I am trying to solve the following problem about group homomorphism. I am taking it as already proved that $\phi$ is a group homomorphism and that $\mathbb{C}^{\star}$ is an abelian group under multiplication.
Take $A = B = \mathbb{C}^{\star}$ and $\phi$ given by $z \mapsto z^2$. Show that $\phi$ is surjective. Show that there does not exist a group homomorphism $\psi: B \to A$ such that $\phi \circ \psi = \mathrm{id}_B$.
The hint is:
Hint: Suppose that $\psi$ existed. Consider the element $\psi(-1)$ and arrive at a contradiction.
I think I'm ok on the first part:
Given $z \in \mathbb{C}^{\star}$, write $z = r e^{i \theta + 2\pi k}$ for $r \in \mathbb{R}$, $\theta \in [0, 2\pi)$, and $k \in \{0,1\}$. For each $k$, let $z_k = r^{1/2} e^{i \left(\frac{\theta}{2} + \pi k \right)}$. For any such $k$, we have \begin{align*} \phi\left(z_k\right) = z_k^2 = \left(r^{1/2} e^{i \left(\frac{\theta}{2} + k \pi \right)} \right)^2 = \left(r^{1/2}\right)^2 e^{i \left(\frac{\theta}{2} + k\pi \right)2} = r e^{i \left(\theta + 2k \pi \right)} = z, \end{align*} so $\phi$ is surjective.
On the second part, I'm less certain, but here is my attempt.
Suppose such a $\psi$ existed, so $\phi \circ \psi = \mathrm{id}_B$. in particular, $(\phi \circ \psi)(-1) = \mathrm{id}_B (-1) = -1$. But then \begin{align*} -1 = (\phi \circ \psi(-1 ) = \phi(\psi(-1)) = [\psi(-1)]^2 = \psi\left((-1)^2\right) = \psi(1). \end{align*} But $\psi$ is a homomorphism, and must carry the identity to the identity, so we must have $\psi(1) = 1$. Having reached a contradiction, we conclude that no such $\psi$ exists.
How do these proofs look?