
By Bukhshtaber
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Example text
By definition, in the given basis φ and φ ⊗Z I|K have the same matrix with integral coefficients. 20) di ⊗Z I|K Ci (f)⊗Z I|K di ⊗Z I|K Ci−1 (X; Z) ⊗Z K / Ci (X; Z) ⊗Z K Ci−1 (f)⊗Z I|K / Ci−1 (X; Z) ⊗Z K , which follows from the corresponding diagram for Ci(f). Denoting di ⊗Z I|K by di , shortly d, and Ci(f)⊗Z I|K by Ci (f), for a given map f: X → Y we have a complex with gradation C∗ (X; K ) and preserving gradation linear map C∗ (f) := Ci (f), which commutes with d. Consequently, Zi (X; K ) := d−1 (0), and respectively i Bi (X; K ) := Im di+1 , called the cycles and boundaries correspondingly, are linear subspaces of Ci(X; K ), and Bi (X; K ) ⊂ Zi (X; K ).
F(x) − f(x0 ) > x − x0 for 0 < x − x0 ≤ ε for an ε > 0. Then x0 is an isolated fixed point and ind (f; x0 ) = deg (x0 − f(x)). 29) Lemma (Hopf Lemma for the fixed point index). Let B be an open ball in the Euclidean space E and let f: cl B → E be a map with no fixed point on the boundary. Then ind (f) = 0 implies a homotopy ft : cl B → E from f0 = f to a fixed point free map. Moreover, the homotopy ft may be constant on the boundary. Proof. Let us notice that F : cl B → E given by F (x) = x−f(x) is a d-compact map and deg (F ) = ind (f) = 0.
Proof. We notice that the composition i i ∗ ∗ Hn (E, E \ 0) ←− Hn (U , U \ 0) −→ Hn (E, E \ 0) is the identity map (by excision) if 0 ∈ U and is zero if 0 ∈ /U since then Hn (U , U \ 0) = 0 . 20) Lemma (Additivity). If U 1 , U 2 ⊂ U are open subsets such that the restrictions f|U1 , f|U2 are compactly fixed and U 1 ∩ U 2 is disjoint from f −1 (0), then deg (f) = deg (f|U1 ) + deg (f|U2 ). Proof. Since (f|U1 )−1 (0), (f|U2 )−1 (0) are compact and disjoint, there exist open disjoint subsets U i satisfying (f|Ui )−1 (0) ⊂ U i ⊂ U i (i = 1, 2).