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If $f, g: S^1\to \mathbb C$ are two functions, what is a homotopy from $f=\frac{g}{\vert g\vert}$ to $g$?

I just want to check whether my homotopy $H(x,t): (1-t)f+tg$ where $x \in S^1, t \in [0,1]$ is correct.

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If you want to check whether your homotopy is right, I suggest you post it. –  msh210 Jun 11 '13 at 18:38
    
$(1-t)f+tg$ but this would work for any two functions, that's why I'm unsure –  topa Jun 11 '13 at 18:39
    
Yes, that works assuming $g$ is never zero. –  Vectk Jun 11 '13 at 18:40
    
This homotopy always works for any $f,g$ if the codomain is convex –  Stefan Hamcke Jun 11 '13 at 18:41
    
And $\mathbb{C}$ is convex, right? –  topa Jun 11 '13 at 18:43

1 Answer 1

Lemma: If $f, g : X \to Y$ continuous where $Y$ is convex, then $f \simeq g$.

Proof: Define $G : X \times I \to Y$ by $G(x, t) = (1 - x)f(x) + t g(x)$ (the straight-line homotopy).

The first thing we need to ask is: is this well-defined? It is since $Y$ is convex.

$G$ is continuous since sum/product of continuous functions is continuous.

Observe that

  • $G(x, 0) = f(x)$
  • $G(x, 1) = g(x)$.

Thus $G$ is a homotopy between $f$ and $g$, so $f \simeq g$.


Given $f, g : S^1 \to \Bbb C$, where $\displaystyle f = \frac{g}{\vert g \vert}$ where $g \neq 0$ anywhere, then $f$ is well-defined and continuous. The homotopy you defined is exactly the straight line homotopy used in the Lemma (and this works since $\mathbb C$ is convex), which yields the homotopy you were looking for.

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