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Manifold degree is functorial and detected in top cohomology
Statement
Let and be continuous maps between nonempty connected closed integrally oriented -manifolds. Then Reversing exactly one of the source or target orientations changes the degree sign, while reversing both leaves it unchanged. These assertions are choice-free.
Assume AC for the following top-cohomology assertion. There is a unique with , and similarly for . These generate their top cohomology groups, and Thus the action on top cohomology detects exactly the degree, including degree zero. AC is inherited only from Poincaré duality's atlas selection and local UCT.
Facts & Assumptions
Degree of a map between oriented closed manifolds gives the unique integer coefficient of in , including its orientation sign conventions.
Singular chains and singular homology are covariantly functorial gives induced homomorphisms, identity and composition on homology.
Singular cohomology is contravariantly functorial gives pullback by cochain precomposition with .
Poincaré duality for oriented topological manifolds gives by fundamental cap under AC.
Zero-th singular homology is free on path components identifies with the free group on path components, sending a point to its component generator.
Cap product with cohomology written first says top-degree cap evaluates a cochain on a simplex and keeps its final vertex.
The Axiom of Choice is assumed for the use of [F4] only.
Proof
Given: as in the statement, and their integral fundamental classes. AC is not used until step 1.3.
By [F2], , so uniqueness in [F1] gives degree one. For the composite, linearity and functoriality give The coefficient is unique by [F1], proving the product formula in . If either degree is zero, this same equality gives zero for the composite.
Negating the source fundamental class negates its image by [F2], hence its coordinate by [F1]. If the target class is replaced by , the same image has coordinate relative to the new class. Applying both changes gives again. These statements include , where negation has no effect.
Now assume [F7]. A nonempty connected manifold is path-connected: the points reachable by a path from a fixed point form an open set, because each point has a path-connected coordinate-ball neighborhood and paths can be extended inside it. Every other reachability class is open for the same reason, so the first class has open complement. Connectedness forces it to be the whole manifold. Thus [F5] identifies with by total-coefficient augmentation. For an -cocycle and fundamental cycle , [F6] gives Hence [F4] followed by this augmentation is precisely the evaluation map , . It is an isomorphism, so exists uniquely and generates the entire group, not just a quotient modulo torsion. The same argument applies to .
Let be a cocycle representing and a cycle representing . By [F3], evaluation of on is . The chain represents by [F1] and [F2]. Cocycle evaluation ignores a boundary since , so this value is . Thus . Injectivity of in step 1.3 proves the asserted equality. In particular degree zero gives zero pullback on the generator, and if the pullback is zero its evaluation makes the degree zero.
The degree identities are steps 1.1 and 1.2, and cohomology detection is step 2.1. For , the manifolds are points, and [F1] gives degree ; the normalized zero-cochain has value on the source point, so pullback value equals , verifying the formula explicitly. Empty manifolds are excluded by [F1]'s scalar-degree hypothesis. Coefficients are , so zero-ring ambiguity does not arise. Degenerate simplices obey the evaluation computation in step 1.3. There is no AC in the identity, composition or orientation calculations; [F7] is used only through [F4] for the top-cohomology isomorphism.
Depends on
- Degree of a map between oriented closed manifolds
- Singular chains and singular homology are covariantly functorial
- Singular cohomology is contravariantly functorial
- Poincaré duality for oriented topological manifolds
- Zero-th singular homology is free on path components
- Cap product with cohomology written first
- The Axiom of Choice
Used by
- Degree is multiplicative under composition Proposition
Dependency tree · two levels
30 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- Hatcher, Algebraic Topology, §3.3, pp.258–259 (standard reference, not scraped)