Degree of a circle endomorphism

Circle endomorphisms are homotopic iff they are of equal degree

Let . Then ¹ iff . homotopy

Proof

Without loss of generality we may assume , since and has the same degree. First we will show that implies . Let , where we may assume without loss of generality that . Let be the uniquely defined morphism for each with and . Since is continuous and thus uniformly continuous by the Heine-Cantor theorem, we can divide by with finite so that for all

for all integers . As in the proof of this theorem, we define as follows

which is continuous by properties of the Main branch of the complex logarithm. Then is continuous, and thus a constant map since it is always an integer. Herefore

as required. For the converse, let . Then let be the uniquely defined morphisms with and for . We may extend this to by

which has the property that for all . Then for all . Note that is just the natural projection for the quotient topology, and thus by its universal property there exists a unique continuous such that . This unique defines a homotopy , since

for all and is a monomorphism, implying as required.

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Footnotes

1. Homotopy of maps ↩