continuous. It has no other opens, so it is the topology with the fewest open sets needed

to make i continuous.

Now let X be a set and ∼ an equivalence relation on X . Consider the set Y = X/∼

of equivalence classes of ∼. That is, in Y , if I write x for the class of x, I have x = y

if and only if x ∼ y, so that Y is obtained by identifying or ˜glueing together™ points

x and y when x ∼ y. Every surjective map f : X ’ Y of X to a set Y is obtained in

this way, by just declaring ∼ to be the relation x ∼ y ⇐’ f (x) = f (y).

Now suppose that X is a topological space, and let ∼ and f : X Y = X/∼ be

as before. The quotient topology of Y has open sets de¬ned by

U ‚ Y is open ⇐’ f ’1 (U ) is open in X .

118 TOPOLOGY

It is easy to see that this satis¬es the axioms for a topology. Clearly f is continuous,

and this is the topology with the most open sets for which f is continuous. It often

happens that the quotient topology of Y is not a metric topology, as we see presently.

As above, let X be a topological space, and ∼ an equivalence relation.

The quotient space Y = X/∼ has the following properties.

Proposition

There is a continuous map f : X ’ Y such that

(1)

x ∼ y =’ f (x) = f (y)

(that is, f is constant on equivalence classes of ∼).

Given a space Z and a continuous map g : X ’ Z that is constant on equivalence

(2)

classes of ∼, there exists a unique continuous map h : Y ’ Z such that g = h —¦ f.

(1) comes from the de¬nition as I discussed above.

Proof

(2) Given g, the map h must take f (x) ∈ Y to g(x). In other words, an element of

Y is an equivalence class [x] of elements of X under ∼, so choose x in that class, and

set h([x]) = g(x). This is well de¬ned because of the assumption that g is constant

on equivalence classes. Why is h continuous? For U ‚ Z open, g ’1 (U ) is open in

X , so that f ’1 (h ’1 (U )) is open in X , and h ’1 (U ) is open in Y by de¬nition of the

quotient topology of Y . QED

This property of the topological space Y and the quotient map f : X ’ Y is called

a universal mapping property or UMP. Constructions throughout abstract math can

be speci¬ed in terms of UMPs: you say what you want to do (in this case, ¬nd a

continuous map that is constant on equivalence classes), and then ask for the solution

of a UMP. In the present case, the universal mapping property says that f does not do

anything that is not forced by the conditions that f is constant on equivalence classes

of ∼, and is continuous. In other words, f identi¬es exactly the equivalence classes

of ∼, and makes no more identi¬cations, and Y has the most open sets subject to f

being continuous. It is interesting to analyse the above proof to see that this is exactly

what is required to make h well de¬ned and continuous.

7.6 Standard examples of glueing

The quotient topology on X/∼ provides the de¬nition of ˜glueing™, the space obtained

from X by glueing together points x ∼ y. Here I discuss some basic examples; see

Exercises 7.18“7.19 below for more.

S 1 = [0, 1]/∼ where ∼ glues the endpoints (see Figure 7.2b).

Example 1

Let X be the unit square [0, 1] — [0, 1]. The M¨ bius strip M is de¬ned

o

Example 2

by glueing some of the sides of X as in Figure 7.6a. More formally, consider the

7.6 STANDARD EXAMPLES OF GLUEING 119

glue

…

The M¨ bius strip M .

Figure 7.6a o

glue

…

The cylinder S 1 — [0, 1].

Figure 7.6b

following equivalence relations on X :

±

either (x, y) = (x , y )

(x, y) ∼ (x , y ) ⇐’ or x = 0, x = 1 and y = 1 ’ y

or vice versa,

and de¬ne the M¨ bius strip M by M = X/∼, with the quotient topology. By de¬nition

o

of the quotient topology, a point on the glued line has a neighbourhood obtained from

neighbourhoods of its two inverse images in X .

The cylinder S 1 — [0, 1] is obtained by glueing the unit square [0, 1] —

Example 3

[0, 1] as in Figure 7.6b.

S 1 — S 1 is obtained from the unit square [0, 1] — [0, 1]

The torus T

Example 4

by the glueing of Figure 7.6c. By de¬nition of the quotient topology, the four corners

of the square correspond to a point of the torus, and a neighbourhood of it is obtained

from neighbourhoods of the four corners in X . You can regard this as a surface of

rotation in R3 , or the surface in R4 given by x1 + y1 = x2 + y2 = 1.

2 2 2 2

The surface with g handles. The picture is as in Figure 7.6d: you get it

Example 5

by starting from S 2 , marking 2g distinct points on S 2 , cutting out small discs around

these, and glueing back in g small cylinders. See Exercise 7.19 as well as 9.4 for

further discussion.

Notice that all these spaces can easily be made into metric spaces, but you do not

really gain anything by doing so.

The M¨ bius strip M, the cylinder N = S 1 — [0, 1] and the torus T

o

Proposition

are not homeomorphic.

120 TOPOLOGY

…

…

Figure 7.6c The torus.

…

glue

Surface with g handles.

Figure 7.6d

I can almost prove this now, though I relegate one crucial statement to the end of

the chapter. The proof consists of the following steps.

Points of the boundary ‚ M ‚ M and ‚ N ‚ N are distin-

Step 1. Main claim

guished from points of the interior by their topological properties.

Therefore, if there exists a homeomorphism • : M ’ N , it must map ‚ M

Step 2

to ‚ N , and the restriction must de¬ne a homeomorphism ‚ M ‚ N .

‚ M is path connected, whereas ‚ N is disconnected; hence a homeo-

Step 3

morphism M N as in Step 2 cannot exist. In the same way, ‚ T = …, so that T M

and T N.

Given the main claim, Steps 2“3 are obvious, and the point is therefore to under-

stand Step 1. How do I distinguish points of the interior of a surface from points on

the boundary? The point is that every small neighbourhood U \ P of an interior point

P contains a small punctured disc D — about P; the punctured disc is the topological

7.7 TOPOLOGY OF Pn 121

R

P P

U U

U \P U \P

Figure 7.6e Boundary and interior points.

space {0 < x 2 + y 2 < 1} ‚ R2 . On the other hand, if P is a boundary point, it has

an arbitrarily small neighbourhood homeomorphic to a closed half-disc, that can be

written in polar coordinates

(r, θ ) 0 ¤ r < 1, θ ∈ [’π/2, π/2]

U

with P at the centre of the half-disc. Hence U \ P is homeomorphic to

U\P (r, θ ) 0 < r < 1, θ ∈ [’π/2, π/2]

which in turn is homeomorphic to a closed disc with parts of the boundary removed,

as in Figure 7.6e. Hence the essential content of telling interior and boundary points

apart consists in showing that the punctured disc D — is not homeomorphic to the

disc D. Think it through yourself to see whether you ¬nd this statement intuitive;

see 7.15.4, Corollary 1 for the proof.

Topology of Pn

7.7 R

Recall 5.2: projective n-space, as a set, is de¬ned to be the set of lines of Rn+1 through

the origin, or in other words, the quotient of Rn+1 \ {0} by the equivalence relation

which identi¬es x with »x for » = 0. The topology of Pn is the quotient topology of

Rn+1 \ {0}. This section considers various ways of looking at this topology.

Write S n = {x ∈ Rn+1 | xi2 = 1} ‚ Rn+1 for the n-sphere. Obviously S n meets

every line of Rn+1 through 0 in a pair of antipodal points. Therefore, as a set, Pn = R

S /±, where ± is the equivalence relation identifying antipodal points of the sphere

n

(that is, pairs ±x of opposite points). The topology of Pn coincides with the quotient

topology of S n /±; indeed, a subset of the lines through 0 is open in Rn+1 \ 0 if and only

if its intersection with S n is open in the subspace topology of S n . Note that S n ‚ Rn+1

is closed and bounded hence compact (Example 7.4.2); thus Pn , being the continuous

122 TOPOLOGY

disk

Möbius strip

Topology of P2 : M¨ bius strip with a disc glued in.

Figure 7.7 o

R

image of a compact space, is also compact by the tautological Proposition 7.4.3. This

was one of the motivations for constructing projective space discussed in 5.1.4.

There are many ways of understanding the quotient, by choosing a closed subset

of S n that picks out just one of each pair of antipodal points for a big open subset

and then glueing around the boundary: for example, the closed northern hemisphere

of S n contains one of each pair of antipodal points, except that I still have to identify

antipodal points of the equatorial sphere S n’1 .

In the case n = 2, we can do the following: view S 2 as the union of 3 pieces, a

cap around the north pole, a band around the equator, and a cap around the south

pole (see Figure 7.7). Every point in the southern cap is equivalent to a point in the

northern cap, so the southern cap is not needed. Now cut the equatorial band into its

front and back halves; as before, every point in the back half is equivalent to a point

in the front half, so this piece is also not needed. Now ± glues together the left and

right intervals of the front half to give a M¨ bius strip; this glueing is the same as in

o

Figure 7.6a. The northern cap is a disc, with boundary a circle; the M¨ bius strip also

o

has boundary a circle, and P is obtained by glueing these two pieces together along

2

their boundaries. Note that this is an abstract construction: you cannot do it in R3

without allowing self-crossing.

It is an interesting exercise to see the components of this construction as the result

of cutting P2 along a line and along a conic. See Exercise 7.17(a).

7.8 Nonmetric quotient topologies

X = {P, Q} is a space with only 2 points

Example 1 (The mousetrap topology)

and open sets

T X = …, {P}, X .

Here P is an open point, but not Q. Every neighbourhood of Q (there is only one)

contains P. In terms of convergence, the constant sequence P, P, . . . converges both

to P and to Q (please check this as an instant exercise; refer back to 7.4.2 for the

7.8 NONMETRIC QUOTIENT TOPOLOGIES 123

Q

P

Figure 7.8a The mousetrap topology.

de¬nition of convergence if needed). This implies, of course, that the topology of X

is not metric. X is a quotient topology: introduce the equivalence relation ∼ on C

de¬ned by

x ∼ y ⇐’ x = »y with » ∈ C, » = 0.

Then there are only two equivalence classes, Q = [0] and P = [» ∈ C \ {0}]; {P} is

obviously open while {Q} is not. The point is that if you are at 0 then any arbitrarily

small perturbation takes you into a nonzero number; that is, viewed from Q, the point

P is in¬nitely close. But if you are at a nonzero number », all the points in a small

neighbourhood are also nonzero, so viewed from P, the point Q is far away. Being

zero is an unstable, or closed condition; being nonzero is a stable or open condition.

I call this the mousetrap topology (Figure 7.8a) because if you are at Q (outside

the trap), it is no distance at all to get into the trap. But if you are at P (inside the

trap), then it is a long way out. Thus the content of the topology is more logical than

geometric.

There are many equivalence relations of interest with this kind of behaviour. One

example is the equivalence relation on R with

{x ∈ R | x > 0}, {0}, {x ∈ R | x < 0}

as its 3 equivalence classes.

A similar but more substantial example: consider

Example 2 (Quadratic forms)

quadratic forms q(x, y) = ax + 2bx y + cy 2 on R2 . There is a coordinate change

2

that puts q(x, y) in one of the 6 normal forms:

q1 = x 2 + y 2 , q2 = x 2 ’ y 2 , q3 = ’x 2 ’ y 2 , q4 = x 2 , q5 = ’x 2 , or q0 = 0.

All the quadratic forms on R2 are parametrised by (a, b, c) ∈ R3 , corresponding

to the symmetric matrix A = a b . Now introduce the equivalence relation on R3

bc

124 TOPOLOGY

q1

q2 q4

q0

q5

ab

= ac ’ b2 =

bc

q3

Equivalence classes of quadratic forms ax 2 + 2bx y + c y 2 .

Figure 7.8b

corresponding to a coordinate change:

A ∼ B ⇐’ ∃M ∈ GL(2, R) such that A = tM B M.

(Here GL(2, R) is the group of 2 — 2 invertible matrixes.) This means exactly that I

consider quadratic forms up to change of basis. So there are exactly the 6 classes, the

strata of Figure 7.8b. The quotient topology on the set

X = R3 /∼ = q1 , q2 , q3 , q4 , q5 , q0

has open sets

{q1 }, {q2 }, {q3 }, {q4 , q1 , q2 }, {q5 , q2 , q3 }, X

and their unions. For example, every neighbourhood of q4 contains q1 , q2 .

7.9 Basis for a topology

This is a formal idea for constructing topologies. Let B be a collection of subsets of

X . Then B is a basis for a topology if it satis¬es the three axioms

1. ¬nite intersections:

U1 , . . . , Un ∈ B =’ U1 © · · · © Un ∈ B;

involves every point: for all x ∈ X there exists U ∈ B such that x ∈ U ;

2.

empty convention: … ∈ B.

3.

7.9 BASIS FOR A TOPOLOGY 125

If B is a basis for a topology, the family of subsets

Construction

T= U» : U» ∈ B, arbitrary index set

»∈

of X is a topology on X , the topology generated by B.

This is entirely formal. X ∈ T using axiom 2 and the construction. T is

Proof

closed under arbitrary unions by construction. To show that T is closed under ¬nite

intersections, note that

U» © = U» © Uµ . QED

Uµ

»∈ »,µ

µ∈M

I can save time by listing only a basis for the topology, rather than by saying what

all the open sets are. The idea here is that a topology is speci¬ed by the neighbour-

hoods of each point (because an open set is determined by the condition that it is a

neighbourhood of each of its points). In turn it is enough to specify any system of

suf¬ciently small neighbourhoods of each point.

In 7.8, Example 2, I described the quotient topology on X = R3 /∼

Example 1

by telling you that its open sets are unions of

{q1 }, {q2 }, {q3 }, {q4 , q1 , q2 }, {q5 , q2 , q3 }, {q0 , q1 , q2 , q3 , q4 , q5 }.

Let X, d be a metric space, and

Example 2

B = B(x1 , µ1 ) © · · · © B(xn , µn )

be the set of ¬nite intersection of open balls B(x, µ) = {y | d(x, y) < µ}. Then B is

a basis for a topology T , the usual metric topology.

Another more substantial ex-

Example 3. Profinite topology of an infinite group

ample. Take any group G; recall that a subgroup H ‚ G is normal (written H G

) if g H = H g for every g ∈ G, that is, its right and left cosets coincide. A normal

subgroup H G of ¬nite index n is the kernel of a surjective homomorphism G ’

to a ¬nite group of order n. For example, if G = Z then every normal subgroup of

¬nite index is just nZ for some integer n.

Let G be a group, with e ∈ G the identity element. Then there is a topology on G

such that:

(a) normal subgroups H G of ¬nite index form a set of suf¬ciently small neighbour-

hoods of e;

the right translation maps r g : G ’ G de¬ned by f ’ f g are homeomorphisms.

(b)

It follows from (a) and (b) that a set of suf¬ciently small neighbourhoods of any

g ∈ G are given by cosets g H , where the H are as in (a). So take

B = {…} ∪ {cosets of normal subgroups of ¬nite index}

126 TOPOLOGY

|δx|, |δy| < µ |δx| + |δy| < µ δx2 + δy2 < µ

Figure 7.10 Balls for product metrics.

as a basis for a topology. I check that this is a basis by going through the three

axioms. Indeed, …, G ∈ B. Also if H1 , . . . , Hn are normal subgroups of ¬nite index

then so is H1 © · · · © Hn , clearly, and if g1 H1 , . . . , gn Hn are their cosets then either

g1 H1 © · · · © gn Hn = …, or ∃g ∈ g1 H1 © · · · © gn Hn , in which case

g1 H1 © · · · © gn Hn = g H1 © · · · © g Hn = g(H1 © · · · © Hn ).

The topology generated by this basis is called the pro¬nite topology of G. Note that

if H G is a normal subgroup of ¬nite index then its cosets form a partition of G

by ¬nitely many disjoint open sets. Therefore any of these cosets is also closed.

Pro¬nite topologies on groups have lots of applications in algebra and

Remark

number theory. For example, in number theory, you may want to solve an equation

f (x, y) = 0 in Z, knowing that you can solve it modulo all N . Another example

occurs in Galois theory. The idea is that if k ‚ L is an in¬nite Galois ¬eld extension,

the ¬nite extension ¬elds k ‚ K ‚ L correspond to subgroups of ¬nite index in the

in¬nite Galois group Gal(L/k). The Galois group Gal(L/k) is automatically pro¬nite,

in the sense that it is de¬ned by its ¬nite quotient groups.

7.10 Product topology

Let X and Y be topological spaces; I show how to put a topology on X — Y . Take the

set of subsets

B = {U — V ‚ X — Y } with U ‚ X and V ‚ Y open.

Then

(U1 — V1 ) © (U2 — V2 ) = (U1 © U2 ) — (V1 © V2 )

gives the ¬nite intersection property; the other two axioms are obvious, so B is a

basis for a topology on X — Y . The product topology on X — Y is de¬ned to be the

topology generated by B.

If X and Y are metric spaces, it is easy to see that the product topology on X — Y

is the topology de¬ned by any of the metrics max(d X , dY ), d X + dY , d X + dY , etc.

2 2

(see Figure 7.10). It follows that for n, m positive integers, the product topology on

Rn — Rm is the same as the metric topology on Rn+m . For example, on R2 = R — R,

7.11 THE HAUSDORFF PROPERTY 127

the sets (a1 , b1 ) — (a2 , b2 ) provide arbitrarily small open sets, but obviously not all

open sets are of this form.

7.11 The Hausdorff property

A topological space is Hausdorff 1 if for all x = y ∈ X , there exist disjoint open sets

U, V ‚ X with x ∈ U , y ∈ V . (See Figure 7.2a.) This is clearly another topological

property.

If X is Hausdorff then every point x ∈ X is closed: for if x = y there exists an open

set containing y and not x, and therefore X \ x is open. This is a weaker separation

axiom,

∀x = y ∈ X, ∃ an open set U containing y and not x

called Hausdorff™s T1 condition. (The Hausdorff condition on X introduced here is

sometimes also called T2 .)

A metric space X is automatically Hausdorff: just choose µ > 0 with

Example 1

µ< y) and set U = B(x, µ), V = B(y, µ).

1

d(x,

2

Examples 1 and 2 of 7.8 are clearly not Hausdorff. The co¬nite topol-

Example 2

ogy of an in¬nite set X (7.1 Example 2) is not Hausdorff either: a nonempty open

set is the complement of a ¬nite set, so the intersection of any two open sets is again

the complement of a ¬nite set, so nonempty. Thus these are certainly not metric

topologies.

A topology on a ¬nite set X is Hausdorff if and only if it is the discrete

Example 3

topology. Indeed, if X is Hausdorff then any point x ∈ X is closed, so every subset

of X is closed.

A topological space X is Hausdorff if and only if the diagonal

Proposition

= {(x, x) | x ∈ X } ‚ X — X

X

is closed in the product topology of X .

Note ¬rst that for any subsets U, V ‚ X ,

Proof

U —V © = {(x, x) | x ∈ U © V },

X

in other words, U — V © X is just the diagonal embedding of U © V into X — X .

A point of X — X \ X is just a pair (x, y) with x = y. Consider the problem

of ¬nding an open neighbourhood W of (x, y) in the product topology such that

1 Felix Hausdorff (1868“1942) was the originator of many of the basic ideas of metric and topological

spaces, and the author of a famous and in¬‚uential book Grundz¨ ge der Mengenlehre. He was Professor at

u

the University of Bonn until he was forced out as a Jew in 1935. He committed suicide in January 1942,

together with several members of his family, to avoid being sent to a Nazi internment camp.

128 TOPOLOGY

Figure 7.12 Separating a point from a compact subset.

W © X = …. By de¬nition of the product topology, an arbitrary small neighbourhood

of (x, y) is U — V with U, V ‚ X open and x ∈ U , y ∈ V . Now by the ¬rst remark,

U — V © X = … if and only if U © V = ….

Since X is closed if and only if X — X \ X is open, this happens if and only if

for every (x, y) with x = y there exist open sets U, V ‚ X open, with x ∈ U , y ∈ V

and U © V = …. QED

7.12 Compact versus closed

Let X be a topological space, and Y ‚ X a subset with the subspace

Proposition

topology.

If X is a compact topological space and Y ‚ X is closed, then Y is also compact.

(i)

If X is Hausdorff and Y ‚ X is compact, then Y is closed.

(ii)

In particular, if X is compact and Hausdorff, then Y ‚ X is compact if and only if it

(iii)

is closed.

(i) Suppose that V» for » ∈ are open subsets of Y , in the subspace

Proof

topology, such that Y = V» . Then by de¬nition of the subspace topology 7.5, for

each » there exists an open set U» of X such that V» = Y © U» . Now also X \ Y is open,

by the assumption that Y is closed. Therefore X = U» ∪ (X \ Y ) is an open cover of

n

X . By de¬nition of compactness, a ¬nite cover will do, say X = i=1 U»i ∪ (X \ Y ),

n

and then obviously Y = i=1 V»i .

(ii) Fix x ∈ X \ Y . For every y ∈ Y , using the Hausdorff assumption on X , choose

disjoint open sets U y and Vy with x ∈ U y and y ∈ Vy . By construction, y ∈ Vy , so

that Y ‚ Vy , or equivalently Y = (Y © Vy ). But since Y is compact, a ¬nite

number of the open sets Y © Vy cover it, and hence there is a ¬nite set of Vyi with Y ‚

n n

i=1 Vyi . Set U = i=1 U yi , which is a ¬nite intersection of opens, therefore open.

n

Since U y © Vy = … for each y, it follows that U © i=1 Vyi = …, and in particular

U © Y = …. (See Figure 7.12.) This proves that for any x ∈ Y , there exists an open

/

set U containing x disjoint from Y , and therefore Y is closed. QED

7.13 CLOSED MAPS 129

V © I — [a, b]

f: R — [a, b] ’ R

V

f(V)

Figure 7.13a Closed map.

xy = 1

f: R2 ’ R

f

)(

f(V) = R \ {0}

Figure 7.13b Nonclosed map.

7.13 Closed maps

A map f : X ’ Y between topological spaces is closed, if f (V ) ‚ Y is closed for

every closed set V ‚ X .

Consider the closed interval [a, b] ‚ R. Then the second projection

Example 1

π : [a, b] — R ’ R is a closed map (Figure 7.13a).

Start with a closed set V ‚ [a, b] — R and a point x ∈ π (V ) of the closure

Proof

of π (V ). Take a closed interval I containing x, and restrict attention to the second

projection

B = [a, b] — I ’ I.

Then B is closed and bounded in R2 , so compact (see Proposition 7.4.2); hence

V © B is compact by Proposition 7.12 (i). Therefore by Proposition 7.4.3, f (V © B)

is a compact subset of I , therefore closed in I . Therefore x ∈ π (V ), and π(V ) is

closed. QED

The projection to the x-axis R2 ’ R is not closed. For consider the

Example 2

hyperbola C : (x y = 1); it is closed in R2 , but its image in R is R \ 0 (Figure 7.13b).

If X is compact and Y Hausdorff then any continuous map f : X ’

Proposition

Y is closed.

130 TOPOLOGY

V ‚ X closed implies V compact by Proposition 7.12 (i). Therefore

Proof

f (V ) is compact by Proposition 7.4.3, and f (V ) ‚ Y is closed by Proposi-

tion 7.12 (ii). QED

7.14 A criterion for homeomorphism

Let X and Y be topological spaces and f : X ’ Y a map. I claim that

f is a homeomorphism ⇐’ f is bijective, continuous, and closed.

=’ is of course clear. If f is bijective, then f closed means exactly that f ’1 is

continuous: for U ‚ X open gives X \ U closed, which implies that f (X \ U ) is

closed; but f (X \ U ) = Y \ f (U ) because f is bijective, so f (U ) is open, that is,

f ’1 is continuous.

Theorem (™ = 0) If X is compact and Y Hausdorff, then a continuous bijective

map f : X ’ Y is a homeomorphism.

f is closed by Proposition 7.13. QED

Proof

A simple closed curve in R2 is a continuous map f : [0, 1] ’ R2 that is

Example

one-to-one except for f (0) = f (1). Write ∼ for the equivalence relation that glues the

endpoints of the interval as in Figure 7.2b. Clearly f de¬nes a continuous one-to-one

map f : [0, 1]/∼ = S 1 ’ R2 . I claim that f : S 1 ’ f (S 1 ) is a homeomorphism.

Indeed, it is a continuous one-to-one map from a compact space S 1 to a Hausdorff

space f (S 1 ) ‚ R2 . This proves that ™ = 0.

7.15 Loops and the winding number

Let D = (x, y) ∈ R2 x 2 + y 2 < 1 be the unit disc in R2 and D — = D \ (0, 0) the

punctured disc. This ¬nal section will answer the following question, left open in the

proof of Proposition 7.6.

How can we tell that D — is not homeomorphic to D?

Question

D is simply connected: any loop in D (starting and ending at P0 , say)

Answer

can be contracted in D to the constant loop; on the other hand, a loop in D — has a

winding number n around the puncture (0, 0), and the loop can be contracted if and

only if n = 0.

The intuitive picture is clear: think of taking a dog on a long lead for a walk in a

park having a tall pole in the middle. In classical math, the winding number n is the

ambiguity of 2πn in the functions arcsin x and arccos x and the ambiguity of n(2πi)

in the complex function log z. The content of the following sections is the ¬rst step in

the theory of the fundamental group π1 (X, P0 ) in algebraic topology; Theorem 7.15.3

on the winding number is closely related to the statement that π1 (D — , P0 ) = Z.

7.15 LOOPS AND THE WINDING NUMBER 131

•s

t’

‘

s

Figure 7.15a Continuous family of paths.

Recall that a path in a topological space X is a continuous map • : [0, 1] ’ X , written

7.15.1

t ’ •(t). Fix a base point P0 ∈ X . A loop in X based at P0 is a path starting and

Paths, loops

ending at P0 ; in other words, a continuous map f : [0, 1] ’ X such that f (0) =

and families

f (1) = P0 . These are called based loops (as opposed to free loops where we insist

that f (0) = f (1), but allow this to be any point in X ). A loop is allowed to cross over

itself any number of times, or even to stop for a while or go back along itself.

A family of paths (or loops) (• (s) ) depending on a parameter s ∈ [0, 1] is just an

indexed family of paths (or loops), one for each s ∈ [0, 1]. Write It for the interval

[0, 1] of the path parameter t, and Is for the interval [0, 1] of the family parameter s.

Let X be a metric space. A family of paths (• (s) ) is continuous

Tentative definition

at s if for every µ > 0, there exists a δ such that

|s ’ s | < δ =’ d(• (s) (t), • (s ) (t)) < µ for all t ∈ [0, 1].

We say that (• (s) ) is a continuous family of paths if it is continuous at all s ∈ [0, 1].

The de¬nition applies in exactly the same way to a family of based loops, except that

I insist that • (s) (0) = • (s) (1) = P0 for every s.

Note that the continuity assumption is uniform in t (the same δ is supposed to

guarantee closeness for all t). The hard thing is to understand why the de¬nition just

given is the right one. The point is that to say that the path • (s) moves just a little, we

have to guarantee that every step • (s) (t) for ¬xed t should move just a little, bounded

in t (compare Exercise 7.20).

Corresponding to a family of paths (• (s) ), consider the map

Lemma

: Is — It = [0, 1] — [0, 1] ’ X (s, t) = • (s) (t).

given by

Then (• (s) ) is a continuous family of paths if and only if is continuous. See

Figure 7.15a.

Notice that continuous is a topological property. The point of the

Remark

lemma is that it makes the notion of continuous family of paths purely topological.

If X is a topological space, the ˜uniform™ de¬nition of a continuous family of paths

is not applicable (it depends on the metric in X ); in the De¬nition below I de¬ne a

family of paths • (s) to be continuous by the property that is continuous.

132 TOPOLOGY

=’ A standard ˜divide the µ in two™ argument. Suppose we are given

Proof

(s0 , t0 ) ∈ Is — It and µ > 0. First, because • (s0 ) is continuous, there exists δ such that

d(t, t0 ) < δ =’ d(• (s0 ) (t), • (s0 ) (t0 )) < µ/2.

Next, because • (s) is a continuous family of paths at s0 , there exists a δ such that

d(s, s0 ) < δ =’ d(• (s) (t), • (s0 ) (t)) < µ/2 for all t.

Therefore max{d(s, s0 ), d(t, t0 )} < δ implies both of these inequalities, so that

d((s, t), (s0 , t0 )) = max{d(s, s0 ), d(t, t0 )} < δ =’

d( (s, t), (s0 , t0 )) ¤ d(• (s0 ) (t), • (s0 ) (t0 )) + d(• (s) (t), • (s0 ) (t)) < µ.

This proves is continuous as a function of (s, t).

⇐= In this direction, I have to use compactness of It to get uniformity in t. If

is continuous, each • (s) : It ’ X is obviously continuous. I ¬x some s0 ∈ Is , and try

to prove that (• (s) ) is a continuous family of paths at s0 . Suppose given µ > 0. Start

by working in a neighbourhood of a ¬xed t ∈ It .

Then because is continuous at (s0 , t), there exists some δ (possibly depending

on t) such that

d((s, t ), (s0 , t)) < δ =’ d(• (s) (t ), • (s0 ) (t)) < µ/2.

Therefore d(s, s0 ) < δ and d(t , t) < δ implies that • (s) (t ) is close to • (s0 ) (t) is close

to • (s0 ) (t ). In other words, for all t , there is a δ neighbourhood of t,

d(s, s0 ) < δ =’ d(• (s) (t ), • (s0 ) (t )) < µ.

Now I have proved that every point of the t-interval has a δ neighbourhood with

this property; by compactness the t-interval is covered by ¬nitely many of these, and

by taking δ to be the minimum of ¬nitely many δi I get • (s) (t) close to • (s0 ) (t) for all

t and all s close to s0 . QED

Let X be a topological space and P0 ∈ X a base point. A family of

Definition

loops • (s)

in X based at P0 is continuous, if the map

: [0, 1] — [0, 1] de¬ned by (s, t) = • (s) (t)

is continuous. A loop • : [0, 1] ’ X based at P0 is contractible in X , if there is a

continuous family of loops joining • to the constant loop •0 (de¬ned by •0 (t) = P0

for all t). A path connected space X is simply connected if every loop in X (with

every possible base point, though see Exercise 7.21) is contractible.

A homeomorphism f : X ’ Y takes paths and continuous families of paths in X

into paths and continuous families of paths in Y . In particular, being simply connected

is a topological property.

7.15 LOOPS AND THE WINDING NUMBER 133

Every loop in the unit disc D ∈ R2 is contractible. This is obvious on

Example

a sheet of paper; formally, it is best to use vector notation: if x0 is the base point,

and •(t) = xt is the loop then (s, t) = x0 + s(xt ’ x0 ) gives a continuous family of

paths connecting • to the constant path at x0 . The point is just that D is convex; the

same argument gives the same conclusion for any convex subset of Rn .

7.15.2 To discuss the winding number formally, I use ordinary Cartesian coordinates (x, y)

on the disc D, and polar coordinates (r, θ ) on the punctured disc D — . Note that r > 0,

The winding

number and that polar coordinates do not really work at the origin. The two coordinate systems

are related by the usual rules x = r sin θ, y = r cos θ.

What values do we allow for θ? Since sin and cos are periodic with period

2π, the right answer is an equivalence class of R modulo 2πZ. Note that every

equivalence class of R/2πZ has a unique representative θ ∈ [0, 2π ); in applications

θ ∈ (’π, π] may be more convenient. If you want θ to be unique, you should insist

that (x, y) = (0, 0), and choose the representative θ ∈ [0, 2π). But if you want θ to

vary continuously with (x, y), you should arrange that (x, y) stays well away from

(0, 0) and choose θ ∈ R.

Suppose that the base point P0 is in the x-axis (so that θ = 0 is

Proposition

a possible choice). Let • : [0, 1] ’ D — be a path with •(0) = P0 . Then there exist

unique continuous functions r : [0, 1] ’ R+ and : [0, 1] ’ R such that

•(t) = (r (t), (t)) for all t ∈ [0, 1].

If • is a loop, then the end point is •(1) = P0 ; hence the value (1) is of the form

2πn for some integer n.

(1) = 2π n is the winding

The integer number n in the expression

Definition

number of the loop •, written n = ν(•).

Write •(t) = (x(t), y(t)) and set r (t) = x(t)2 + y(t)2 for t ∈ [0, 1].

Proof

Clearly r (t) is continuous and strictly positive. Since [0, 1] is compact, r (t) is bounded

above and below by some R, ρ > 0. De¬ne

x(t) y(t)

•1 : [0, 1] ’ S 1 •1 (t) = , .

by

r (t) r (t)

Then •1 is continuous, because x, y and r are, and r (t) is bounded away from 0.

Now •1 (t) ∈ S 1 is certainly of the form (sin θ, cos θ) for some θ = θ(t) ∈ R. The

problem is that θ(t) is determined up to addition of multiples of 2π, and we have to

choose the value for each t to make the function continuous.

Clearly the map e : R ’ S 1 de¬ned by

e : θ ’ (sin θ, cos θ)

134 TOPOLOGY

∆+

∆“

D — covered by overlapping open radial sectors.

Figure 7.15b

¦

[() ) ( ) ) ]

ai bi ¦ bi +1

0 a1 b1 b2 1

Figure 7.15c Overlapping intervals.

de¬nes a homeomorphism of any open interval (a, b) ‚ R of length b ’ a < 2π onto

an open sector of the circle S 1 (similarly for closed). To prove the proposition, it is

enough to chop up [0, 1] into ¬nitely many short intervals Ui so that •1 maps each

Ui into such a sector, then take a suitable branch of e’1 on each of these.

To do this very explicitly, cover D — by a number of overlapping open radial sectors.

To be de¬nite, say, the ˜top™ and ˜bottom™ 200—¦ sectors

: ’10—¦ < θ < 190—¦ , : 170—¦ < θ < 370—¦ ,

+ ’

as in Figure 7.15b (or make your own choice). Let me write µ = 10—¦ = π/18, so that

the sector intervals are (0 ’ µ, π + µ) and (π ’ µ, 2π + µ). Then R is divided up into

countably many intervals

I+ = (2lπ ’ µ, (2l + 1)π + µ) I’ = ((2l ’ 1)π ’ µ, 2lπ + µ)

l l

and

for l ∈ Z, in such a way that the restriction of e to each interval I± is a homeomorphism

l

e± : I ± ’ ± .

l l

For every t ∈ [0, 1], the image •1 (t) ∈ D — is in one of the ± . Since •1 is con-

’1

tinuous, •1 ( ± ) is open, so there exists a neighbourhood U (t) ‚ [0, 1] of t with

•1 (U (t)) ‚ ± . I can assume that each of the U (t) is an open interval of [0, 1] (ex-

cept the ¬rst and last, which are half-open intervals). The U (t) form an open cover of

[0, 1], so by compactness it has a ¬nite subcover. It follows that I can choose a cover

7.15 LOOPS AND THE WINDING NUMBER 135

of [0, 1] by a ¬nite number of overlapping open intervals (Figure 7.15c)

m

with U0 = [0, b1 ), Ui = (ai , bi+1 ), Un = (am , 1], and

[0, 1] = Ui ,

0 < a1 < b1 < a2 < · · · < bm’1 < an < bm < 1,

i=0

such that •1 (Ui ) ‚ ± . (For each Ui , if there is any doubt, make the choice of ± at

the outset.)

Now since e± : I± ’ ± is a homeomorphism, we clearly de¬ne over Ui ‚ ±

l l

to be (e± )’1 —¦ •1 , and the only remaining question is the choice of l. First, •(0) = P0

l

has θ = 0 by assumption, so that either U0 ‚ + or U0 ‚ ’ . In the ¬rst case, choose

I+ , in the second choose I’ . These are forced by the requirement that (0) = 0. Next,

0 0

suppose by induction that is de¬ned and continuous on U0 ∪ U1 ∪ · · · ∪ Ui’1 . The

initial point ai of Ui is in the overlap with Ui’1 , so that is already de¬ned there.

l

This determines the choice of I± . QED

Let (• (s) ) be a continuous family of loops • (s) : [0, 1] ’ D — . Then

7.15.3 Theorem

the winding number of the loop • (s) is constant (independent of s). In particular

Winding

ν(• (0) ) = ν(• (1) ).

number is

constant in

Write ν(•) for the winding number of a loop •. The point is to show that

a family Proof

ν(•) depends continuously on the path • : [0, 1] ’ D — .

For some value s, suppose that ν(• (s) ) = n. I claim that there is a neighbourhood

Vs = (s ’ δ, s + δ) such that ν(• (s ) ) = n for all s ∈ Vs . In other words, the subset

= s ν(• (s) ) = n ‚ [0, 1]

n

is open. This claim proves the theorem, because the interval [0, 1] is connected, and

is a disjoint union of the open sets n , therefore only one value of n occurs.

First, as in the proof of Proposition 7.15.2, I normalise all the paths by dividing by

the factor r (s) (t), so that each • (s) maps to S 1 . The normalisation factor is bounded

away from 0 because Is — It = [0, 1] — [0, 1] is compact and : Is — It ’ D — is

continuous. Thus I assume from now on that • (s) : [0, 1] ’ S 1 .

Recall the construction of Proposition 7.15.2 for • (s) . There is a cover of [0, 1] = It

(s)

by a ¬nite chain of overlapping open intervals Ui = (ai , bi+1 ) such that •1 (Ui ) ‚

n

± . After this, the map just lifts ± to I± , where the value of n is determined

inductively by the already known value of the starting point (ai ).

Now I choose slightly bigger ˜top™ and ˜bottom™ sectors ± of S 1 ; to be explicit,

choose

: ’20—¦ < θ < 200—¦ , : 160—¦ < θ < 380—¦ ,

+ ’

or in the previous notation + = (0 ’ 2µ, π + 2µ), etc. As far as • (s) is concerned,

(s)

nothing has changed: I still have •1 (Ui ) ‚ ± ‚ ± , and the construction of can

be made equally well with the bigger intervals.

136 TOPOLOGY

However, by the de¬nition of continuous family of loops, there exists a small

neighbourhood s ∈ Vs ‚ [0, 1] such that also • (s ) (Ui ) ‚ ± for all s ∈ Vs . Thus I

can use the same collection of intervals Ui to construct the argument function (s )

of • (s ) for all s ∈ Vs .

Then (s ) (t) on Vs — Ui is equal to the composite e± —¦ (s , t), and hence it is

n

a continuous function of (s , t) ∈ Vs — Ui . It follows that (s ) (t) is a continuous

function of s ∈ Vs for any t. In particular, (s ) (1) is a continuous function of s ∈ Vs .

However, it is an integer multiple of 2π. Therefore it is constant for s ∈ Vs . This

proves the claim. QED

The punctured disc D — is not homeomorphic to the disc D.

Corollary 1

7.15.4

Applications

By Theorem 7.15.3, a loop • in D — of winding number = 0 is not con-

Proof

of the

tractible. On the other hand, every loop in D is contractible (Example 7.15.1). The

winding

property that a loop is contractible is a topological property, so is preserved by

number

homeomorphism. Therefore there does not exist a homeomorphism between D and

D — . QED

The same proof shows that the punctured disc D — is not homeomorphic

Remark

to the disc D with some of its boundary added, since loops in the latter are still

contractible. This concludes the proof of the main claim in Proposition 7.6: a boundary

point of a surface is topologically different from an interior point.

Let

Corollary 2 (˜Fundamental theorem of algebra™)

f (z) = z n + an’1 z n’1 + · · · + a1 z + a0

be a polynomial of degree n ≥ 1 in z, with complex coef¬cients ai ∈ C. Then there

exists a complex number ζ such that f (ζ ) = 0. In other words, C is algebraically

closed.

Write C— = C \ {0}. Obviously C— is homeomorphic to D — , so that the

Proof

de¬nition and properties of the winding number apply also to C— .

I ¬rst give the proof forgetting the small detail of the base point P0 , then explain

how to patch this up. For K ∈ R, K ≥ 0, de¬ne

• K : [0, 1] ’ C by t ’ f (K exp(2πit)).

If • K (t) = 0 for some K and some t then f (ζ ) = 0 for ζ = K exp(2πit). Assume by

contradiction that this never happens. Then • K : [0, 1] ’ C— is a continuous family

of loops in C— . When K = 0 it is the constant loop: •0 (t) = a0 for all t. When K 0

n’1

it has winding number n. Indeed, if K > 1 + i=0 |ai | the term z in f (z) is bigger

n

than all the other terms put together, so that the loop looks like K n (sin nt + i cos nt)

plus a smaller error term that does not allow the path to reach to the origin.

EXERCISES 137

However, by Theorem 7.15.3, if we assume that • K maps [0, 1] to C— , the wind-

ing number must be constant, independent of K . This is a contradiction. Therefore,

sometimes f (z) = 0.

The proof just given does not work as it stands, because Theorem 7.15.3 dealt only

in based loops. There are several ways of dealing with this; one method would be

to reprove Theorem 7.15.3 without base points, or to prove that the winding number

does not depend on the choice of a base point.

An easy ad hoc method is to de¬ne a new family of paths • K starting from the base

point P0 = a0 in the following way: we spend the ¬rst 1/3 of the time in the interval

[0, 1] plodding out from f (0) = a0 to f (K ) = • K (0) along the path f (R); then we

pursue the loop • K at 3 times the original speed, returning to f (K ) = • K (1) at time t =

2/3; then we spend the ¬nal 1/3 of the time returning from f (K ) to f (0) by retracing

our steps along the same path f (R). The new path has the same winding number as

the old, because any change in the argument θ made in plodding out to f (K ) is exactly

cancelled when we retraced our steps. The details are easy to work out. QED

Exercises

7.1 Let (X, d) be a metric space. Check that De¬nition 7.2 does indeed de¬ne a topology

on X ; in other words, check that the set T of open sets in the metric sense is a topology.

[Hint: use the triangle inequality.]

Questions on point-set topology.

X, Y, Z are topological spaces and f : X ’ Y , g : Y ’ Z continuous maps. Prove

7.2

that g —¦ f is continuous. Count the lines of your proof, and compare with the same

proof in a standard analysis or metric spaces course.

X is a metric space with metric topology T X . Prove that a sequence of points ai ∈ X

7.3

converges to l in the sense of the metric if and only if it converges in the sense of

topology as in 7.4.2.

By de¬nition, a sequence of points {xi }i=1,2,... converges to x ∈ X in a topological

7.4

space if every neighbourhood U of x contains all but ¬nitely many of the xi . Let X, Y

be topological spaces and f : X ’ Y continuous.

(a) Prove that {xi } converge to x implies { f (xi )} converge to f (x). That is, ˜continuity

implies sequential continuity™ for topological spaces.

(b) Conversely, prove that for a metric space X , this convergence for all sequences

implies that f is continuous. In other words, ˜sequential continuity implies con-

tinuity™ for metric spaces.

(c) Now let X be a topological space, not necessarily metric, in which every point x ∈

X has a countable basis of neighbourhoods (referred to in 7.2). Prove sequential

continuity implies continuity.

(d) Prove that if X is an uncountable set with the co¬nite topology (7.1 Example 2),

then there does not exist a countable basis for the neighbourhoods of x ∈ X .

(e) (Harder) Find a topological space and a map f : Y ’ X which is sequentially

continuous but not continuous.

138 TOPOLOGY

X is a metric space, x, y ∈ X and a1 , a2 , . . . a sequence of points of X . Which of the

7.5

following are topological properties?

(a) X \ x is disconnected.

(b) ai ’ x as i ’ ∞.

(c) x is in the closure of {y}.

(d) ai is a Cauchy sequence.

(e) The ball B(x, 1) is compact.

(f) Every neighbourhood of x is a countable set.

(g) The closure of the ball B(x, 1) is connected.

(h) For every compact subset V ‚ X , the complement X \ V is disconnected.

For each statement, give a proof or a counterexample, or both.

7.6 How many capital letters of the alphabet are there up to homeomorphism in a typeface

without knobs on, such as

ABCDEFGHIJKLMNOPQRSTUVWXYZ?

Scrabble players do it with K and Q.

X and Y are topological spaces and f : X ’ Y a continuous surjective map. Prove

7.7

that if X is sequentially compact, so is Y . [Hint: consider a sequence in Y and use the

stated properties of f and X . Compare the proof of Proposition 7.4.2.]

Prove that a continuous function f : X ’ R on a compact space X is bounded, and

7.8

achieves its bounds. [Hint: to get bounded, just say balls, lots of balls, . . . as before.

Let K = sup f (X ) ∈ R, which exists by the completeness axiom. By contradiction

assume that f (x) = K for all x ∈ X ; consider the open sets Uµ ‚ X de¬ned by

Uµ = x f (x) ¤ K ’ µ .]

Prove that a continuous function f : [a, b] ’ R is uniformly continuous. [Hint: for

7.9

a given µ, the de¬nition of continuity gives balls B(x, δx ), . . . ]

X is a topological space and Y ‚ X a subset with the subspace topology; prove that

7.10

every closed subset of Y is of the form Y © V with V closed in X .

X is a metric space and Y ‚ X a subset. Prove that the following two topologies on

7.11

Y are identical.

(a) Take the metric topology T X and the subspace topology TY,1 on Y .

(b) Restrict the metric d X to Y to get a metric dY , then take the metric topology TY,2

on Y corresponding to dY .

Find all the possible topologies on a set {x, y} with two points.

7.12

7.13 Study the possible topologies on a ¬nite set.

(a) If a topological space is not T1 (see 7.11) then there exist x = y such that the

constant sequence y, y, . . . converges to x. That is, x is in the closure of the set

{y}.

(b) Write x C y if x is in the closure of y, and think of this as a relation between x

and y. Prove that C is a transitive relation.

(c) De¬ne the relation x R y by

x R y ⇐’ x C y and y C x.

EXERCISES 139

Prove that R is an equivalence relation.

(d) Let Y ‚ X be an equivalence class of R; prove that the subspace topology on Y

is the indiscrete topology (no opens other than … and X ).

(e) (Harder) Use steps (a)“(d) to describe all possible topologies on a ¬nite set Y .

Let X be a topological space, ∼ an equivalence relation on X and Y = X/∼ the

7.14

quotient topological space. Think of the relation ∼ as the subset

Z (∼) = (x, y) x ∼ y ‚ X — X

where X — X is given the product topology.

(a) By imitating the proof of Proposition 7.11, prove that Y is Hausdorff if and only

if Z (∼) ‚ X — X is closed.

(b) Let Z ‚ X — X be the closure of the diagonal, considered as a relation (x ∼ y if

and only if (x, y) ∈ Z ); describe what x ∼ y means in terms of neighbourhoods

of x and y, and prove that ∼ is an equivalence relation.

(c) Prove that X has a continuous map f : X ’ X to a Hausdorff space which has

the UMP for such maps.

Exercises on surfaces.

7.15 Write down equations for a torus, a solid torus and a M¨ bius strip in terms of Cartesian

o

coordinates (x, y, z) or cylindrical polar coordinates (r, θ, z) for R3 . [Hint: you get

a torus by rotating a circle about an axis outside it, and a M¨ bius strip by letting a

o

diameter of the circle rotate simultaneously to get 1, 3, 5, . . . half-twists.]

Prove that S 2 \ {2 points} is homeomorphic to the cylinder S 1 — R. [Hint: let the two

7.16

points be the poles N and S, and think of Mercator™s projection.]

7.17 Using Figure 7.7, prove the following statements.

(a) If L = P1 is the line obtained from the equatorial circle, then P2 \ L is topologi-

cally a disc (the upper half-sphere), and a neighbourhood of L in P2 is a M¨ bius

o

strip.

(b) If Q = {x 2 + y 2 = z 2 } ‚ P2 is a conic curve, then P2 \ Q consists of two pieces,

one a M¨ bius strip and the other a disc; a neighbourhood of Q in P2 is a

o

cylinder.

Draw pictures illustrating the following statement: cutting P2 along a line is like

cutting a M¨ bius strip along its central curve, whereas cutting P2 along a conic is like

o

cutting a M¨ bius strip along the curve trisecting the width of the strip.

o

7.18 In 7.6, I obtained the M¨ bius strip, the cylinder and the torus from a square by glueing

o

its edges in a particular fashion. In Figure 7.16a, I give two other glueing rules.

(a) Show that the ¬rst pattern builds a surface homeomorphic to the projective plane

P2 .

(b) Show that the second pattern corresponds to a surface that you can build in two

steps, ¬rst glueing a cylinder as in Figure 7.6c and then identifying the circles at

the ends, carefully remembering their orientation. This surface is called the Klein

bottle. It shares with P2 the property that it cannot be embedded in R3 without

self-crossing.

7.19 The top panel of Figure 7.16b shows a surface with two handles, with a set of circles

marked on its surface, in analogy with the last panel of Figure 7.6c.

140 TOPOLOGY

Figure 7.16a Glueing patterns on the square.

Figure 7.16b The surface with two handles and the 12-gon.

(a) Verify that cutting the surface along the marked circles leads to the 12-gon on

the bottom panel of Figure 7.16b, with the edges identi¬ed as shown. Hence

conversely, glueing the 12-gon with the given pattern leads to a surface with two

handles!

(b) Triangulate the surface by triangulating the 12-gon. Compute the Euler number

˜faces ’ edges + vertexes™. Compare 9.4.

Exercises on loops

7.20 Draw the graph of the function

±

4t/s for 0 ¤ t ¤ s/4

f (t) = 2 ’ 4t/s

(s)

for s/4 ¤ t ¤ s/2

0 for s/2 ¤ t ¤ 1.

Here s ∈ (0, 1]. Have you seen anything like this before? Set f (0) (t) = 0, and prove

the following:

EXERCISES 141

(a) for any ¬xed s ∈ [0, 1] the formula • (s) (t) = ( f (s) (t), t) de¬nes a path

• (s) : [0, 1] ’ R2 (i.e. it is continuous);

(b) for ¬xed t ∈ [0, 1] the map s ’ • (s) (t) is continuous;

(c) • (s) is not a continuous family of paths in R2 in the sense of De¬nition 7.15.1;

(d) • (s) is something you would not do to a dog lead;

(e) (s, t) = f (s) (t) is not a continuous function of s, t near (0, 0).

The point of the question is to justify the tentative de¬nition in 7.15.1, in particular

to convince you of the requirement for uniformity in t.

Suppose that X is a path connected topological space and pick two points P0 , Q 0 ∈ X .

7.21

Prove that all loops in X based at P0 are contractible if and only if all loops in X

based at Q 0 are contractible. [Hint: compare the end of 7.15.4.]

8 Quaternions, rotations and the geometry

of transformation groups

Chapters 1“ 5 discussed transformations that depend continuously on parameters: for

example, Euclidean rotations in the plane that depend on the centre and the angle

of rotation. I stressed that composition of transformations is a natural operation, an

idea that led in Chapter 6 to the de¬nition of a geometric transformation group. Here

I focus on groups with a continuous family of elements, especially some examples

arising in geometry where the group of transformations has an interesting geometry of

its own. The discussion is a ¬rst introduction to some of the basic ideas of ˜continuous

transformation groups™. The formal de¬nition and a detailed treatment of this type

of ˜group-manifold™ (or Lie group) is beyond the scope of this book, but see 8.8 and

Segal [22].

As an example, recall that a rotation of E2 around a ¬xed point P is given by

the matrix cos θ ’ sinθθ , and so depends continuously on the real parameter θ. This

θ

sin cos

parameter takes values in a circle. Thus the group of rotations of E2 around a ¬xed

point has a geometry of its own, that of the circle, as shown in Figure 8.0. The relation

between rotations in the plane and the circle can be conveniently expressed in terms of

complex numbers, with the action of rotation by θ on the column vector x written

y

as multiplication of the complex number x + iy by the complex number exp(iθ) of

absolute value 1. On the other hand, the set of unit complex numbers is the circle S 1

in the complex plane.

A highlight of this chapter is Corollary 8.5.3, which applies the homeomorphism

criterion Theorem 7.14 (one of the main results of Chapter 7) to give a description in

similar terms of the topology of the groups of rotations of E3 and E4 around a ¬xed

point. The algebra of complex numbers is replaced by the algebra of quaternions

H = a + bi + cj + dk with a, b, c, d ∈ R,

where i, j, k all square to ’1 and multiply together wisely. Corollary 8.5.3 describes

the topology of the group of three- and four-dimensional rotations in terms of the

sphere S 3 of unit quaternions.

The group of three dimensional rotations is of basic importance in many areas of

mechanics and physics, describing symmetries of Euclidean space E3 , a space that

old-fashioned empiricists believe we inhabit. The quantum mechanical treatment of

142

8.1 TOPOLOGY ON GROUPS 143

θ ’ sin θ

(cos θ )

cos θ

sin

θ

S1

Figure 8.0 The geometry of the group of planar rotations.

the spin of the electron is a pretty illustration of my treatment of the topology of

the group of three-dimensional rotations. As most ingredients are at hand already, I

cannot resist the temptation to include a section on this, cribbed more or less directly

from Feynman [7]. The discussion puts together in a very satisfactory way ideas from

algebra (groups, algebra of quaternions), analysis (topology, compactness), geometry

(rotations of E3 ) and quantum physics (wave function, spin of the electron).

8.1 Topology on groups

A group G is a topological group if it has a topology de¬ned on it so that multiplication

and inverse are continuous. In more detail, a topological group is an object G having

two quite different structures: a collection of open subsets satisfying the axioms for

a topology, and a multiplication map with identity and inverse satisfying the group

axioms. I require the group structure to respect the topological structure in the sense

that

mult : G — G ’ G inv : G ’ G

and

g ’ g ’1

(g, h) ’ gh

are both continuous maps of topological spaces; here G — G has the product topology

of 7.10.

Any ¬nite group G is a topological group under the discrete topology.

Example 1

The groups (R, +) and (R— , —) are topological groups with respect

Example 2

to the usual topology of R. This is just a fancy way of restating the fact, used all

over the place in a ¬rst analysis course, that the four operations addition, subtraction,

multiplication and division are continuous on the reals.

A substantial generalisation of the previous example brings us back

Example 3

to the linear geometries of Chapters 1“5. Recall the general linear group GL(n, R)

of n — n real invertible matrixes. Note that GL(n, R) is a subset of the set of real

2

matrixes M(n — n, R) = Rn . This latter is a metric space, and therefore has a natural

metric topology. Moreover, it is an easy fact that matrix multiplication and inverse

144 GEOMETRY OF TRANSFORMATION GROUPS

are continuous. Hence GL(n, R) is a topological group. As a consequence, af¬ne

transformations Aff(n, R) (compare 4.5) also form a topological group.

The group R— of constant diagonal matrixes is a subgroup of GL(n + 1, R), the

centre of GL(n + 1, R), that is, the subgroup of elements commuting with every

element g ∈ GL(n + 1, R) (see 5.5 and Exercise 6.3). The quotient

PGL(n + 1, R) = GL(n + 1, R)/R—

is a topological group with the quotient topology. This is of course the group of

projective linear transformations of Pn familiar from 5.5, the projective linear group.

The orthogonal group

Example 4

O(n) = A ∈ GL(n, R) A A = 1n ,