## Week by Week

(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 nth 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: T1 is finer than T2 (both topologies on the same set X) if every T2-open set is also T1-open; equivalently, if the identity map id: (X,T1) --> (X,T2) 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 Br(x) = {y : d(x,y) < r} forms a basis for the metric topology on Rn. The set of balls with rational radii about points with rational coordinates also forms a basis for the metric topology on Rn. 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.

Week 2
1. Proof that R is connected. Proposition: If X is the union of connected subsets Ca, all of which intersect the connected subset C, then X is connected. Class exercise: prove that the circle is connected. Proposition: Rn 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 Bd(x), x in X,
d > 0, form the basis of a topology. In a compact metric space, every infinite sequence has a convergent subsequence.

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 = Pia Xa (each Xa 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 Xa has the property that it cannot be covered by open sets whose inverse image belongs to R (here is where compactness of Xa is used), so has a point xa contained in no such set. The point x=(xa) lies in some open set V of R; and is contained in the intersection of some finite number of inverse images of Vi (open in Xi), 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: T0, T1, T2: 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.

Week 4
1. More separation axioms: Regular: T1 and T3; Normal: T1 and T4. 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 T1 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:

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 {fn} 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. |fn| < = Mn, and \Sum1inftyMn < \infty) then \Sum1inftyfn converges to a continuous function.

Next a useful fact:
(*) Given a space X, a closed subset C and a continuous ga:C --> [-a,a], let Ha = { ga > = a/3} and Ka =
{ ga < = -a/3}. (These sets are closed.) Then the function ha: X --> [-a/3,a/3] equal to a/3 on Ha and -a/3 on Ka (given by the Urysohn Lemma) satisfies |ga(x) - ha(x)| < = 2/3 for x in C. Proof: Class exercise.

Now the proof of the TET. Step 1: apply (*) to the function ga =f', and call f1 the ha it produces. Here a=1. By (*), the function f'-f1 maps C into [-2/3,2/3]. Notice that |f1| < = 1/3.

Step 2: apply (*) to ga = f'-f1, with a=2/3, and call f2 the ha it produces. By (*), the function f'-f1-f2 maps C into [-4/9,4/9]. Notice that |f2| < = 2/9.

Step n: apply (*) to ga = f'-f1-...-fn-1, with a= (2/3)n-1, and call fn the ha it produces. By (*), the function f'-f1-f2-...-fn maps C into [-(2/3)n,(2/3)n]. Notice that |fn| < = (1/2)(2/3)n.

Wrap-up: Apply the Lemma to the sequence f1, f2, ... , with Mn = (1/2)(2/3)n. Since the sum of the Mn is 1, this sequence converges to a continuous function f: X-->[-1,1]. On the other hand, for x in C, the difference |f'-f1-f2-...-fn| is less than or equal to (2/3)n. In the limit this goes to 0.

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 S1 --> S1 does *not* extend to D2 (the closed unit disc in R2) even though S1 is closed and D2 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 S1 in D2 is a neighborhood retract), absolute neighborhood retract. The sphere Sn 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.

Week 7
1. Hilbert Space: the space of all square-summable sequences of real numbers: H = {y = (yi), i=1..\infty | \sum yi2 < \infty} with distance function dist(x,y) = \sqrt(\sum(xi - yi)2). Class exercise: check the triangle inequality for this metric.

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 x1, x2, etc. and define f(x1) = (2,0,0,0,...), f(x2) = (0,2,0,0,...), etc.; for non-isolated points the f-values we construct will have all components bounded by 1; so the images of the isolated points are all distinct and isolated from the rest of the image. This part of the embedding is done.

We delete the isolated points from the rest of the construction. Let B1, B2, ... be the elements of the countable basis. There is therefore a countable number of pairs (Bi, Bj) and in particular a countable number P1, P2, ... of pairs such that closure(Bi) is contained in Bj, and Bi is different from Bj. For each Pn, let hn be the function : X--> [0,1] given (X is normal) by the Urysohn Lemma, equal to 0 on closure(Bi) and to 1 on X-Bj. Let f = (h1, (1/2)h2, (1/3)h3, ...). Clearly f maps X to H.

* 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 Rn.

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 ft(x) = F(x,t), f1 is the identity and f0 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 Cinfty 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 C0, 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 RN. 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 RN, N finite.) For x in M, let Tx be the tangent space to M at x, and Px: RN --> Tx 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, Py: U --> Uy = Py(U) in Ty is a diffeomorphism.

(b) For any y in the closure of U, the map Py (which as a projection must be distance-decreasing) does not shrink distances by more than 1/2. I.e. if z1 and z2 are in the closure of U, then d(Py(z1), Py(z2)) > = (1/2)d(z1,z2).

(c) For any y in U, and any z0 in U, the real-valued function defined on Uy by Py(z)--> d(z,z0)^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(tA + (1-t)B) < = tF(A) + (1-t)F(B) for A and B in its domain, and 0 < = t < = 1.)

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 BR(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 BR(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 BR(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 BR(x) is relatively compact.
Claim 1: For every x in K the projection Px(BR(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(BR(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 = Px(zi). By relative compactness, a subsequence of the zi converges to some point z, with Px(z)=z' by continuity. By property (a) z cannot be an interior point of the ball, or else z' would be an interior point of the image. So d(x,z) > = R; by property (b) d(Px(x), Px(z)) = d(x,z') > = R/2, justifying the claim.
Claim 2: Choose x in K and a positive r < = R/4. Then for any y in Br(x), the projection Px: Br(y) --> Tx is a diffeomorphism onto a convex set.

2. Proof of claim: Let z1'and z2' be two points in Px(Br(y)); we must show that the segment S between them also lies in Px(Br(y)). By hypothesis z1'= Px(z1), z2'= Px(z2), with d(z1,y), d(z2,y) < r. It follows that d(z1,x), d(z>2,x) < 2r, since d(x,y) < r. So d(z1',x), d(z2',x) < 2r since the projection Px cannot increase distances, and Px(x) = x. It follows by an elementary plane geometry argument that d(z',x) < 2r for every point of S. Since by Claim 1 all points at distance < 2r of x in Tx lie in the image Px(BR(x)), it follows that all of S lies in Px(BR(x)). Now BR(x) is contained in one of the U's from Lemma 2, and so inherits property (c). Interpreting property (c) replacing y by x and z0 by y, it follows that on Px(BR(x)) the function which takes z' to d(z,y)^2 is convex. Since at each end of S this function has value < r^2, this inequality must hold for every z' = tz1' + (1-t)z2' in S. So S is contained in Px(Br(y)), as claimed.

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 BRi(xia) = Uia where a ranges over a finite set of indices; similarly {closure of Wj} can be covered by a finite number of BRj(xjb) = Ujb, 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 Uia, Ujb, ... (a finite number!) and suppose that Ri is the largest of the corresponding R's. Then each of the Ukc is a set of the form Bs(y) for y in Br(x), and some s < r. It follows from Claim 2 that Px(Ukc) 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.

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 * * * * * *

forming a matrix of rank, clearly, n.
A smooth embedding has a well-defined tangent space at each point. By definition this is the set of velocity vectors of smooth curves passing through that point. Notice that if M is an abstract manifold, the tangent vectors at x exist abstractly, and manifest themselves by acting as directional derivations at x, by the rule c'(0) f = (foc)'(0); whereas once M is smootly embedded in R^n, these tangent vectors become concrete "arrows" based at x and filling out an affine n-dimensional space tangent to M at x. Class exercise: suppose H(x) = 0 in R^n (suppressing the chart-index) and let ci(t) = H^(-1)(0...t...0), t in the i-th position. Then the vectors ci'(0) form a basis for the tangent space at x, and their images (F o ci)'(0) span the tangent n-plane to F(M) at F(x).

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 fog is homotopic to the identity map of Y, and gof 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 y0 is a point of a topological space Y, let L(Y,y0) (the loops based at y0) be the space of maps f: I --> Y with f(0)=f(1)=y0 (compact-open topology). Definition of "two loops homotopic relative to y0." This homotopy is an equivalence relation. The set of homotopy classes is called the fundamental group of Y based at y0, denoted pi1(Y,y0). 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(2x) if x <= 1/2, and = g(2x-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.

Week 12
1. Elementary properties of the fundamental group.
* Independence of basepoint if space is arcwise connected. Suppose x0 and x1 are two points of the arcwise-connected space X. So there is a path p: [0,1] --> X, with p(0) = x0, p(1) = x1. Let q be the reverse path: q(x) = p(1-x). The path p defines a homomorphism P: pi1(X,x1) --> pi1(X,x0) as follows: if f is in L(X,x1), then the concatenation p f q is in L(X,x0) 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 p1 and p2 are two paths with reverse paths q1 and q2. Then the isomorphisms P1 and P2 are related by conjugation by the homotopy class of the loop p2 q1: P2[f} = [p2 q1] P1[f] [p1 q2]. So if the fundamental group is abelian, the isomorphism is path-independent.
* Functoriality. A basepoint-preserving map h: X,x0 --> Y,y0 induces a homomorphism h*: pi1(X,x0) --> pi1(Y,y0); the identity map induces the identity isomorphismn, and if k: Y,y0 --> Z,z0 is also b.p.-p., then (koh)* = k*oh*. In other words, pi1 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 pi1(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 pi1. The values for the groups are entered in the third diagram (pi1(D^2) = 0 because the disk is contractible). Obviously the completion is impossible.

          id                    id* = id                 id
S^1 ------> S^1    pi(S^1) ------> pi1(S^1)    Z -------> Z
\         .          \               .           \       .
\       .            \             .             \     .
\     .              \           .               \   .
\   .                \         .                 \ .
D^2                  pi1(D^2)                   0


* The homomorphisn h* only depends on the (basepoint-preserving) homotopy class of h. In particular, a homotopy-equivalence induces an isomorphism of fundamental groups.

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 = (x1,x2) is a point on the circle, then {x1 > 0}, {x1 < 0}, {x2 > 0}, {x2 < 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(x1) and arcsin(x2).
Path-lifting theorem: Given a covering p: Y --> X, a path or curve c: [0,1] --> X, and a point y0 lying over c(0). Then there exists a unique curve c': [0,1] --> Y with c'(0) = y0 and p(c'(t))= c(t). If two curves c1 and c2 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.

Week 13
Corollary of Path-Lifting Theorem: If p: Y,y0 --> X,x0 is a covering map, then the induced homomorphism p*: pi1(Y,y0) --> pi1(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 pi1(S^1,b0) --> Z by

D([a])=a'(1)

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.

Week 14
1. Proposition: Let p: E --> B be a covering map, with E path- connected. Let e0 and e1 be two basepoints for E, both projecting to b0 in B. Then the subgroups p*pi1(E,e0) and p*pi1(E,e1) are conjugate in pi1(B,b0). Furthermore any subgroup H conjugate to p*pi1(E,e0) is the image of the homotopy group of E for some basepoint e'.
Homotopy Lifting Theorem. Let p:E,e0 --> B,b0 be a covering map, with E path=connected, and let f: Y,y0 --> B,b0 be continuous. Then there exists a continuous lifting F: Y,y0 --> E,e0 (lifting means p o F = f) if and only if f*pi1(Y,y0) is contained in p*pi1(E,e0).
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 y0 to y, and (f o c)' is the unique lift of f o c (a curve starting at b0) to a curve starting at e0. 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,e0 --> B,bo and q: E',e'0 --> B,b0 are equivalent if and only if the subgroups p*pi1(E,e0) and q*pi1(E', e'0) are conjugate in pi1(B,b0). (In the proof, the "only if" is straightforward; the "if" follows from the Homotopy Lifting Theorem.)

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 pi1(E) = {1}. (A Universal cover of B'').