M8130 Algebraic topology, tutorial 11, 2020 6. 1. 2020
Exercise 1. If (X, A) is relative CW-complex such that there are no cells in dimension
≤ n in X \ A, then (X, A) is n-connected.
Solution. Recall the denition of n-connectness of a pair. For [f] ∈ πi(X, A, x0), i ≤ n,
use cell approximation of f: There is a cell map q: (Di
, Si−1
, s0) → (X, A, x0), such that
q ∼ f relatively Si−1
and q(Di
) ⊆ X(i)
= A since X(−1)
= X(i)
= · · · = X(n)
= A. Note
the following very useful criterion:
[f] = 0 in πi(X, A, x0) ⇐⇒ f ∼ q relatively Si−1
, g(Di
) = A.
Thus [f] = 0 in our case, and we are done.
Exercise 2. Let [X, Y ] denote a set of homotopy classes of maps from X to Y . If (X, x0)
is a CW-complex and Y is simply connected, then [X, Y ] ∼= [(X, x0), (Y, y0)].
Exercise 3. Show that the Hurewicz homomorphism is natural.
Exercise 4. Show that the Hurewicz homomorphism commutes with connecting homomorphisms.
It means: Let (X, A) be a pair. Show that the following diagram commutes:
πn(X, A, x0)
∂ //
h

πn−1(A, x0)
h

Hn(X, A) ∂∗
// Hn−1(A)
where ∂ is the boundary homomorphism, h is the Hurewicz homomorphism and ∂∗ is the
connecting homomorphism.
Solution. Take [f] ∈ πn(X, A, x0), that is f : (Dn
, Dn−1
, s0) → (X, A, x0). Then ∂[f] =
[f/Sn−1
] and h∂[f] = h[f/Sn−1
] = (f/Sn−1
)∗(b), where b is a generator in Hn−1(Sn−1
). (We
recall the denition of the Hurewicz homomorphism: if g: Sn−1
→ A, g∗ : Hn−1(Sn−1
) →
Hn−1(A) and h[g] = g∗(b) ∈ Hn−1(A).)
Let a ∈ Hn(Dn
, Sn−1
) be a generator such that ∂∗a = b. We proceed using commutativity
of the following diagram:
Hn(Dn
, Sn−1
)
∂∗
∼=
//
f∗

Hn−1(Sn−1
)
(f/Sn−1)∗

Hn(X, A) ∂∗
// Hn−1(A)
Now ∂∗h[f] = ∂∗(f∗a) = (f/Sn−1
)∗(∂∗a) = h∂[f] which concludes the proof.
Exercise 5. Show that the Hurewicz homomorphism h : πn(Sn
) → Hn(Sn
) is an isomor-
phism.
Exercise 6. Use Hopf bration S1
→ S3
→ S2
to compute π3(S2
).
M8130 Algebraic topology, tutorial 11, 2020 6. 1. 2020
Exercise 7. (application) We know that deg(f) is an invariant of [Sn
, Sn
] = πn(Sn
).
Study [S2n−1
, Sn
] ∼= π2n−1(Sn
) and describe its co called Hopf invariant H(f).
Solution. Have f : ∂D2n
= S2n−1
→ Sn
and Sn
∪f D2n
. For f ∼ g we have Sn
∪f D2n
Sn
∪gD2n
, moreover Sn
∪f D2n
= Cf (the cylinder of f). For n ≥ 2 we have Cf = e0
∪en
∪e2n
.
Using cohomology: H∗
(Cf ) = Z for ∗ ∈ {0, n, 2n} and 0 elsewhere. Take α ∈ Hn
(Cf )
generator, we have cup product. Then α ∪ α ∈ H2n
(Cf ) and for β ∈ H2n
(Cf ) we have
α ∪ α = H(f)β, where H(f) is the Hopf invariant.
Exercise 8. What can we say in this case about Hopf inviariant for n odd and for n even?
Thanks.
Solution. Knowing α ∪ β = (−1)|α||β|
β ∪ α we see that α ∪ α = 0. So for n odd Hopf
invariant is zero.
For n even consider the Hopf bration S1
→ S3
→ S2
= CP1
. For CP2
= D4
∪f CP1
(recall how CPn
is built up from CPn−1
) we have Cf = CP2
and H∗
(CP2
) = Z[α]/ α3
,
with α ∈ H2
. The generator of H4
is α2
. We get that H(f) = 1.
Exercise 9. Show that H(f) = 1 for the Hopf bration f : S3
→ S2
.
Exercise 10. Find a map f with Hopf invariant H(f) = 2.
Solution. We study a space X with a basepoint e. Denote construction J2(X) = X×X/ ∼,
where (x, e) ∼ (e, x). Apply this idea to Sn
. We get a projection p: Sn
×Sn
→ J2(Sn
). On
the left we have one 0-cell, two n-cells and one 2n-cell, while on the right we have one of
each. We get that J2(Sn
) has to be a space of the form Cf , so Hn
(J2) = Z given by a and
H2n
(J2) = Z given by b and Hn
(Sn
×Sn
) = Z⊕Z (generators a1, a2) and H2n
(Sn
×Sn
) = Z
(with b0). Now, p∗
: Hi
(J2) → Hi
(Sn
× Sn
) and p∗
(a) = a1 + a2, p∗
(b) = b0.
a2
= H(f)b
p∗
(a2
) = H(f)p∗
(b)
(a1 + a2)2
= H(f)b0
(a2
1 + a1a2 + a2a1 + a2
2) = H(f)a1a2
2a1a2 = H(f)a1a2
H(f) = 2
because b0 = a1a2 and by evenness of the dimension a1a2 = a2a1.