattern
in order to obtain a workable theory is the fundamental observation in the subject. It
is responsible for the power of the method, and on the other hand for the technical
complexity of the subject.
Proof. Reflexivity is evident since f ' f by the constant homotopy H(x, t) = f (x).
Symmetry is also easy since if f0 ' f1 via H, then f1 ' f0 via the inverse homotopy
H 0 (x, t) = H(x, 1 − t). For transitivity, if f0 ' f1 via F and if f1 = g0 with g0 ' g1
via G, then f0 ' g1 via the homotopy H such that H(s, t) = F (s, 2t) for 0 ≤ t ≤ 21
and H(s, t) = G(s, 2t − 1) for 21 ≤ t ≤ 1. Since H is continuous on [0, 1] × [0, 21 ] and on
[0, 1] × [ 12 , 1], it is continuous on [0, 1] × [0, 1].
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Example 6.2. Any two paths f, g : [0, 1] → C, where C is convex, are path homotopic.
The path homotopy H : [0, 1] × [0, 1] → C from f to g is given as
H(x, t) = tg(x) + (1 − t)f (x).
Definition 6.3. Let (X, TX ) be a topological space, and let x0 ∈ X be a point. A path
f : [0, 1] → X is called a loop based at x0 if it starts and ends at the point x0 , in the sense
that f (0) = f (1) = x0 . We define
π1 (X, x0 ) = {[f ] | f : I → X is a path with f (0) = f (1) = x0 }
by considering all loops based at x0 , up to path homotopy. In fact, π1 (X, x0 ) is not just a
set, but a group; the group operation is given by composition by paths. Given two paths
f, g : I → X such that f (1) = g(0), there is a composition or product path f · g that
traverses first f and then g, defined by the formula
f (2s),
0 ≤ s ≤ 21
f · g(s) =
g(2s − 1), 12 ≤ s ≤ 1
This is called the fundamental group of X.
Proof.
Associativity Note that p · (q · r) 6= (p · q) · r, even if all products are defined, but
we can reparametrize to make them equal so they are certainly equivalent. Hence
[p] · ([q] · [r]) = ([p] · [q]) · [r]).
Neutral Element Write 1x (s) = x for the constant path at x ∈ X. Clearly, [1x ]·[p] = [p]
and [p] · [1x ] = [p].
Inverse Element Let p−1 (s) = p(1 − s) be the same path but going backwards. Then
every path in [p] · [p−1 ] is equivalent to 1x and every path in [p−1 ] · [p] is equivalent
to 1y .
Similarly, one can in a natural way impose the structure of a group on the set of
equivalence classes of based maps from S n to X. The resulting group is denoted by
πn (X, x0 ). Applied to the example above, with a single obstacle in the plane, this group
π1 is a single copy of the integers with addition as the operation. The integer assigned to
a given loop is the so-called winding number of the loop, which counts how many times
the loop wraps around the obstacle, with orientation taken into account as a sign.
Theorem 6.4. Suppose that f : (X, TX ) → (Y, TY ) is a homeomorphism. Then πn (X, x0 )
is isomorphic to π(Y, f (x0 )) for n ≥ 1.
There is a notion weaker than that of a homeomorphism, namely that of homotopy
equivalence.
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Definition 6.5. Two maps f, g : (X, TX ) → (Y, TY ) are homotopic, we write f ' g, if
there is a continuous map H : X × [0, 1] → Y so that H(x, 0) = f (x) and H(x, 1) = g(x)
for all x ∈ X.
Spaces (X, TX ) and (Y, TY ) are homotopy equivalent if f : (X, TX ) → (Y, TY ) and
g : (Y, TY ) → (X, TX ) exist such that g ◦ f ' IdX and f ◦ g ' IdY .
Homotopy groups of homotopy equivalent spaces are isomorphic.
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Compactness
Compactness is a sort of finiteness property that some spaces have and others do not. The
rough idea is that spaces which are ‘infinitely large’ such as R or [0, ∞) are not compact.
However, we want compactness to depend just on the topology on a space, so it will have
to be defined purely in terms of open sets. This means that any space homeomorphic to
a noncompact space will also be noncompact, so finite intervals (a, b) and [a, b) will also
be noncompact in spite of their ‘finiteness’. On the other hand, closed intervals [a, b] will
be compact - they cannot be stretched to be ‘infinitely large’.
How can this idea be expressed just in terms of open sets rather than in some numerical
measure of size? This would seem to be difficult since open sets themselves can be large
or small. But large open sets can be expressed as unions of small open sets, so perhaps
we should think about counting how many small open sets are needed when a large open
set in a space X, such as the whole space X itself, is expressed as a union of small open
sets. The most basic question in this situation is whether the number of small open sets
needed is finite or infinite. For example, if X is a metric space, then X is the union of
all its balls B (x) of fixed radius > 0, so we could ask whether X is in fact the union of
a finite collection of these balls B (x) of fixed radius. To generalize this idea to arbitrary
spaces which need not have a metric, we replace balls by arbitrary open sets, and this
leads to the following general definition:
Definition 7.1. A space X is compact if for each collection of open sets Uα in X whose
union is X, there exist a finite subset U1 , U2 , . . . , Un whose union is X. More concisely,
one says that every open cover of X has a finite subcover, where an open cover of X
is a collection of open sets in X whose union is X , and a finite subcover is a finite
subcollection whose union is still X.
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Example 7.2. R is not compact because the cover by the open intervals (−n, n) for
n = 1, 2, . . . has no finite subcover, since infinitely many of these intervals are needed
to cover all of R. Another open cover which has no finite subcover is the collection of
intervals (n − 1, n + 1) for n ∈ Z. In a similar vein, the interval (0, 1) fails to be compact
since the cover by the open intervals ( n1 , 1) for n ≥ 1 has no finite subcover.
Proposition 7.3. If f : X → Y is continuous and onto, and X is compact, then Y is
also compact.
Proof. Let a cover of Y by open sets Uα be given. Then the sets f −1 (Uα ) form an
open cover of X. If X is compact, this cover has a finite subcover. Call this finite
subcover f −1 (U1 ), . . . , f −1 (Un ).The corresponding sets U1 , . . . , Un then cover Y since for
each y ∈ f (X) there exists x ∈ X with f (x) = y, and this x will be in some set f −1 (Ui )
of the finite cover of X, so y will be in the corresponding set Ui .
Theorem 7.4 (Heine-Borel Theorem). A subspace X ⊂ Rn is compact if and only if it
is closed and bounded.
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Constructing new topological spaces from existing
ones
Our next objective is to describe a general procedure for building complicated spaces out
of simpler spaces. This is the topological analog of how all sorts of objects are made in
the real world. Just think of all the different words there are in the English language
for putting things together: gluing, pasting, taping, stapling, stitching, sewing, welding,
riveting, soldering, brazing, bonding, nailing, bolting, clamping, etc.
8.1
Product Spaces
Definition 8.1. If (X, TX ) and (Y, TY ) are topological spaces, then the collection BX×Y
of subsets of the form U × V ⊂ X × Y , U ∈ TX , V ∈ TY forms a basis of a topology. The
topology generated by BX×Y is called product topology on X × Y .
More generally one can define the product X1 × . . . × Xn to consist of all ordered
n-tuples (x1 , . . . , xn ) with xi ∈ Xi for each i. A basis for the product topology on
X1 × . . . × Xn consists of all products U1 × . . . × Un as each Ui ranges over open sets
in Xi , or just over a basis for the topology on Xi .
8.2
Quotients
Let (X, TX ) be a topological space and ∼ an equivalence relation on X. For every x ∈ X,
denote by [x] its equivalence class. The quotient space of X modulo ∼ is given by the set
X/∼ = {[x] : x ∈ X}.
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We have the projection map
p : X → X/∼ , x 7→ [x]
and we equip X/∼ by the topology: U ⊂ X/∼ is open iff p−1 (U ) is an open subset of X.
In practice the partition of X is often specified by saying which points we wish to glue
together.
Example 8.2.
1. Let X = [0, 1] ∪ [2, 3]. We define an equivalence relation: 1 ∼ 2. Then [1] = [2] =
{1, 2}, while [x] = {x}, ∀x ∈ X \ {1, 2}. Then X/∼ is homeomorphic to a closed
interval.
2. Let X = [0, 1] and ∼ an equivalence relation on X such that 0 ∼ 1 and [x] = {x},
∀x ∈ X \ {0, 1}. Then X/∼ is homeomorphic to S 1 .
For example, to obtain the cylinder from the rectangle [0, 1] × [0, 1] we make the
identifications (0, t) ∼ (1, t) for t ∈ [0, 1], and to obtain the Mobius band we instead
identify (0, t) ∼ (1, 1 − t).
Figure 2: Cylinder, Mobius band.
Example 8.3. The torus S 1 × S 1 can be realized as a quotient space of a rectangle
[0, 1]×[0, 1] by identifying both pairs of opposite edges via (0, t) ∼ (1, t) and (s, 0) ∼ (s, 1).
If we first identify the top and bottom edges of the rectangle we obtain a cylinder, and then
identifying the two ends of the cylinder produces a torus. Notice that the identifications
of opposite edges of the rectangle force all four corners to be identified to a single point.
In particular, we are making the identification (0, 0) ∼ (1, 1) even though this is not part
of either of the original identifications (0, t) ∼ (1, t) and (s, 0) ∼ (s, 1), but follows since
(0, 0) ∼ (0, 1) ∼ (1, 1).
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Figure 3: Torus.
Figure 4: Klein Bottle.
Example 8.4. Let us construct a surface, commonly called the Klein bottle, by modifying
the earlier construction of the torus from a rectangle by reversing the orientation of one
of the edges of the rectangle: By definition, the Klein bottle is the quotient space of the
rectangle obtained by identifying opposite edges according to the orientations shown. We
will use the notation K 2 for this space, the superscript indicating that it is a 2-dimensional
surface. Identifying the top and bottom edges of the rectangle gives a cylinder, but to
identify the two ends of the cylinder by deforming the cylinder in R3 requires that the
cylinder pass through itself in order to make the orientations match, as in the third figure.
Thus we have a map from K 2 into R3 which is not one-to-one because two circles in K 2
have the same image circle C in R3 . In fact, it is a theorem that there is no embedding
of K 2 into R3 , i.e., there is no subspace of R3 homeomorphic to K 2 . However, there is an
embedding into R4 . An ordinary (three-dimensional) bottle has a crease or fold around
the opening where the inside and outside of the bottle meet. A sphere doesn’t have this
crease or fold, but it has no opening. A Klein bottle has an opening but no crease: like a
Mobius band, it is a continuous one-sided structure. Because it has no crease or fold, it
has no verifiable definition of where it’s inside and outside begin. Therefore, the volume
of a Klein bottle is considered to be zero, and the bottle has no real contents – except
itself! As the joke goes: ‘In topological hell the beer is packed in Klein bottles.’ Take a
coin, slide it across the surface of a Klein bottle until it returns to its starting point, and
the coin, as if by magic, will be flipped over. This is because, unlike a sphere or a regular
bottle, a Klein bottle is non-orientable.
https : //www.youtube.com/watch?v = GGlmppx − 2M 8
https : //www.youtube.com/watch?v = yaeyN jU P V qs
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