Suppose that $f:S^1 \to S^1$ is continuous and has closed degree $n$, how would you show that $f$ is homotopic to the map $z \mapsto z^n$?

I know that by definition of closed degree, we have $\deg(f \circ \exp) = n$.

And the closed degree of $g(z) : = z^n$ is also $n$, so $\deg (g \circ \exp) = n$ and therefore we have $g \circ \exp (t) = \exp(nt)$ but I can't really see how this helps?

And a theorem states that every loop is homotopic to a constant speed loop of the same degree therefore $g \circ \exp (t) = \exp(nt)$ is homotopic to $f \circ \exp$ but I don't see how this can imply that $g$ and $f$ are homotopic! I would appreciate any help!

  • $\begingroup$ This reduces to proving that a degree $0$ map is null homotopic. I think it is easier to approach this. $\endgroup$ – Julien Mar 25 '13 at 21:18
  • $\begingroup$ Seems given the homotopy you've proved existence of, it should not be hard to construct the homotopy over $S^1$ in terms of the first. $\endgroup$ – Brady Trainor Mar 25 '13 at 22:05

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