Lagmental Vicfred

The Degree of a Sphere Map Is Its Action on Top Homology by Vicfred

Last updated: Sat 01 March 2014

tags: mathematics, cohomology, cup products, Poincare duality

A continuous map from S^n to itself multiplies the fundamental class by an integer. This is a compact note, but the quantifiers and hypotheses stay on the page.

Notation

Cohomology applies \(\operatorname{Hom}(-,R)\) to chains and reverses arrows. The cup product makes \(H^\ast(X;R)\) a graded ring rather than only a graded group.

$$ f:S^n\to S^n $$

The formulas should not be merged too early. The datum \(\mathsf D\), the conclusion \(\mathsf C\), and the bridge \(\Longrightarrow\) have three different logical jobs.

$$ f_\ast([S^n])=\deg(f)[S^n] $$

Stress the formula

This is the algebraic core of the note. Once this line is correct, the surrounding interpretation has something solid to refer to.

$$ \deg(f\circ g)=\deg(f)\deg(g),\qquad\deg(\operatorname{id})=1,\qquad\deg(\text{constant})=0 $$

Interpretation

What survives the example is not its particular numbers but the relation encoded by the two rows below. That relation is the part worth transporting to a new setting.

$$ \begin{aligned} \mathsf{D}\;&:\quad f:S^n\to S^n,\\[5pt] \mathsf{C}\;&:\quad f_\ast([S^n])=\deg(f)[S^n]. \end{aligned} $$

Limit of the argument

Isomorphic cohomology groups do not guarantee isomorphic cohomology rings. Products can distinguish spaces that additive invariants cannot.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] f_\ast([S^n])=\deg(f)[S^n] \end{gathered}} $$

With the dependency made explicit, the same pattern can be recognised safely in nearby problems. A changed hypothesis should now be easy to spot.

This article was posted on Sun 28 July 2013. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.