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