## Cobordisms in problems of algebraic topology by Bukhshtaber By Bukhshtaber

Best geometry and topology books

Low-dimensional geometry: From Euclidean surfaces to hyperbolic knots

The research of third-dimensional areas brings jointly components from a number of components of arithmetic. the main outstanding are topology and geometry, yet parts of quantity idea and research additionally make appearances. long ago 30 years, there were awesome advancements within the arithmetic of three-d manifolds.

Additional info for Cobordisms in problems of algebraic topology

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).