October 9th
Today I learned that every character $\chi:\QQ\to S^1$ can indeed by expressed as $\Psi_a$ for some adele $a\in\AA_\QQ.$ This is very classic local-global principle: we're going to decompose $\chi$ into a product of character-like things $\chi_\infty$ and $\chi_p$ for all of our places, and then we'll finish by showing our character-like things are actually characters. This will let us identify $\chi$ with some adele, finishing.
For each $\psi_p,$ we have that $\psi_p(1a_p)=e^{2\pi i\{a_p\}_p}$ is a root of unity, so any non-root-of-unity contribution to $\chi$ must come from $\chi_\infty.$ In particular, let $\chi(1)=e^{-2\pi ia_\infty}$ so that $\chi_\infty(q)=e^{-2\pi ia_\infty q}$ satisfies $\chi(1)=\chi_\infty(1).$ Then for any $q=\frac mn,$ we have that\[\frac\chi{\chi_\infty}(q)^n=\frac\chi{\chi_\infty}(nq)=\frac\chi{\chi_\infty}(m)=\frac\chi{\chi_\infty}(1)^m=1,\]so indeed, $\frac\chi{\chi_\infty}$ has image only equal to roots of unity, dictated by the denominator. With that in mind, we define $\frac\chi{\chi_\infty}(q)=e^{2\pi i\theta}$ for $\theta\in\QQ$ so that $\chi_p(q)=e^{2\pi i\{\theta\}_p}.$ In particular, $-\theta+\sum_p\{\theta\}_p\in\ZZ$ implies that\[\frac\chi{\chi_\infty}(q)=e^{2\pi i\theta q}=\prod_pe^{2\pi i\{q\}_p}=\prod_p\chi_p(q).\]Namely,\[\chi(q)=\chi_\infty(q)\prod_p\chi_p(q).\]So we have decomposed $\chi$ into character-like $\chi_\bullet$s. This ends the global step.
We defined $\chi_\infty(q)=e^{2\pi ia_\infty q}=\psi_\infty(a_\infty q),$ so we'll finish by showing that $\chi_p(q)=\psi_p(a_pq)$ for some $a_p\in\ZZ_p.$ This will tell us that each character $\chi$ is associated with some $\Psi_a$ for $a\in[0,1)\times\prod_p\ZZ_p\subsetneq\AA_\QQ,$ completing the proof of the surjection of $\Psi_\bullet.$ We begin by studying $\chi_p(1)$ in order to construct $a_p.$ For any $\bullet\in\NN,$ recall that\[\chi_p\left(1/p^\bullet\right)^{p^\bullet}=\chi_p(1)=1,\]so it follows $\chi_p\left(1/p^\bullet\right)=e^{2\pi ic_\bullet/p^\bullet}$ for some $c_\bullet\in\ZZ/p^\bullet\ZZ.$ We'd like to have $\{c_\bullet/p^\bullet\}_p=\{1a_p/p^\bullet\}_p,$ so we had better have $a_p\equiv c_\bullet\pmod{p^\bullet}.$ Because we're constructing an element of $\ZZ_p,$ this modular information is enough to construct $a_p$ as a Cauchy sequence, but we need to show that this is really a Cauchy sequence. Well, $\chi_p\left(1/p^{\bullet+1}\right)^p=\chi_p\left(1/p^\bullet\right),$ so $c_{\bullet+1}\equiv c_\bullet\pmod{p^\bullet}.$ Thus,\[a_p=\{c_1,c_2,\ldots\}\in\ZZ_p\]is indeed a Cauchy sequence in $\ZZ_p,$ which we'll say converges to $a_p.$ Now we have to extend this $a_p$ to work for all $q=\frac mn.$ To control the denominator, fix $n=p^\bullet n'$ with $(n',p)=1.$ This gives\[\chi_p(q)^{n'}=\chi_p\left(m/np^\bullet\right)^{n'}=\chi_p\left(1/p^\bullet\right)^m=e^{2\pi ia_pm/p^\bullet}.\]We would like this exponent to have a $\{a_pq\}_p$ in it, and indeed\[\frac{a_pm}{p^\bullet}\equiv m\left\{a_p/p^\bullet\right\}_p=\left\{a_pm/p^\bullet\right\}_p=\{a_pqn'\}_p\equiv n'\{a_pq\}_p\pmod\ZZ\]Here we have used $\{x\}_p+\{y\}_p=\{x+y\}_p$ repeatedly. So we see that $\chi_p(q)^{n'}=\left(e^{2\pi i\{a_pq\}_p}\right)^{n'}.$ But $\chi_p(q)$ is a $p$th root of unity, so the quotient is a $p$th root of unity as well as an $n'$th root of unity, forcing the quotient to be $1$ because $(n',p)=1.$ It follows $\chi_p(q)=e^{2\pi i\{a_pq\}_p},$ which is what we wanted.