M8130 Algebraic topology, tutorial 11, 2019 6. 5. 2019
Note: This tutorial originates in 20171
.
Exercise 57. Recall denitions of n-connectedness and n-equivalence. Prove the following
lemma: An inclusion A → X is n-equivalence if and only if (X, A) is n-connected.
Solution. "⇐" Take long exact sequence:
→ πn(A, x0)
fn
→ πn(X, x0) → πn(X, A, x0) → πn−1(A, x0)
fn−1
→ πn−1(X) → πn−1(X, A) →
and use the assumption that πi(X, A, x0) = 0 for i ≤ n. Then we get that fn is epimorphism
and fn−1 is isomorphism.
"⇒" Reasoning is the same as in the other direction, the only thing we need to realize
is π0(A, x0)
∼=
−→ π0(X, x0).
Exercise 58. Show πk(S∞
) = 0 for all k, where S∞
is colim Sn
.
Solution. We have S1
⊂ S2
⊂ S3
⊂ · · · ⊂ Sn
⊂ · · · ⊂ S∞
. Take element in πk(S∞
), that
is f : Sk
→ S∞
. We know that f(Sk
) is compact in S∞
. Consider other CW-complex
structure than e0
∪ ek
for Sk
that is Sk
= k
i=0 ei
1 ∪ ei
2 (two hemispheres). Then the
following holds: S∞
= ∞
i=0 ei
1 ∪ ei
2. So f(Sk
) ⊆ (S∞
)(N)
= SN
for some N where (S∞
)(N)
is N-skeleton of S∞
. Now,
f : Sk
→ SN
→ SN+1
→ S∞
,
so the composition Sk
→ SN+1
is a map that is not onto and therefore f is homotopic
to constant map. (map into a disc is homotopic to constant map, disc is contractible)
Then we have [f] = 0. Thus we have proved πk(S∞
) = 0. One can conclude that S∞
is
contractible by the Whitehead theorem as S∞
→ ∗ is a weak equivalence.
Remark. Another approach is to dene the map T : S∞
→ S∞
by the assignment
T(x1, x2, x3, . . . ) = (0, x1, x2, x3, . . . )
and show that T ∼ id via the ane homotopy
H(x, t) =
x(1 − t) + tT(x)
x(1 − t) + tT(x)
(x(1 − t) + tT(x) = 0  realize that location of leading non-zero element for id and T is
dierent). Similarly, there is an ane homotopy T ∼ C where C(x) = (1, 0, 0, . . . ) is a
constant map. Thus we obtain id = πk(id) = πk(T) = πk(C) = 0: πk(S∞
) → πk(S∞
) what
implies πk(S∞
) = 0.
1see https://is.muni.cz/el/sci/jaro2019/M8130/um/68045051/
M8130 Algebraic topology, tutorial 11, 2019 6. 5. 2019
Remark. In general, the functor πn : Top∗ → Ab for n > 1 doesn't preserve colimits.
Consider diagram Dn
← Sn−1
→ Dn
where both maps are inclusions of a boundary. A
pushout of the diagram is Sn
as we identied two hemispheres on their boundaries (equator).
However pushout of the induced diagram 0 = πn(Dn
) ← πn(Sn−1
) → πn(Dn
) = 0 is the
trivial group which doesn't agree with πn(Sn
) = Z.
Remark. A solution of the previous exercise can be generalized to the following statement.
Let X be a topological space such that X = i≥0 Xi satisfying X0 ⊆ X1 ⊆ . . . where Xi
are Hausdor. Then the functor πn commutes with colimit
lim
−→
πn(Xi, x0) ∼= πn(X, x0).
Sketch of proof: Every continuous map from a compact space to X factors through some
Xi. As the target space Xi is Hausdor such map is continuous as well as original one.
Exercise 59. Compute homotopy groups of RP∞
.
Solution. Surprisingly use previous exercise: We can view RP∞
as lines going through
origin in S∞
, or...just take S∞
/Z/2, where the action is x → −x. So we work with the
following bration (we don't write the distinguished points as they are not needed) Z/2 →
S∞
→ RP∞
and the long exact sequence
πn(Z/2) → πn(S∞
) → πn(RP∞
)
∂
→ πn−1(Z/2) → πn−1(S∞
),
where for all n ≥ 2 we have all zeroes, for n = 1 consider 0 → π1(RP∞
)
∂
→ π0(Z/2) →
π0(S∞
). Since π0(S∞
) = 0 and π0(Z/2) = Z/2, we get that the homomorphism ∂ (it
is homomorphism, really, we did it in previous tutorial, but it's still a homomorphism
independently on whether we did it or not) is an isomorphism of groups. By connectivity
we also know the π0 group. So the nal results are:
πn(RP∞
) = 0 for n ≥ 2, π1(RP∞
) = Z/2, π0(RP∞
) = 0.
Exercise 60. Show that the spaces S2
× RP∞
and RP2
have the same homotopy groups
but they are not homotopy equivalent.
Solution. Here, also use previous exercise. It is known that πn(X × Y ) = πn(X) × πn(Y ).
With this we can compute
π0(S2
× RP∞
) = 0, π1(S2
× RP∞
) = π1(S2
) × π1(RP∞
) = 0 × Z/2 ∼= Z/2,
for n ≥ 2 πn(S2
× RP∞
) = πn(S2
) × 0 ∼= πn(S2
).
M8130 Algebraic topology, tutorial 11, 2019 6. 5. 2019
Now consider RP2
as S2
/Z/2, work with the bration (scheme as follows)
Z/2 // S2 // RP2
πn(Z/2) // πn(S2
) // πn(RP2
) // πn−1(Z/2)
n ≥ 2 0 πn(S2
) ∼= πn(RP2
) 0
n = 1 0 π1(RP2
) ∼= Z/2 0
and π0(RP2
) = 0. Thus we showed that these two spaces have the same homotopy groups.
Marvellous. How to show, that they are not homotopy equivalent? Use cohomology group!
That's right. It is well known (or we should already know) that
H∗
(RP2
; Z/2) = Z/2[α]/ α3
, α, γ ∈ H1
and
H∗
(S2
× RP∞
; Z/2) = H∗
(S2
; Z/2) ⊗ H∗
(RP∞
; Z/2) = Z/2[β]/ β2
⊗ Z/2[γ], β ∈ H2
.
The former space obviously has no non-zero elements of order 4, while the latter has a
non-zero element of order 4. This is impossible for homotopy equivalent spaces. We are
done.
Exercise 61. Extension lemma: Let (X, A) be a pair of CW-complexes, Y a path-connected
space with πn−1(Y ) = 0 whenever there is a cell of dimension n in X −A. Then every map
f : A → Y can be extended to a map F : X → Y .
Solution. Set X−1 = A, X0 = X(0)
∪A, Xk = X(k)
∪A, and f = f−1 : X−1 → Y, f0 : X0 → Y
which extends f−1 to 0-cells in an arbitrary way.
We have fk−1 : Xk−1 → Y together with πk−1(Y ) = 0 for some k ≥ 1 and we want to
extend fk−1 to g: Dk
∪ Xk−1 → Y for any k-cell in X − A such that the following diagram
commutes:
Sk−1
ϕ|Sk−1
//

Xk−1

fk−1
// Y
Dk ϕ
//
g
##
Xk fk
//
fk

Y
Y
The class [g|Sk−1 ] = [fk−1 ◦ ϕ|Sk−1 ] is in πk−1(Y ) = 0 so it is homotopic to a constant
map via h: Sk−1
× I → Y such that h(−, 1) = const. Now, use HEP to complete the
diagram:
Dk
× {1} ∪ Sk−1
× I
const ∪h//

Y
Dk
× I
H
77
M8130 Algebraic topology, tutorial 11, 2019 6. 5. 2019
Put g = H(−, 0) and do the same procedure for the remaining k-cells in X − A. We
extended fk−1 to fk : Xk → Y and we proceed to innity (and beyond), as we always
do.
Exercise 62. Compare πn(X, x0) and πn(X, x1), when distinguished points (are / are not)
path connected. Use proof with pillows.
Solution. First, the case where X = S1
S2
and x0 ∈ S1
and x1 ∈ S2
. Then π1(X, x0) =
π1(S1
) = Z but π1(X, x1) = π1(S2
) = 0. Thus if distinguished points are not path connected,
homotopy groups can be dierent.
Now consider a curve ω: I → X connecting x0 and x1 i.e. ω(0) = x0 and ω(1) = x1.
Dene a map πn(X, x0) → πn(X, x1) by the assignment [f] → [ω · f] where g = ω · f is
determined by Figure 1.
We verify that the map is well-dened and an isomorphism.
• A homotopy invariance. If f1 ∼ f2 then ω ·f1 ∼ ω ·f2. Similarly, Figure 2 shows that
ω1 ∼ ω2 ⇒ ω1 · f ∼ ω2 · f for curves ω1, ω2 connecting points x0 and x1. You can
imagine this situation by adding the third dimension to Figure 1 which represents a
time component of a homotopy.
• A group homomorphism i.e. ω · (f + g) ∼ (ω · f) + (ω · g). This is proved on the
M8130 Algebraic topology, tutorial 11, 2019 6. 5. 2019
following picture.
• A bijection i.e. ω−1
· (ω · f) ∼ (ω−1
∗ ω) · f ∼ f ∼ ω · (ω−1
· f) ∼ (ω ∗ ω−1
) · f. This
follows from transitivity of the action.
We get that if x0, x1 are in the same path component then πn(X, x0) → πn(X, x1) is an
isomorphism. In particular, if X is simply connected, then every curve induces the same
isomorphism.