Simplicial Structures in Topology

III.3 Some Applications 121
has no fixed points and so Λ( f )=0; but f is homotopic to the identity 1 S 1 through
H(e 2πit ,s)=e 2πi(t+s 1 12 )
and therefore Λ(1 S 1)=0 (because the Lefschetz Number is invariant up to homotopy);
however, 1 S 1 has only fixed points.
We now consider an important corollary to the Lefschetz Fixed Point Theorem,
known as the Brouwer Fixed Point Theorem. Aself-map of a space X is a map of
X into itself.
(III.3.4) Corollary. Every nonconstant self-map of a connected acyclic polyhedron
has a fixed point.
Proof. Let |K| be such a polyhedron 3 ; by hypothesis, the homology of |K| is all
trivial, except for H0(K;Q) ∼ = Q. Let us suppose that f is simplicially approximated
by g: K (r) → K. Letx0 be a vertex of |K| and let us suppose that {x0} + B0(K,Q)
is a generator of H0(K,Q). On the other hand,
C0(g)ℵ0({x0})=g(x0)
and g(x0) is a vertex of K because g is simplicial. Since |K| is connected, g(x0)
is homological to {x0} and therefore Λ( f )=1. Lefschetz theorem allows us to
conclude that f must have a fixed point. �
Before stating the next corollary, let us look into the definition of the degree of
a self-map of a sphere. It is easy to see that the n-sphere S n (n ≥ 1) is the geometric
realization of a simplicial complex Σ n .Letf : S n → S n be a given map and
g: (Σ n ) (r) → Σ n be a simplicial approximation of f . Σ n has trivial homology in all
dimensions except for 0 and n, when it is isomorphic to the Abelian group Z. We
recall that Hn((Σ n ) (r) ) ∼ = Zn((Σ n ) (r) ) has only two possible generators (differing by
their orientation); let z be one of them. Then, there exists an integer d such that
Hn(g)(z)=dz. This number, which is obviously independent from the homotopy
class of the map f ,thecyclez, and the simplicial approximation g of f ,isthedegree
of f (notation: gr ( f )). 4
(III.3.5) Lemma. For every map f : S n → S n ,
Λ( f )=1 +(−1) n gr ( f ) .
Proof. The Lefschetz number is defined for the homology with rational coefficients.
Since Σ n is connected, Tr H0( f ,Q)=1; we only need to prove that Tr Hn( f ,Q)=
gr ( f ), to which end it is sufficient noting that
Zn((Σ n ) (r) ,Q) ∼ = Zn((Σ n ) (r) ) ⊗ Q . �
3 See examples of acyclic complexes in Sect. II.4.
4 Case n = 1 is particularly interesting: when we move along the cycle z, its image g(z) wraps the
corresponding generating cycle of Z1(S 1 ) d times around S 1 .