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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!

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  • $\begingroup$ I assume you've already looked into / tried to use $\hat{\Bbb Z} \simeq \prod \Bbb Z_p$? $\endgroup$ Commented Jun 22, 2018 at 16:30
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    $\begingroup$ Dear @TorstenSchoeneberg : thank you for your comment. Yes, this is why I mentioned the non-existence of a surjective morphism $\Bbb Z_p \to \Bbb Z$. Defining morphisms from products (or projective limits) is a bit tricky. $\endgroup$
    – Watson
    Commented Jun 22, 2018 at 16:33
  • $\begingroup$ (On the other hand, there are many continuous morphisms $\hat{\Bbb Z} \to \Bbb C^{\times}$ — which must all factor through $S^1$. Indeed, we have $\Bbb Q / \Bbb Z = \varinjlim_n (1/n \Bbb Z)/\Bbb Z$ and so $$\mathrm{Hom}_{\text{cont}}(\Bbb Q / \Bbb Z, S^1) = \varprojlim_n \mathrm{Hom}_{\text{cont}}((1/n \Bbb Z)/\Bbb Z, S^1) \cong \varprojlim_n \Bbb Z / n \Bbb Z = \hat{\Bbb Z},$$ and Pontryagin duality finishes the proof.) $\endgroup$
    – Watson
    Commented Jun 22, 2018 at 16:50

2 Answers 2

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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.

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  • $\begingroup$ Dear @Minseon Shin, in view of your answer here, you might be possibly interested by the question there. $\endgroup$
    – Watson
    Commented Jul 13, 2018 at 12:42
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Little improvement: every nonzero homomorphism $\widehat{\mathbf{Z}}\to\mathbf{Q}$ is surjective.

More precisely, let us show that every torsion-free quotient of $\widehat{\mathbf{Z}}$ with cardinal $<2^{\aleph_0}$ is divisible.

First, this property that every torsion-free proper 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 of cardinal $<2^{\aleph_0}$. 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.

[Edit May 2023, I fixed the above statement from this June 2018 answer, since I first claimed it without the cardinal restriction, which is obviously incorrect: I need the quotient map to be non-injective on each $\mathbf{Z}_p$.]

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  • $\begingroup$ Dear @YCor, in view of your answer here, you might be possibly interested by the question there. $\endgroup$
    – Watson
    Commented Jul 13, 2018 at 12:41

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