Let $F:B^2 \to B^2$ be a continuous map w/ $F(S^1)\subseteq S^1$, and let $f:S^1\to S^1$ be $f(z)=F(z)$. Prove $F$ is onto or $f$ is nullhomotopic.

Question

This is an exercise from "Fundamental Groups & Covering Spaces" by Elon Lages Lima.

Let $F:B^2 \to B^2$ be a continuous map such that $F(S^1)\subseteq S^1$. Let $f:S^1 \to S^1$ be a map given by $f(z)=F(z)$ for all $z\in S^1$. Prove that either $F$ is surjective, or $f$ is homotopic to a constant map.

My attempt: Suppose that $F$ is surjective, i.e., $F(S^1)=S^1$. Then by definition we have $f$ is surjective, so also $f(S^1)=S^1$. In other words, $f$ is homotopic to the identity map on $S^1$. Alternatively, we must have that $F(S^1)\neq S^1$. However, since $F$ is continuous then $F(S^1)$ must remain connected. This means that $F(S^1)$ is an arc on the circle (or equal to a single point), so the image $f(S^1)$ is homeomorphic to an interval (or a single point), which is nullhomotopic. Therefore $f$ is homotopic to a constant map. Since these are the only two options, we are done. $\Box$

Is this correct?

Answer 1

No, that is not correct:

1. $F$ being surjective doesn't mean $F(S^1)=S^1$, it means $F(B^2)=B^2$.
2. $f(S^1)=S^1$ doesn't mean $f$ is homotopic to the identity map. In fact, any map $S^1\to S^1$ is homotopic to a surjective map. Its the other way around: if $f$ is not surjective then it is nullhomotopic.

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So how to prove this?

Assume that $F$ is not surjective. We just need to show that $f$ is nullhomotopic.

Pick $\gamma\in B^2\backslash im(F)$.

We will now construct $G:B^2\to S^1$ such that $G(v)=F(v)=f(v)$ for $v\in S^1$. Indeed, for any $v,w\in B^2$, $v\neq w$ define $L(v,w)$ to be the projection of $w$ onto $S^1$ by a straight line $L$ starting at $v$ and going through $w$ and finally hitting $S^1$ at $L(v,w)$. More formally $L(v,w)$ arises from the unique solution to $\lVert (1-t)\cdot v+t\cdot w\lVert=1$ over $t> 0$. For such solution $t_0$ we have $L(v,w)=(1-t_0)\cdot v+t_0\cdot w$. This calculation of $t_0$ is important because it also shows that $L(v,w)$ is continuous. Note that if $w\in S^1$ then $L(v,w)=w$.

Then $G(v)$ is given by $L\big(\gamma, F(v)\big)$. It is well defined because for any $v$ we have $F(v)\neq \gamma$. It is also continuous by previous remark.

In other words we showed that $f:S^1\to S^1$ extends to $G:B^2\to S^1$. But then $B^2$ is contractible and so $G$ is homotopic to some constant map through some homotopy $H:B^2\times I\to S^1$. This homotopy gives us the nullhomotopy of $f$ by simply restricting it to $S^1\times I$.