INTRODUCTION TO MANIFOLDS - PART 1/3 Contents 1. What is Algebraic Topology?

INTRODUCTION TO MANIFOLDS - PART 1/3
T.K.SUBRAHMONIAN MOOTHATHU
Contents
1. What is Algebraic Topology?
1
2. A warm-up with quotient maps, etc.
3
3. Homotopy and the fundamental group
6
4. Computations and applications of the fundamental group
11
5. Galois theory of covering spaces
19
We plan to discuss some introductory topics from Algebraic Topology and Differential Topology.
The student may complement this by studying the basics of Differential Geometry from other
sources. Topology, Basic Algebra, and Multivariable Calculus are the pre-requisites.
1. What is Algebraic Topology?
Here is a loose introduction to Algebraic topology. Checking whether two topological spaces are
homeomorphic is a difficult problem in general. In Algebraic Topology, we make a compromise and
consider the simpler problem of checking the equivalence of spaces under a weaker notion called
homotopy. Roughly speaking, two topological spaces are homotopic if one space can be continuously
deformed to the other by stretching and shrinking but without tearing and pasting.
For example, consider a line segment, a circle and a compact annulus on the Euclidean plane.
No two of the three spaces are homeomorphic (∵ line segment minus one interior point is not
connected, circle minus one point is connected but circle minus two points is not connected, and
1
2
T.K.SUBRAHMONIAN MOOTHATHU
annulus minus two points is connected). But, the annulus can be continuously shrunk to the circle
and the circle can be continuously stretched to get the annulus so that the circle and the annulus
are homotopic. However it is not possible to deform the line segment continuously to get the circle
or the annulus since we are not allowed to paste together the end points of the line segment; hence
the line segment is not homotopic to either the circle or the annulus. What distinguishes the circle
(or the annulus) from the line segment is the presence of a ‘hole’ in the middle.
Though we will not cover much, some aspects of Algebraic topology can be summarized as below:
(i) Topological spaces that look locally like nice geometric objects in Rn may have holes of different
dimensions, where by the dimension of a hole we mean the dimension of the boundary of the hole1.
A circle and annulus in R2 have 1-dimensional holes, a sphere in R3 has a 2-dimensional hole, and
a torus in R3 has two 1-dimensional holes. Continuous deformation can neither create nor destroy
holes in a topological space. For two spaces to be homotopic, it is necessary that both should have
the same number of holes in each dimension.
(ii) Algebraic objects (groups, rings, modules, vector spaces) can be attached to many topological
spaces in such a way that the information about holes gets stored in these algebraic objects. If
two topological spaces have different sets of algebraic objects attached to them, we can conclude
that the two spaces are not homotopic (and in particular, not homeomorphic). Moreover, these
attached algebraic objects are often of finite nature (for example, finitely generated groups) so that
certain topological problems finally get reduced to elementary combinatorial problems.
(iii) Continuous maps which look locally like projections appear in various contexts. They are
useful in transferring theories up and down between the top space and the base space.
Topological spaces that we consider are assumed to be Hausdorff unless we use the phrase ‘general
topological space’. Often we will have to consider a topological space together with a fixed base
point in it. We will use the following brief expressions for convenience:
1The notion of a hole of a particular dimension is subject to one’s interpretation: compare homotopy groups and
homology groups.
INTRODUCTION TO MANIFOLDS - PART 1/3
Expression
Meaning
X∈T
X is a topological space.
(X, x0 ) ∈ T∗
X ∈ T and x0 ∈ X.
f ∈ C(X, Y )
X, Y ∈ T and f : X → Y is continuous.
3
f : (X, x0 ) → (Y, y0 ) is continuous (X, x0 ), (Y, y0 ) ∈ T∗ , f ∈ C(X, Y ), and f (x0 ) = y0 .
2. A warm-up with quotient maps, etc.
Since quotient spaces occur at many places, some points about them are worth noting.
Definition: Let X, Y ∈ T and q : X → Y be a continuous surjection. Then q induces an equivalence
relation on X by the condition that x1 ∼ x2 iff q(x1 ) = q(x2 ), and Y as a set can be identified
with the collection of equivalence classes. The quotient topology on Y induced by q is {V ⊂ Y :
q −1 (V ) is open in X}, which is finer than the existing topology of Y . If the two topologies on Y
coincide (i.e., V ⊂ Y is open in Y ⇔ q −1 (V ) is open in X), then q is called a quotient map.
Some elementary observations: Let the spaces mentioned below be general topological spaces.
(i) Let q : X → Y be a quotient map. For D ⊂ Y , we have D is open (closed) in Y ⇔ q −1 (D) is
open (closed) in X. It follows that a map f : Y → Z is continuous ⇔ f ◦ q : X → Z is continuous.
e be a quotient space of X and q : X → X
e be the quotient map. Let f : X → X be
(ii) Let X
continuous, and suppose f (x1 ) ∼ f (x2 ) whenever x1 ∼ x2 , i.e., f (q(x)) = q(f (x)) for every x.
e → X
e by the rule fe(q(x)) = q(f (x)). And fe is continuous by (i)
Then f induces a map fe : X
since fe ◦ q = q ◦ f is continuous. Similarly if g : X → Z is continuous and g is constant on each
e → Z by the
equivalence class, i.e., if g(x1 ) = g(x2 ) whenever x1 ∼ x2 , then g induces a map ge : X
rule ge(q(x)) = g(x). Again, ge is continuous by (i) since ge ◦ q = g is continuous.
(iii) If q1 : X → Y and q2 : Y → Z are quotient maps, then q2 ◦ q1 : X → Z is a quotient map.
(iv) Let q : X → Y be a continuous surjection. If q is either an open map or a closed map, then q
is a quotient map. If X is compact, then q is a closed map and hence a quotient map.
(v) A quotient map need not be open or closed. Let q : R2 → [0, ∞) be the quotient map sending
(x, y) to 0 if x < 0 and (x, y) to x if x ≥ 0. The image of the open set {(x, y) : x < 0} is {0}, which
is not open. The image of the closed set {(x, y) : x > 0 and xy = 1} is (0, ∞), which is not closed.
(vi) The product of two open maps is open, but the product of two closed maps need not be closed;
if we consider I × 0 : R × R → R × R, then the image of the closed set {(x, y) : x > 0 and xy = 1}
is (0, ∞) × {0}, which is not closed.
4
T.K.SUBRAHMONIAN MOOTHATHU
(vii) [Hausdorff but not regular; this is for Exercise-2] Let X = R, with topology generated by
{U ⊂ R : U is Euclidean open} ∪ {Q}. Note that W ⊂ X is open iff there are Euclidean open sets
U, V ⊂ R with W = U ∪ (Q ∩ V ). Since the topology on X is finer than Euclidean topology, X is
Hausdorff. But X is not regular since the closed set R \ Q cannot be separated from 0 (verify).
Exercise-1: Let the spaces mentioned below be general topological spaces.
(i) Let {x} × K ⊂ W ⊂ X × Y , where K is compact and W is open. Then there exists an open
neighborhood U ⊂ X of x with U × K ⊂ W .
(ii) [Whitehead’s theorem] If q : X → Y is a quotient map and Z is locally compact, then q × IZ :
X × Z → Y × Z is a quotient map.
(iii) Let X1 , X2 be locally compact, and q1 : X1 → Y1 , q2 : X2 → Y2 be quotient maps. Then
q1 × q2 : X1 × X2 → Y1 × Y2 is a quotient map.
(iv) In general, the product of two quotient maps need not be a quotient map.
∪
[Hint: (i) For each y ∈ K, we have (x, y) ∈ Uy × Vy ⊂ W for some open Uy × Vy . If K ⊂ kj=1 Vyj ,
∩
take U = kj=1 Uyj . (ii) Let g = (q × IZ ), and W ⊂ Y × Z be such that g −1 (W ) is open in
X × Z. To show W is open in Y × Z, consider (y0 , z0 ) ∈ W . There is open V ⊂ Z containing
z0 with V compact such that {y0 } × V ⊂ W . If q(x0 ) = y0 , then {x0 } × V ⊂ g −1 (W ). Let
U = {x ∈ X : {x} × V ⊂ g −1 (W )}, which is open in X by part (i). Check that q −1 (q(U )) = U and
hence q(U ) × V is an open neighborhood of (y0 , z0 ) contained in W . (iii) q1 × q2 = (q1 × I) ◦ (I × q2 ).
(iv) q × I : Q × Q → Q/Z × Q is not a quotient map; see p.111 of Brown, Topology and Groupoids.]
e of R obtained by
In general, a quotient space may not be Hausdorff. The quotient space X
e intersect.
collapsing Q to a singleton is not Hausdorff since any two nonempty open sets in X
Good to know some conditions ensuring that the quotient is Hausdorff.
Exercise-2: Let X, Y be general topological spaces with X Hausdorff and q : X → Y be a quotient
map. Let R ⊂ X 2 be the equivalence relation R = {(x1 , x2 ) : q(x1 ) = q(x2 )} ⊂ X 2 . Then,
(i) If Y is Hausdorff, then R is closed in X 2 .
(ii) If R is closed in X 2 , then Y need not be Hausdorff.
(iii) If q is an open map and R is closed in X 2 , then Y is Hausdorff.
(iv) If X is locally compact and R is closed in X 2 , then Y is Hausdorff.
[Hint: (i) Y is Hausdorff iff its diagonal ∆Y is closed in Y 2 . Note (q × q)−1 (∆Y ) = R. (ii) Suppose
X is Hausdorff but not regular, and A ⊂ X be a closed set which cannot be separated from some
b ∈ X \ A. Let Y be the quotient space obtained by collapsing A to a singleton. Here R = A2 ∪ ∆X ,
which is closed in X 2 . If Y is Hausdorff, the pre-images of open sets separating q(A) and q(b) would
INTRODUCTION TO MANIFOLDS - PART 1/3
5
separate A and b, a contradiction. (iii) q × q is an open map, and Y 2 \ ∆Y = (q × q)(X 2 \ R). (iv)
q × q is a quotient map by Exercise-1 and (q × q)−1 (∆Y ) = R so that ∆Y is closed.]
Definition: X ∈ T is called a topological group if there is a group structure on X such that the maps
x 7→ x−1 from X to X and (x, y) 7→ xy from X 2 to X are continuous; equivalently, if (x, y) 7→ xy −1
from X 2 to X is continuous. For example, (Rn , +), (S1 , ·), (GL(n, R), ·) are topological groups.
Exercise-3: (i) Let X, Y ∈ T and f : X → Y be continuous. Define x1 ∼ x2 iff f (x1 ) = f (x2 ).
e of X is Hausdorff.
Then the quotient space X
(ii) Let X be a topological group and Y ⊂ X be a closed subgroup. Then the quotient map
q : X → X/Y is an open map, and the quotient space X/Y (space of left cosets) is Hausdorff.
e be the quotient map. If q(a) ̸= q(b), then there are disjoint open sets
[Hint: (i) Let q : X → X
V1 , V2 ⊂ Y separating f (a) and f (b). Let Uj = f −1 (Vj ) ⊂ X and note q −1 (q(Uj )) = Uj so that
e (ii) If U ⊂ X is open, then q −1 (q(U )) = U Y = ∪
q(Uj ) is open in X.
y∈Y U y, which is open in
X so that q(U ) is open. Now aY = bY iff b = ay for some y ∈ Y . So the equivalence relation is
R = {(a, ay) : a ∈ X, y ∈ Y }. Assume X is metrizable for simplicity. If (an , an yn ) 7→ (b, z), then
(yn ) = (an−1 an yn ) → b−1 z ∈ Y and hence (b, z) = (b, bb−1 z) ∈ R, which shows R is closed.]
In Algebraic Topology, path connected spaces are more relevant than connected spaces.
Exercise-4: Let X ∈ T be path connected, and locally path connected (i.e., every point has a base
consisting of path connected open sets). Then each connected component of X is path connected
and clopen in X. [Hint: The hypothesis implies each path component is open in X. Hence each
path component, being the complement of the union of the other path components, is also closed.]
Some common spaces: (i) [n-sphere] Sn = {x ∈ Rn+1 : |x| = 1} for n ≥ 0. In particular, S0 =
{−1, 1} and S1 is the unit circle.
(ii) [n-disc] Dn = {x ∈ Rn : |x| ≤ 1} for n ≥ 0. Observe that ∂Dn+1 = Sn .
(iii) [n-torus] Tn = Rn /Zn = S1 × · · · × S1 (product of n copies of S1 ).
(iv) [real projective n-space] RPn = Sn /(x ∼ −x), i.e., the quotient space of Sn obtained by
identifying each x ∈ Sn with its antipodal pair −x. By Exercise-2(iv), RPn is Hausdorff. We
may also think of RPn as a quotient of Rn+1 \ {0} by the equivalence relation R = {(x, rx) : x ∈
Rn+1 \ {0} and r ∈ R \ {0}}, i.e., RPn is the space of lines passing through the origin in Rn+1 .
Remark: For n ≥ 1, the spaces Sn , Dn , Tn and RPn are all compact, path connected, and locally
path connected, where the case of RPn is justified by the fact that RPn is a quotient of Sn .
6
T.K.SUBRAHMONIAN MOOTHATHU
In a finite group, there may exist non-trivial relations among different elements. Loosely speaking, a group in which there are no non-trivial relations among the generators is called a free group
(we will define it precisely later). For the moment, let us see what a free abelian group is.
Definition: The free abelian group G generated by a set A is G :=
⊕
a∈A Z
(direct sum of copies of
Z, with one copy for each a ∈ A). An element of G will be of the form (na )a∈A , where na ∈ Z and
na ̸= 0 for only finitely many a ∈ A. Equivalently, an element of G can be thought of as a finite
∑
formal sum
na a with na ∈ Z and a ∈ A. The rank of G is defined as the cardinality of A.
Exercise-5: (i) Two free abelian groups of finite rank are isomorphic iff they have the same rank.
(ii) Every abelian group is the quotient of a free abelian group.
⊕
[Hint: (ii) If A is an abelian group, consider the free abelian group G = a∈A Z generated by A.
∑
Check that q : G → A defined as (na )a∈A 7→ a∈A na a is a surjective group homomorphism and
apply the First isomorphism theorem of groups to identify A with G/ker(q).]
3. Homotopy and the fundamental group
The notion of homotopy captures the idea of continuous deformation. We will attach a group
called the fundamental group to a topological space X by defining a group operation on the homotopy classes of closed paths at a point. This group contains data about 1-dimensional holes.
Definition: Let X, Y ∈ T . (i) Let f, g ∈ C(X, Y ). We say f is homotopic to g if there is continuous
h : X × [0, 1] → Y such that h(x, 0) = f (x) and h(x, 1) = g(x) for every x ∈ X, and we write f ∼ g
(via h), or just f ∼ g. In this case h is said to be a homotopy from f to g. We say f ∈ C(X, Y )
is a homotopy equivalence between X and Y if there is g ∈ C(Y, X) such that f ◦ g ∼ IY and
g ◦ f ∼ IX . Spaces X, Y are said to be homotopic (or, homotopically equivalent, or, of the same
homotopy type) if there is a homotopy equivalence f ∈ C(X, Y ), and in this case we write X ∼ Y .
Example: (i) Let X ∈ T , Y ⊂ Rn be convex, and f, g ∈ C(X, Y ). Then f ∼ g (via h), where
h : X × [0, 1] → Y is h(x, t) = (1 − t)f (x) + tg(x).
INTRODUCTION TO MANIFOLDS - PART 1/3
7
(ii) Let X, Y ∈ T and suppose Y is path connected. Let f, g ∈ C(X, Y ) be constant maps, say
f ≡ c and g ≡ d, where c, d ∈ Y . If α : [0, 1] → Y is a path with α(0) = c and α(1) = d, then f ∼ g
(via h), where h : X × [0, 1] → Y is h(x, t) = α(t).
(iii) Let X = R2 \ {0}, Y = S1 , let f : X → Y be f (x) = x/|x|, and g : Y → X be the inclusion
map g(y) = y. Note that f ◦ g = IY , and g ◦ f = f ∼ IX (via h), where h : X × [0, 1] → [0, 1] is
h(x, t) = tx + (1 − t)f (x). Hence X is homotopic to Y .
Exercise-6: (i) [Pasting lemma] Let X, Y ∈ T and let X1 , . . . , Xk ⊂ X be closed subspaces with
∪
X = kj=1 Xj . If fj : Xj → Y is continuous for 1 ≤ j ≤ k and fi = fj on Xi ∩ Xj , then f : X → Y
defined as f (x) = fj (x) for x ∈ Xj is continuous.
(ii) If X, Y ∈ T , then homotopy is an equivalence relation on C(X, Y ).
(iii) Let X, Y, Z ∈ T . If X ∼ Y and Y ∼ Z, then X ∼ Z.
∪
[Hint: (i) If Z ⊂ Y is closed, then check that f −1 (Z) = kj=1 fj−1 (Z) is closed in X. (ii) If f ∼ g
(via h1 ), then g ∼ f (via h2 ), where h2 (x, t) = h1 (x, 1 − t). If f1 ∼ f2 (via h1 ) and f2 ∼ f3 (via
h2 ), then f1 ∼ f3 (via h3 ), where h3 (x, t) = h1 (x, 2t) for 0 ≤ t ≤ 1/2 and h3 (x, t) = h2 (x, 2t − 1)
for 1/2 ≤ t ≤ 1 (use pasting lemma to see h3 is continuous).]
Definition: Let X, Y ∈ T . We say f ∈ C(X, Y ) is null-homotopic if f is homotopic to a constant
map. A topological space X is contractible if IX is null-homotopic. Every convex subset of a
Euclidean space is contractible and thus Rn , Dn , Rk × [0, 1]n−k are contractible.
Exercise-7: If X ∈ T is contractible, then X is path connected. [Hint: Let c ∈ X and suppose IX
is homotopic to the constant map c via a homotopy h. Then for any x1 , x2 ∈ X, t 7→ h(xj , t) is a
path from xj to c for j = 1, 2.]
[101] Let X ∈ T be path connected. Then the following are equivalent:
(i) X is contractible.
(ii) IX is homotopic to each constant map from X into X.
(iii) X is homotopic to a singleton.
(iv) Every f ∈ C(X, X) is null-homotopic.
(v) Any two f, g ∈ C(X, X) are homotopic to each other.
(vi) Any two f, g ∈ C(Y, X) are homotopic for each Y ∈ T .
(vii) Any two f, g ∈ C(X, Y ) are homotopic for each path connected space Y .
Proof. As X is path connected, any two constant maps of X are homotopic, giving the equivalence
of (i), (ii), and (iii). The implications (vi) ⇒ (v) ⇒ (iv) ⇒ (i) and (vii) ⇒ (i) are trivial.
8
T.K.SUBRAHMONIAN MOOTHATHU
(i) ⇒ (vi): Let c ∈ X and h : X × [0, 1] → X be a continuous map with h(x, 0) = x and h(x, 1) = c.
Then f ∼ c (via h1 ), where h1 (y, t) = h(f (y), t), and g ∼ c (via h2 ), where h2 (y, t) = h(g(y), t).
Also recall that homotopy is an equivalence relation.
(i) ⇒ (vii): Let h be as above with h(x, 0) = x and h(x, 1) = c. Let c1 = f (c) and c2 = g(c).
Then f ∼ c1 (via h1 ), where h1 (x, t) = f (h(x, t)), and g ∼ c2 (via h2 ), where h2 (x, t) = g(h(x, t)).
Moreover, the constant maps c1 and c2 on X are homotopic since Y is path connected.
[102] Let Y ∈ T , and f : Sn → Y be continuous. Then the following are equivalent:
(i) There is a continuous map fe : Dn+1 → Y extending f .
(ii) f is null-homotopic.
Proof. (i) ⇒ (ii): Let c = fe(0) ∈ Y . Then c ∼ f (via h), where h(x, t) = fe(tx).
(ii) ⇒ (ii): If c ∈ Y and c ∼ f (via h), define fe : Dn+1 → Y as fe(tx) = h(x, t) for x ∈ Sn and
t ∈ [0, 1], where a little work is needed to show fe is continuous. Or define fe(x) = c for 0 ≤ |x| ≤ 1/2
and fe(x) = h(x/|x|, 2|x|−2) for 1/2 ≤ |x| ≤ 1; here pasting lemma ensures that fe is continuous.
Any two paths in a path-connected space are homotopic by [101](vii) since [0, 1] is contractible.
To get a non-trivial equivalence relation on paths, and to define the fundamental group, we will
now consider a restricted type of homotopy that preserves the end-points of paths.
Definition: Let X ∈ T and x0 , x1 ∈ X. Let P(X, x0 , x1 ) = {α ∈ C([0, 1], X) : α(0) = x0 and α(1) =
x1 }, the collection of all paths in X from x0 to x1 . If x0 = x1 , we just write P(X, x0 ) and any
member of P(X, x0 ) is called a closed path or loop in X at x0 . We say α, β ∈ P(X, x0 , x1 ) are
path-homotopic if α ∼ β (via h), where h : [0, 1]2 → X has the additional property that h(0, t) = x0
and h(1, t) = x1 for every t ∈ [0, 1], i.e., h(·, t) ∈ P(X, x0 , x1 ) for each t ∈ [0, 1]. In this case h is
called a path-homotopy. Note that path-homotopy is an equivalence relation on P(X, x0 , x1 ). We
will continue to use the notation α ∼ β for path-homotopy also.
Exercise-8: Let X ∈ T . (i) If paths α, β in X are path-homotopic, and f ∈ C(X, Y ), then f ◦ α is
path-homotopic to f ◦ β. (ii) If α is a path in X, and ϕ : [0, 1] → [0, 1] is a continuous map with
ϕ(0) = 0 and ϕ(1) = 1, then α is path-homotopic to α ◦ ϕ. [Hint: If α ∼ β (via h), then f ◦ h is a
path-homotopy from f ◦ α to f ◦ β. (ii) Since [0, 1] is convex, I[0,1] and ϕ are path-homotopic. So
the composition of α with them are also path-homotopic by part (i). Specifically, h : [0, 1]2 → X
given by h(s, t) = α((1 − t)s + tϕ(s)) is a path-homotopy from α to α ◦ ϕ.]
Definition: (i) If a < b and c < d are reals, the unique continuous map ϕ : [a, b] → [c, d] having a
linear graph and satisfying ϕ(a) = c, ϕ(b) = d, is called the canonical map from [a, b] to [c, d].
INTRODUCTION TO MANIFOLDS - PART 1/3
9
(ii) [Inverse of a path] Let X ∈ T . If α ∈ P(X, x0 , x1 ), then the inverse of α is the path α ∈
P(X, x1 , x0 ) defined as α(s) = α(1 − s) for s ∈ [0, 1]. Clearly, α = α.
(iii) [Product of paths] Let P = {0 = a0 < a1 < · · · < ak−1 < ak = 1} be a partition of [0, 1],
let X ∈ T , x0 , x1 , . . . , xk ∈ X, and let αj ∈ P(X, xj−1 , xj ) for 1 ≤ j ≤ k. Then their product
(α1 · · · αk )P w.r.to the partition P is defined by the condition that (α1 · · · αk )P restricted to [aj−1 , aj ]
is αj ◦ ϕj , where ϕj : [aj−1 , aj ] → [0, 1] is the canonical map. By pasting lemma, this product is
continuous. If P is the uniform partition P = {0 < 1/k < 2/k < · · · < (k − 1)/k < 1}, then we
write α1 · · · αk for (α1 · · · αk )P . Note that (α1 · · · αk )(s) = αj (ks − (j − 1)) if s ∈ [(j − 1)/k, j/k].
Let (X, x0 ) ∈ T . Our next aim is to give a group structure to the the space P(X, x0 )/ ∼ of
path-homotopy classes of loops in X at x0 by defining a binary operation as [α] ∗ [β] := [αβ]. The
result [103] below establishes the necessary properties for this binary operation.
[103] Let X ∈ T . (i) Let αj ∈ P(X, xj−1 , xj ) for 1 ≤ j ≤ k, let P = {0 = a0 < a1 < · · · < ak = 1}
and Q = {0 = b0 < b1 < · · · < bk = 1}. Then (α1 · · · αk )P is path-homotopic to (α1 · · · αk )Q .
Consequently, α1 (α2 α3 ) is path-homotopic to (α1 α2 )α3 .
(ii) If α1 , α2 ∈ P(X, x0 , x1 ) are path-homotopic, and β1 , β2 ∈ P(X, x1 , x2 ) are path-homotopic,
then the products α1 β1 , α2 β2 ∈ P(X, x0 , x2 ) are path-homotopic.
(iii) Let ex0 ∈ P(X, x0 ) denote the constant path at x0 , let α ∈ P(X, x0 , x1 ) and β ∈ P(X, x1 , x0 ).
Then we have the following path-homotopies: ex0 α ∼ α, βex0 ∼ β, and ex0 ∼ αα.
(iv) Let α ∈ P(X, x0 , x1 ), β ∈ P(X, x1 , x2 ) and γ ∈ P(X, x0 , x2 ). If αβ is path-homotopic to γ,
then β is path-homotopic to αγ, and α is path-homotopic to γβ.
Proof. (i) Let ϕ : [0, 1] → [0, 1] be the homeomorphism defined by the condition that ϕ restricted
to [aj−1 , aj ] is the canonical map from [aj−1 , aj ] to [bj−1 , bj ]. Then (α1 · · · αk )P = (α1 · · · αk )Q ◦ ϕ
and hence the first assertion follows by Exercise-8(ii). Next observe that α1 (α2 α3 ) = (α1 α2 α3 )P
and (α1 α2 )α3 = (α1 α2 α3 )Q , where P = {0 < 1/2 < 3/4 < 1} and Q = {0 < 1/4 < 1/2 < 1}.
10
T.K.SUBRAHMONIAN MOOTHATHU
(ii) If we have path-homotopies α1 ∼ α2 (via h1 ) and β1 ∼ β2 (via h2 ), then verify that α1 β1 ∼ α2 β2
(via h), where h(s, t) = h1 (2s, t) for 0 ≤ s ≤ 1/2 and h(s, t) = h2 (2s − 1, t) for 1/2 ≤ s ≤ 1.
(iii) Let ϕ : [0, 1] → [0, 1] be the continuous map defined by the condition that ϕ([0, 1/2]) = {0}
and ϕ restricted to [1/2, 1] is the canonical map from [1/2, 1] to [0, 1]. Then α is path-homotopic
to α ◦ ϕ = ex0 α by Exercise-8(ii). Similarly, β is path-homotopic to βex0 . A path-homotopy from
ex0 to αα is h(s, t) = (αt αt )(s), where αt ∈ P(X, x0 , α(t)) is defined as αt (s) = α(st).
(iv) Path-homotopies: αγ ∼ α(αβ) ∼ (αα)β ∼ ex1 β ∼ β, and similarly γβ ∼ (αβ)β ∼ α.
Definition: Let (X, x0 ) ∈ T∗ and π1 (X, x0 ) := P(X, x0 )/ ∼ be the collection of path-homotopy
classes of loops in X at x0 . Define a binary operation ∗ on π1 (X, x0 ) as [α] ∗ [β] = [αβ]. By
[103], the binary operation ∗ is well-defined, and (π1 (X, x0 ), ∗) is a group with identity [ex0 ], where
[α]−1 = [α]. This group is called the fundamental group of X at x0 . A loop in X can also be
thought of as a continuous map from S1 to X. More generally, one can define the nth homotopy
group πn (X, x0 ) by considering C(Sn , X), etc. However, we will study only π1 (X, x0 ).
[104] [Fundamental group is independent of the base point for path connected spaces] If X ∈ T is
path connected and α ∈ P(X, x0 , x1 ), then [β] 7→ [αβα] is a group isomorphism from π1 (X, x0 ) to
π1 (X, x1 ). Thus we write π1 (X) (up to isomorphism), and call it the fundamental group of X.
Proof. The map [β] 7→ [αβα] is well-defined by [103](ii), and has inverse [γ] 7→ [αγα]. The map is
a group homomorphism since [αβ1 α] ∗ [αβ2 α] = [αβ1 ααβ2 α] = [αβ1 β2 α], also by [103].
[105] [Homomorphism induced by a continuous map] (i) Let f : (X, x0 ) → (Y, y0 ) be continuous.
Then f∗ : π1 (X, x0 ) → π1 (Y, y0 ) given by f∗ ([α]) = [f ◦ α] is a well-defined group homomorphism.
(ii) If (X, x0 ) ∈ T∗ , then (IX )∗ is the identity map of π1 (X, x0 ).
(iii) If f : (X, x0 ) → (Y, y0 ) and g : (Y, y0 ) → (Z, z0 ) are continuous, then (g ◦ f )∗ = g∗ ◦ f∗ .
(iv) If f : (X, x0 ) → (Y, y0 ) is continuous and null-homotopic, then f∗ is the trivial homomorphism.
(v) Let f : (X, x0 ) → (Y, y0 ) and g : (X, x0 ) → (Y, y1 ) be continuous, and suppose f ∼ g (via h).
Let α be the path in Y from y0 to y1 given by α(t) = h(x0 , t), and let ϕ : π1 (Y, y0 ) → π1 (Y, y1 ) be
the isomorphism given by ϕ([γ]) = [αγα]. Then g∗ = ϕ ◦ f∗ .
(vi) [Homotopy invariance of the fundamental group] Let X, Y be path connected and of the same
homotopy type. Then π1 (X) ∼
= π1 (Y ). In fact, if f : X → Y is a homotopy equivalence, then
f∗ : π1 (X, x0 ) → π1 (Y, f (x0 )) is an isomorphism for any x0 ∈ X. In particular, if X is contractible
(for example, Rn or Dn ), then π1 (X) = {0}.
Proof. Statements (i), (ii), (iii) are easy to verify and left to the student.
INTRODUCTION TO MANIFOLDS - PART 1/3
11
(iv) Suppose f ∼ c (via h). Then t 7→ h(x0 , t) is a path from y0 to c, and consequently f is
homotopic to the constant y0 . So, assume c = y0 . Then for [α] ∈ π1 (X, x0 ), (s, t) 7→ h(α(s), t) from
[0, 1]2 to Y gives the homotopy f ◦ α ∼ y0 .
(v) We know ϕ is an isomorphism by the proof of [104]. Now consider [β] ∈ π1 (X, x0 ). We need to
show [g ◦ β] = [α(f ◦ β)α], or equivalently [α(g ◦ β)α] = [f ◦ β]. Let αt : [0, 1] → Y be αt (s) = α(st).
Note that αt is a path in Y from y0 to α(t), that α0 = ey0 and α1 = α. Let ft : X → Y be
ft (x) = h(x, t). Note that f0 = f , f1 = g, and ft (x0 ) = α(t). Define h1 (s, t) = (αt (ft ◦ β)αt )(s) and
check that f ◦ β and α(g ◦ β)α are path-homotopic via h1 .
(vi) Let f ∈ C(X, Y ) and g ∈ C(Y, X) be such that f ◦ g ∼ IY and g ◦ f ∼ IX . Let x0 ∈
X, y0 = f (x0 ), x1 = g(y0 ), and y1 = f (x1 ). Let f∗0 = f∗ : π1 (X, x0 ) → π1 (Y, y0 ) and f∗1 :
π1 (X, x1 ) → π1 (Y, y1 ). Applying part (v) to the homotopic pairs g ◦ f ∼ IX and f ◦ g ∼ IY ,
we can find isomorphisms ϕ : π1 (X, x0 ) → π1 (X, x1 ) and ψ : π1 (Y, y0 ) → π1 (Y, y1 ) such that
g∗ ◦ f∗0 = (g ◦ f )∗ = ϕ ◦ (IX )∗ = ϕ ◦ IX = ϕ, and f∗1 ◦ g∗ = (f ◦ g)∗ = ψ ◦ (IY )∗ = ψ ◦ IY = ψ. The
first expression says g∗ is surjective while the second expression says g∗ is injective. Thus g∗ is an
isomorphism, and consequently f∗0 , f∗1 are also isomorphisms.
Remark: It follows from [105](vi) that if X, Y are path connected spaces with π1 (X) ̸= π1 (Y ), then
X is not homotopic (hence, not homeomorphic) to Y .
4. Computations and applications of the fundamental group
[106] π1 (Sn ) = {0} for n ≥ 2.
Proof. Fix x0 ∈ Sn , and α be a loop in Sn at x0 . Choose x1 ∈ Sn \ {x0 } such that the compact set
α−1 (x1 ) is nowhere dense in [0, 1]. Such a choice is possible since Sn \ {x0 } is uncountable and [0, 1]
is separable. Let Ui = Sn \ {xi } for i = 0, 1, and note that {U0 , U1 } is an open cover for the compact
12
T.K.SUBRAHMONIAN MOOTHATHU
set Sn . By Lebesgue number property, we may find a partition P = {0 = a0 < a1 < · · · ak−1 <
ak = 1} with α(aj ) ̸= x1 for 0 ≤ j ≤ k such that for each j ∈ {1, . . . , k}, there is i ∈ {0, 1} with
α([aj−1 , aj ]) ⊂ Ui . Since Ui is homeomorphic to Rn and n ≥ 2, on each interval [aj−1 , aj ] we may
modify α through a path-homotopy in Ui to another path that avoids the point x1 . Thus we get a
path β : [0, 1] → Sn \ {x1 } such that β(aj ) = α(aj ) for 0 ≤ j ≤ k and such that α is path-homotopic
to β. Since Sn \ {x1 } is homeomorphic to the contractible space Rn , we have [α] = [β] = [ex0 ]. Definition: A path connected space X is simply connected if π1 (X) = {0}. Note that a path
connected space X is simply connected ⇔ any two paths α, β in X with α(0) = β(0) and α(1) = β(1)
are path-homotopic; see [103](iv).
Remark: Fundamental group can detect only 1-dimensional holes. Simply connected means the
space has no 1-dimensional holes, and contractible means the space has no holes in any dimension.
Thus, contractible spaces (for example, Rn , Dn ) are simply connected. But, a space having holes
of dimension ≥ 2 but having no 1-dimensional hole will turn out to be simply connected but not
contractible. For example, Sn is simply connected for n ≥ 2 by [106], and it will be seen later that
Sn is not contractible (∵ it has an n-dimensional hole).
Next we would like to compute the fundamental group of S1 , Tn , S1 × D2 , etc. Two tools helpful
for this task are: (i) the notion of covering spaces, and (ii) van Kampen’s theorem. We will take
up van Kampen’s theorem a little later. But let us explain briefly how the first tool is going to
work. Given a topological space X, it may be possible to find a ‘simpler’ topological space Z and a
continuous surjection p : Z → X that looks locally like a projection (to be defined precisely below).
In this case the paths and path-homotopies in X can be lifted to paths and path-homotopies in
the ‘simpler’ space Z, and we can do better analysis of the situation on Z. The knowledge that we
obtain by our study on Z can then be pushed down to X through the map p : Z → X.
Definition: Let X, Z ∈ T (in general assumed to be path connected) and p : Z → X be a continuous
surjection. A nonempty open set U ⊂ X is evenly covered by p if p−1 (U ) can be written as a disjoint
⊔
union p−1 (U ) = j∈J Vj , where Vj ⊂ Z is open and p|Vj : Vj → U is a homeomorphism for each
j ∈ J. In this case, we say Vj ’s are the p-slices of U in Z. If each x ∈ X has an open neighborhood
evenly covered by p, then we say p is a covering map (or a covering projection).
Example: The map p : R → S1 ⊂ C given by p(x) = exp(2πix) and p : S1 → S1 given by p(x) = xn
(where n ∈ N) are covering maps. The map p : [−1, 1] → [0, 1] given by p(x) = |x| and the
projection p : R2 → R, (x, y) 7→ x, are not covering maps. The projection P : X × Y → X is a
covering map iff Y is discrete. The quotient map q : Sn → RPn is a covering map, but the quotient
INTRODUCTION TO MANIFOLDS - PART 1/3
13
map q : [0, 1] → S1 given by q(s) = exp(2πis) (which identifies the two end points of [0,1]) is not a
covering map.
Exercise-9: Let p : Z → X be a covering map of path connected spaces. Then the cardinality of
p−1 (x) is constant on X. If this constant is k ∈ N ∪ {∞}, then p is said to be a k-fold covering. For
example, x 7→ xn of S1 is an n-fold covering. [Hint: Cardinality of p−1 (x) is constant on any open
set U ⊂ X evenly covered by p. Also any path in X is covered by finitely many such open sets.]
Path-lifting problem: Let X, Z be path connected and p : Z → X be a continuous surjection. Let
α be a path in X, let x0 = α(0) and z0 ∈ p−1 (x0 ). Is there a path α
b in Z with α
b(0) = z0 and
p◦α
b = α? If so, is α
b unique? In general the answer is negative. If we consider p : [0, 1] → S1 given
by p(s) = exp(2πis) and if α is a path in S1 parametrizing a small closed arc containing 1 ∈ S1 in
the interior, then α has no lift to [0, 1]. For the projection p : R2 → R given by (x, y) 7→ x, it can
be seen that any path in R has infinitely many lifts with any prescribed starting point.
The path-lifting problem has a positive answer for covering maps. In fact, we will see below that
even path-homotopies have unique lifts.
[107] Let p : Z → X be a covering map of path connected spaces, let x0 ∈ X and z0 ∈ p−1 (x0 ).
(i) [Unique path lifting property] If α : [0, 1] → X is a path with α(0) = x0 , then there is a unique
path α
b : [0, 1] → Z such that p ◦ α
b = α and α
b(0) = z0 .
(ii) [Homotopy lifting property] Let α, β : [0, 1] → X be paths with α(0) = β(0) = x0 and
b
α(1) = β(1) = x1 (say). Let α
b, βb be the unique lifts of α, β with α
b(0) = β(0)
= z0 , given by (i).
If we have a path-homotopy α ∼ β (via h), then there is a unique continuous map b
h : [0, 1]2 → Z
b
with p ◦ b
h = h such that α
b is path-homotopic to βb via b
h, and in particular α
b(1) = β(1).
Proof. (i) Let U be a finite open cover for the compact set α([0, 1]) ⊂ X such that each U ∈ U is
evenly covered by p. Using the Lebesgue number property of the open cover {α−1 (U ) : U ∈ U} of
[0, 1], we may find k ∈ N such that for each j ∈ {1, . . . , k}, there is U ∈ U with α([(j − 1)/k, j/k]) ⊂
14
T.K.SUBRAHMONIAN MOOTHATHU
U . We will define α
b on the subintervals [(j − 1)/k, j/k] inductively. Let U0 ∈ U be with x0 ∈ U0 ,
and let V0 ⊂ Z be a p-slice of U0 with z0 ∈ V0 . Define α
b = (p|V0 )−1 ◦ α on [0, 1/k].
Now assume that we have defined α
b on [0, j/k] for some j < k. We describe how to define α
b on
[j/k, (j + 1)/k]. Put xj = α(j/k) ∈ X, zj = α
b(j/k) ∈ Z, and note p(zj ) = xj . Let Uj ∈ U be with
xj ∈ Uj , and let Vj ⊂ Z be a p-slice of Uj with zj ∈ Vj . Define α
b = (p|Vj )−1 ◦ α on [j/k, (j + 1)/k].
In this way, α
b gets defined on the whole of [0, 1], and its continuity is ensured by the pasting lemma.
Let γ : [0, 1] → Z be another lift of α (i.e., p ◦ γ = α) with γ(0) = z0 . Now V0 is one of the p-slices
of U0 . Since γ(0) ∈ V0 and since γ([0, 1/k]) is connected, we must have γ([0, 1/k]) ⊂ V0 . Since
p|V0 : V0 → U0 is a homeomorphism, we obtain that γ = α
b on [0, 1/k]. Similarly, it may also by
verified inductively that γ agrees with α
b on [(j − 1)/k, j/k] for 2 ≤ j ≤ k.
(ii) Let U be a finite open cover for the compact set h([0, 1]2 ) ⊂ X such that each U ∈ U is evenly
covered by p. Using the Lebesgue number property of the open cover {h−1 (U ) : U ∈ U} of [0, 1]2 ,
we may find k ∈ N such that for each (i, j) ∈ {1, . . . , k}2 , there is U ∈ U with h(A(i, j)) ⊂ U , where
A(i, j) = [(i−1)/k, i/k]×[(j −1)/k, j/k]. Define b
h inductively on A(i, j)’s in the lexicographic order
of the pairs (i, j), starting with A(1, 1), by imitating the argument given for part (i). When two
squares intersect, say A(i, j) and A(i, j + 1), the intersection is a line segment and h restricted to
this line segment is a path in X. This path has unique lift by part (i) and this will ensure that the
definitions of b
h on two intersecting A(i, j)’s agree on their intersection; see Lemma 54.2 of Munkres,
Topology, for finer details. The uniqueness of b
h is also proved as in part (i), by working with A(i, j)’s
inductively. Since b
h(·, 0) is a lift of α and b
h(·, 1) is a lift of β, we get by the uniqueness in part (i)
b in other words, α
that b
h(·, 0) = α
b and b
h(·, 1) = β;
b ∼ βb (via b
h). Since h is a path-homotopy, we have
h({0} × [0, 1]) = {x0 } and h([0, 1] × {1}) = {x1 }. By the uniqueness of path-lifting, b
h({0} × [0, 1])
INTRODUCTION TO MANIFOLDS - PART 1/3
15
b
b
and h([0, 1] × {1}) should be singletons, necessarily equal to {b
α(0)} = {β(0)}
and {b
α(1)} = {β(1)}
respectively. This proves that b
h is a path-homotopy.
Remark: If p : Z → X is a covering map, any path-homotopy in Z can be pushed-down to X by
composing with p. In combination with the lifting property of [107], we conclude:
If p : Z → X is a covering map, then path-homotopy in X is equivalent to path-homotopy in Z.
Hence we may study path-homotopy in X by lifting everything to a possibly simpler space Z.
Remark: If p : R → S1 is p(y) = exp(2πiy) and α : [0, 1] → S1 is α(s) = exp(2πis), then the unique
lift α
b of α to R with α
b(0) = 0 is α
b(s) = s. Note that the lift α
b is not a loop even though α is a
loop. The Exercise below throws a little more light on the possibilities of α
b(1) for the lift of a loop.
Exercise-10: Let p : Z → X be a covering map of path connected spaces, let x0 ∈ X and z0 ∈
p−1 (x0 ). Define ψ : π1 (X, x0 ) → p−1 (x0 ) as ψ([α]) = α
b(1), where α
b is the unique lift of α with
α
b(0) = z0 , given by [107]. Then, ψ is well-defined and surjective. If Z is simply connected, then ψ
is also injective. [Hint: Well-definedness is assured by [107](ii). Given z1 ∈ p−1 (x0 ), there is path
α
b in Z from z0 to z1 . If α = p ◦ α
b, then ψ([α]) = z1 . If Z is simply connected and ψ([α]) = ψ([β]),
b Now α is path-homotopic to β via p ◦ b
then there is a path-homotopy b
h between α
b and β.
h.]
[108] π1 (S1 ) = (Z, +). In particular S1 is not simply connected, and hence not contractible.
Proof. First we give an intuitive idea. Since y 7→ exp(2πiy) from R to S1 is a covering map, pathhomotopy in S1 is equivalent to path-homotopy in R. Let [α], [β] ∈ π1 (S1 , 1) and α
b, βb be the lifts
b
to R with starting point 0. Since R is simply connected, α
b is path-homotopic to βb iff α
b(1) = β(1).
Thus path-homotopy in R is determined by the value α
b(1). This value must be an integer, and any
integer can be α
b(1) for some [α] ∈ π1 (S1 , 1). Thus π1 (S1 , 1) is bijective with Z and this bijective
correspondence is actually a group homomorphism. The rigorous proof is given below.
Let α : [0, 1] → S1 ⊂ C be α(t) = exp(2πit). Then α is a loop in S1 at x0 = 1. Define
ϕ : Z → π1 (S1 , x0 ) as ϕ(n) = [α]n , which is clearly a group homomorphism. We claim that there
is a bijection ψ : π1 (S1 , x0 ) → Z such that ψ ◦ ϕ = IZ . If the claim is proved, then ϕ = ψ −1 is
also a bijection, and we will be through. Let p : R → S1 be the covering map p(y) = exp(2πiy).
b
Let ψ : π1 (S1 , x0 ) → p−1 (x0 ) = Z be ψ([β]) = β(1)
and note that ψ is a bijection by Exercise-10
since R is simply connected. Fix n ∈ Z and let γ
b : [0, 1] → R be the path γ
b(t) = nt. Then γ
b is the
unique lift of αn , and therefore ψ(ϕ(n)) = ψ([α]n ) = ψ([αn ]) = γ
b(1) = n.
Remark: (i) Let f ∈ C(S1 , S1 ). Since π1 (S1 ) = Z, the induced homomorphism f∗ : π1 (S1 , 1) →
π1 (S1 , f (1)) has the form n 7→ kn. This integer k is defined as the degree of f , and it counts how
16
T.K.SUBRAHMONIAN MOOTHATHU
many times f winds around S1 in the anti-clockwise direction. If f (1) = 1 and fb : S1 → R is the
unique lift of f with fb(1) = 0 w.r.to the exponential map, then it may be seen that deg(f ) = fb(1).
Since a loop in S1 can be thought of as a continuous map from S1 to S1 , the map ψ appearing in
the above proof is precisely the isomorphism [α] 7→ deg(α) from π1 (S1 , 1) to Z.
Exercise-11: If X, Y are path connected, then π1 (X ×Y ) = π1 (X)×π1 (Y ). [Hint: If p : X ×Y → X
and q : X ×Y → Y are the projections, consider the homomorphism p∗ ×q∗ : π1 ((X ×Y ), (x0 , y0 )) →
π1 (X, x0 ) × π1 (Y, y0 ). It is a bijection since it has the inverse [α] × [β] 7→ [α × β].]
[109] (i) π1 (Tn ) = (Zn , +), where Tn is the n-torus Tn = S1 × · · · × S1 (n-times).
(ii) π1 (Tk × Dm ) = (Zk , +). In particular, if X is the solid torus S1 × D2 , then π1 (X) = (Z, +).
(iii) π1 (RPn ) = (Z2 , +2 ) for n ≥ 2; and RP1 is homeomorphic to S1 so that π1 (RP1 ) = (Z, +).
Proof. Statements (i) and (ii) follow from [108], Exercise-11 and the fact that π1 (Dn ) = {0}.
(iii) Fix n ≥ 2 and x0 ∈ RPn . Recall that RPn = Sn /(x ∼ −x). Now the quotient map q : Sn → RPn
is a 2-fold covering map. Since Sn is simply connected by [106], there is bijection ψ : π1 (RPn , x0 ) →
q −1 (x0 ) by Exercise-10 and hence |π1 (RPn , x0 )| = |q −1 (x0 )| = 2.
Remark: By the results above, the following spaces have distinct fundamental groups: S1 × S1 × S1 ,
S1 × S1 × D1 , S1 × D2 , D3 , S1 × RP2 , and RP3 . It follows that no two of these spaces are homotopic
to each other, and hence no two of them are homeomorphic to each other. To really appreciate
this, try to prove this using elementary Topology!
Definition: Let A ⊂ X ∈ T . We say A is a retract of X if there is a continuous map r : X → A
(called a retraction) such that r|A = IA , or more formally r ◦ i = IA , where i : A → X is the
inclusion map. For example, x 7→ x/|x| is a retraction from C \ {0} to S1 .
[110]: (i) If r : X → A is a retraction, i : A → X is the inclusion map, and a0 ∈ A, then
r∗ : π1 (X, a0 ) → π1 (A, a0 ) is surjective and i∗ : π1 (A, a0 ) → π1 (X, a0 ) is injective.
(ii) S1 is not a retract of D2 .
(iii) [Brouwer’s fixed point theorem for D2 ] If f : D2 → D2 is continuous, then f has a fixed point.
Proof. (i) r ◦ i = IA implies r∗ ◦ i∗ = (r ◦ i)∗ = (IA )∗ = I.
(ii) No surjection exists from π1 (D2 ) = {0} onto π1 (S1 ) = Z, and use part (i).
(iii) Suppose f has no fixed points. Define r : D2 → S1 by the condition that r(x) is the unique
point on S1 where the line starting from f (x) and passing through x intersects S1 \ {f (x)}. It may
be checked that r is a retraction from D2 onto S1 ; this contradicts part (ii).
INTRODUCTION TO MANIFOLDS - PART 1/3
17
Exercise-12: [Borsuk-Ulam Theorem for S1 ] If f ∈ C(S1 , R), then ∃ x ∈ S1 with f (x) = f (−x).
[Hint: Let g(x) = f (x) − f (−x), compare g(x), g(−x) and apply Intermediate value property.]
[111] (i) If g ∈ C(S1 , S1 ) satisfies g(−x) = −g(x) for every x ∈ S1 , then g is not null-homotopic.
(ii) [Borsuk-Ulam theorem for S2 ] If f ∈ C(S2 , R2 ), then there is x ∈ S2 such that f (x) = f (−x).
Proof. (i) Replacing g with g/g(1), we may assume g(1) = 1. In view of [105](iv), it suffices to show
g∗ : π1 (S1 , 1) → π1 (S1 , 1) is not trivial. Let β = g ◦ α, where α : [0, 1] → S1 is α(s) = exp(2πis).
We will show that [β] = g∗ ([α]) is non-trivial in π1 (S1 , 1). Consider the covering map p : R → S1
b
given by p(y) = exp(2πiy), and let βb : [0, 1] → R be the unique lift of β with β(0)
= 0. From the
proof of [108] we know that [γ] 7→ γ
b(1) is an isomorphism from π1 (S1 , 1) to Z. So now it suffices to
b
show deg(β) = β(1)
̸= 0. If s ∈ [0, 1/2], then β(s + 1/2) = g(α(s + 1/2)) = g(−α(s)) = −g(α(s)) =
b + 1/2)) = exp(πi + 2πiβ(s))
b
b
exp(πi)β(s) and therefore exp(2πiβ(s
= exp(2πi(1/2 + β(s))).
Hence
b + 1/2) = b
for each s ∈ [0, 1/2], there is ms ∈ Z such that β(s
1/2 + β(s) + ms . Since ms is
b
integer-valued and depends continuously on s, it must be a constant, say m ∈ Z. Then β(1)
=
b
b + m + 1/2) + m + 1/2 = 2m + 1 ̸= 0.
β(1/2)
+ m + 1/2 = (β(0)
(ii) If f (x) ̸= f (−x) for every x ∈ S2 , define g : S2 → S1 as g(x) = (f (x) − f (−x))/|f (x) − f (−x)|
and observe that the continuous map g satisfies g(−x) = −g(x). The restriction of g to the equator
of S2 must be null-homotopic by [102] since it has a continuous extension to the upper half-sphere
∼
= D2 . This contradicts part (i).
Remark: See p.358 of Munkres, Topology, for an interesting application of Borsuk-Ulam Theorem.
Definition: Let A ⊂ X ∈ T and i : A → X be the inclusion map. We say A is a deformation retract
of X if there is a retraction r : X → A such that r and i are homotopy inverses of each other,
i.e., if r = i ◦ r ∼ IX (note: r ◦ i = IA since r is a retraction). If X is path connected and A is a
18
T.K.SUBRAHMONIAN MOOTHATHU
deformation retract of X, then X is homotopic to A by definition, and hence π1 (X) = π1 (A) by
[105].
Example: (i) S1 is a deformation retract of C \ {0} through the retraction x 7→ x/|x|. Hence
π1 (C \ {0}) = Z. (ii) If X ∈ T , then every singleton {x0 } ⊂ X is a retract of X, but {x0 } is a
deformation retract of X iff X is contractible. Thus a retract need not be a deformation retract;
for example, consider {1} ⊂ S1 .
Remark: If α is a loop in C and if a ∈ C is a point not on the image of α, then α can be considered
as a continuous map from S1 to C \ {a}. Since a circle centered at a is a deformation retract of
C \ {a}, we have π1 (C \ {a}) = Z. Therefore, the induced homomorphism α∗ : π1 (S1 ) → π1 (C \ {a})
has the form n 7→ kn. This integer k is the winding number of α about the point a.
Definition: Let X, Y ∈ T and f : X → Y be continuous. The mapping cylinder Mf ∈ T of f is the
⊔
quotient of the disjoint union (X × [0, 1]) Y by the identification (x, 1) ∼ f (x) for x ∈ X. Note
that X = X × {0} and Y can be thought of as subspaces of Mf .
Example: (i) If X = D1 = [−1, 1], Y is a singleton, and f : X → Y is the constant map, then
Mf can be identified with a solid triangle and hence with D2 . More generally, if X = Dn , Y
is a singleton, and f : X → Y is the constant map, then Mf can be identified with Dn+1 . (ii)
X = [−1, 1], Y = S1 , and f : X → Y is f (x) = exp(2πix), then Mf can be identified with the
compact space bounded by two tangential circles, one inside the other.
[112] Let X, Y ∈ T . (i) If f : X → Y is continuous, then Y is a deformation retract of Mf .
(ii) f ∈ C(X, Y ) is a homotopy equivalence ⇔ X is also a deformation retract of Mf .
Proof. (i) For points in Mf , we will continue to use the notations (x, s) ∈ X × [0, 1] and y ∈ Y ,
by keeping in mind the identification (x, 1) ∼ f (x). The map r : Mf → Y sending (x, s) to
f (x), and y to y is a retraction. If i : Y → Mf is the inclusion, then IMf ∼ i ◦ r (via h), where
h : Mf × [0, 1] → Mf sends ((x, s), t) to (x, (1 − t)s + t), and (y, t) to y.
INTRODUCTION TO MANIFOLDS - PART 1/3
(ii) This is not easy; see Proposition 0.19 and Corollary 0.21 of Hatcher, Algebraic Topology.
19
Remark: By [112], two spaces X, Y are homotopic iff both are deformation retracts of a third space.
Seminar Topic: Other constructions - cone, suspension, etc. - involving topological spaces.
5. Galois theory of covering spaces
In this section we will see that if X is a nice topological space, then there is a 1-1 correspondence
between (equivalence classes) of covering spaces of X and conjugacy classes of subgroups of π1 (X).
This correspondence is order-reversing: bigger the cover, smaller the subgroup.
Convention and Definition: Let X ∈ T be path connected and locally simply connected throughout
this section, and let Cov(X) = {(Z, p) : Z is path connected and p : Z → X is a covering map}.
Note that if (Z, p) ∈ Cov(X), then Z is also locally simply connected. We say (Z1 , p1 ), (Z1 , p2 ) ∈
Cov(X) are isomorphic if there is a homeomorphism f : Z1 → Z2 with p2 ◦ f = p1 . The notion of
isomorphism is an equivalence relation on Cov(X).
Exercise-13: Let (Z, p) ∈ Cov(X), x0 ∈ X and z0 ∈ p−1 (x0 ). Then,
(i) p∗ : π1 (Z, z0 ) → π1 (X, x0 ) is injective; hence π1 (Z) may be identified with a subgroup of π1 (X).
(ii) For [α] ∈ π1 (X, x0 ), we have [α] ∈ p∗ (π1 (Z, z0 )) ⇔ the lift α
b of α is a loop in Z at z0 .
[Hint: (i) If α
b ∈ ker(p∗ ), then p ◦ α
b and ex0 are path-homotopic, and hence by [107] their lifts
b = [α], then [α] = [p ◦ β].
b Hence α
α
b and ez0 are path-homotopic in Z. (ii) ⇒: If p∗ ([β])
b, βb are
b
path-homotopic by [107], and in particular α
b(1) = β(1)
= z0 .]
[113] [Lifting criterion] Let (Z, p) ∈ Cov(X), x0 ∈ X, and z0 ∈ p−1 (x0 ). Let Y ∈ T be path
connected and locally path connected, and f : (Y, y0 ) → (X, x0 ) be continuous. Then there is a
continuous map fb : (Y, y0 ) → (Z, z0 ) with p ◦ fb = f ⇔ f∗ (π1 (Y, y0 )) ⊂ p∗ (π1 (Z, z0 )). Also, if the
lift fb exists, then it is unique.
Proof. ⇒: If p ◦ fb = f , then f∗ (π1 (Y, y0 )) = p∗ (fb∗ (π1 (Y, y0 ))) ⊂ p∗ (π1 (Z, z0 )).
⇐: Let y1 ∈ Y , let α be a path in Y from y0 to y1 and define fb(y1 ) = f[
◦ α(1), where f[
◦ α is the
unique lift to Z with starting point z0 , of the path f ◦ α in X. Assuming fb is well-defined (this is
proved below), note that p ◦ fb(y1 ) = p ◦ f[
◦ α(1) = f ◦ α(1) = f (y1 ) so that fb is indeed a lift of f .
fb is well-defined : If α, β are two paths in Y from y0 to y1 , then αβ is a loop in Y at y0 . Hence
[(f ◦ α)(f ◦ β)] = [f ◦ (αβ)] ∈ f∗ (π1 (Y, y0 )) ⊂ p∗ (π1 (Z, z0 )). So there is a loop γ
b in Z at z0 such
that (f ◦ α)(f ◦ β) is path-homotopic to p ◦ γ
b. Hence f ◦ α is path-homotopic to (p ◦ γ
b)(f ◦ β). Since
the loop γ
b is the lift of p ◦ γ
b, we get fb ◦ α(1) = γ
b f[
◦ β(1) = f[
◦ β(1) by [107].
20
T.K.SUBRAHMONIAN MOOTHATHU
fb is continuous: Let y1 ∈ Y and V ⊂ Z be an open neighborhood of fb(y1 ). We need to find
an open neighborhood W ⊂ Y of y1 such that fb(W ) ⊂ V . Replacing V with a smaller open
set we may assume that V is a p-slice of the open neighborhood U := p(V ) ⊂ X of the point
p(fb(y1 )) = f (y1 ). Since Y is locally path connected, we may find a path connected open set
W ⊂ Y with y1 ∈ W ⊂ f −1 (U ). We claim fb(W ) ⊂ V . Let α be a path in Y from y0 to y1 . If
y2 ∈ W , find a path β from y1 to y2 in W . Since f ◦β ⊂ f (W ) ⊂ U , we have that γ
b := (p|V )−1 ◦f ◦β
is a lift of f ◦ β with starting point fb(y1 ). Hence fb(y2 ) = f[
◦ αγ
b(1) = γ
b(1) ∈ V , proving the claim.
The uniqueness of fb follows essentially by the uniqueness of path-lifting.
Exercise-14: [Conjugating a subgroup below is equivalent to changing the base point above] Let
(Z, p) ∈ Cov(X), x0 ∈ X, z0 ∈ p−1 (z0 ) ∈ Z, and let H = p∗ (π1 (Z, z0 )) ⊂ π1 (X, x0 ). Then a
subgroup H1 of π1 (X, x0 ) is conjugate to H ⇔ H1 = p∗ (π1 (Z, z1 )) for some z1 ∈ p−1 (x0 ). [Hint:
Let [α] ∈ π1 (X, x0 ) and H1 = [α]H[α]. If α
b is the unique lift of α to Z with starting point z0 , take
z1 = α
b(1). Then p(z1 ) = p(b
α(1)) = α(1) = x0 . Check that p∗ (π1 (Z, z1 )) = [α]H[α]. Conversely, if
H1 = p∗ (π1 (Z, z1 )) for some z1 ∈ p−1 (x0 ), let α
b be a path in Z from z0 to z1 , let α = p ◦ α
b, and
show H1 = [α]H[α].]
[114] [Comparing two covers] Let (Z1 , p1 ), (Z2 , p2 ) ∈ Cov(X).
(i) If q : Z1 → Z2 is a continuous map with p2 ◦ q = p1 , then q is a covering map.
(ii) Let x0 ∈ X, let zk ∈ p−1
k (x0 ) ∈ Zk , and Hk = (pk )∗ (π1 (Zk , zk )) ⊂ π1 (X, x0 ) for k = 1, 2.
(a) H1 = H2 ⇔ there is a homeomorphism f : (Z1 , z1 ) → (Z2 , z2 ) such that p2 ◦ f = p1 .
(b) H1 is conjugate to H2 ⇔ there is a homeomorphism f : Z1 → Z2 such that p2 ◦ f = p1 .
Proof. (i) First we show q is surjective. Let z2 ∈ Z2 , let x = p2 (z2 ) ∈ X, and z1 ∈ p−1
1 (x) ∈ Z1 . If
α
b2 is a path in Z2 from q(z1 ) to z2 , then α := p2 ◦ α2 is a loop in X at x. Let the path α
b1 in Z1
INTRODUCTION TO MANIFOLDS - PART 1/3
21
be the unique lift of α with starting point z1 . Then both q ◦ α
b1 and α
b2 are lifts of α to Z2 with
starting point q(z1 ) and hence q ◦ α
b1 = α
b2 by [107]. In particular, z2 = α
b2 (1) = q(b
α1 (1)).
To prove the covering property, consider z ∈ Z2 . Let x = p2 (z) ∈ X, and choose a simply
connected open neighborhood U ⊂ X of x such that U is evenly covered by both p1 and p2 . Let
Vj ⊂ Z1 be the p1 -slices of U for j ∈ J, and let W ⊂ Z2 be the p2 -slice of U containing z. Since
⊔
⊔
−1 −1
p1 = p2 ◦ q and q is surjective, we have q( j∈J Vj ) = p−1
2 (U ) and
j∈J Vj = q (p2 (U )). Let
J0 = {j ∈ J : W ∩ q(Vj ) ̸= ∅}. Since each Vj is simply connected (being homeomorphic to U ), and
in particular connected, and since the p2 -slices of U are disjoint open sets, it follows that q(Vj ) ⊂ W
⊔
for each j ∈ J0 and that q −1 (W ) = j∈J0 Vi . Thus W is evenly covered by q.
(iia) ⇒: If H1 = H2 , then by [113] we can find a lift pb1 : (Z1 , z1 ) → (Z2 , z2 ) of p1 w.r.to p2 , and
similarly a lift pb2 : (Z2 , z2 ) → (Z1 , z1 ) of p2 w.r.to p1 . That is, p1 = p2 ◦ pb1 , and p2 = p1 ◦ pb2 . Then
p1 = p1 ◦ pb2 ◦ pb1 and p2 = p2 ◦ pb1 ◦ pb2 . Since pb2 ◦ pb1 , IZ1 : (Z1 , z1 ) → (Z1 , z1 ) are both lifts of p1 w.r.to
p1 , we get pb2 ◦ pb1 = IZ1 by the uniqueness of lift in [113]. Similarly, pb1 ◦ pb2 = IZ2 . Take f = pb1 .
⇐: H1 ⊂ H2 since (p1 )∗ = (p2 )∗ ◦ f∗ , and H2 ⊂ H1 since (p2 )∗ = (p1 )∗ ◦ (f −1 )∗ .
(iib) This follows from (iia) and Exercise-14.
Definition: (Z, p) ∈ Cov(X) is said to be a universal cover for X if Z is simply connected. For
example, (R, y 7→ exp(2πiy)) is a universal cover of S1 . If (Z1 , p1 ), (Z2 , p2 ) ∈ Cov(X), write
(Z2 , p2 ) ≤ (Z1 , p1 ) if there is a covering map q : Z1 → Z2 with p2 ◦ q = p1 .
[115] [Universal cover] X has a universal cover (Z, p), and any two universal covers of X are
isomorphic. Moreover, (Z2 , p2 ) ≤ (Z, p) for any (Z2 , p2 ) ∈ Cov(X).
22
T.K.SUBRAHMONIAN MOOTHATHU
Proof. Fix x0 ∈ X. Let P = {all paths α in X starting at x0 } and P/ ∼= {[α] : α ∈ P}, where
the equivalence relation is path-homotopy. To make a guess about the construction of Z, observe
that if (Z, p) is a universal cover for X, and z0 ∈ p−1 (x0 ), then [α] 7→ α
b(1) must be a bijection from
P/ ∼ to Z (compare with Exercise-10), where α
b is the lift of α to Z with α
b(0) = z0 . Motivated by
this observation, we put Z = P/ ∼, and define p : Z → X as p([α]) = α(1).
Now we need to topologize Z. Let U = {∅ ̸= U ⊂ X : U is open and simply conncted}, and
note that U is a base for the topology of X since X is locally simply connected. If α ∈ P,
and α(1) ∈ U ∈ U , let [αU ] = {[αγ] : γ is a path in U starting at α(1)}, which is going to be a
neighborhood of [α] in Z. We verify various properties step by step. Let U, V ∈ U.
(a) [αU ] = [βU ] ⇔ [α] ∈ [βU ] ⇔ [β] ∈ [αU ] (∵ [αγ] = [β] ⇔ [α] = [βγ]).
(b) If [δ] ∈ [αU ] ∩ [βV ], then [δU ] = [αU ] and [δV ] = [βV ] so that [δ(U ∩ V )] ⊂ [αU ] ∩ [βV ].
(c) [αU ] and [βU ] are either identical or disjoint (∵ take V = U in (b)).
(d) Sets of the form [αU ] (with both α and U varying) form a base for a topology on Z by (b).
(e) p([αU ]) = U , and hence p : Z → X is a surjective open map.
∪
(f) p−1 (U ) = α(1)∈U [αU ], and hence p is continuous.
(g) p restricted to [αU ] is injective, and thus p|[αU ] : [αU ] → U is a homeomorphism. Proof :
Suppose [β], [γ] ∈ [αU ] and β(1) = p([β]) = p([γ]) = γ(1). By (a), we have [α] ∈ [βU ] ∩ [γU ] and
hence there are paths β1 , γ1 in U such that [α] = [ββ1 ] = [γγ1 ]. Then [β] = [γγ1 β1 ] = [γ] since γ1 β1
is a loop in U at β(1) = γ(1) and since U is simply connected.
(h) Z is Hausdorff. Proof : Consider [α] ̸= [β]. If α(1) ̸= β(1), and if U, V ∈ U separate α(1) and
β(1) in X, then [αU ] and [βV ] separate [α] and [β] in Z. So assume α(1) = β(1) ∈ U . If [β] ∈
/ [αU ],
then [αU ] and [βU ] are disjoint open sets separating [α] and [β] by (a) and (c). If β ∈ [αU ], then
use the facts that [αU ] is homeomorphic to U by (g) and U is Hausdorff.
(i) Z is path connected. If αt (s) = α(st), then t 7→ [αt ] is a path in Z from z0 := [ex0 ] to [α].
(j) p : Z → X is a covering map and the p-slices of U are the disjoint sets of the form [αU ]. This
follows from (c), (e), (f) and (g).
(k) Z is simply connected. Proof : We need to show π1 (Z, z0 ) = {0}, where z0 = [ex0 ]. Since p∗
is injective, it suffices to show p∗ (π1 (Z, z0 )) = {0}. Let [α] ∈ π1 (X, x0 ) and note that its lift α
b
to Z starting at z0 is given as α
b(t) = [αt ], where αt (s) = α(st). In particular, α
b(1) = [α]. If
[α] ∈ p∗ (π1 (Z, z0 )), then the lift α
b must be a loop by Exercise-13, and thus [ex0 ] = z0 = α
b(1) = [α].
INTRODUCTION TO MANIFOLDS - PART 1/3
23
(l) If (Zk , pk ) are universal covers of X for k = 1, 2, then (p1 )∗ (π1 (Z1 )) = {0} = (p2 )∗ (π1 (Z2 )) so
that the two universal covers are isomorphic by [114](ii).
(m) Let z0 ∈ p−1 (x0 ) and z2 ∈ p−1
2 (x0 ). Now p∗ (π1 (Z, z0 )) = {0} ⊂ (p2 )∗ (π1 (Z2 , z2 )) so that by
[113], p has a lift q : Z → Z2 w.r.to the covering map p2 . By [114](i), q is a covering map.
Remark: (i) If (Zk , pk ) ∈ Cov(Xk ) for k = 1, 2, then it is easy to see that (Z1 × Z2 , p1 × p2 ) ∈
Cov(X1 × X2 ). We may deduce that the universal cover of the torus S1 × S1 is (R2 , p), where
p(y1 , y2 ) = (exp(2πiy1 ), exp(2πiy2 )). (ii) Recall that Sn is simply connected for n ≥ 2, and hence
the universal cover of RPn for n ≥ 2 is (Sn , p) where p : Sn → RPn is the quotient map identifying
x and −x (which is a covering map).
[116] [Galois correspondence] (i) If H ⊂ π1 (X, x0 ) is a subgroup, then there exists (Z, p) ∈ Cov(X)
and z0 ∈ p−1 (x0 ) such that p∗ (π1 (Z, z0 )) = H.
(ii) The map (Z, p) 7→ p∗ (π1 (Z)) induces a bijection between isomorphic classes of members of
Cov(X) and conjugacy classes of subgroups of π1 (X). This correspondence is order-reversing in
the sense that if (Z2 , p2 ) ≤ (Z1 , p1 ), then (p1 )∗ (π1 (Z1 )) ⊂ (p2 )∗ (π2 (Z2 )).
Proof. We indicate how to modify the construction in [115](i). Fix x0 ∈ X and define an equivalence
relation on paths in X starting at x0 by the condition α ∼H β iff α(1) = β(1) and [αβ] ∈ H. Let
[α]H denote the equivalence class of α, and Z be the collection of all such equivalence classes. Let
z0 = [ex0 ]H , and p : Z → X be p([α]H ) = α(1). See Theorem 82.1 of Munkres, Topology, or
Proposition 1.36 of Hatcher, Algebraic Topology, for the details. For instance, if α is a loop in X
at x0 , and if α
b is its lift to Z with starting point z0 , then [α] ∈ p∗ (π1 (Z, z0 ) ⇔ α
b is a loop in Z at
z0 ⇔ [ex0 ]H = z0 = α
b(1) = [α]H ⇔ α ∼H ex0 ⇔ [α] ∈ H, and hence p∗ (π1 (Z, z0 )) = H. This gives
(i); and the assertion (ii) follows by part (i), [114], and [115].
Example: We know π1 (S1 ) = Z, which is abelian. The universal cover of S1 is (R, exp) which
corresponds to the subgroup {0}. The covering space (S1 , x 7→ xn ) of S1 corresponds to the
subgroup nZ of Z. We have listed all subgroups of Z and hence by [116] all covering spaces of S1 .
Definition: The covering group Gp of (Z, p) ∈ Cov(X) is the collection of all homeomorphisms
f : Z → Z satisfying p ◦ f = p; indeed Gp is a group under composition. Any f ∈ Gp is called a deck
transformation or covering transformation. Note that any f ∈ Gp is a lift of p w.r.to p. If X = S1
and (Z, p) = (R, y 7→ exp(2πiy)), then Gp = {fk : k ∈ Z} ∼
= Z = π1 (S1 ), where fk (x) = x + k.
Exercise-15: Let (Z, p) ∈ Cov(X) and f ∈ Gp . Then,
(i) For each x ∈ X, f |p−1 (x) : p−1 (x) → p−1 (x) is a bijection.
24
T.K.SUBRAHMONIAN MOOTHATHU
(ii) If U ⊂ X is a connected open set evenly covered by p, then f permutes the p-slices of U .
(iii) If g ∈ Gp and f (z0 ) = g(z0 ) for some z0 ∈ Z, then f = g. In particular, If f ̸= IZ , then f has
no fixed points.
[Hint: (ii) f restricted to p−1 (U ) is a homeomorphism of p−1 (U ) to itself and the p-slices of U are
disjoint open connected sets. (iii) The set A = {z ∈ Z : f (z) = g(z)} is closed. If z ∈ A and if V
is a p-slice of an open connected neighborhood of f (z) with z ∈ V , then check that f |V = g|V by
using (ii). So V ⊂ A and thus A is also open.]
[117] Let (Z, p) ∈ Cov(X), x0 ∈ X, z0 ∈ p−1 (x0 ), and let H = p∗ (π1 (Z, z0 )). Let N (H) be the
normalizer of H in π1 (X, x0 ). Then,
(i) Let [α] ∈ π1 (X, x0 ) and z1 = α
b(1) ∈ p−1 (x0 ), where α
b is the lift of α to Z with starting point
z0 . Then [α] ∈ N (H) ⇔ there is a unique f ∈ Gp with f (z0 ) = z1 .
(ii) Gp is isomorphic to the quotient group N (H)/H.
(iii) H is normal ⇔ Gp acts transitively on p−1 (x0 ). If this happens, (Z, p) is called normal.
(iv) If (Z, p) is a normal (in particular, when π1 (X, x0 ) is abelian), Gp is isomorphic to π1 (X, x0 )/H.
(v) If (Z, p) is the universal cover of X, then Gp ∼
= π1 (X).
Proof. (i) [α] ∈ N (H) ⇔ H = [α]H[α] = p∗ (π1 (Z, z1 )) ⇔ (by [114] and Exercise-15) there is a
unique f ∈ Gp with f (z0 ) = z1 .
(ii) Define ψ : N (H) → Gp as ψ([α]) = f , where f ∈ Gp is the unique deck transformation with
f (z0 ) = α
b(1). Then ψ is surjective by (i), and it may be checked that ψ is a homomorphism. Also
[α] ∈ ker(ψ) ⇔ ψ([α]) = IZ ⇔ α
b(1) = z0 ⇔ [α] ∈ H by Exercise-13.
Now, statements (iii), (iv) and (iv) are corollaries.
*****