I am trying to solve.
Take $A = B = \mathbb{Z}$ and $\phi$ given by $n \mapsto 2 \cdot n$. Show that $\phi$ is injective. Show that there does not exist a group homomorphism $\psi: B \to A$ such that $\psi \circ \phi = \mathrm{id}_A$.
The proof of injectivity seems straightforward, but I'm unsure if my approach is valid.
Suppose $\phi(n) = \phi(m)$ for $n,m \in \mathbb{Z}$. Then. $2n = 2m$, so $n = m$.
The problem here is: I'm working in the world of the integers, but I've multiplied both sides by $\frac{1}{2}$, which is not an integer. Is this allowed? Alternatively, I could rearrange and prove that $2(n-m) = 0$ and argue that because $\mathbb{Z}$ has no zero divisors, $n-m = 0$, so $n=m$. But I haven't proven that $\mathbb{Z}$ has no zero divisors and am not sure how to prove that from first principles.
As for the second condition:
Suppose for the sake of contradiction that there exists such a homomorphism $\psi$. Then $(\psi \circ \phi)(1) = 1$. Define $\psi(1) = g$. Then $$ 1 = (\psi \circ \phi)(1) = \psi(\phi(1)) = \psi(2) = \psi(1 + 1) = \psi(1) + \psi(1) = g + g = 2g. $$ But $2g = 1$, so $g = \frac{1}{2} \not \in \mathbb{Z}$, so we have a contradiction.
My worry again is that I've multiplied by a non-integer.