# 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!

• I assume you've already looked into / tried to use $\hat{\Bbb Z} \simeq \prod \Bbb Z_p$? Commented Jun 22, 2018 at 16:30
• 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. Commented Jun 22, 2018 at 16:33
• (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.) Commented Jun 22, 2018 at 16:50

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.

• Dear @Minseon Shin, in view of your answer here, you might be possibly interested by the question there. Commented Jul 13, 2018 at 12:42

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 $$\mathbf{Z}_p$$.]

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