I am trying to understand the below problem.
Take $A = B = \mathbb{Z}$ and $\phi$ given by $n \longmapsto 2 \cdot n$. Show that there there does not exist a group homomorphism $\psi: B \to A$ such that $\psi \circ \phi = \mathrm{id}_A$. Hint: suppose that $\psi$ existed. Consider the element $\psi(1)$ and arrive at a contradiction.
I've written up the question verbatim, but there are two ambiguities that I cannot solve (and don't have the intuition to discern). First is the domain/codomain of $\phi$, though I believe they are $A$ and $B$, respectively. Second, $A$ and $B$ are clearly groups, I assume under multiplication (or, at the very least, $B$ is, since $\phi$ maps to a product of two integers.
Injectivity isn't hard to prove. If $\phi(a) = \phi(b)$, then $2a = 2b$, so $2(a - b) = 0$, so $a - b = 0$, hence $a = b$.
Proving that no such $\psi$ existed is harder. That the problem asks me to consider $\psi(1)$ suggests to me that $1$ is the identity in $A$, in which case $A$ is a multiplicative group of integers. Homomorphisms have to carry identities to identity, so assuming $B$ is also multiplicative, we have $\psi(1) = 1$. I don't know how this yields a contradiction, however.
Any hints?
Let me rephrase the problem: $\DeclareMathOperator{id}{id}$
Take $G = \mathbb{Z}$, the group of integers under the $+$ operation. Define the group homomorphism $\phi : G \to G$ by $\phi(n) = 2 \cdot n$.
Show that there is no homomorphism $\psi : G \to G$ such that $\psi \circ \phi = \id_G$. Hint: consider the value of $\psi(1)$ and derive a contradiction.
If you want an even bigger clue, consider $\psi(1) + \psi(1)$.