Let $f : S^1 \rightarrow S^1$ be a continuous map such that $\operatorname{deg}(f) \ne 0.$

[Note : I have the degree defined as in this question.]

I'm trying to prove that the hypotheses above implie that $f(S^1)=S^1.$

What I've tried is the following,

Suppose that $\exists \ x_0 \in S^1 \setminus f(S^1).$

Having in mind that $S^1 \setminus \{x_0\}$ is a subspace of $S^1$ that is contractible, we obtain a diagram:

$$S^1 \overset{f}{\longrightarrow} S^1 \setminus \{x_0\} \overset{r}{\longrightarrow} \{f(1)\},$$

where $r$ is a strong deformation retraction of $S^1 \setminus \{x_0\}$ onto $\{ f(1) \}.$

Hence, $r \circ f = c_{f(1)}$, the constant map from $S^1$ to $f(1).$ Writing $i : \{f(1)\} \rightarrow S^1$ for the inclusion map, we have that $i \circ r \simeq Id_{S^1 \setminus \{x_0\}}$, that is, $i \circ r$ is homotopic to the identity relatively to the point $f(1).$

Hence $f = Id_{S^1 \setminus \{x_0\}} \circ f \simeq (i \circ r) \circ f = i \circ c_{f(1)} = c_{f(1)}$ and we get that f is homotopic to the constant map relatively to the point $f(1)$ and this contradicts the fact that $\operatorname{deg}(f) \ne 0.$

I don't know if my proof is correct or if it can be improved.

Thanks for your time!


Let $\bar f:I\rightarrow \mathbb{R}$ a lift of $f$, you have $|\bar f(0)-\bar f(1)|<1$ otherwise, you have $\bar f(I)$ is an interval (since $f$ is continuous) of diameter superior or equal to $1$, this would implies that $f$ is surjective. You deduce that $|\bar f(1)-\bar f(0)|<1$ and is an integer so it is zero.

  • $\begingroup$ I considered an alternative solution to the problem in terms of lifts, and is exactly the way you did it! $\endgroup$ – DrinkingDonuts Apr 28 at 14:39

I would have done it essentially the same way. If $f$ is homotopic to a constant (which it is if it isn't surjective), $f_*(1)=0$, where $f_*:H_1(S^1)\to H_1(S^1)$ is the induced homomorphism.

This is one definition of degree. Remember $H_1(S^1)\cong\Bbb Z$.

In other words, the winding number is $0$.


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