How is this homomorphism from Algebraic Number Theory surjective? This is on page 72 of Cassels-Fröhlich's Algebraic Number Theory. 
Take $S$ to be the archimedian valuations of $k$, a finite extension of $\mathbb{Q}$, and define $J_S \subset V_k $ as the subgroup of idèles $\boldsymbol\alpha=\{\alpha_v\}$ with $|\alpha_v|_v=1$ for $v \not\in S$. For convenience write $s := |S|$.
Now define the map $$\lambda : J_S \rightarrow \mathbb{R}^+\;\; \oplus\;\; ...\; \oplus\;\; \mathbb{R}^+$$ where there are $s$ copies of $\mathbb{R}^+$ on the right, and  $$\boldsymbol\alpha \mapsto (\textrm{log}(|\alpha_1|_1,\textrm{log}(|\alpha_2|_2,...,\textrm{log}(|\alpha_s|_s)$$
(Where $\{|\cdot|_i\}$ are the valuations in $S$). This is clearly a continuous homomorphism, but more dangerously and far less clearly, Cassels goes on to state that this is also surjective - and this isn't just a typo of some sort, it's essential to the proof.
I don't see how it could possibly be surjective; as what seems like a simple counterexample, take $k = \mathbb{Q}$; then $s=1$, and since $|\alpha_v|_v=1$ for $v$ non-archimedean, our set $J_S$ is just a copy of $\mathbb{Q}$; but then we've now got a 'surjective' map from a countable set to an uncountable set! What am I missing here?

In this case $J_S$ is not just $\mathbb{Q}$. The ideles are the invertible adeles, and the adeles are $\mathbb{R}\times \prod ' \mathbb{Q}_p$. Requiring all but the first valuation to be $1$ still leaves you with $\mathbb{R}^\times$. Thus in this case, all the theorem is saying is that $\log |\cdot |: \mathbb{R}^\times\to \mathbb{R}^+$ is surjective.
