The following proposition is from Algebraic Number Theory by Neukirch (Proposition 1.11, Chapter 3, p.191), but I doubt that exat sequence. Let $K$ be a number field, $O$ be the ring of integers. And $\mathfrak{p}\mid \infty$ means $\mathfrak{p}$ is a infinite place.
Let $\Gamma=\lambda(O^{\star})$ denote the complete lattice of units in trace-zero space, $H=\{(v_{\mathfrak{p}}) \in \prod_{\mathfrak{p}|\infty} \mathbb{R} \mid \sum_{\mathfrak{p}|\infty}{ v_{\mathfrak{p}}}=0 \}.$ There is an exact sequence $ 0\rightarrow H/ {\Gamma} \rightarrow CH^{1}(\bar{O})^{0} \rightarrow CH^{1}(O) \rightarrow 0\quad.$
Let $Div(\bar{O})$ be a replete divisor (Arakelov divisor), $CH^{1}(\bar{O})=Div(\bar{O})/P(\bar{O})$ be a replete divisor class group. Let $CH^{1}(\bar{O})^{0}$ be the kernel of $\deg: CH^{1}(\bar{O}) \rightarrow \mathbb{R}$, which is $\deg(\sum_{\mathfrak{p}}{v_{\mathfrak{p}} \mathfrak{p} })=\log(\prod_{\mathfrak{p}}\mathfrak{N}(\mathfrak{p})^{v_{\mathfrak{p}}}).$
My question comes from the exactness of $0\rightarrow H\stackrel{\alpha}{\longrightarrow} Div(\bar{O})^{0} \stackrel{\beta}{\longrightarrow} Div(O)\rightarrow 0,$ which is crucial in the proof. $\beta$ is the projection map, mapping all the finite place part of $Div(\bar{O})^{0}$ to $Div(O)$. It is surjective. $\alpha$ is defined to be $\quad \alpha((v_{\mathfrak{p}}))=\sum_{v_{\mathfrak{p}}\mid\infty}{\frac{v_{\mathfrak{p}}}{f_{\mathfrak{p}}} \mathfrak{p}}\quad$ (where $f_{\mathfrak{p}}=[K_{\mathfrak{p}}: \mathbb{R}]$, $K_{\mathfrak{p}}$ is the completion of $K$ with respect to place $\mathfrak{p}$, actually one can show: for $\mathfrak{p}\mid \infty$, $f_{\mathfrak{p}}=1$ if $\mathfrak{p}$ is a real embedding, and $f_{\mathfrak{p}}=2$ if $ \mathfrak{p}$ is a complex embedding.)
The point is that $Ker(\beta)$ is larger than $Im(\alpha)$. Say, one can pick $\mathfrak{p}_{1}$ as a finite place with $v_{\mathfrak{p}_{1}}=1$, and some infinite place $\mathfrak{p}_{2}$, such that $v_{\mathfrak{p}_{2}}=\log(p^{f_{\mathfrak{p}_{1}}})$. And all the other $v_{\mathfrak{p}}=0$. This is in the $Ker(\beta)$, but not in $Im(\alpha)$!