Is there a surjective morphism $\widehat{\Bbb Z} \to \Bbb Z$? Consider the profinite completion $\widehat{\Bbb Z}$ of the additive group of $\Bbb Z$.

Is there a surjective group morphism $s : \widehat{\Bbb Z} \to \Bbb Z$ ?
  If so, can we moreover assume that $s \circ i = \rm{id}_{\Bbb Z}$, where $i : \Bbb Z \to \widehat{\Bbb Z}$ is the canonical embedding?

Clearly, there is no continuous map $\widehat{\Bbb Z} \to \Bbb Z$, otherwise $\Bbb Z$ would be compact and discrete... ; so $s$ can't be continuous.
I also noticed that there is no surjective group morphism $\Bbb Z_p \to \Bbb Z$ (i.e. $\Bbb Z$ is not a quotient of the $p$-adic integers), for any prime $p$, since the former is $q$-divisible for any prime $q \neq p$, while the latter is not. 
Thank you!
 A: In fact it seems there are no nontrivial group homomorphisms $\widehat{\mathbb{Z}} \to \mathbb{Z}$. 
First let's note that there are no nontrivial group homomorphisms $\mathbb{Z}_{p} \to \mathbb{Z}$, since, as you say, the image of any such map must be $q$-divisible for $q \ne p$.
Write $\widehat{\mathbb{Z}} = \prod \mathbb{Z}_{p} = \mathbb{Z}_{2} \times \prod_{p \ne 2} \mathbb{Z}_{p}$. There are no nontrivial maps $\mathbb{Z}_{2} \to \mathbb{Z}$ by the above, and there are no nontrivial maps $\prod_{p \ne 2} \mathbb{Z}_{p} \to \mathbb{Z}$, since $\prod_{p \ne 2} \mathbb{Z}_{p}$ is $2$-divisible.
A: Little improvement: every nonzero homomorphism $\widehat{\mathbf{Z}}\to\mathbf{Q}$ is surjective. More precisely, let us show that every torsion-free quotient of $\widehat{\mathbf{Z}}$ is divisible.
First, this property (that every torsion-free quotient is divisible) is satisfied by $\mathbf{Z}_p$ for every prime $p$, see the simple argument in my MathSE answer here. 
Let $f:\mathbf{\widehat{Z}}\to A$ be a surjective homomorphism, $A$ torsion-free abelian group. Then $f(\mathbf{Z}_p)$ is divisible by the above reference, for every $p$. So $C=f(\bigoplus_p\mathbf{Z}_p)$ is divisible. Since $\mathbf{\widehat{Z}}/\bigoplus_p\mathbf{Z}_p$ is divisible, its homomorphic image $A/C$ is also divisible. Hence $A$ is divisible.
