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MATH 151A: TOPOLOGY I
0. Introduction
The first topic we need to discuss is the philosophy of algebraic topology. In a nutshell:
(1) We associate algebraic (usually discrete) objects to each space and apply these
to study the topology and homotopy type of the space.
(2) We usually associate a group or ring to each space. The size of the group should
indicate the complexity of the space.
(3) We explicitly avoid pathological examples which contradict our ideas about how
this should work.
An easy example is the connectivity of the space.
0.1. Connectivity. Recall the definition:
Definition 0.1.1.
` A topological space X is connected if it cannot be written as a disjoint
union X = U V of two nonempty open subsets.
Example 0.1.2. The set of rational numbers Q, with the usual metric topology, is not
connected since it is the union of two nonempty open sets:
na
o ` na
o
Q=
2 Q | a2 < 2b2
2 Q | a2 > 2b2
b
b
We rephrase this in the spirit of algebraic topology:
Proposition 0.1.3. A topological space X is connected i↵ every continuous mapping:
is constant.
f : X ! {0, 1}
Here {0, 1} has the discrete topology (all subsets are open). This set is also written
as @I where I = [0, 1] ⇢ R with the usual metric topology.
Proof. If there is a nonconstant (surjective) mapping f : X ! {0, 1}, then X is the
disjoint`union of the open subsets U = f 1 (0) and V = f 1 (1). Conversely, given
X = U V , f : X ⇣ {0, 1} can be given by
(
0 if x 2 U
f (x) =
1 otherwise
This map is continuous since U, V are open subsets of X.
⇤
Philosophical point: In algebraic topology, we look for more structure behind concepts. Mappings have more structure than subsets. In this spirit, we put more structure
on the set {0, 1}. Namely, this is a ring:
{0, 1} = Z/2Z
(It is also a field {0, 1} = F2 .)
This leads us to the first algebraic invariant of spaces:
Definition 0.1.4. For any topological space X, let C(X) be the set of all continuous
mappings
X ! Z/2Z
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(1) The set C(X) is a ring (with unit) under pointwise addition and multiplication.
(2) For any continuous mapping ' : X ! Y we get an induced mapping:
'⇤ : C(Y ) ! C(X)
given by '⇤ (g) = g '. This mapping is a (unital) ring homomorphism.
We can rephrase Proposition 0.1.3 in terms of the ring C(X):
Theorem 0.1.5. X is connected i↵ C(X) ⇠
= Z/2Z.
One student made the interesting comment that X is connected if and only if C(X)
is a field!
Another fascinating idea is: What happens when X = Z/2Z and we consider the
third operation on C(Z/2Z) given by composition?
Next introductory topic: What are the central problems of Algebraic Topology?
0.2. Homotopy. The main di↵erence between topology and geometry is the lack of
rigidity. In geometry we are interested in real number invariants of metric spaces such
as: diameter (supremum of the distance between points), volume (given a measure on
the space), curvature (given a Riemannian manifold). These are all real numbers.
In topology we interested in properties which are invariant under continous deformation. The famous saying is: “A topologist is someone who cannot tell the di↵erence
between a donut and a co↵ee cup” since these two shapes can be deformed into each
other. Deformation invariants are usually discrete, often integers, not real numbers. For
example, the number of path components of a space.
Definition 0.2.1. A path in a space X is defined to be a continuous mapping: : I ! X
where by I is the standard symbol for the unit interval I = [0, 1] with the metric topology.
We say that is a path in X from x = (0) to y = (1). We also say that x, y are
connected by a path in X. This is an equivalence relation and two elements of X are in
the same path component if there is a path in X connecting them.
The main properties of the interval I = [0, 1] are that is compact and connected. This
implies:
Proposition 0.2.2. Every path connected space is connected.
⇤
Definition 0.2.3. A space X is locally path connected if for every open set U in X and
every x 2 U there is a path connected open nbh V of x in U . [The concept of “locally
xxx” is defined analogously.]
Proposition 0.2.4. A locally path connected space X is connected i↵ it is path connected.
⇤
A path is a deformation of one point in X. A “deformation of X” refers to a deformation of all the points. This is a continuous mapping
h:X ⇥I !X
where X ⇥ I has the product topology so that h(x, 0) = x for all x 2 X. If h(X ⇥ 1) is
contained in a subspace A of X we say that h is a deformation of X into A.
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MATH 151A: TOPOLOGY I
We say that h is a deformation retraction of X to A ✓ X if, in addition, h(a, t) = a
for all t 2 I.
Example 0.2.5. Take X = Rn with the metric topology. Let h be the deformation of
X given by h(x, t) = (1 t)x. Then h is a deformation retraction of X = Rn onto the
origin A = 0 where we use the standard shorthand 0 for {0}. (This also works for any
convex subset of Rn containing the origin.)
There are two ways to think about a deformation h : X ⇥ I ! X:
(1) As a family of paths x : I ! X for all x 2 X given by x (t) = h(x, t),
(2) As a one-parameter family of mappings ht : X ! X, t 2 I, given by ht (x) =
h(x, t). (t is the parameter.)
We take the second point of view. Then h0 = idX and h1 (X) ⇢ A. For a deformation
retraction we have: ht |A = idA for all t 2 I. This is a special case of a “homotopy”.
Definition 0.2.6. A homotopy is a one-parameter family of continuous maps ft : X !
Y between two spaces X, Y so that the adjoint map F : X ⇥ I ! Y given by F (x, t) =
ft (x) is continuous.1 We say that F is a homotopy from f0 to f1 and we write:
F : f0 ' f1
One major problem in algebraic topology is to determine when two mappings f, g :
X ! Y are homotopic. The philosophy is to convert this into an algebraic problem! For
example, here is one necessary condition:
Theorem 0.2.7. If f, g : X ! Y are homotopic (notation: f ' g) they induce the same
ring homomorphism f ⇤ = g ⇤ : C(Y ) ! C(X).
Proof. Suppose that : Y ! Z/2 is an element of C(Y ). Then for any x 2 X we f (x)
and g(x) are connected by a path in Y . Therefore (f (x)) = (g(x)). So,
f ⇤ ( )(x) = (f (x)) = (g(x)) = g ⇤ ( )(x)
for all x 2 X. So, f ⇤ ( ) = g ⇤ ( ). So, f ⇤ = g ⇤ .
⇤
Another major problem in algebraic topology is to determine when two spaces X, Y
are “homotopy equivalent”.
Definition 0.2.8. A homotopy equivalence X ' Y is defined to be a mapping f :
X ! Y so that, for some mapping g : Y ! X we have g f ' idX : X ! X and
f f ' idY : Y ! Y . In that case, we say that X, Y are homotopy equivalent or that
they have the same homotopy type. g is called the homotopy inverse of f but it is not
unique.
Proposition 0.2.9. Homotopy equivalence is an equivalence relation. I.e.,
(1) X ' X
(2) X ' Y implies Y ' X
(3) X ' Y ' Z implies X ' Z.
1This
⇤
is our first example of “adjunction”: F and {ft } are adjoint to each other and we view them
as equivalent since they contain the same information.
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However, Brandon pointed out that the collection of all topological spaces does not
form a set. So, strictly speaking, we do not have an equivalence relation on “the set of
all topological spaces”. However, (1), (2), (3) make sense and are clearly true.
The homotopy type of X is the collection of all Y ' X. This is not a set and we
should be careful.
Corollary 0.2.10. If X, Y are homotopy equivalent then C(X) ⇠
= C(Y ).
Proof. Suppose that f : X ! Y is a homotopy equivalence with homotopy inverse
g : Y ! X. Then we get induced ring mappings f ⇤ : C(Y ) ! C(X) and g ⇤ : C(X) !
C(Y ). By the Theorem we have:
f ⇤ g ⇤ = (g f )⇤ = id⇤X = idC(X) : C(X) ! C(X)
and, similarly, g ⇤ f ⇤ = idC(Y ) . Therefore, f ⇤ and g ⇤ are isomorphisms of rings.
⇤
This corollary is a “meta-corollary”. It did not use any properties of C(X) except
for what is stated in the Theorem: If we can replace “C” with another “functor” in the
Theorem, we can replace “C” with the functor in the Corollary and in its proof. The
concept of a “functor” lies in two steps in the proof:
Lemma 0.2.11.
(1) If f : X ! Y, g : Y ! Z are maps then (g
C(Z) ! C(X).
(2) (idX )⇤ = idC(X) .
f )⇤ = f ⇤
g⇤ :
Proof. These are both “obvious” since, by definition, f ⇤ is composition with f . For any
' : Z ! Z/2,
(g f )⇤ (') = ' (g f ) = (' g) f = (g ⇤ (') f = f ⇤ (g ⇤ (')
which proves (1). And (2) is even easier.
⇤
Proposition 0.2.12. If A is a deformation retract of X then A ' X.
Proof. In this proof, which I did in detail in class, I emphasized one point: The homotopy
ft : X ! X ends with a mapping f1 : X ! X whose image is A. So, f1 = j f10 where
j : A ! X is the inclusion and f10 : X ! A is the mapping f1 with codomain restricted
to A. So, the homotopy equivalence A ' X is given by inclusion j : A ! X and
retraction f10 : X ! A and ft is the homotopy idX ' j f10 = f1 and f10 j = idA .
⇤
A very important special case is when A is single point.
Definition 0.2.13. A space X is called contractible if it is homotopy equivalent to one
point: X ' ⇤.
Theorem 0.2.14. A space X is contractible i↵ the identity mapping on X is nullhomotopic (homotopic to a constant map):
X ' ⇤ () idX ' cx0 : X ! X
where cx0 is the map cx0 (x) = x0 for all x 2 X.
Proof. ()) The definition of X ' ⇤ gives maps X ! ⇤, ⇤ ! X. The second map gives
the point x0 . The composition X ! ⇤ ! X is cx0 . The definition of X ' ⇤ also gives
the homotopy idX ' cx0 .
⇤
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MATH 151A: TOPOLOGY I
Example 0.2.15. Some standard examples of contractible space:
(1) CX, the cone of X. This is the quotient space:
CX = X ⇥ I/X ⇥ 1
where the notation means the top of the cylinder X ⇥ 1 ⇢ X ⇥ I is “smashed to
a point”. One important special case is X = ; in which case this point must be
added: C; = ⇤. So, the correct words are: Add a point ⇤ and identify all points
of X ⇥ 1 with ⇤.
The cone CX has its “cone point” as deformation retract with deformation:
ft [x, s] = [x, (1
t)s + t]
(2) Comb. This is the subspace of the plane R2 given by
1
[
1
[0, 1] ⇥ 0 [ 0 ⇥ [0, 1] [
⇥ [0, 1
n
n=1
1/n]
This space is contractible with (0, 0) as deformation retract but (0, 1) is not a
deformation retract. (See HW problem.)
(3) House with two rooms. (See the book.)
The homotopy in (1) is an example of a linear homotopy f ' g which has the following
general formula:
ft = (1 t)f + tg
In algebraic topology, we like to rephrase concepts in terms of mapping rather than
spaces.
Corollary 0.2.16. A space X is contractible if and only if it is a retract of its cone
CX.
This is expressed in terms of the following diagram:
X
⇢
CX
/
=
✏ }
X
This is an example of the extension problem.
0.3. Extension Problem. Another main problem in algebraic topology is to determine
when maps can be extended: Given A ⇢ X and a mapping f : A ! Y , when does there
exist a mapping f : X ! Y so that f |A = f ? f is called an extension of f to X.
A
⇢
f
✏ ~
Y
f
/
X
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We usually convert this diagram into a diagram of groups where it becomes a lifting
⇤
problem. f , if it exists, is called a lifting of f ⇤ to F (X).
F (A) o
j⇤
F (X)
O
;
f⇤
f
⇤
F (Y )
For example, F (X) = the ring C(X). If f = idA : A ! Y = A has more components
than X, then the identity map f ⇤ : F (Y ) ⇠
= F (A) cannot factor through F (X).
1
Another example is: F (X) = [X, S ]. The notation is
[X, Y ] = the set of homotopy classes of maps f : X ! Y . The homotopy class of f
is denoted [f ].
S 1 = multiplicative topological group of unit complex numbers. S 1 is isomorphic to
the additive group R/Z.
Lemma 0.3.1. [X, S 1 ] is a group for any X.
Proof. Mappings f, g : X ! S 1 are multiplied pointwise: (f · g)(x) = f (x)g(x). This
makes the set of all mappings X ! S 1 into a group. The group is commutative since
S 1 is commutative. The identity is the constant map 1(x) = 1 for all x 2 X. The group
inverse of f : X ! S 1 is given by composition c f where c : S 1 ! S 1 is the inverse
map (given by complex conjugation!)
Let N be the set of all mappings f : X ! S 1 which are null-homotopic (homotopic
to 1).
Claim: N is a subgroup of M ap(X, S 1 ).
This follows from the following general conditions:
(1) If f ' g. Then c f ' c g.
(2) If f ' g and f 0 ' g 0 then f · f 0 ' g · g 0 .
These imply the claim since f 2 N i↵ f ' 1 i↵ c f ' c 1 = 1. So, c f 2 N . And
f, g 2 N implies f · g ' 1 · 1 = 1 ) f · g 2 N .
Since M ap(X, S 1 ) is an abelian group, N is normal. The quotient group is equal to
[X, S 1 ]. (The homotopy classes are equal to the cosets: [f ] = f · N .)
⇤
Theorem 0.3.2. If the extension problem for spaces has a solution, the corresponding
lifting problem for groups has a solution.
S1
⇢
/
=
D2
F (S 1 ) = Z o
O
F (D2 ) = 0
7
=
✏ 
S1
f
F (S 1 ) = Z
Since the identity homomorphism on the group Z does not factor through the trivial
group F (D2 ) = [D2 , S 1 ] = 0, there is no way to fill in the dotted line on the right. So,
there is no way to fill in the dotted line on the left. So, X = S 1 is not contractible. This
argument uses a theorem that we have not yet proven:
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MATH 151A: TOPOLOGY I
Theorem 0.3.3. [S 1 , S 1 ] ⇠
= Z.
This isomorphism is given by assigning to each mapping S 1 ! S 1 its “winding number” which is an integer. We will prove this later using the theory of covering spaces.
0.4. Cell complexes. Cell complexes are examples of “mapping cones” which look like
this:
C(f ) :
CA
Y
f (A)
Given a map f : A ! Y , the cone of f is the space obtained from CA
[a, 0] to f (a) 2 Y for all a 2 A:
`
C(f ) = CA Y /([a, 0] ⇠ f (a) 8a 2 A)
`
Y by pasting
We say that the mapping cone C(f ) is obtained by “attaching CA to Y along a map f :
A ! Y ”. We give C(f ) the quotient topology. In other words, a mapping g : C(F ) ! Z
is continuous if and only if both g|Y : Y ! Z and the composition A ⇥ I ! CA !
C(f ) ! Z are continuous.
For a cell complex, we take A to be a sphere A = S n 1 , the unit sphere in Rn and we
take, instead of CS n 1 the n-dimensional disk
Dn = {x 2 Rn | ||x||  1}
Exercise: Show that Dn is homeomorphic to the cone Dn ⌘ CS n 1 .
A CW complex is constructed inductively by: X 1 = ; and X n is obtained from
n 1
X
by attaching a collection of n-disks D↵n to X n 1 along maps ⌘↵ : S n 1 ! X n 1 .
Xn :
en↵
en
⌘↵ (S n 1 )
⌘ (S n 1 )
Xn = Xn
1
Xn
1
[ en↵ [ en [ · · ·
0.5. Homotopy extension property. One of the important properties of a CWcomplex is that it has the “homotopy extension property” with respect to “subcomplexes”. Today, I will explain what this means. For full details, please read this last
section of in the introductory chapter in Hatcher’s book.
Terminology: A pair of spaces (X, A) is simply a space X and a subspace A.
Definition 0.5.1. A pair of spaces (X, A) has the homotopy extension property (HEP)
if, for any mapping f of X into any space Z, any homotopy F of f |A extends to X. I.e.,
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there exists a homotopy F : X ⇥ I ! Z of f whose restriction to A ⇥ I is equal to F .
This is indicated by the following diagram.
A
✓
A⇥I
/
✓
✏
X
/
F
✏
9F
X ⇥I
'/
9
Z
f
This diagram can be rewritten as an extension problem:
X [A⇥I
f [F
✏ x
✓
/
X ⇥I
F
Z
Lemma 0.5.2. A pair (X, A) has the property that (1) any map f : A ! Z (for any
Z) extends to X if and only if (2) A is a retract of X.
The proof is left as an easy exercise.
Theorem 0.5.3. A pair (X, A) has the HEP if and only if X [ A ⇥ I is a retract of
X ⇥ I.
The main example of a pair having the HEP is a mapping cone:
Theorem 0.5.4. For any map f : A ! Y , the pair (Y, C(f )) has the HEP.
Proof. By the theorem we need to construct a retraction of C(f ) ⇥ I to the subspace
Y ⇥ I [ C(f ). This is easy to visualize and to covert into a simple formula.
Y ⇥I
C(f ) ⇥ 0
The retraction R : C(f ) ⇥ I ! C(f ) ⇥ 0 [ Y ⇥ I is given by R = id on Y ⇥ I and on
the image of CX ⇥ I by
( ⇥
⇤
t
a, 2s
,0
if 2s > t
2
t
R([a, s], t) =
([a, 0], t 2s) if 2s  t
The key point is that the cone points ([a, 1], t) go to the cone point ([a, 1], 0) and the
points in Y ⇥ I (where s = 0) go to themselves.
⇤
Lemma 0.5.5. If A ✓ B and B ✓ C have the HEP then so does A ✓ C.
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MATH 151A: TOPOLOGY I
Proof. This is “obvious”: Given f : C ! Z and a homotopy F of f |A, we first observe
that F extends to F : B ⇥I ! Z since (B, A) has the HEP. Then this homotopy extends
to C since (C, B) has the HEP.
⇤
Definition 0.5.6. A subcomplex of a CW-complex X is defined to be a closed subset
A which is a union of cells. X is called a finite CW-complex if it has a finite number of
cells.
Theorem 0.5.7. (X, A) has the HEP for every finite CW-complex X and subcomplex
A.
Proof. X is constructed from A by attaching a finite number of cells. Each step is the
mapping cone of the attaching map ⌘↵ : S n 1 ! A [ (previous cells). Now apply Lemma
0.5.5.
⇤
Note that the same proof works if X is obtained from an arbitrary space A by attaching a finite number of cells. Such a pair (X, A) is called a CW-pair provided that the
cells are attached in increasing order of dimension and X has the quotient topology (is
an iterated mapping cone). Later we will see that the same statement holds for infinite
cell complexes (having infinitely many cells). In your homework, please assume that all
CW-complexes under consideration are finite.
Added in the lecture:
Definition 0.5.8. A mapping f : A ⇢ X is called a cofibration if it is a homeomorphism
onto its image f (A) = B ✓ X and (X, B) has the HEP.
A very important fact is: “Every map is equivalent to a cofibration.” This is not very
difficult to prove and we will do that later.
0.6. Mapping spaces. Today we will talk about the topology of the mapping space
M ap(X, Y ) and the adjunction theorem which is proved using the hot dog lemma.
Definition 0.6.1. The set M ap(X, Y ) of all continuous mappings f : X ! Y is a
topological space using the compact-open topology whose subbasic open sets are:
hK, U i := {f : X ! Y : f (K) ✓ U }
for all compact K ✓ X and open U ✓ Y .
Exercise 0.6.2. If X, Y are metric spaces and X is compact then show that the compactopen topology on M ap(X, Y ) is equivalent to the uniform metric topology given by
d(f, g) = sup d(f (x), g(x)) (which is finite since X is compact).
The main theorem is:
Theorem 0.6.3. If Z is a locally compact Hausdor↵ space then we have a bijection
: M ap(X ⇥ Z, Y ) ⇠
= M ap(X, M ap(Z, Y ))
given by adjunction:
(F )(x)(z) = F (x, z).
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The main application of this is when Z = I. Then the statement is that a homotopy
F : X ⇥ I ! Y is equivalent to a continuous map X ! M ap(I, Y ).
If F : X ⇥ Z ! Y is continuous then we need to show that (F ) is also continuous.
The key step in the proof is given by:
Lemma 0.6.4 (Hot Dog Lemma). Let W be an open subset of X ⇥ Y . If W contains
K ⇥ y0 where K is a compact subset of X and y0 2 Y then there are open subsets U ⇢ X
and V ⇢ Y so that K ⇢ U, y0 2 V and U ⇥ V ✓ W .
Proof. For each point x 2 K, W is a nbh of (x, y0 ). So, there exist open nbhs Ux ✓ X
of x and Vx ✓ Y of y0 so that Ux ⇥ Vx ✓ W . The open sets Ux form a covering
of
S
K. Since T
K is compact, there is a finite subcovering Uxi . Then K ⇢ U = Uxi and
y0 2 V = Vxi and U ⇥ V ⇢ W .
⇤
Take a subbasic open set hK, U i ✓ M ap(Z, Y ) and x0 2 X so that (F )(x0 ) 2 hK, U i.
Then F (x0 ⇥ K) ✓ U . By the Hot Dog Lemma, there are open nbhs V, W of x0 , K in
X, Z resp. so that F (V ⇥ W ) ✓ U . This implies that (F )(V ) ✓ hK, U i. Therefore,
(F ) : X ! M ap(Z, Y ) is continuous (without any assumptions on X, Y, Z).
Proof of Theorem 0.6.3. It is obvious that is 1-1. So, it suffices to show it is onto
assuming Z is locally compact Hausdor↵.
Suppose that
: X ! M ap(Z, Y ) is continuous. Let F : X ⇥ Z ! Y be the
adjoint mapping. We want to show it is continuous. So, let F (x0 , z0 ) = y0 2 Y and
let U be an open nbh of y0 in Y . We are given that (x0 ) : Z ! Y is continuous and
sends z0 to y0 . So, there is a compact nbh K of z0 in Z so that (x0 )(K) ✓ U . So,
(x0 ) 2 hK, U i. Since is assumed to be continuous, there is an open nbh V of x0 in
X so that (V ) ✓ hK, U i. Then F (V ⇥ K) ✓ U showing that F is continuous. So, is
onto.
⇤
Exercise 0.6.5. Show that, if X, Z are both locally compact Hausdor↵ then
homeomorphism.
is a
Contents
0. Introduction
0.1. Connectivity
0.2. Homotopy
0.3. Extension Problem
0.4. Cell complexes
0.5. Homotopy extension property
0.6. Mapping spaces
References
References
[Ha] Allen Hatcher, Algebraic Topology, Cambridge Univ. Press, Cambridge, 2002
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