Download HOMOTOPY THEORY 1. Homotopy Let X and Y be two topological

HOMOTOPY THEORY
JESPER M. MØLLER
Abstract. This note contains comments to Chapter 0 in Allan Hatcher’s book [5].
1. Homotopy
Let X and Y be two topological spaces and f0 , f1 : X → Y be two continuous maps. (In the
following, ‘map’ will mean ‘continuous map’.)
1.1. Definition. The map f0 is homotopic to the map f1 , f0 ' f1 , if there exists a map (a
homotopy) f : X × I → Y such that f0 (x) = f (x, 0) and f1 (x) = f (x, 1) for all x ∈ X.
It is easily seen that homotopy is an equivalence relation. The fundamental problem of algebraic
topology is to calculate [X, Y ], the set of homotopy classes of maps from X to Y , for given spaces
X and Y .
1.2. Definition. The map f : X → Y is a homotopy equivalence if there exists a map g : Y → X
such that the two compositions are homotopic to the respective identity maps. Two spaces are homotopy equivalent if there exists a homotopy equivalence between them. A space is contractible
if it is homotopy equivalent to a point.
A space is contractible if and only if the identity map is homotopic to a constant map. The
House with Two Rooms and the House with One Room are examples of contractible spaces.
There is also a notion of relative homotopy. If the two maps f0 , f1 : X → Y agree on some
subspace A ⊂ X, then we say that f0 is homotopic to f1 relative to A, f0 ' f1 rel A, if there exists
a homotopy f : X × I → Y from f0 to f1 such that f0 (a) = f (a, t) = f1 (a) for all a ∈ A and all
t ∈ I.
1.3. Definition. Let X be a space and A ⊂ X a subspace.
• A is a retract of X if there exists a continuous map (a retraction) r : X → A such that
r(a) = a for all a ∈ A.
• A is a deformation retract of X if the identity map of X is homotopic rel A to a
retraction r : X → A of X onto A. A deformation retraction is such a homotopy X ×
I → X rel A between the identity on X and the retraction.
Thus the subspace A is a deformation retract of X if one of the following equivalent conditions
hold
• there exists a map, the deformation retraction, R : X × I → X such that R(x, 0) = x and
R(x, 1) ∈ A for all x ∈ X while R(a, t) = a for all a ∈ A, t ∈ I
• there exists a map r : X → A such that ri = 1A and ir ' 1X rel A (where i is the inclusion
map)
For instance S 2 is a retract of S 2 ∨ S 1 and a deformation retract of S 2 ∨ I. There are several
other good examples of (deformation) retractions in [5]. A retract is a left inverse to the inclusion
map. Any retract A of a Hausdorff space X is closed for A = {x ∈ X|r(x) = x} is the equalizer of
two continuous maps [7, Ex 31.5].
If A is a deformation retract of X, then the inclusion map i : A ,→ X is a homotopy equivalence.
Conversely, if the inclusion map is a homotopy equivalence, there exists a map r : X → A such
that ri ' 1A and ir ' 1X . This is not quite the same as saying that A is a deformation retract of
X since r may not fix the points of A and, even if it does, the points in A may not be fixed under
the homotopy from ri to the identity of A. However, surprisingly enough, the converse does hold
if the pair (X, A) is sufficiently well behaved (3.20).
Date: 17th August 2005.
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2
J.M. MØLLER
1.4. Example (Mapping cylinders and mapping cones). Let f : X → Y be a map. The mapping
cylinder of f
(X × I) q Y
Mf =
(x, 1) ∼ f (x)
is obtained by gluing one end of the cylinder on X onto Y by means of the map f . The mapping cylinder deformation retracts onto its subspace Y . What is the mapping cylinder of z → z 2 : S 1 → S 1 ?
The cone on X
X ×I
CX =
X ×0
is obtained by collapsing one end of the cylinder on X. The cone is always contractible. What is
the cone on the n-sphere S n ?
The mapping cone on f
CX q Y
Cf =
(x, 1) ∼ f (x)
is obtained by gluing the cone on X onto Y by means of the map f . What is the mapping cone of
z → z2 : S 1 → S 1?
2. CW-complexes
The CW-complexes are a class of spaces that are particulaly well suited for the methods of
algebraic topology. CW-complexes
are built inductively out of cells.
` n−1
Let X be a space and φ :
Sα → X a map from a disjoint union of spheres into X. The
adjunction space [7, Exercise 35.8] (solution) (the mapping cone on φ)
`
φ
Sαn−1
_
` n
Dα
φ
/ X
_
/ X ∪φ
`
Dαn
is called the n-cellular extension of X with attaching map φ and characteristic map φ.
2.1. Definition. A CW-complex is a space X with a sequence of subspaces (called skeleta)
[
∅ = X −1 ⊂ X 0 ⊂ X 1 ⊂ · · · ⊂ X n−1 ⊂ X n ⊂ · · · ⊂ X =
Xn
such that
• X 0 is a discrete set of points
• X n is (homeomorphic to) an n-cellular extension of X n−1 for n ≥ 1
• The topology on X is coherent with the filtration in the sense that
A is closed (open) ⇐⇒ A ∩ X n is closed (open) for all n
for any subset A of X.
Any CW-complex is a normal topological space [7, Exercise 35.8] (solution). CW-decompositions
are not unique; there are generally many CW-decompositions of a given space X – consider for
instance X = S 2 .
2.2. Example. The closed orientable surface Mg of genus g is a CW-complex
_
Mg =
S 1 ai ∨ S 1 bi ∪Q[ai ,bi ] D2
1≤i≤g
with 1 0-cell, 2g 1-cells, and 1 2-cell.
The closed nonorientable surface Ng of genus g is a CW-complex
_
Ng =
S 1 ai ∪Q a2i D2
1≤i≤g
with 1 0-cell, g 1-cells, and 1 2-cell.
HOMOTOPY THEORY
3
The n-sphere S n can be obtained from S n−1 by attaching two n-cells (the Northern and Southern
hemisphere). Thus S n is a finite CW-complex with two cells in each dimension 0 through n. The
infinite sphere S ∞ is an infinite dimensional CW-complex
∞
[
S ◦ ⊂ S 1 ⊂ · · · ⊂ S n−1 ⊂ S n ⊂ · · · ⊂ S ∞ =
Sn
n=0
with two cells in each dimension.
Real projective n-space RP n can be obtained from RP n−1 by attaching one n-cell along the
canonical quotient map pn−1 : S n−1 → RP n−1 . Thus RP n is a finite CW-complex
with one cell
S∞
in each dimension 0 through n. The infinite real projective space RP ∞ = n=0 RP n is an infinite
dimensional CW-complex
RP ◦ ⊂ RP 1 ⊂ · · · ⊂ RP n−1 ⊂ RP n ⊂ · · · ⊂ RP ∞
when equipped with the coherent topology.
3. The Homotopy Extension Property
The Homotopy Extension Property will be very important to us.
3.1. Definition. [1, VII.1] Let X be a space with a subspace A ⊂ X. The pair (X, A) has the
Homotopy Extension Property (HEP) if it is always possible to complete the diagram
X × {0} ∪
_ A×I
/8 Y
incl
X ×I
for any space Y and any partial homotopy of a map X → Y .
Thus (X, A) has the HEP if any partial homotopy on A of a map X → Y can be extended to a
homotopy of the map. The pair (X, ∅) always has the HEP. A nondegenerate base point is a
point x0 ∈ X such that (X, {x0 }) has the HEP.
3.2. Proposition. [5, p 14, Ex 0.26] Let X a space and A ⊂ X a subspace. The following three
conditions are equivalent
(1) (X, A) has the HEP
(2) The partial cylinder X × {0} ∪ A × I is a retract of the cylinder X × I.
(3) The partial cylinder X × {0} ∪ A × I is a deformation retract of the cylinder X × I.
Proof. If (X, A) has the HEP then the identity map of the partial cylinder X × {0} ∪ A × I extends
to a retraction of the cylinder X × I onto the partial cylinder. Conversely, if the inclusion of the
partial cylinder into the cylinder has a left inverse r then it is very easy
h /
X × {0} ∪ A × I
8Y
_O
qqq
q
q
q
r
incl
qqqhr
qqq
X ×I
to find an extension of any partial homotopy h. This shows that (1) ⇐⇒ (2). It is clear that
(3) =⇒ (2). To prove that (2) =⇒ (3) let r : X × I → X × {0} ∪ A × I be a retraction. Define a
homotopy [2, p 329] H : X × I × I → X × I by
H(x, t, s) = (π1 r(x, st), (1 − s)t + π2 r(x, t))
π1
π
2
where (as usual) X ←− X × I −→
I are the projections. Then H(x, t, 0) = (x, t), H(x, 1) = r(x, t),
H(x, 0, s) = (x, 0), and H(a, t, s) = (a, t) for all a ∈ A. Thus H is a deformation retraction of the
cylinder X × I onto the partial cylinder X × {0} ∪ A × I.
3.3. Example. The pair (I, ∂I) has the HEP, for radial projection from ( 21 , 2) ∈ R2 gives a
(deformation) retraction of I × I ⊆ R2 onto I × {0} ∪ ∂I × I. More generally, the pair (Dn , ∂Dn ) =
(Dn , S n−1 ) has the HEP for all n ≥ 1.
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J.M. MØLLER
3.4. Proposition. If (X, A) has the HEP and X is Hausdorff, then A is a closed subspace of X.
Proof. X × {0} ∪ A × I is a closed subspace of X × I since it is a retract. Now look at X at level
1
2 inside the cylinder X × I.
See [1, VII.1.5] for a necessary condition for an inclusion to have the HEP.
3.5. Example. (A closed subspace that does not have the HEP) (General topology exam, Problem
3). (I, A) where A = {0} ∪ { n1 |n = 1, 2, . . .} does not have the HEP since I × {0} ∪ A × I is not a
retract of I × I. (The partial homotopy which is the identity map of I × {0} ∪ A × I does not extend
to a homotopy.) Indeed, assume that r : I × I → A × I is a retraction. For each n ∈ Z+ , the map
1
1
t 7→ r(t × 1), n+1
≤ t ≤ n1 , is a path in A from n+1
× 1 to n1 × 1. Such a path must pass through
1
1
1
1
all points of ( n+1 , n ) × {0} ⊂ I × {0} because π1 r([ n+1
, n1 ] × {1}) ⊃ [ n+1
, n1 ] by connectedness.
1
1
1
1
Thus there is a point tn × 1 ∈ ( n+1 , n ) × {1} such that r(tn × 1) ∈ ( n+1 , n ) × {0}. This contradicts
continuity of r for tn × 1 converges to 0 × 1 and r(t1 × 1) converges to 0 × 0 6= r(0 × 1).
The following two results show the usefulness of the HEP.
3.6. Proposition. Suppose that (X, A) has the HEP. Any homotopy left inverse r : X → A to the
inclusion map i : A → X is homotopic to a retract of X onto A.
Proof. Assume that r : X → A is a map such that ri ' 1A . We must change r on A so that it
actually fixes points of A. Let f : A × I → X be a homotopy with f0 = ri and f1 = i. Using the
HEP we may complete the commuative diagram
r∪f
X × {0} ∪
_ A×I
X ×I
/7 X
h
and get a homotopy h : X × I → X. The (corestriction of the) end-value of this homotopy is a
map h1 : X → A such that h1 i = 1A (a retract).
3.7. Proposition. [5, Proposition 0.17] [1, VII.4.5] Suppose that (X, A) has the HEP and that A
is contractible. Then the quotient map q : X → X/A is a homotopy equivalence.
Proof. Let c : A × I → A ⊂ X be a contraction of A, a homotopy of the identity map to a constant
map. Use the HEP to extend the contraction of A and the identity on X
1X ∪c
X × {0} ∪
_ A×I
X ×I
/7 X
h
to a homotopy h : X × I → X such that h0 is the identity map, ht sends A to A for all t ∈ I, and
h1 sends A to a point of A. Then h1 induces a map
h1
/X
XC
CC
{=
{
CC
{{
C
{{
q CC
!
{{ h1
X/A
such that h1 = h1 q. We claim that X o
/ X/A are homotopy inverse to each other. Certainly,
q
h1
h1 q = h1 ' h0 = 1X .
The universal property of quotient maps provides a factorization of qh
/ X/A
/X
X × IJ
JJJ
u:
u
u
JJJ
uu
uu qh
q×1 JJJ
u
%
u
X/A × I
h
q
HOMOTOPY THEORY
5
showing that qh1 = qh1 ' qh0 = 1X/A . Note here that the product map q × 1 : X × A → X/A × I
is quotient since I is locally compact Hausdorff [7, Ex 29.11].
3.8. Proposition. The HEP property is preserved under some constructions.
(1) (Transitivity) If X0 ⊆ X1 ⊆ X2 and both pairs (X2 , X1 ) and (X0 , X1 ) have the HEP, then
(X2 , X0 ) has the HEP.
(2) If (X, A) has the HEP then Y × (X, A) = (Y × X, Y × A) has the HEP for all spaces Y .
(3) If (X, A) has the HEP then (Y ∪f X, Y ∪f A) has the HEP for all spaces Y and all maps
f : B → Y defined on a subspace B of A. In particular [5, Ex 0.28], (Y ∪f X, Y ) has the
HEP.
` n
(4) The n-cellular
` n−1extension (X ∪φ D , X) of any space X has the HEP for any attaching
map φ :
S
→ X.
Proof. (1) Using the HEP twice we may extend
k/5 Z
kkwkww;
k
k
k
kkk ww
kkk www
k
k
kk
ww
ww
X2 × {0} ∪
 _ X1 × Iwww
ww
ww
w
w
ww
X2 × I
X2 × {0} ∪
 _ X0 × I
any given map X2 × {0} ∪ X0 × I → Z to a homotopy X2 × I → Z.
(2) We use 3.2. Let r : X × I → X × I be a retraction onto X × {0} ∪ A × I. Then the product
map 1Y × r is a retraction of (Y × X) × I onto (Y × X) × {0} ∪ (Y × A) × I.
(3) We use 3.2 again. Let r : X × I → X × I be a retraction onto X × {0} ∪ A × I. The universal
property of quotient maps provides a factorization
(Y q X) × I
1Y ×I qr
q×1I
q×1I
(Y ∪f X) × I
/ (Y q X) × I
1Y ×I qr
/ (Y ∪f X) × I
of Y ∪f X onto Y ∪f A. Note here that the left vertical map is a quotient map since I is locally
compact Hausdorff [7, Ex 29.11]. If we`take B
`= A, we see that (Y ∪f X, X) has the HEP.
(4) This is a special case of (3) since ( Dn , S n−1 ) has the HEP (3.3).
We shall need, but not prove, the following variant of 3.8.(2).
3.9. Proposition. [2, 7.5 p. 330] If (X, A) has the HEP then also (X × I, X × ∂I ∪ A × I) has
the HEP
3.10. Example. (HEP for mapping cylinders.) Let f : X → Y be a map. We apply (3.8) in
connection with the mapping cylinder Mf = Y ∪f (X × I).
3.8.(2)
3.8.(3)
(I, ∂I) has the HEP =⇒ (X × I, X × ∂I) has the HEP =⇒ (Mf , X ∪ Y ) has the HEP
3.8.(2)
3.8.(3)
(I, {0}) has the HEP =⇒ (X × I, X × {0}) has the HEP =⇒ (Mf , Y ) has the HEP
The fact that (Mf , X ∪ Y ) has the HEP implies that also (Mf , X) has the HEP (simply take a
constant homotopy on Y ). See 3.11 below for another application.
3.11. Example. (HEP for subspaces with mapping cylinder neighborhoods [5, Example 0.15]) For
another application of (3.8), suppose that the subspace A ⊆ X has a mapping cylinder neighborhood. This means that A has a closed neighborhood N containing a subspace B (thought of as
the boundary of N ) such that N − B is an open neighborhood of A and (N, A ∪ B) is homeomorphic to (Mf , A ∪ B) for some map f : B → A. Then (X, A) has the HEP. To see this, let
h : X × {0} ∪ A × I → Y by a partial homotopy of a map X → Y . Extend it to a partial homotopy on X × {0} ∪ (A ∪ B) × I by using the constant homotopy on B × I. Since (N, A ∪ B) has the
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J.M. MØLLER
HEP, we can extend further to a partial homotopy defined on X × {0} ∪ N × I. Finally, extend to
X × I by using a constant homotopy on X − (N − B) × I. In this way we get extensions
/
jjj5 > C Y
j
j
j
jjj
jjjj
j
j
j
jj
X × {0} ∪ (A
∪
B)
×I
_
X × {0} ∪
_ A×I
h
X × {0} ∪
_ N ×I
X ×I
The final map is continuous since it restricts to continuous maps on the closed subspaces X − (N −
B) × I and N × I with union X × I.
For instance, the subspace ABC ⊆ R2 consisting of the three thin letters in the figure on [5,
p. 1] is a subspace with a mapping cylinder neighborhood, namely the three thick letters. Thus
(R2 , ABC) has the HEP.
Transitivity may be applied infinitely many times as in the following situation.
3.12. Proposition. [5, Proposition 0.16] [1, VII.1.4] Any CW-pair (X, A) has the HEP.
Proof. In the filtration
A = A ∪ X −1 ⊂ A ∪ X 0 ⊂ · · · ⊂ A ∪ X n−1 ⊂ A ∪ X n ⊂ · · · ⊂ X
each pair (X n ∪ A, X n−1 ∪ A) has the HEP for X n ∪ A is an n-cellular extension of X n−1 ∪ A
by the n-cells of X that are not already in A. By transitivity (applied perhaps infinitely many
times), (X, X n−1 ∪ A) has the HEP for all n ≥ 0. (It is here necessary to observe that the topology
on X × I is coherent with the filtration (X n ∪ A) × I.) In particular for n = 0, (X, A) has the
HEP.
3.13. Example. Let C be the quasi-circle [5, Ex 1.3.7]. Collapsing the interval [−1, 1] ⊂ C lying
on the vertical axis gives a quotient map C → S 1 [6, §28] which is not a homotopy equivalence
(since π1 (S 1 ) = Z and π1 (C) is trivial) even though [−1, 1] is contractible.
3.14. Proposition. [5, Proposition 0.18, Ex 0.26] Suppose that (X, A) has the HEP. Let Y be
any space and φ : A → Y any map. Then the homotopy type of the adjunction space Y ∪φ X only
depends on the homotopy class of the attaching map φ : A → Y .
Proof. Let f : A × I → Y be a homotopy of f0 to f1 . Form the adjunction space Y ∪f (A × I). We
intend to show that the inclusions
Y ∪f0 X = Y ∪f (X × {0} ∪ A × I) ⊂ Y ∪f (X × I) ⊃ Y ∪f (X × {1} ∪ A × I) = Y ∪f1 X
are homotopy equivalences rel Y . (The equality signs are there because all points of A × I have
been identified to points in Y .) The left inclusion is a homotopy equivalence because the subspace
is a deformation retract of the big space. The deformation retraction h rel Y of Y ∪f (X × I) onto
Y ∪f0 X is induced as in the diagram
/
A × I fM
kk5 Y
MMM
kkk
k
k
MMM
kk
kkk
π1 ×π2 MMM
kkk π1
k
k
f ×1
/ Y ×I
A×I ×I
X ×I ×I
ss
h sss
s
ss
y ss
s
X ×I
/ Y ∪f (X × I) × I
RRR
RRR h
RRR
RRR
R(
/ Y ∪f (X × I)
HOMOTOPY THEORY
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from a deformation retraction h : X × I × I → X × I of X × I onto X × {0} ∪ A × I. (The inner
square is just the push-out diagram for Y ∪f (X × I) crossed with the unit interval.) By symmetry,a
the right inclusion is also a homotopy equivalence (rel Y ), of course.
As an application, show that the dunce cap [3] is contractible.
The proof of [5, Proposition 0.19] uses the homotopy commutative mapping cylinder diagram
Mf
|> O
|
|| ' j
||
||
?
X f /Y
i
which shows that any map is an inclusion (satisfying the HEP), up to a homotopy equivalence.
3.16. Example. The unreduced suspension SX and the reduced suspension ΣX are homotopy
equivalent for all CW-complexes X based at a 0-cell.
3.17. Example. [4, 21.21] (Homotopic maps have homotopy equivalent mapping cones). The pair
(CX, X) clearly has the HEP for X has a mapping cylinder neighborhood in CX for any space X.
Suppose that f ' g : X → Y are homotopic maps. Then 3.14 shows that
Cf = Y ∪f CX ' Y ∪g CX = Cg
In particular, the homotopy type of an n-cellular extension only depends on the homotopy class of
the attaching map.
3.18. Example. Suppose that (X, A) has the HEP. Then also (X ∪CA, CA) has the HEP (3.8.(3)).
The cone on A is always contractible. Hence
X ∪ CA → X ∪ CA/CA = X/A
is a homotopy equivalence.
3.19. Example. Suppose that (X, A) has the HEP and that the inclusion map A ,→ X is homotopic
to the constant map 0 : A → X, ie. if A is contractible in X. Then
3.18
3.17
X/A ' X ∪ CA = Ci ' C0 = X ∨ SA
For instance, S n /S i ' S n ∨ S i+1 for all i ≤ 0 < n. (The inclusion S i → S n , 0 < i < n, is
nullhomotopic since it factors through the contractible space S n − ∗ = Rn .) See [5, Exmp 0.8] for
an illustration of S 2 /S 0 ' S 2 ∨ S 1 .
3.20. Proposition. Suppose that (X, A) has the HEP. If the inclusion map is a homotopy equivalence, then A is a deformation retract of X.
Proof. Let i : A → X be the inclusion map. The assumption is that there exists a map r : X → A
such that ri ' 1A and ir ' 1X . By 3.6 we can assume that ri = 1A , ie that A is a retract of X.
Let G : X × I → X be a homotopy with start value G0 = 1X and end value G1 = ir. We want to
modify G into a deformation retraction, that is a homotopy 1X ' ir rel A. Since (X, A) has the
HEP so does (X, A) × (I, ∂I) = (X × I, A × I ∪ X × ∂I) (3.9). Let H : X × I × I → X × I be an
extension (a homotopy of homotopies) of the map X × I × {0} ∪ A × I × I ∪ X × ∂I × I given by
H(x, t, 0) = G(x, t)
H(a, t, s) = G(a, t(1 − s)) for a ∈ A
H(x, 0, s) = x
H(x, 1, s) = G(ir(x), 1 − s)
Note that H is well-defined since the first line, H(x, 1, 0) = G(x, 1) = ir(x), and the fourth
line, H(x, 1, 0) = G(ir(x), 1) = irir(x) = ir(x), yield the same result. The end value of H,
(x, t) 7→ H(x, t, 1), is a homotopy rel A of H(x, 0, 1) = x to H(x, 1, 1) = G(ir(x), 0) = ir(x). This
is a homotopy rel A since H(a, t, 1) = G(a, 0) = a for all a ∈ A.
@
@ a
@
I
@
@
a
@
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J.M. MØLLER
References
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[2] James Dugundji, Topology, Allyn and Bacon Inc., Boston, Mass., 1966. MR 33 #1824
[3] George K. Francis, A topological picturebook, Springer-Verlag, New York, 1987. MR 88a:57002
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[5] Allen Hatcher, Algebraic topology, Cambridge University Press, Cambridge, 2002. MR 2002k:55001
[6] Jesper M. Møller, General topology, http://www.math.ku.dk/~moller/e03/3gt/notes/gtnotes.pdf.
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Matematisk Institut, Universitetsparken 5, DK–2100 København
E-mail address: [email protected]
URL: http://www.math.ku.dk/~moller