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Is there any example of a smooth map $f:S^1\to S^1$ that has degree zero that is not the constant map?

Either the map would have no regular values or every regular value would have an even number of pre-images with cancelling degrees, but I'm having a hard time seeing what the regular values should be.

Dennis Gulko
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user114326
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3 Answers3

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Here's a specific map which is actually surjective. Considering $S^1$ as $\{(x, y): x^2 + y^2 = 1\}$ in the Euclidean plane, take the projection $\pi_x: \mathbb{R}^2 \to \mathbb{R}, (x, y) \mapsto x$. Define $f: \mathbb{R} \to \mathbb{R}^2, t \mapsto (\cos \pi t, \sin \pi t)$. Finally define $g$ as the composition restricted to $S^1$: $f \circ (\pi_x | S^1): S^1 \to S^1$. $\pi_x$ and $f$ are both smooth, so $g$ is too. $\pi_x$ is nullhomotopic, so $g$ is nullhomotopic and therefore has degree $0$. And it's clear that (roughly speaking) $g$ goes through all angles in the range $[-\pi, \pi]$ and is therefore onto.

Edit: fixed scaling mistake, clarified domain and range.

Hew Wolff
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  • It generalises more nicely if you let $\pi_x$ denote the projection map from $S^1 \to D^1$ and let $f$ be the map the natural map from $D^1 \to S^1$ via the quotient $q : D^1 \to D^1/S^{0}$. Replacing all 1s by $n$s if one feels the need to generalise. –  Sep 27 '16 at 06:33
  • Don't you have to define $t \mapsto (\cos(2\pi t),\sin(2\pi t))$ in order for $g=f\circ \pi_x$ to wrap around $S^1$ twice because Image($\pi_x)=[-1,1]$ if we restrict to your parameterisation of the circle? – exchange May 26 '17 at 02:37
  • @exchange, yes, thank you, the map as given is not onto. Fixing now. – Hew Wolff May 29 '17 at 01:59
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Yes, identify $S^1$ with $[-\pi,\pi]$ and consider the map: $$f(t) = \cos(t):S^1\rightarrow S^1$$ It is obviously smooth and not surjective, and thus has degree zero.

Daniel Robert-Nicoud
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Degree is a homotopy invariant, which is to say that if $f$ and $g$ are homotopic maps from $S^1$ to $S^1$ then they have the same degree. Then any map $f:S^1\to S^1$ which is null-homotopic (homotopic to a constant map) will have degree zero, even if it itself is not constant. An example of such a map is given by flattening $S^1$ down onto the unit interval (say, by projecting it onto the $x$-axis, thinking of $S^1$ as embedded in $\mathbb{R}^2$) and then identifying this interval with a hemisphere of $S^1$. Contracting this hemisphere to a point gives the desired null-homotopy.

bradhd
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