The continuous function $f: S^1 \rightarrow S^1, f(z) = z^3,$ has no sections

Question

A section of a continuous function $f : X → Y$ is a continuous function $g : Y → X$ such that $f ◦ g$ is the identity on $Y$ . Let $S^1$ to be the unit circle in $\mathbb C$. Prove that the map $f : S^1 → S^1$ given by $f(z) = z^3$ has no section.

If such a $g: S^1 \rightarrow S^1$ exists, then $f\circ g = id$ gives us that $g(z)^3 = z$. I am really unsure on how to use this information to further proceed into solving this problem. The only idea I had so far was to utilize the induced homormophism between fundamental groups to obtain $(g_z)_* \circ (g_{g(z)})_* \circ (g_{g^2(z)})_* = id_{\pi_1(S^1,\,z)}$. However, the group homomorphisms $(g_z)_*, (g_{g(z)})_*$ and $(g_{g^2(z)})_*$ are not the same, so we dont get something like $h^3(u) = u$, where $h: \mathbb Z \rightarrow \mathbb Z$ is a homomorphism of groups, which cannot exist.

Any help will be very much appreciated.

Answer 1

If $f\circ g = \text{id}_{S^1}$, then $f_* \circ g_* = \text{id}_{\pi_1(S^1)}$. Now $f_*$, viewed as a homomorphism $\mathbb{Z} \to \mathbb{Z}$, is $n\mapsto 3n$; the map $g_* : \mathbb{Z} \to \mathbb{Z}$ is given by $n\mapsto dn$ for some integer $d$. So $f_*\circ g_*$ is the map $n\mapsto (3d)n$. Therefore $3d = 1$, a contradiction.

Answer 2

Here is an alternative argument, which only uses the fact that $f$ is not injective. Suppose there exists a section $g\colon S^1\to S^1$. I claim $g$ is not surjective. Indeed, since $f(1)=f(\omega)=f(\omega^2)$, where $\omega$ is a primitive third root of unity, at least one of $\{1,\omega,\omega^2\}$ is not in the image of $g$.

Thus, $g$ can be viewed as a map $S^1\to S^1\setminus\{pt\}\simeq(0,1)$, the unit interval. Since $(0,1)$ is contractible, $g$ must be nullhomotopic.

Now, $f\circ g=id_{S^1}$ must also be nullhomotopic, and so $S^1$ must be contractible. This is a contradiction.

Answer 3

Note that we can extend $f(z) = z^3: S^1 \to S^1$ to a continuous function $$ \tilde f: S^1 \times \mathbf{R}_{>0} \to S^1 \times \mathbf{R}_{>0} $$ by $$ (z,r) \mapsto (z^3,\, r^3), $$ if $f$ has a continuous section $g$ then $\tilde f$ has \begin{align} \tilde f: S^1 \times \mathbf{R}_{>0} &\to S^1 \times \mathbf{R}_{>0} \\ (r,z) &\mapsto(g(z),\, r^{1/3}) \end{align} for it, by conjugating the homeomorphism \begin{align} S^1 \times \mathbf R_{>0} \xrightarrow{\sim} \mathbf C^* \\ (r,z) \mapsto rz, \end{align} may suppose $\tilde f$ is just $\mathbf{C}^* \to \mathbf{C}^*,\, z \mapsto z^3$. Note also $\tilde g$ induces a continuous section of $\tilde f$. Since $\tilde f$ is a holomorphic function with $\partial f \neq 0$, using the complex chain rule $$ \bar\partial \bigl( f \circ g(z) \bigr) = \partial f\, \bar\partial g + \bar\partial f\, \bar\partial \bar g, $$ where $\partial := \frac{\partial}{\partial z}$, $\,\bar\partial:=\frac{\partial}{\partial \bar z}$, we see $\tilde g$ is holomorphic function on $\mathbf{C}^*$ with $(\tilde g)^3(z) = z$, it has a removable singularity at $0$ so we may suppose $g$ defined on $\mathbf{C}$ with $g(0)=0$.

Now everything left is just to show that there is no holomorphic function on $\mathbf{C}$ such that $g^3(z)=z$ using the fact that $\mathbf{C}$ is not simply-connected on which the complex logarithm cannot be well-defined. If there is such $g$, then by Cauchy integration formula $$ \int_{S^1} \frac{g^3(z)}{z^2} \, dz = 2\pi i \cdot 3g^2(0)\, g'(0) = 0 $$ but $$ \int_{S^1} \frac{g^3(z)}{z^2}\, g'(z)\, dz = \int_{S^1} \frac{1}{z}\, dz = 2\pi i. $$

Answer 4

This is a variation of a degree theory argument, but achieves the result via the Argument Principle.

Let $f_1:\mathbb C \longrightarrow \mathbb C$ be given by $f_1(z)=z^3$.
Then for any closed curve $\gamma: [0,1]\longrightarrow S^1$ and the inclusion map $i:S^1\longrightarrow \mathbb C$
$n\big(f_1\circ i \circ \gamma,0\big)= n\big(i\circ f\circ \gamma,0\big)$
i.e. the winding numbers are the same because $f_1\circ i \circ \gamma(t) = i\circ f\circ \gamma(t)$ for $t\in[0,1]$.

with $\gamma(t) = \exp(2\pi i t)$ set $\gamma^* := i \circ g \circ \gamma$
letting $D$ be the open disc with radius 2 and applying the Argument Principle for a disc (I suggest chp 8, section 5 of Beardon's Complex Analysis) we get

$1= n\big(i\circ \gamma,0\big)= n\big(i\circ f\circ g\circ\gamma,0\big) = n\big(f_1\circ \gamma^*,0\big)=v_{f_1}(0)\cdot n\big(\gamma^*, 0\big)=3 \cdot n\big(\gamma^*, 0\big)$
$\implies n\big(\gamma^*, 0\big) =\frac{1}{3}$
where $v_{f_1}$ is the valency of the zero but the winding number for a closed curve is integer valued, which is a contradition.