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, acircle and a square are topologically equivalent. Physically, a rubber band can be stretchedinto the form of either a circle or a square, as well as many other shapes which are alsoviewed as being topologically equivalent. On the other hand, a figure eight curve formedby two circles touching at a point is to be regarded as topologically distinct from a circleor square. A qualitative property that distinguishes the circle from the figure eight is thenumber of connected pieces that remain when a single point is removed: when a pointis removed from a circle what remains is a single arc, whereas for a figure eight if oneremoves the point of contact of its two circles, what remains is two separate arcs, twoseparate pieces. The term used to describe two geometric objects that are topologicallyequivalent is homeomorphic. One of the basic problems of topology is to determine whentwo 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 studiesin topology. These are called topological spaces. The definition turns out to be extremelygeneral, so that many objects that are topological spaces are not very geometric at all.

1 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 (fortopological spaces). The main impetus for this work was to provide a framework inwhich to discuss continuous functions, with the goal of examining their attributes morethoroughly 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 collectionof 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 setswith respect to T .

It is always possible to construct at least two topologies on every set X by choosingthe 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 discretetopology ;

• 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 ′ on a set X, withevery set that is open in the T topology is also open in the T ′ topology, so T ⊂ T ′. Inthis situation we say that the topology T ′ is finer than T and that T is coarser than T ′.Thus the discrete topology on X is finer than any other topology and the trivial topologyis coarser than any other topology. Of course, given two topologies on a set X, it neednot 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 astructure on a set, we often try to understand what the ‘manageable’ data is needed tospecify the structure. The notion of a topological basis will help us to describe differenttopologies 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 ofX such that

(B1) For every x ∈ X, there is an element B ∈ B such that x ∈ B.

(B2) If x ∈ B1∩B2 whereB1, B2 are in B, then there isB3 in B such that x ∈ B3 ⊂ B1 ∩B2.

Lemma 1.3. (Generating a topology). Let B be a topological basis on X. Define TB tobe 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 abasis of T . Note that T itself is a basis of the topology T . So there is always a basis fora 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 :=⋃i∈I

Ui ∈ TB.

By the definition of the union, for each x ∈ U , there is Ui such that x ∈ Ui. SinceUi ∈ TB, there is B ∈ B such that x ∈ B ⊂ Ui. For this B ∈ B, x ∈ B ⊂ U . ThusU ∈ 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 thedefinition of the intersection, x ∈ Ui for i = 1, 2. For i = 1, 2, Bi ∈ TB exists suchthat 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 openintervals:

(a, b) := {x ∈ R | a < x < b}

Then B is a basis of a topology and the topology generated by B is called the standardtopology of R (usual, Euclidean).

So far we have three different topologies on R, the standard topology, the discretetopology and the trivial topology. Here are two more, the first with fewer open sets thanthe usual topology, the second with more open sets:

• Let T consist of the empty set together with all subsets of R whose complementis finite. The axioms (1) − (3) are easily verified (homework). Every set in T isopen in the usual topology, but not vice versa. This topology is called the Cofinitetopology on R.

• Let T consist of all sets T such that for each x ∈ T there is an interval [a, b) withx ∈ [a, b) ⊂ T . Intervals [a, b) are certainly in T so this topology is different fromthe usual topology on R. Every interval (a, b) is in T since it can be expressed as aunion of an increasing sequence of intervals [an, b) in T . It follows that T containsall sets that are open in the usual topology since these can be expressed as unionsof 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 distancefunction in the plane it says that length of one side of a triangle is always less than orequal 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 fora topology on X.

A space whose topology can be obtained in this way via a basis of open balls withrespect to a metric is called a metric space.

Proof. First a preliminary observation: For a point y ∈ Br(x) the ball Bs(y) is con-tained in Br(x) if s ≤ r − d(x, y), since for z ∈ Bs(y) we have d(z, y) < s and henced(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 pointy ∈ Br1(x1) ∩ Br2(x2). Then the observation in the preceding paragraph implies thatBs(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 thestandard distance function

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 ballBε(x1, y1) coincides with the set of all points inside (but not on) the circle of radius εcentered at (x1, y1).

Just as there are many different topologies on R, so there are many different types ofdistance functions on R2.

Different metrics on the same set X can give rise to different bases for the same topol-ogy. For example, if we instead take d∞(x, y) = max{|x1−y1|, |x2−y2|} then the ‘balls’ aresquares. 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 pre-vious 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 isopen.

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 itsboundary circle, is a closed subset.

Given a subset A of a topological space X , then for each point x ∈ X exactly one ofthe 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 setcalled 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 arecalled 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 beinga disjoint union. We will use the symbol t to denote union of disjoint subsets when wewant to emphasize the disjointness, so A = int(A)t∂A and X = int(A)t∂At 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 usualtopology, if A is the union of an open disk B1(0, 0) with any subset of its boundary circleS1 then int(A) = B1(0, 0), A = B1(0, 0) ∪ S1, and ∂A = S1. For a somewhat differenttype 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 subspacetopology. 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 collectionTY := {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 atopological space. In particular, geometric figures such as circles and polygons can nowbe viewed as topological spaces. Likewise, geometric figures in R3 such as spheres andpolyhedra become topological spaces, with the subspace topology from the usual topologyon R3.

Example 3.2. Let Y = [0, 1). The set [0, 12) is open in the subspace topology, since

[0, 12) = [0, 1) ∩ (−1

2, 12) and (−1

2, 12) is open.

4 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 inthe everyday sense of the word is in fact a function from the points on the map to thepoints 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 continuousif 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,suppose f−1(B) is open for each B ∈ B. Then any open set U in Y is a union

⋃αBα of

basis sets Bα, hence f−1(U) = f−1(⋃αBα) =

⋃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 compositiong ◦ 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 restric-tion 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 inA 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 one-to-one and onto, and its inverse function f−1 : Y → X is also continuous.

Example 4.6. Stereographic ProjectionStereographic projection is an important homeomorphism between the plane R2 and

the sphere S2 minus a point. The 2-sphere S2 is the set of points (x, y, z) ∈ R3 such thatx2 + y2 + z2 = 1. Let S2 \ {N} denote the 2-sphere minus its north pole, i.e. the point(0, 0, 1).

There exists a homeomorphism f : S2 \ {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 thesegment Nf(p) and S2 is p. In coordinates, this map is precisely

f(x, y, z) = (x

1− z,

y

1− z)

We can also compute that

f−1(u, v) = (2u

u2 + v2 + 1,

2v

u2 + v2 + 1,u2 + v2 − 1

u2 + v2 + 1).

Via similarly defined maps, one can show that that the n-sphere minus a point ishomeomorphic 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 7iw635iJ8https : //www.youtube.com/watch?v = dwrhCSORERAA 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 finda continuous bijection between them whose inverse is also continuous; this comes down tobeing able to construct continuous functions. On the other hand, to show that X and Yare not homeomorphic, we have to prove that there does not exist any homeomorphismbetween them. This can be difficult or even impossible. In practice, one tries to dothis with the help of certain ‘invariants’: for example, topological properties such asconnectedness or compactness that (as it will turn out) are the same for homeomorphicspaces.

5 Connectedness

Some spaces are in a sense ‘disconnected’, being the union of two or more completelyseparate 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 whichis the union of two disjoint circles A and B. As these examples show, it is reasonable tointerpret the idea of A and B being ‘completely separate’ as saying not only that they aredisjoint, 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 Aand B each contain all their limit points. In other words, A and B are both closed subsetsof X . Since each of A and B is the complement of the other, it would be equivalent to saythat 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 bothopen 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 opensets 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 aseparation.

Example 5.3. The trivial topology Ttrivial on a set X is connected since in this topologythe only open sets are ∅ and X. If X is a set with at least two distinct elements then thediscrete 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 openand 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] isdecomposed as the disjoint union of sets U and V that are open in [a, b] with the subspacetopology, so they are closed in [a, b]. After possibly changing notation we may assumethat a is in U . Since U is open in [a, b] there is an interval [a, a+ε) contained in U for someε > 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 propertyof R is that any set that is bounded above has a least upper bound.) We know thatL > a by the earlier observation that there is an interval [a, c] ⊂ U with c > a. Sinceno number smaller than L is an upper bound for C, there exist intervals [a, c] ⊂ A withc ≤ L and c arbitrarily close to L. These numbers c are in U , hence L must also be in Usince U is closed. Thus we have [a, L] ⊂ U . Now if we assume that L < b we can derivea contradiction in the following way. Since U is open and contains [a, L], it follows thatA contains [a, L + ε] for some ε > 0. But this means that C contains numbers biggerthan L, contradicting the fact that L was an upper bound for C. Thus the assumptionL < 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 mustbe empty, and we have shown that it is impossible to decompose [a, b] into two disjointnonempty 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] → S1 be given by ϕ 7→ (sinϕ, cosϕ). This is a continuousfunction. Since images of continuous functions are continuous, so is S1.

There is a more refined way to apply this fact that connectedness is preserved byhomeomorphisms, using the following idea. In a connected space X, a point x ∈ X iscalled a cut point if removing x from X produces a disconnected space X \ {x}. Cutpoints can also be used to show that R is not homeomorphic to Rn for n > 1, since thelatter 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 canbe 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 homeomorphismbetween 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 disjointconnected subspaces that are maximal in the sense that none of them is contained in anylarger connected subspace. These maximal connected subspaces are called the connectedcomponents 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 connectedsubspace of X containing the point x. Any two connected components of X are eitherequal 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 aconnected subspace, then Y ⊂ C or Y ⊂ D.

Proof. We have Y = (Y ∩C)∪ (Y ∩D), and both sets are open in the subspace topologyand disjoint. Since Y is connected, one of them must be empty; but then we either haveY = 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 connectedsubspace Y has to lie entirely in one of the two sets; without loss of generality, we mayassume that Y ⊂ C. Now C = Z ∩ A for some closed set A⊂ X, because C is closedin the subspace topology of Z. Since Y ⊂ A, we also have Z ⊂ Y ⊂ A, and thereforeZ = C. This contradicts the fact that D is nonempty.

If we join together any number of connected subspaces at a common point, the resultis again connected.

Proposition 5.10. Let {Yi}i∈I be a collection of connected subspaces of a topological spaceX. If the intersection ∩iYi is nonempty, then the union Y = ∪iYi is again connected.

Proof. We argue by contradiction. Suppose that Y = C ∪D was a separation. Choose apoint x ∈ ∩iYi; without loss of generality x ∈ C. Since each Yi is connected, and since Yiand C have the point x in common, the previous lemma tells us that Yi ⊂ C. This beingtrue 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 thereforeconnected 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 shallargue 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). Hencex ∈ C(z), and for the same reason, C(z) ⊂ C(x).

• Let Y ⊂ X be nonempty connected subspace. Choose a point x ∈ Y ; thenY ⊂ 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 ∈ Xthere exists a continuous function f from the unit interval [0, 1] to X with f(0) = x andf(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

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 wayas 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).

6 Homotopy Groups

The fundamental idea of algebraic topology is that one should develop methods for count-ing the occurrences of geometric patterns in a topological space in order to distinguishit from other spaces, or to suggest similarities between different space. Trying to figureout how many ‘pieces’ a topological space is made of is only the first step. Other natu-ral questions include: ‘Are there any loops or higher dimensional voids in a topologicalspace?’

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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 interestingquantity to consider, from an intuitive point of view. However, it appears difficult to makeprecise mathematics out of this intuition. It is reasonably easy to make sense of whatone means by a loop in a space X, i.e. a continuous map f : S1 → X. So in this case,the pattern associated to a loop is the circle itself, and an occurrence of the pattern is acontinuous map from the circle S1 to X. However, there are almost always infinitely manyloops in a space. For example, any loop can always be reparametrized by precomposingwith any self homeomorphism of the circle. Another difficulty, though, is the situationillustrated by the following subset of the plane.

The interesting feature is the hole in the center, and both the loops (as well as aninfinity of others) capture that feature, in the sense that they ‘go around’ the hole. Thismakes for an even larger set of loops, and the idea here is to create a kind of count whichcaptures the feature using the presence of loops around it, rather than producing aninfinity of loops. The key insight to be had here is that the idea of counting occurrencesof patterns directly is unworkable, but that counting equivalence classes of occurrences ofpatterns under an equivalence relation is workable.

Definition 6.1. Two paths f, g : [0, 1] → Y are homotopic, we write f ' g, if there isa continuous map H : [0, 1] × [0, 1] → Y so that H(x, 0) = f(x) and H(x, 1) = g(x) forall x ∈ [0, 1], and H(0, t) = x0, H(1, t) = x1 for all t ∈ [0, 1]. The relationship of beingpath homotopic is an equivalence relation. We denote the equivalence class of paths pathhomotopic to f by [f ].

Remark. The fact that one most choose equivalence classes of occurrences of a patternin order to obtain a workable theory is the fundamental observation in the subject. Itis responsible for the power of the method, and on the other hand for the technicalcomplexity 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 homotopyH ′(x, t) = H(x, 1 − t). For transitivity, if f0 ' f1 via F and if f1 = g0 with g0 ' g1via G, then f0 ' g1 via the homotopy H such that H(s, t) = F (s, 2t) for 0 ≤ t ≤ 1

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and H(s, t) = G(s, 2t − 1) for 12≤ t ≤ 1. Since H is continuous on [0, 1] × [0, 1

2] 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 pathf : [0, 1]→ X is called a loop based at x0 if it starts and ends at the point x0, in the sensethat 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 aset, but a group; the group operation is given by composition by paths. Given two pathsf, g : I → X such that f(1) = g(0), there is a composition or product path f · g thattraverses first f and then g, defined by the formula

f · g(s) =

{f(2s), 0 ≤ s ≤ 1

2

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, butwe 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. Thenevery path in [p] · [p−1] is equivalent to 1x and every path in [p−1] · [p] is equivalentto 1y.

Similarly, one can in a natural way impose the structure of a group on the set ofequivalence classes of based maps from Sn to X. The resulting group is denoted byπn(X, x0). Applied to the example above, with a single obstacle in the plane, this groupπ1 is a single copy of the integers with addition as the operation. The integer assigned toa given loop is the so-called winding number of the loop, which counts how many timesthe 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 homotopyequivalence.

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Definition 6.5. Two maps f, g : (X, TX) → (Y, TY ) are homotopic, we write f ' g, ifthere 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 ) andg : (Y, TY )→ (X, TX) exist such that g ◦ f ' IdX and f ◦ g ' IdY .

Homotopy groups of homotopy equivalent spaces are isomorphic.

7 Compactness

Compactness is a sort of finiteness property that some spaces have and others do not. Therough 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 haveto be defined purely in terms of open sets. This means that any space homeomorphic toa noncompact space will also be noncompact, so finite intervals (a, b) and [a, b) will alsobe noncompact in spite of their ‘finiteness’. On the other hand, closed intervals [a, b] willbe 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 numericalmeasure of size? This would seem to be difficult since open sets themselves can be largeor small. But large open sets can be expressed as unions of small open sets, so perhapswe should think about counting how many small open sets are needed when a large openset in a space X, such as the whole space X itself, is expressed as a union of small opensets. The most basic question in this situation is whether the number of small open setsneeded is finite or infinite. For example, if X is a metric space, then X is the union ofall its balls Bε(x) of fixed radius ε > 0, so we could ask whether X is in fact the union ofa finite collection of these balls Bε(x) of fixed radius. To generalize this idea to arbitraryspaces which need not have a metric, we replace balls by arbitrary open sets, and thisleads to the following general definition:

Definition 7.1. A space X is compact if for each collection of open sets Uα in X whoseunion 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 Xis a collection of open sets in X whose union is X , and a finite subcover is a finitesubcollection 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) forn = 1, 2, . . . has no finite subcover, since infinitely many of these intervals are neededto cover all of R. Another open cover which has no finite subcover is the collection ofintervals (n− 1, n+ 1) for n ∈ Z. In a similar vein, the interval (0, 1) fails to be compactsince the cover by the open intervals ( 1

n, 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 isalso compact.

Proof. Let a cover of Y by open sets Uα be given. Then the sets f−1(Uα) form anopen cover of X. If X is compact, this cover has a finite subcover. Call this finitesubcover f−1(U1), . . . , f

−1(Un).The corresponding sets U1, . . . , Un then cover Y since foreach 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 itis closed and bounded.

8 Constructing new topological spaces from existing

ones

Our next objective is to describe a general procedure for building complicated spaces outof simpler spaces. This is the topological analog of how all sorts of objects are made inthe real world. Just think of all the different words there are in the English languagefor 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×Yof subsets of the form U × V ⊂ X × Y , U ∈ TX , V ∈ TY forms a basis of a topology. Thetopology generated by BX×Y is called product topology on X × Y .

More generally one can define the product X1 × . . . × Xn to consist of all orderedn-tuples (x1, . . . , xn) with xi ∈ Xi for each i. A basis for the product topology onX1 × . . .×Xn consists of all products U1 × . . . × Un as each Ui ranges over open setsin 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 mapp : 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 closedinterval.

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 S1.

For example, to obtain the cylinder from the rectangle [0, 1] × [0, 1] we make theidentifications (0, t) ∼ (1, t) for t ∈ [0, 1], and to obtain the Mobius band we insteadidentify (0, t) ∼ (1, 1− t).

Figure 2: Cylinder, Mobius band.

Example 8.3. The torus S1 × S1 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 thenidentifying the two ends of the cylinder produces a torus. Notice that the identificationsof 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 partof 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 modifyingthe earlier construction of the torus from a rectangle by reversing the orientation of oneof the edges of the rectangle: By definition, the Klein bottle is the quotient space of therectangle obtained by identifying opposite edges according to the orientations shown. Wewill use the notation K2 for this space, the superscript indicating that it is a 2-dimensionalsurface. Identifying the top and bottom edges of the rectangle gives a cylinder, but toidentify the two ends of the cylinder by deforming the cylinder in R3 requires that thecylinder pass through itself in order to make the orientations match, as in the third figure.Thus we have a map from K2 into R3 which is not one-to-one because two circles in K2

have the same image circle C in R3 . In fact, it is a theorem that there is no embeddingof K2 into R3, i.e., there is no subspace of R3 homeomorphic to K2. However, there is anembedding into R4. An ordinary (three-dimensional) bottle has a crease or fold aroundthe opening where the inside and outside of the bottle meet. A sphere doesn’t have thiscrease or fold, but it has no opening. A Klein bottle has an opening but no crease: like aMobius band, it is a continuous one-sided structure. Because it has no crease or fold, ithas no verifiable definition of where it’s inside and outside begin. Therefore, the volumeof a Klein bottle is considered to be zero, and the bottle has no real contents – exceptitself! As the joke goes: ‘In topological hell the beer is packed in Klein bottles.’ Take acoin, slide it across the surface of a Klein bottle until it returns to its starting point, andthe coin, as if by magic, will be flipped over. This is because, unlike a sphere or a regularbottle, a Klein bottle is non-orientable.

https : //www.youtube.com/watch?v = GGlmppx− 2M8https : //www.youtube.com/watch?v = yaeyNjUPV qs

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