MATH 753 WEEK 2
NOTES BY MEGAN MAGUIRE
1. Relative homotopy groups
Let x0 ∈ A ⊂ X, and n ∈ Z with n ≥ 1. Let
I n−1 = {(s1 , . . . , sn ) ∈ I n : sn = 0}.
Define J n−1 by
J n−1 := ∂I n − I n−1 .
Then we define the nth homotopy group of the pair (X, A) with basepoint x0 to be
πn (X, A, x0 ) = {f : (I n , ∂I n , J n−1 ) → (X, A, x0 )}/{homotopy equivalence}.
We have the following commutative triangle
f
(I n , ∂I n , J n−1 )
/
(X, A, x0 )
6
g
(Dn , S n−1 , s0 )
where g is obtained by collapsing J n−1 .
Special case: n = 1. Then
π1 (X, A, x0 ) = {f : (I, {0, 1}, {1}) → (X, A, x0 )}/{homotopy equivalence}.
In words, π1 (X, A, x0 ) is the set of paths that start anywhere in A and end at x0 (up to
homotopy equivalence). We cannot define a group structure on this set.
Remark 1. πn (X, x0 , x0 ) = πn (X, x0 ).
Proposition 1. For n ≥ 2, πn (X, A, x0 ) is a group with the usual + operation as in the
absolute case. Moreover, πn (X, A, x0 ) is abelian if n ≥ 3.
Proof. The same as in the absolute case.
Remark 2. Maps of pairs ϕ : (X, A, x0 ) −→ (Y, B, y0 ) induce (by composition) group homomorphisms
ϕ∗ : πn (X, A, x0 ) −→ πn (Y, B, y0 )
[f ] 7−→ [ϕ ◦ f ].
Proposition 2. The relative homotopy groups fit into a long exact sequence:
...
/
πn (A, x0 )
/
...
i∗
/
j∗
/
πn (X, x0 )
/
πn (X, A, x0 )
/
π0 (X, x0 )
1
0
∂n
/
πn−1 (A, x0 )
2
NOTES BY MEGAN MAGUIRE
where i∗ and j∗ are
induced by inclusions of pairs/triples, i : A ,→ X, j : (X, x0 ) ,→ (X, A),
and ∂ n ([f ]) := [f ]
.
n−1
I
Proof. Exercise.
Remark 3. Towards the end of the sequence exactness still makes sense even though no
group structures are present.
Lemma 1. Let [f ] ∈ πn (X, A, x0 ). Then [f ] = 0 if and only if f ' g so that im(g) ⊆ A.
Sketch of Proof. It is clear that if [f ] = 0 then f ' g with im(g) ⊆ A, since f is homotopic
to the constant map at x0 . Conversely suppose that f ' g with im(g) ⊆ A. Deform I n to
J n−1 by pushing I n−1 inside. Then we obtain g ' cx0 .
2. (In)dependence of basepoints.
Proposition 3. If A is path connected, then
βγ : πn (X, A, x0 ) −→ πn (X, A, x1 )
is an isomorphism where γ is a path in A from x0 to x1 .
Proof. Similar to the absolute case.
Remark 4. π1 (A) acts on πn (X, A, x0 ) (just take x0 = x1 ).
We say that (X, A) is n-connected if πi (X, A) = 0 for all i ≤ n. We say that X is
n-connected if πi (X) = 0 for all i ≤ n.
Observe that X is 0-connected if and only if X is connected and X is 1-connected if and
only if X is simply connected.
Hurewicz Theorem. Given any connected topological space X there exists a map
π1 (X) −→ H1 (X) = π1 (X)ab
f : S 1 → X 7−→ f∗ ([S 1 ]),
where [S 1 ] denotes the fundamental class of S 1 . More generally there is a map of
πn (X) −→ Hn (X)
[f : S → X] 7−→ f∗ ([S n ]).
n
Theorem 1. If n ≥ 2 and πi (X) = 0 for all i < n then Hi (X) = 0 for all i < n and
πn (X) ∼
= Hn (X) (and onto on πn+1 ).
Proof. Will be proven later.
3. Homotopy groups of spheres.
What is πi (S n )?
For i < n, πi (S n ) = 0. For i = n, n + 1, πi (S n ) = Z.
For i = n, f∗ ([S n ]) 7→ deg(f ) · [S n ] is an isomorphism. To show that πi (S n ) = 0 for i < n
use cellular approximation.
Theorem 2. Given f : X −→ Y a map of CW complexes, one can find a cellular map
g : X −→ Y (i.e. g(Xi ) ⊆ Yi for all i) such that f ' g. Similarly for maps of pairs.
MATH 753 WEEK 2
3
Corollary 1. For i < n, πi (S n ) = 0.
Proof. Can assume f : S i −→ S n is cellular, which implies that
f (S i )i ⊆ (S n )i = pt, so f is homotopic to a constant map.
Corollary 2. Let A ⊆ X be CW complexes and suppose all cells in X − A have dimension
greater than n. Then πi (X, A) = 0 for all i ≤ n.
Corollary 3. For all i ≤ n, πi (X, Xn ) = 0.
From the above corollary and the long exact sequence of relative homotopy groups we
obtain the following corollary:
Corollary 4. For all i ≤ n − 1, πi (X) ∼
= πi (Xn ).
Proof of Corollary 2. For [f ] ∈ πi (X, A), the map
f : (Di , S i−1 ) −→ (X, A)
is homotopic to g such that g(Di ) ⊆ Xi ⊆ A for all i ≤ n. Thus [f ] = 0 by Lemma 1.
Suspension Theorem. Let f : S i −→ S n be a map. Consider its suspension Σf : ΣS i '
S i+1 −→ ΣS n ' S n+1 . The assignment
πi (S n ) −→ πi+1 (S n+1 )
[f ] 7−→ [Σf ]
is an isomorphism for i < 2n − 1 and onto for i = 2n − 1.
Corollary 5. We have πn (S n ) is either Z or a finite quotient of Z, generated by the degree
map.
Proof. By the Suspension Theorem
Z∼
= π1 (S 1 ) π2 (S 2 ) ∼
= ... ∼
= πn (S n ) ∼
= ....
To get that π1 (S 1 ) π2 (S 2 ) is an isomorphism, use the long exact sequence of πi for the
Hopf fibration, or alternatively use Hurewicz.
Recall that if X is a finite CW complex, then H∗ (X) and H ∗ (X) are finitely generated
abelian groups. This is not true for homotopy groups.
Example 1. Let X = S 1 ∨ S 2 . Then π2 (X) = π2 (X̃), for X̃ the universal cover of X. Recall
that the universal cover of X is the real line, with spheres attached
! at the integer points, which
_
_
is homotopy equivalent to
Sk2 . We claim that π2
Sk2 is generated by inclusions of
k∈Z
k∈Z
W
factors [Sk2 → k∈Z Sk2 ]. Assuming this, π1 (X) = Z, which implies that Deck(X̃/X) = Z.
Since π1 (X) acts on π2 (X), we have that π2 (X) is a Z[π1 ]-module. Now
(
)
X
Z[π1 ] =
ni γi : ni ∈ Z, γi ∈ πi ∼
= Z[Z] ∼
= Z[t, t−1 ]
i
(where this last isomorphism is obtained via the map n 7→ tn . Thus we see that π2 (X) is a
free Z[π1 ]-module of rank 1 with generator any of [Sk2 ,→ X̃]. So π2 (X) = Z[π1 ] ∼
= Z[t, t−1 ]
which is infinitely generated as a Z-module by {tk : k ∈ Z}.
4
NOTES BY MEGAN MAGUIRE
W
W
Lemma 2. πn ( α Sαn ) is generated (over Z) by the inclusions of factors [Sαn ,→ α Sαn ].
Proof. If there are only finitely many Sαn , then
Y
_
πn ( Sαn ) ∼
= πn ( Sαn ) (remains to be proven)
α
α
∼
=
Y
∼
=
Y
πn (Sαn )
α
Z
α
∼
=
M
Z.
α
Q
W
of α Sαn . Now Sαn =
To prove the W
first isomorphism, we claim
that α Sαn is the n-skeleton
Q
Q
e0α ∪ enα and α Sαn = e0 ∪ (enα )α . In α Sαn there is one 0-cell, α e0α := e0 and n-cells of
Q n
W n
S Q
0
n
S . So
the form α
α Sα is indeed the n-skeleton of
β6=α eβ × eα . Thus we see that
Q n Wα αn
Q n W n
2, ( α Sα , Q
α Sα ) is
α Sα only has cell of dimension at least 2n. From CorollaryW
α Sα −
(2n − 1)-connected. Since n ≥W2, and thus n < 2n − 1, we have that πn ( α Sαn ) = πn ( α Sαn ).
In general, any f : S n −→ α Sαn hasL
compact image, and
W thus touched only finitely many
spheres, Sαn . From this we obtain that α πn (Sαn ) −→ πn ( α Sαn ) is an isomorphism.
Whitehead’s Theorem. If f : X −→ Y is a map of CW complexes which induces isomorphisms on all πn , then f is a homotopy equivalence. Moreover, if f is an inclusion of a
subcomplex, then X is a deformation retract of Y .
Corollary 6. If f : X −→ Y induces isomorphisms on H∗ (−, Z) and π1 (X) = π1 (Y ) = 0,
then f is a homotopy equivalence.
Before we can proceed, we need the following lemma:
Lemma 3. In homotopy theory “any map is an inclusion”. That is, any f : X −→ Y can
be factored as X 
i
/
Mf
r
/
Y , where r is a deformation retract.
Proof. Define the mapping cylinder of f by
Mf := (X × [0, 1])
a
Y /((x, 1) ∼ f (x)).
Note that X ⊆ Mf and Y ⊆ Mf . Moreover, we have a deformation retract r : Mf −→ Y.
Thus in the above theorem and corollary one can replace f by X ,→ Mf , provided that Mf
has a CW structure. If f is cellular then Mf has a CW structure (exercise). If f is not
cellular then f ' g with g cellular, and one can work with X ,→ Mg .
Proof of Corollary 6. From the above lemma, we can assume that f : X −→ Y is an inclusion.
From the long exact sequence of homology groups for (Y, X) we obtain that Hi (Y, X) = 0
for all i. By the relative version of the Hurewicz Theorem, we get that πi (Y, X) = 0 for
all i. The long exact sequence of relative homotopy groups for the pair (Y, X) tells us that
f∗ : πi (X) −→ πi (Y ) is an isomorphism. Thus from Whitehead’s Theorem we see that f is
a homotopy equivalence.
Remark/Example 1. The simply connected assumptions of the above corollary are necessary. Let X = S 1 and Y = (S 1 ∨ S n )∪en+1 where the attaching map ∂en+1 = S n −→ S 1 ∨S n
MATH 753 WEEK 2
5
corresponds to the element 2t − 1 ∈ Z[t, t−1 ] ∼
= πn (S 1 ∨ S n ). So we have that π1 (X) = Z 6= 0,
f∗ : Hi (X) −→ Hi (Y ) is an isomorphism for all i, and πi (X) = 0 for all i > 1. However,
πn (Y ) ∼
= πn (S 1 ∨ S n )/(2t − 1) 6= 0
∼
= Z[t, t−1 ]/(2t − 1)
1
∼
= Z[ ].
2
Exercise: Find spaces X and Y with the same homotopy groups so the isomorphisms cannot
be induced by any map.