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Suppose $f,g: X\to S^1$ are such that $f(x)\neq -g(x)$ for any $x\in X$. I need to construct a homotopy between these two functions. Now, the fact that $f(x)\neq -g(x)$ guarantees that there is always a unique shortest path between $f(x)$ and $g(x)$ for any $x$, suppose we call it $\omega_{f(x),g(x)}: I\to S^1$. In fact, we can think of $\omega$ as a function $$ \omega: S^1\times S^1\times I\to X $$ Then a homotopy will look like $$ h: X\times I\to S^1 $$ $$ h(x,t)=\omega(f(x), g(x), t) $$ so that indeed $h(x,0)=f(x), h(x,1)=g(x)$ for any $x\in X$. The homotopy is continuous because it is a combination of continuous functions in each variable. Have I constructed the right homotopy and is there maybe a way to find an explicit formula for $\omega$? I tried doing that but it turns out to be really complicated.

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You're correct broadly speaking, but you haven't said how you know that $\omega$ is a continuous function of it two arguments. I suggest constructing the homotopy $\omega$ explicitly as follows:

For each $x$, the straight line from $f(x)$ to $g(x)$ passes through the unit disk, but doesn't hit the origin (because $f(x) \ne -g(x)$). You could write $$ k(x, t) = (1-t) f(x) + t g(x). $$

Evidently $k$ is a continuous function of $x$ and $t$.

Now since $k(x, t)$ is not the origin for any $(x, t)$ pair, you can radially project it out to the unit circle; that gives you the homotopy you're looking for.

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  • $\begingroup$ $f(x),g(x)\in S^1$, and hence $(1-t)f(x)+tg(x)$ does not in general belong to $S^1$! $\endgroup$ Jan 19, 2014 at 18:34
  • $\begingroup$ I think you missed the part where I said "radially project $k(x, t)$ onto the circle (by which I meant "Let K(x, t) = \frac{k(x, t)}{\|k(x, t) \|}$"); that gives a point that is on the unit circle. And since radial projection is continuous away from the origin, we're done. $\endgroup$ Jan 19, 2014 at 19:05

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