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Let $\gamma$ be a smooth, simple, closed curve and let $f : \gamma \to S^1$ assign to each $x \in \gamma$ the unit normal vector there. We can find a diffeomorphism $g: \gamma \to S^1$ and define the map $f': S^1 \to S^1$ assign to each point the unit normal. Is $f' \circ g$ (smoothly) homotopic to $f$?

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1 Answer 1

Not as you've stated it- the homotopy classes could differ up to sign.

For example, let $\gamma$ be the circle, $S^1$. Viewed as an element of the complex plane, the map you've just described is given by multiplication by $i$, which is homotopic to the identity. If we had accidentally chosen $g$ to be the map $S^1 \rightarrow S^1$ given by $z \mapsto 1/z$, then the resulting map, at the end of the day, would not be homotopic to the identity.

But that's the worst that can happen: you can always choose a diffeomorphism to make what you say true.

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