Simplicial Structures in Topology

226 VI Homotopy Groups
(VI.3.16) Remark. Let S 0 be the 0-dimensional sphere, that is to say, the pair
{−1,1} of points of R with the discrete topology and the base point (1). Forany
based space (Y,y0), letπ0(Y ) be the set [S 0 ,Y ]∗; it is easily seen that, if Y is pathconnected,
then π0(Y) =0: in fact, two based maps f ,g: S 0 → Y are always homotopic,
the homotopy being given by a path that joins f (1) and g(1); thus, there
exists only one homotopy class in [S 0 ,Y ]∗, that of the constant map.
In particular, since for each n ≥ 1, S n is path-connected, π0(S n )=0.
We know that π1(S1 ,e0) ∼ = Z; what can we tell about the groups πn(Sn ,e0)? We
have the following result:
(VI.3.17) Theorem. For every n ≥ 1, πn(Sn ,e0) ∼ = Z.
Before proving the theorem, we recall that the degree of a map f : S n → S n (see
p. 121) is homotopy invariant and so, it induces a function
d : πn(S n ) −→ Z.
(VI.3.18) Lemma. For every n ≥ 1, the degree function
is a group isomorphism.
d : πn(S n ,e0) −→ Z
Proof. Let [ f ],[g] ∈ πn(Sn ,e0) be given arbitrarily; we wish to determine the degree
of the function
σ( f ∨ g)νn : S n −→ S n .
First of all we note that, by Theorem (III.4.3),
Hn(S n ∨ S n ;Z) ∼ = Hn(S n ;Z) ⊕ Hn(S n ;Z);
we now identify the components {e0}×S n and S n ×{e0} of S n ∨ S n with S n (in
other words, we interpret S n ∨ S n as a “union” of two spheres S n ) and consider the
“projections”
p1 : S n ∨ S n → S n and p2 : S n ∨ S n → S n ;
it is easy to verify that p1νn ∼ 1Sn and p2νn ∼ 1Sn; hence,
Hn(νn): Hn(S n ,Z) −→ Hn(S n ;Z) ⊕ Hn(S n ;Z)
is such that Hn(νn)({z})={z}⊕{z}. Moreover,
Hn(σ): Hn(S n ;Z) ⊕ Hn(S n ;Z) −→ Hn(S n ;Z) , x ⊕ y ↦→ x + y
for every x ⊕ y ∈ Hn(S n ;Z) ⊕ Hn(S n ;Z). Therefore,
Hn(σ( f ∨ g)νn)({z})=Hn(σ)(Hn( f )({z}) ⊕ Hn(g)({z}))
= Hn( f )({z})+Hn(g)({z})=d( f )+d(g).