(Skip to Week 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 )

**Week 1**

1. Categories; examples: *Top, Gr, R-Vect*. Functors;
example (from later in the course) the *n*th homology
group as a functor *Top --> Gr*. Topological space:
definition *via* axioms for open sets. Group exercise:
check that *R* is a topological space when ``open''
is defined in terms of epsilon-neighborhoods. Continuous
map. Group exercise: check consistency with delta-epsilon
definition of continuity for maps *f*:*R*-->*R*.

2. Comparison of topologies: *T*_{1} is finer than
*T*_{2} (both topologies on the same set *X*)
if every *T*_{2}-open set is also *T*_{1}-open;
equivalently, if the identity map *id:* (*X*,*T*_{1}) -->
(*X*,*T*_{2}) is continuous. The discrete and the coarse
topologies. Putting a finer topology on *X* makes it
``easier'' for a map *from X* to be continuous, and
``harder'' for a map *to X* to be continuous.

Bases and Sub-bases for a topology. Group exercise:
The set of balls *B*_{r}(*x*) =
{*y* : d(*x*,*y*) < *r*}
forms
a basis for the metric topology on R^{n}. The set of balls with
rational radii about points with rational coordinates also forms
a basis for the metric topology on R^{n}. This basis is
countable.

Connectedness. The rationals in the subspace topology are
separated, or disconnected. Connectedness of the reals
*via* the Least Upper Bound Axiom.

(Back to Top)

**Week 2**

1. Proof that R is connected. Proposition: If *X*
is the union of connected subsets *C*_{a},
all of which intersect the connected subset *C*,
then *X* is connected. Class exercise: prove
that the circle is connected. Proposition: R^{n} minus
the origin is connected

(*n* > one). Proposition:
The continuous
image of a connected set is connected. Class exercise:
give the 2-line proof. Proposition: The *n*-sphere
is connected.

2. Compactness: every covering has a finite sub-cover. Class exercise: R is not compact. Proof that [0,1] is compact, using the L.U.B. axiom. The continuous image of a compact set is compact. Class exercise: prove it.

Limit point of a set. In a compact space, every infinite subset has a limit point.

Metric on a space *X*. The metric topology: check
that the set of balls *B*_{d}(*x*),
*x* in *X*,

*d* > 0, form the
basis of a topology. In a compact metric space, every
infinite sequence has a convergent subsequence.

(Back to Top)

**Week 3**

1. Closed sets and the open/closed, union/intersection duality. Class
exercise: a closed set (=the complement of an open set) contains all
its limit points. Proof that compactness (defined in terms of open covers)
is equivalent to the Finite Intersection Property. Product spaces and
the product (or "Tychonoff") topology. Preparation for Tychonoff's
Theorem: the Maximal Principle (in a partially ordered set, every
totally ordered subset (chain) is contained in a maximal chain).

2. Proof that this Maximal Principle implies the ``Other Maximal
Principle:'' Let **R* be a nonempty collection of sets with the property
(*)q for every chain in **R*, there is an element *A* in
**R* which
contains every element of the chain; then **R* has a maximal element.

Proof of Tychonoff Theorem, following Nachbin:
Suppose *X* = Pi_{a} *X*_{a}
(each *X*_{a} compact) has an open
covering with no finite subcover. Let **R* be the set of all
such coverings. Check that **R* satisfies (*) and so there is a
maximal such covering *R*. Maximality means that if any single
other open set *V* is added to *R*, then there is a finite
*R'* contained in *R* such that the sets in *R'*,
together with *V*, cover *X*. Each *X*_{a}
has the
property that it cannot be covered by open sets whose inverse image
belongs to *R* (here is where compactness of *X*_{a}
is used),
so has a point *x*_{a} contained in no such set.
The point *x*=(*x*_{a}) lies in some
open set *V* of *R*;
and is contained in the intersection of some finite number of
inverse images of *V*_{i} (open in *X*_{i}), with the
intersection contained in *V*, since such intersections form a
basis for the product topology. None of these inverse images belongs
to *R*, so each of them, together with a finite subset
*R'*_{i},
covers *X*; then their intersection, together with the union of
the *R'*_{i}, also covers; so *V*, together with that union,
i.e. a finite subset of *R*, also covers, a contradiction.

Separation axioms: T_{0}, T_{1}, T_{2}: Hausdorff. Class exercise: in a
Hausdorff space, every compact set is closed. In a Hausdorff space,
any 2 compact sets can be separated by open sets.

(Back to Top)

**Week 4**

1. More separation axioms: Regular: T_{1} and T_{3}; Normal: T_{1} and T_{4}.
Class exercise: a closed subset of a compact Hausdorff space is
compact. A compact Hausdorff space is regular. Proposition: *X* regular
is equivalent to: *X* is T_{1} and if *p* belongs to an open
set *U*, then
there is an "interpolated" open set *V*
with *p* in *V* and *V*-closure
contained in *U*. A metric space is regular. A compact Hausdorff space
is normal. Definition of "completely normal:"
if *H* and *K* are completely
disjoint sets (neither intersects the closure of the other) then
they can be separated by open sets (there exist disjoint,
open *U* and *V*
with H contained in *U* and *K* in *V*).
Proposition: a metric space is
completely normal.

2. The Urysohn Lemma: If *A*, *B* are disjoint, closed subsets of a
normal space *X*, then there exists a continuous
*f*: *X*-->[0,1] with
*f*(*A*)=0, *f*(*B*)=1.
* Proof:*

(Back to Top)

**Week 5**

The Tietze Extension Theorem: Any continuous function *f'* defined on a
closed subset *C* of a normal space *X*
and with values in an interval
(say, [-1,1]) can be extended to a continuous function *f*
defined on all of *X*.

*Proof* (following H&Y)

First a Lemma: If {*f _{n}*} is a sequence of continuous,
real-valued functions defined on a topological space X, individually
uniformly bounded in absolute value by the terms of a convergent series
(i.e. |

Next a useful fact:

(*) Given a space *X*, a closed subset *C*
and a continuous *g _{a}:C * --> [

Now the proof of the TET.
Step 1: apply (*) to the function *g _{a}
=f'*, and call

Step 2: apply (*) to *g _{a} = f'-f*

Step *n*: apply (*) to *g _{a} = f'-f*

Wrap-up: Apply the Lemma to the sequence *f*_{1}, *f*_{2}, ... ,
with *M _{n}* = (1/2)(2/3)

(Back to Top)

**Week 6**

1. Retracts. Turn the Tietze Extension Theorem around and interpret
it as a statement about the topology of [0,1]. Note that the
identity map *S*^{1} --> *S*^{1} does *not* extend to
*D*^{2}
(the closed unit disc in *R*^{2})
even though *S*^{1} is closed and *D*^{2} is normal
(proof later this term). Extension problems in terms of
completion of commutative diagrams.
Definitions of *retract, absolute
retract* (Proof that [0,1] is an AR),
*neighborhood retract* (Class exercise: Prove that
*S*^{1} in *D*^{2} is a neighborhood retract),*
absolute neighborhood retract*. The sphere *S*^{n} is
an ANR. Brief allusion to smooth n-dimensional manifolds,
which are also ANR's by essentially the same argument
(proof in a differential topology course).
For later use: proof of the homotopy extension theorem.

{A *commutative diagram* is a convenient and
visual way of organizing many problems in topology and in algebra.
Students should familiarize themselves with this schema, which
is not used in H&Y. Another archaism in this week's material is
the use of ``throwing *X* into *Y*''
in the sense of ``mapping *X* into *Y*'' or
``defined on *X* with values in *Y*.'' This
would certainly raise a few eyebrows if uttered today.}

2. Separability. Definitions of separable and completely
separable (CS; also, ``second countable''). Examples: Euclidean spaces. CS => separable,
and a separable metric space is CS. Class exercise: the
cartesian product of two separable spaces is separable,
and same for CS. A subspace of a CS space is CS. Think of
CS as an analogue of compactness, as in the following three
propositions. In a
CS space, every uncountable set *X*
has a limit point. In fact *X*
contains uncountably many limit points of itself.
In a CS space, every open cover contains a countable
subcover (Lindeloef's Theorem). A regular CS space
is (completely) normal.

(Back to Top)

**Week 7**

1. Hilbert Space: the space of all square-summable sequences of
real numbers: *H* = {*y* = (*y _{i}*),

Prop.: A completely separable, normal space *X* can be
embedded in *H*.

Proof: (after H&Y) we construct a map *f* = (*fi*): *X*
--> *H*, and show
that it is continuous, 1-1 and open. It follows that *f* is a
homeomorphism onto its image, i.e. an embedding.

*Construction of *f*. The isolated points of *X* are disposed of
first. By separability there can be at most countably many of
them (each one constitutes an open set in *X*); number them
*x _{1}*,

We delete the isolated points from the rest of the construction.
Let *B _{1}*,

* *f* is continuous. Just as in H&Y. Compare with the proof that a
uniform limit of continuous functions is continuous. In fact,
let *f*{*N*} = *f* with all components past *N*
set to 0. Then *f*{*N*}
is continuous because each component is continuous, as usual.
On the other hand dist(*f*{*N*}(*x*),*f*(*x*))
< = \sqrt(1/(*N*+1)^{2} +
1/(*N*+2)^{2} + ...) < \sqrt(1/*N*), which goes to zero as
*N* --> \infty independently of *x*.

* *f* is 1-1.
Just as in H&Y, noting that the isolated points have been
disposed of separately.

* *f* is open. As in H&Y.

2. Paracompactness. Definition. A compact Hausdorff space is paracompact. Class exercise: The real line is paracompact. Reading assignment: local compactness, the one-point compactification of a locally compact space.

**Week 8**

1. Partitions of unity. (Following Munkres, Chapter 4,
section 7). A normal space covered by a finite collection
of open sets admits a partition of unity subordinate to
(also, "fitting," "dominated by") the covering. Definition
of *m*-dimensional manifold. A compact *m*-manifold can be
embedded in R^{n}.

Definition of differentiable *m*-dimensional
manifold of class *C*^*k*. Mention of
Whitney's theorem:
Every *C*^*k*
manifold, *k* > = 1, is *C*^*k*-diffeomorphic to
a *C*^{infty} manifold. Definition of (differentiable, or "smooth")
deformation retraction:
*A* is contained in *X* as (smooth) deformation retract if there
is a (smooth) continuous *F*: *X* times [0,1] --> *X* such
that, setting *f*_{t}(*x*) =
*F*(*x*,*t*),
*f*_{1} is the identity and *f*_{0} is a retraction of *X*
onto *A*. A space is *(smoothly) contractible* if it has a point as
(smooth) deformation retract.

Introduction to Andre Weil's proof
of the following theorem: A *C*^{infty} paracompact manifold
admits a *simple covering*, i.e. a locally finite covering
such that each set of the covering, and each intersection of two
or more sets of the covering, is differentiably contractible.

Library exercises: 1. Locate that theorem of Whitney's. 2. Is there
a proof of the simple covering theorem in the textbook literature?
3. What is the status of that theorem for *C*^{0}, i.e.
continuous, manifolds and maps. Does a topological manifold admit
a simple covering?

2. Statements of Lemmas necesary for Weil's proof.

*Lemma 1.* A smooth manifold with a locally finite open covering admits
a smooth partition of unity fitting that covering.

Application: A smooth compact *m*-manifold can be smoothly
embedded in R^{N}. If the manifold is covered by *n*
smooth coordinate charts, we can take *N* = *nm* + *n*.
If the manifold is paracompact, we can use a partition of unity
fitting a locally finite covering by smooth coordinate charts to
embed it in Hilbert space; restricted to any compact portion of the
manifold, this map embeds that portion in a finite-dimensional
subspace.

Notation for Lemma 2. The manifold *M* is identified with its
image in euclidean space. (We will work on compact parts of *M*,
so the euclidean space can be some R^{N}, *N* finite.)
For *x* in *M*, let *T*_{x} be the tangent
space to *M* at *x*, and
*P*_{x}: R^{N} -->
*T*_{x}
orthogonal projection onto that (affine)
subspace.

*Lemma 2.* Any *x* in *M* has a neighborhod *U*
with the following
three properties.

(a) For any *y* in *U*, *P*_{y}: *U* -->
*U*_{y} = *P*_{y}(*U*)
in *T*_{y}
is a diffeomorphism.

(b) For any *y* in the closure of *U*, the map
*P*_{y}
(which as a projection
must be distance-decreasing) does not shrink distances by more than
1/2. I.e. if *z*1 and *z*2 are in the closure of *U*,
then *d*(*P*_{y}(*z*1),
*P*_{y}(*z*2))
> = (1/2)*d*(*z*1,*z*2).

(c) For any *y* in *U*, and any *z*0 in *U*,
the real-valued function
defined on *U**y* by *P*_{y}(*z*)-->
*d*(*z*,*z*0)^2 is a *convex function*.
(This is a function defined on an open set of R^*m*. Such a function
*F* is convex if
its graph is convex upwards, in the sense that *F*(*t**A* +
(1-*t*)*B*) < =
*t**F*(*A*) + (1-*t*)*F*(*B*) for *A*
and *B* in its domain, and 0 < = *t* < = 1.)

(Back to Top)

**Week 9**

1. Existence of simple coverings (continued). Some motivation:
notice that in Euclidean space any convex set is
contractible, and the intersection of two convex sets is convex.
So any locally finite covering by convex sets is automatically simple.
The idea behind this proof is to use the local identification possible
between the manifold and its tangent plane to lift back certain convex
sets in the tangent planes to sets in *M* with the right properties.
The complication comes from having to coordinate sets coming from
nearby, but different tangent planes. Now back to work.

Let *K* be a compact
subset of *M*. Then *K* is covered by
a finite number of neighborhoods
*U* as in Lemma 2. Let *R* = *R*(*K*)
be the Lebesgue number of this cover:
a positive number such that for any *x* in *K*
the ball *B*_{R}(*x*) (= the set
of points in *K* at distance < *R*
from *x*) is contained in at least one
of the *U*'s. (See homework for proof that such a number exists). So
*B*_{R}(*x*)
will inherit properties (a), (b) and (c) given by the Lemma. It
costs nothing to require *R* < 1. This will have the consequence
that every *B*_{R}(*x*)
is contained in the closure of one of the *W*'*i*, because
the function *fi* enters into the computation
of distance. In particular
each *B*_{R}(*x*) is relatively compact.

*Claim 1:* For every *x* in *K* the projection
*Px*(*B*_{R}(*x*)) will
contain all the points in *Tx* at distance
< *R*/2 from *x*. *Proof
of claim:* Suppose *z*' is a boundary point of
*Px*(*B*_{R}(*x*)), i.e.
a point in the closure both of this set and of its complement. We
will show *d*(*x*,*z*') > = *R*/2.
By hypothesis *z*' is the limit of a
sequence *z*'_{i} in the image, i.e.
*z*'_{i} = *P _{x}*(

2. *Proof of claim:* Let z_{1}'and z_{2}'
be two points in
*P _{x}*(

*
Now we can construct the simple cover. Let each {closure of Wi'}
play the part of K in Tuesday's analysis, and let Ri be the
corresponding Lebesgue number. The compact set {closure of Wi}
can be covered by the
B_{Ri}(x_{ia}) = U_{ia} where
a ranges over a finite set
of indices; similarly {closure of Wj} can be covered by a
finite number of
B_{Rj}(x_{jb}) = U_{jb}, etc.
The collection of all the
balls involved covers (since the Wi cover) and is a locally
finite covering, since the Wi are. *

Claim 3: This covering is simple. Proof of claim: suppose x lies in the intersection Z of U_{ia}, U_{jb}, ... (a finite number!) and suppose that Ri is the largest of the corresponding R's. Then each of the U_{kc} is a set of the form B_{s}(y) for y in B_{r}(x), and some s < r. It follows from Claim 2 that Px(U_{kc}) is a convex set in Tx, and therefore so is their intersection. The intersection of the projections is therefore smoothly contractible, and this contraction can be lifted, via the diffeomorphism Px, to a contraction of Z to a point.

*(Back to Top)*

*
Week 10
1. 80-minute Midterm Examination
on Point-Set Topology (through
partitions of unity).*

*
2. Proofs of Lemmas for the Simple Covering Theorem.
Lemma 1: see Whitney, Geometric Integration
Theory, Appendix III.
Lemma 2: first in constructing the smooth embedding into Euclidean
space, it is convenient to use the Urysohn-type functions *before*
their normalization, so the component functions of the embedding are *

*
f1, f1H1, f2, f2H2, etc *

*
with fi identically one on Wi, and zero off
W'i, and Hi the
coordinate chart from Ui to R^n. Claim F
is a smooth
embedding (meaning it is a homeomorphism onto its image and
at each point x of M it has rank n,
where this is defined as the
rank of the matrix of partial derivatives of the components of F
with respect to a (any) set of local coordinates at x. Proof
of claim: The point x must belong to some Wi;
since fi is
then identically one on a neighborhood of x, the partial
derivatives of F with respect to the coordinates given by
Hi are then
*

* * * * * * 1 0 ... 0 * * * * * * * * * * * * 0 1 ... 0 * * * * * * ... * * * * * * 0 0 ... 1 * * * * * *

A smooth embedding has a well-defined

(Back to Top)

**Week 11**

1. (End of proof of Lemma 2 postponed.)

*Homotopy Theory*. The aim of this and other parts of Algebraic
Topology is to assign invariants to spaces and maps. An invariant
for spaces would be, in the most primitive sense, the assignment
of a number *N*(*X*) to each topological space *X*
in such a way that
if *X* and *Y* are homeomorphic, then *N*(*X*)=
*N*(*Y*). *Example* *N*(*X*) =
the number of connected components of *X*. (A subset *A* of *X*
is a
connected component if *A* is connected and any connected subset
containing *A* is equal to *A*.) Class exercise: this number is
an invariant. Definition of arc-connected; the number of
arc-connected components is an invariant.
For topological spaces *X*,*Y* the symbols *C*(*X*,
*Y*) or *Y*^*X* denote
the set of all continuous maps from *X* to *Y*. The most natural
topology to put on this space is the *compact-open topology*.
A sub-basis for this topology is indexed by the pairs (*K*,*U*)
where *K* is compact in *X* and *U* open in *Y*.
The sub-basis element
*B*(*K*,*U*) is the set of all *f* such that
*f*(*K*) is contained in *U*.
Class exercise: the *B*(*K*,*U*)
do form a sub-basis. *Homotopy
classes* of maps from *X* to *Y* are the arc-connected components
of *C*(*X*,*Y*) in the compact-open topology, and two maps are
*homotopic* if they are in the same component. The arc
joining them is called a *homotopy*.

*Proposition:* if *X* is locally compact, and *F*:
*X* x *I* --> *Y* is a continuous map, then the map
f: I --> C(X,Y) defined by *ft*(x) = *F*(*x*,*t*)
is a continuous map with respect to the compact-open topology.

2. Class exercise: prove the converse: If *X* is locally compact,
and *ft* is a
homotopy between *f0* and *f1*, then the map *F*:
*X* x *I* --> *Y*
defined by *F*(*x*,*t*) = *ft*(x) is a continuous map.

Homotopy is an equivalence relation in *C*(*X*,*Y*);
the set of homotopy
classes is denoted [*X*,*Y*].

A map *f*: *X* --> *Y* is a
*homotopy equivalence* if there exists a
map *g*: *Y* --> *X* such that *f*o*g*
is homotopic to the identity map of *Y*,
and *g*o*f* to the identity map of *X*.
The spaces *X* and *Y* are then said
to be *homotopy equivalent*. Check that in that case, for any
space *Z*, [*X*,*Z*]=[*Y*,*Z*] and
[*Z*,*X*]=[*Z*,*Y*].

*The Fundamental Group*. If *y*0 is a point of a topological
space *Y*, let *L*(*Y*,*y*0)
(the *loops based at y*0) be the space
of maps *f*: I --> *Y* with *f*(0)=*f*(1)=*y*0
(compact-open topology).
Definition of "two loops homotopic relative to *y*0." This homotopy
is an equivalence relation. The set of homotopy classes is called
the fundamental group of *Y* based at *y*0,
denoted pi_{1}(*Y*,*y*0). The
group structure is defined by concatenation: the product of two
equivalence classes [*f*] and [*g*]
is defined to be [*f***g*], where
*f***g* is the loop given by *f***g*(*x*)=
*f*(2*x*) if *x* <= 1/2, and
= *g*(2*x*-1) if *x* > 1/2. Four things to check. Product is
well-defined (does not depend on representatives of equivalence
classes) and the three group axioms.

(Back to Top)

**Week 12**

1. Elementary properties of the fundamental group.

* Independence of basepoint if space is arcwise connected. Suppose
*x*0 and *x*1 are two points of the arcwise-connected
space *X*. So
there is a path *p*: [0,1] --> *X*, with
*p*(0) = *x*0, *p*(1) = *x*1. Let
*q* be the reverse path: *q*(*x*) = *p*(1-*x*).
The path *p* defines a
homomorphism *P*: pi_{1}(*X*,*x*1) -->
pi_{1}(*X*,*x*0) as follows: if
*f* is in *L*(*X*,*x*1), then the concatenation
*p* *f* *q* is in *L*(*X*,*x*0)
Define *P*[*f*]=[*p* *f* *q*].
Check: *P*[*f*] is independent of the
representative chosen; *P* is 1-1 and onto, *P*
is a homomorphism.

* Dependence of this isomorphism on path. Suppose *p*1 and *p*2
are two
paths with reverse paths *q*1 and *q*2.
Then the isomorphisms *P*1 and *P*2
are related by conjugation by the homotopy class of the loop
*p*2 *q*1:
*P*2[*f*} = [*p*2 *q*1] *P*1[*f*]
[*p*1 *q*2]. So if the fundamental group is
abelian, the isomorphism is path-independent.

* Functoriality. A basepoint-preserving map *h*: *X*,*x*0
--> *Y*,*y*0
induces a homomorphism *h**: pi_{1}(*X*,*x*0)
--> pi_{1}(*Y*,*y*0);
the identity map induces the identity isomorphismn, and
if *k*: *Y*,*y*0 --> *Z*,*z*0 is also b.p.-p.,
then (*k*o*h*)* = *k**o*h**. In other
words, pi_{1} is a functor from the category *Top.* of
topological-spaces-with-basepoints and basepoint-preserving maps
to the gategory *Gr* of groups and group homomorphisms.

* Importance of functoriality. It means that a commutative
diagram of maps becomes a commutative diagram of groups. For
example, suppose we know that pi_{1}(*S*^1) = Z (since the group
is abelian, we can suppress the basepoint). Then the fact that the
identity map of *S*^1 cannot be extended to the disk *D*^2 can be
proved as follows. The extension would mean completing the
commutative triangle on the left by a dotted line upwards.
(the line on the left is the inclusion map of *S*^1 in *D*^2).
The center triangle is the translation of the problem into
algebra, using the functoriality of pi_{1}. The values for the
groups are entered in the third diagram (pi_{1}(*D*^2) = 0 because
the disk is contractible). Obviously the completion is impossible.

id id* = id id* The homomorphisnS^1 ------>S^1 pi_{(}S^1) ------> pi_{1}(S^1) Z -------> Z \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ .D^2 pi_{1}(D^2) 0

2. The fundamental group and covering spaces.
A map *p*: *Y* --> *X*
is a *covering map* (and *Y* is a *covering space* of *X*)
if each point *x* in *X* has an *evenly covered neighborhood*: a
neighborhood *U* such that *p*^(-1) *U*
is a disjoint union of subsets
{*Vi*} of *Y*, restricted to each of which
*p*: *Vi* --> *U* is a
homeomorphism.

The standard example: *p*: R --> *S*^1 given by
*p*(*t*) =(cos(2 pi *t*),sin(2 pi *t*)).
Check that if *x* = (*x*1,*x*2)
is
a point on the circle, then {*x*1 > 0}, {*x*1 < 0},
{*x*2 > 0}, {*x*2 < 0} can be used as evenly covered
neighborhoods (according to which of the inequalities
*x* satisfies; it must satisfy at least one!) via
translations of the maps arccos(*x*1) and arcsin(*x*2).

*Path-lifting theorem:* Given a covering
*p*: *Y* --> *X*,
a path or curve *c*: [0,1] --> *X*, and a point *y*0
lying over
*c*(0). Then there exists a unique curve *c*': [0,1] --> *Y*
with *c*'(0) = *y*0 and *p*(*c*'(*t*))=
*c*(*t*). If two curves *c*1 and *c*2
are homotopic keeping endpoints fixed, then the lifts
*c*'1 and *c*'2 satisfy *c*'1(1) = *c*'2(1)
and are also homotopic
keeping endpoints fixed.

(Back to Top)

**Week 13**

Corollary of Path-Lifting Theorem: If *p*: *Y*,*y0*
--> *X*,*x0*
is a covering map, then the induced homomorphism
*p**: pi_{1}(*Y*,*y0*)
--> pi_{1}(*X*,*x0*) is injective. (
H & Y, p.189.)

Theorem: The fundamental group of the circle is infinite cyclic.
Following Munkres, p. 340, let p: R
--> S^1 be the standard example as above, so p(0) = (1,0) = b0.
Define a map D from pi_{1}(S^1,b0) --> Z by

where [a] is the homotopy class of the loop a based at b0, and a':[0,1] --> R is the unique lift of a starting at 0. The proof consists then in checking that D is well-defined, a homomorphism, one-one and onto.

(Back to Top)

**Week 14**

1. Proposition: Let *p*: *E* --> *B*
be a covering map, with *E* path-
connected. Let *e*0 and *e*1 be two basepoints for *E*,
both
projecting to *b*0 in *B*. Then the subgroups
*p**pi_{1}(*E*,*e*0) and
*p**pi_{1}(*E*,*e*1) are conjugate in pi_{1}(*B*,*b*0).
Furthermore
any subgroup *H* conjugate to *p**pi_{1}(*E*,*e*0)
is the image of
the homotopy group of *E* for some basepoint *e*'.

Homotopy Lifting Theorem. Let *p*:*E*,*e*0 -->
*B*,*b*0 be a
covering map, with *E* path=connected, and let *f*:
*Y*,*y*0
--> *B*,*b*0 be continuous. Then there exists a continuous
lifting *F*: *Y*,*y*0 --> *E*,*e*0
(lifting means *p* o *F* = *f*) if
and only if *f**pi_{1}(*Y*,*y*0) is contained in
*p**pi_{1}(*E*,*e*0).

The "only if" part is clear. For the "if" define *F*(*y*)
to be (*f* o *c*)'(1) where *c* is a curve in *Y*
from *y*0 to *y*,
and (*f* o *c*)' is the unique lift of *f* o *c* (a curve
starting at *b*0) to a curve starting at *e*0. The algebraic
hypothesis is necessary for proving this map is well-defined.
Then continuity must be checked.

2. Two covering spaces over the same base are *equivalent*
if there is a homeomorphism between them which commutes with
the covering maps.

Class exercise: the "Furthermore..." part of Tuesday's
Proposition.

Theorem: two arc-connected covering spaces *p*: *E*,*e*0
--> *B*,*b*o
and *q*: *E*',*e*'0 --> *B*,*b*0 are
equivalent if and only if the
subgroups *p**pi_{1}(*E*,*e*0) and
*q**pi_{1}(*E*', *e*'0) are conjugate in
pi_{1}(*B*,*b*0). (In the proof, the "only if"
is straightforward; the "if" follows from the Homotopy Lifting
Theorem.)

(Back to Top)

**Week 15**

1. Construction of universal covering spaces.

Theorem. Given a topological space *B*
which is connected, locally arc-connected,
and semi-locally simply connected (each point has a neighborhood *U* such
that any loop in *U* is contractible in *B*),
then there exists a connected
covering
space *p*: *E* --> *B* such that pi_{1}(*E*) = {1}.
(``A Universal cover of *B*'').

Proof following Munkres,
pp 394-397, with the simplification *H*={1}.
Today
we defined *E* and *p* and proved *p* was a covering map.
(Continuity of *p*: class exercise.)

2. (continuation) Proof that E is arc-connected and simply connected.
(Continuity of lifted paths left for students to read in
Munkres.

Final Exam: Wednesday, December 13 8:30 AM in the regular classroom.