By Ian F. Putnam

The writer develops a homology concept for Smale areas, which come with the fundamentals units for an Axiom A diffeomorphism. it really is according to parts. the 1st is a much better model of Bowen's end result that each such method is clone of a shift of finite style below a finite-to-one issue map. the second one is Krieger's size crew invariant for shifts of finite sort. He proves a Lefschetz formulation which relates the variety of periodic issues of the approach for a given interval to track facts from the motion of the dynamics at the homology teams. The lifestyles of the sort of thought used to be proposed by means of Bowen within the Seventies

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**Extra resources for A homology theory for Smale spaces**

**Example text**

ZM ) is actually in ΣL,M . What we have described is a bijection between (ΣL,0 (π))M (ρL, ) and ΣL,M . It is an easy matter to see it is invariant for the actions and a homeomorphism. Its biggest complication would be in writing it explicitly, which we avoid. The same kind of analysis applies to (Σ0,M (π))L (ρ,M ). We have proved the following. 12. For any L, M ≥ 0, we have (ΣL,0 (π))M (ρL, ) = ΣL,M (π) = (Σ0,M (π))L (ρ,M ). This result has the following easy consequence (although it could also be proved directly earlier).

2. For K ≥ 1, as maps deﬁned on ZGK , we have i ◦ t∗ t ◦ i∗ = t∗ ◦ i, = i∗ ◦ t. For K > j, as maps deﬁned on ZGK , we have (γ s )j (γ u )j = ij ◦ tj∗ = tj ◦ ij∗ = tj∗ ◦ ij , = ij∗ ◦ tj . Proof. Let p be in GK . By deﬁnition i ◦ t∗ = t(q)=p i(q), while t∗ ◦ i(p) = t(q)=i(p) q. We claim that i : {q | t(q) = p} → {q | t(q ) = i(p)} is a bijection. Since we suppose K ≥ 1, if q is such that t(q) = p, then t(i(q)) = i(t(q)) = i(p). Moreover, the map sending q 1 · · · q k to q 1 · · · q K pK is the inverse of i and this establishes the claim.

Let (X, ϕ) and (Y, ψ) be Smale spaces and let π : (Y, ψ) → (X, ϕ) be either an s-resolving or a u-resolving map. 5. 12. Proof. From the symmetry of the statement, it suﬃces to consider the case that π is s-resolving. 12. 2, it is easy to see that π is continuous on Y s (y). The same argument covers the case of π on Y u (y). To see the map π on Y s (y) is proper, it suﬃces to consider a sequence yn in s Y (y) such that π(yn ) is convergent in the topology of X s (π(y)), say with limit x, and show that it has a convergent subsequence.