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I understand what is meant by the degree of a continuous map $f$ from $S^1$ to $S^1$. If we let $[S^1, S^1]$ denote the set of homotopy classes of continuous maps from $S^1$ to $S^1$, it turns out that the degree map gives a bijection from $[S^1, S^1]$ to the integers. I am also cool with this.

My problem is I heard that this bijection fact is equivalent to the following: The degree map from $C(S^1,S^1)$ to the integers is a continuous map, and whenever $deg(f_0) = deg(f_1)$, there exists a path in $C(S^1,S^1)$ from $f_0$ to $f_1$.

How are the two notions equivalent? I don't have a very good grasp (or intuition) for continuous maps from $C(S^1,S^1)$ to the integers, and how that is related to modding functions out by homotopy.

(You may assume that I have background knowledge equivalent to Munkres chapter 9)

suncup224
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  • A homotopy between two maps $f,,g \colon X \to Y$ is a path in $C(X,Y)$ connecting $f$ and $g$ (with the right topology on $C(X,Y)$). Whenever $f,,g\colon S^1\to S^1$ are close, they have the same degree, and there exists a homotopy between $f$ and $g$ [the former follows from the latter]. So: $C(S^1,S^1)$ is locally path-connected, and the degree map gives a bijection between the path-components and the integers. – Daniel Fischer Nov 03 '13 at 15:03
  • Am I right to say that whenever $f,g$ are close in $C(X,Y)$, then they are homotopic? How would you show that? – suncup224 Nov 03 '13 at 15:18

2 Answers2

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If you lift the map $f\colon S^1\to S^1$ to a map $\tilde f\colon \Bbb R\to\Bbb R$, then the degree is given by $\dfrac{\tilde f(2\pi)-\tilde f(0)}{2\pi}$, and you can easily construct a path joining two maps of the same degree by taking the straight-line homotopy between their respective lifts.

Ted Shifrin
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Hint:Let $z\in \Bbb Z$ and $deg:C(S^1,S^1)->\Bbb Z$ Then $deg^{-1}$({$z$})={the set of the maps in $C(S^1,S^1)$ that wrap around the $S^1$ $z$ times}. Show that $deg^{-1}$({$z$}) is connected and open => is path connected. Thus if $deg(f_0)=deg(f_1)$ then there is a path from $f_0$ to $f_1$.

Haha
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