# Prove that if $f$ is not surjective, then $f$ is homotopic to a constant via a homotopy that fixes a point

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...

• It suffices to show that you can contract an interval by a homotopy that fixes a point, which should be clear – leibnewtz Apr 29 '19 at 1:51
• @leibnewtz which can be done since a point is a deformation retract of an interval? – Derek Adams Apr 29 '19 at 9:49
• yes, do you see why this is enough? – leibnewtz 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

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.

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.

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.