HOMEWORK 12
This should be a combination of some exposition and actual exercises:
Let Y be a path connected CW–complex. Then πn(Y ) = 0 iff any morphism f :
Sn
→ Y can be extended to F : Dn+1
→ Y .
Sn f
// _

Y
Dn+1
F
<<y
y
y
y
This is a rather easy observation. The direction ⇐ says that any map from a sphere
can be extended to a map F. Then F describes a nullhomotopy of f.
The direction ⇒ says that any map f : Sn
→ Y is nullhomotopic. This means that
there exists a nullhomotopy H : Sn
× I → Y , where H|Sn×{1} = y for some y ∈ Y .
We can imagine that we glue Dn+1
to the top of the cylinder Sn
× I and we have a
map H ∪ {y} : Sn
× I ∪ Dn+1
× {1}. Thus by lifting extension property H ∪ {y} can
be lifted to a map G : Dn+1
× I. We define F = G|Dn+1×{y}.
Exercise 1. Prove the following theorem: Let X, A be a pair of CW–complexes and let
Y be a path–connected space. Then f : A → Y can be extended to a map F : X → Y
if πn−1(Y ) = 0 for all n such that X \ A has cells in dimension n. (Hint: use the
preceding exercise and extend f to skeletons Xi
).
Exercise 2. By presenting a counterexample, disprove the following: Let X, A be a
pair of CW–complexes and let Y be a path–connected space. Then f : A → Y can
be extended to a map F : X → Y if πn(Y ) = 0 for all n such that X \ A has cells in
dimension n. (Hint: Assume (X, A) = (Dn+1
, Sn
))
Let us have the following diagram:
F
i

Sn
f
<<y
y
y
y f
// _

E
p

Dn+1
h
<<y
y
y
y g
// B
where p : E → B is a fibration of path–connected spaces with fibre F such that
πnF = 0. We will prove, that there exists a diagonal h that will make everything
commute. First, we notice that as pf : Sn
→ B can be extended to g : Dn+1
→ B
it must be nullhomotopic and the nullhomotopy is provided by g. Let b0 ∈ B be the
point in the center of the disc g(Dn+1
). Then we have a homotopy ˆg : Sn
× I → B,
which is pf on one end and b0 on the other end.
Date: May 13, 2013.
1
2 HOMEWORK 12
By the lifting property of fibration, we then get a lift ˆh in the diagram
Sn f
// _

E
p

Sn
× I
ˆh
;;w
w
w
w
w ˆg
// B
We can see that ˆh is the homotopy between f and some f , where pf = b0 and f is
therefore a mapping into the fibre F (or its image i(F)). But as πn(F) = 0, we can
see that f is nullhomotopic. Let h be the nullhomotopy of f . We will interpret this
as a map h : Dn+1
→ i(F). Now we can glue the disc h to top of the cylinder Sn
× I.
Thus we have a map h = ˆh ∪ ˜h : Dn+1
(∼ Dn+1
∪ Sn
× I) → E. This map is the lift
we were originally looking for.
Exercise 3. Prove the theorem: Let (X, A) be a CW–pair and p : E → B a fibration
of path connected spaces such that πn(F) = 0 whenever there is a cell of dimenstion
n + 1 in X \ A. Then there exists a lift in the following diagram:
A
f
// _

E
p

X
h
>>~
~
~
~ g
// B
Exercise 4. Let (X, A) be a CW–pair and p : E → B a fibration of path connected
spaces which is also a weak equivalence. Prove, that there exists a lift in the following
diagram:
A
f
// _

E
p

X
h
>>~
~
~
~ g
// B