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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?

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    $\begingroup$ The equation $(x^2−13)(x^2−17)(x^2−221)=0$ has a solution in $\hat{\mathbb{Z}}$ but not in $\mathbb{Z}$. See Page 3 of "Number Theory" by Borevich and Shafarevich. $\endgroup$ – Sidney Raffer May 24 '11 at 13:16
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    $\begingroup$ $\hat{\mathbb Z}/\mathbb Z$ is a rational vector space (isomorphic to $\hat{\mathbb Z}\otimes\mathbb Q/\mathbb Z\otimes \mathbb Q$). $\endgroup$ – Tom Goodwillie May 24 '11 at 13:16
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    $\begingroup$ Perhaps someone should make the category clear in which we talk about homomorphisms (groups? rings?). Because the comment by SJR obviously makes use of the ring structure, but I think that groups are meant. $\endgroup$ – Martin Brandenburg May 24 '11 at 13:26
  • $\begingroup$ @Martin: Indeed, I was thinking of group homomorphisms. But rings are interesting too... $\endgroup$ – Martin Bright May 24 '11 at 13:33
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    $\begingroup$ I would have given the question +1, except for the "We all know from primary school" affectation. I'm sure it's supposed to be funny, but it's a ridiculous exaggeration: probably the majority of research mathematicians have never learned what $\widehat{\mathbb{Z}}$ is. I would be willing to bet that at least one person read that sentence, didn't understand it, and felt at least a little bit bad about themselves because of it. There's really no need for this sort of schoolboy humor. $\endgroup$ – Pete L. Clark May 24 '11 at 20:17
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The answer is that there are no such homomorphisms. See the following preprint of Nik Nikolov http://arxiv.org/abs/0901.0244.

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  • $\begingroup$ A perfect answer! Except that I was hoping to use $H^1(\mathbb{F}_p,\mathbb{Z})$ as an example to show why it's important to use continous cochains... $\endgroup$ – Martin Bright May 24 '11 at 12:38
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    $\begingroup$ I'm not sure an Annals paper, relying on the CFSG is quite necessary here. $\endgroup$ – Junkie May 24 '11 at 13:54
  • $\begingroup$ Junkie you are right as you can see from Neil Strickland's answer. But the paper is of course much more general. $\endgroup$ – Yiftach Barnea May 24 '11 at 15:50
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    $\begingroup$ Although this is an excellent answer, Yiftach, this preprint of Nikolay and Dan Segal supercedes the above one arxiv.org/abs/1102.3037 $\endgroup$ – Jonathan Kiehlmann May 24 '11 at 22:06
  • $\begingroup$ In addition by an old result of R. Alperin, every homomorphism from any compact group to $\mathbf{Z}$ is continuous. Hence $\mathrm{Hom}(K,\mathbf{Z})=\{0\}$ for every compact group $K$. $\endgroup$ – YCor Apr 22 '18 at 9:08
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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.

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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.

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    $\begingroup$ One formalization of this argument: $\hat{\mathbf{Z}}=A\oplus B$ where $A=\mathbf{Z}_2$ and $B=\prod_{p\ge 3}\mathbf{Z}_p$. Then $A$ is 3-divisible and $B$ is 2-divisible. So the images of both $A$ and $B$ in $\mathbf{Z}$ have to be zero. $\endgroup$ – YCor Apr 22 '18 at 9:29

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