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I'm trying to prove that if $f:S^1 \rightarrow S^1$ is not surjective, then $f$ is homotopic to a constant function via a homotopy that fixes a point $\theta \in S^1$. Showing that it is homotopic to a constant function is simple, but showing that there exists a homotopy that fixes a point is proving to be a bit tricky... Is showing that $f$ must have a fixed point enough?

I can extend $f$ to a map on the disk $g:D^2 \rightarrow S^1$. If $i:S^1 \rightarrow D^2$ is the inclusion mapping, then $i \circ g: D^2 \rightarrow D^2$ is a map on the disk that must have a fixed point, so that $g(\theta) = \theta$ for some $\theta \in D^2$. This was my idea at a proof, but I fail to see how it has a connection, if any, to a homotopy...

Connor Malin
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Derek Adams
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    It suffices to show that you can contract an interval by a homotopy that fixes a point, which should be clear – Exit path Apr 29 '19 at 01:51
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    @leibnewtz which can be done since a point is a deformation retract of an interval? – Derek Adams Apr 29 '19 at 09:49
  • yes, do you see why this is enough? – Exit path Apr 29 '19 at 23:01
  • Any fixed point $\theta$ of $i \circ g$ is contained in $S^1$ and therefore a fixed point of $f$: For $x \in D^2 \setminus S^1$ you have $(i \circ g)(x) = g(x) \in S^1$, hence $x$ cannot be a fixed point of $i \circ g$. – Paul Frost Apr 30 '19 at 12:58
  • @leibnewtz It depends on how you interpret the question. If you understand it in the sense that the required homotopy is stationary on some $\theta \in S^1$, then your argument is correct. If you understand it in the sense that the homotopy keeps $\theta$ fixed, then you must first find a fixed point of $f$. – Paul Frost Apr 30 '19 at 13:25

3 Answers3

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More generally consider $f:S^n\to S^n$ and let $P\in S^n$ be such that $P\not\in f(S^n)$. For any $\theta\in S^n$ we have a homotopy

$$H:I\times S^n\to S^n$$ $$H(t, s)=\pi^{-1}\big(t\cdot \pi(f(s))+(1-t)\cdot \pi(f(\theta))\big)$$

where $\pi:S^n\backslash\{P\}\to\mathbb{R}^n$ is the stereographic projection which is a homeomorphism. Note that $H$ is well defined, continuous and we have

$$H(0, s)=f(\theta)$$ $$H(1, s)=f(s)$$ $$H(t,\theta)=f(\theta)$$

Finally since $\theta$ was arbitrary then all you need now is a fixed point $f(\theta)=\theta$. And the existance of such point follows from the Brouwer's fixed point theorem as you've mentioned yourself.

freakish
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As $f:S^1 \rightarrow S^1$ is not surjective then $f(S^1) \cong [0,1] $, where $x$. $f(S)$ must be a simply connected, so our loop $f:S^1\rightarrow S^1$ is contractible to a point, as required.

ben huni
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Suppose the circle function is not continuous, then $f(S^1)$ is a closed subset of $S^1$ and homeomorphic to $[0,1]$. Therefore, let us define our homotopy $H:S^1 \times I \rightarrow I$ by $H(x,t)=(1-t)i(f(x))$, where $i$ is the homoemorphism $f(S^1) \cong [0,1]$.

Hence, it follows clearly if we define our homotopy as $$G(x,t)=i^{-1} \circ H(x,t):S^1 \times I \rightarrow f(S^1) \subset S^1$$ , all conditions will be satisfied, as required.

ben huni
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