Homomorphism from $\hat{\mathbb{Z}}$ to $\mathbb{Z}$ I expect this question has a very simple answer.
We all know from primary school that there are no non-trivial continuous homomorphisms from $\hat{\mathbb{Z}}$ to $\mathbb{Z}$.  What if we forget continuity: can anybody give an explicit example of a homomorphism?
Note that $\hat{\mathbb{Z}}$ is torsion-free, and not divisible (since it's isomorphic to $\prod_p \mathbb{Z}_p$ and $\mathbb{Z}_p$ is not divisible by $p$).  There is the canonical injection $\mathbb{Z} \to \hat{\mathbb{Z}}$; is there some abstract reason why it ought to have a left inverse, and if so can we write it down?
 A: Let $\phi:\hat{\mathbb{Z}}\to\mathbb{Z}$ be a nontrivial homomorphism.  As every nontrivial subgroup of $\mathbb{Z}$ is isomorphic to $\mathbb{Z}$, we may suppose that $\phi$ is surjective, with kernel $K$ say.  Now $\phi$ induces a surjective homomorphism $\phi_n:\hat{\mathbb{Z}}/n\hat{\mathbb{Z}}\to\mathbb{Z}/n\mathbb{Z}$, but it is standard that $\hat{\mathbb{Z}}/n\hat{\mathbb{Z}}$ has order $n$, so $\phi_n$ must be an isomorphism.  This implies that $K\leq n\hat{\mathbb{Z}}$ for all $n$, but $\bigcap_n n\hat{\mathbb{Z}}=0$, so $\phi$ is injective, which is clearly impossible.  
A: The answer is that there are no such homomorphisms. See the following preprint of Nik Nikolov
https://arxiv.org/abs/0901.0244.
A: Let $\phi$ be such a homomorphism, on additive groups $\hat Z\rightarrow Z$. Write $(\vec x,\vec y)\in\hat Z$ for the element that is $x$ on primes that are 1 mod 3, and $y$ on primes that are 2 mod 3. 
Then $\phi(\vec x,\vec 0)=0$ for all $x\in Z$, for $(\vec x,\vec 0)$ is $l$-divisible for any prime $l$ that is 2 mod 3. The symmetrical argument claims $\phi(\vec 0,\vec y)=0$ too.
Without loss of generality, we can assume that a preimage of $1$ is given by $(\vec 1,\vec 1)$.
Next, applying the group law and setting $\alpha=\phi(\vec 1,\vec{-1})$, we derive the system $$1+\alpha=\phi(\vec 1,\vec 1)+\phi(\vec 1,\vec{-1})=\phi(\vec 2,\vec 0)=0$$
$$1-\alpha=\phi(\vec 1,\vec 1)-\phi(\vec 1,\vec{-1})=\phi(\vec 0,\vec 2)=0$$
This is impossible, so $\phi$ does not exist.
A: Here a slightly different proof. Decompose $\hat{\mathbb{Z}}$ into product of the pro-l Sylows. Now a vector in this decomposition having a zero coordinate at the prime p must be divisible by any power of p.
Any other element is the sum of two such elements, hence you are done (or alternatively fix only one prime, apply the above, now you have a map from the p-adics to the integers and it must be 0, as the p-adics are l-divisible for any l coprime to p).
