M8130 Algebraic topology, tutorial 12, 2017 11. 5. 2017
Exercise 63. Using homotopy groups show that RPk
is not a retract of RPn
, n > k ≥ 1.
Solution. Retract: The composition RPk i
→ RPn r
→ RPk
is the identity, so we have the
following diagram and want to use it to obtain a contradiction.
π∗(RPk
)
π∗(i)
//
id &&
π∗(RPn
)
π∗(r)

π∗(RPk
)
From the long exact sequence of the bration, we know that 0 = π1(Sk
) → π1(RPk
)
∼=
→
π0(S0
) → 0 and πi(Sk
) = πi(RPk
) for k ≥ 2. Then considering diagram
πk(Sk
) = Z //
''
πk(Sn
) = 0

πk(Sk
) = Z
we see that we are factoring identity through zero group and that is Mission Impossible:
Factor Zero (in theaters never). Continue with k = 1. We use the knowledge of RP1 ∼= S1
.
Then our diagram is a triangle with Z, Z/2, Z with identity Z → Z, factoring identity
through nite group is Mission Impossible: Group protocol (in theaters maybe one day,
one can only hope) ...and we are done.
Exercise 64. Consider the map q: S1
× S1
× S1
→ S3
dened as a map S1
× S1
× S1
→
D3
/S2
where D3
is a small disk in the triple torus which is the identity in the interior of
D3
and constant on its complement. Further, consider the Hopf map p: S3
→ S2
= CP1
(described in Hopf bration S1
→ S3
→ S2
). Compute q∗ and (pq)∗ in homotopy groups.
Show that pq is not homotopic to a constant map.
Solution. Take following diagram and treat it like your own:
S1

(ber)
S1
× S1
× S1 q
// S3
p

⊆ C2
S2
= CP1
where p(z1, z2) =
z1
z2
. We know that only nontrivial homotopy group of the triple torus
is π1(S1
× S1
× S1
), but π1S3
is is trivial, so q∗ is zero in homotopy groups and the
composition as well. By the following diagram we have H, thanks to homotopy lifting
property, of course. (denote (S1
)3
the triple torus S1
× S1
× S1
)
M8130 Algebraic topology, tutorial 12, 2017 11. 5. 2017
S1
= p−1
(s0)

(S1
)3 q
//
77

S3

(S1
)3
× I
H
77
h
// (S2
, s0)
where h(0, −) is constant map, h(1, −) = pq, p(H(0)) = s0,Im H(0) ⊆ p−1
(s0) = S1
and
H(1) = q ∼ H(0). However, H(0)∗ and q∗ dier in the third homology groups:
H3(S1
) = 0

H3((S1
)3
) q∗=id
//
H(0)∗
77
H3(S3
) = Z
We are trying to factor through zero, a contradiction.
Exercise 65. (detail for the lecture 9. 5. 2017) 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 66. 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 path connected, then [X, Y ] ∼= [(X, x0), (Y, y0)].
Solution. Surely, [(X, x0), (Y, y0)] ⊆ [X, Y ] and denote the class in the left set as (g) and
[g] the class in the [X, Y ]. The map (g) → [g] is well dened and injective. To prove
that it is also surjective take [f] ∈ [X, Y ]. Using HEP for f : X × 0 → Y and a curve
ω : x0 × I → Y which connects f(x0) with y0 we get g : X × 1 → Y such that f ∼ g and
g(x0) = y0, then [f] = [g] and (g) ∈ [(X, x0), (Y, y0)]. We are done.
Exercise 67. (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.
M8130 Algebraic topology, tutorial 12, 2017 11. 5. 2017
Exercise 68. (continuation of previous exercise) For n odd, what can we say in this case
about Hopf inviariant? 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.