Isomorphism of algebraic closure of p-adics with their completion

Consider the following fields:

1) $\mathbb{C}$ the complex numbers

2) $\overline{\mathbb{Q}}_p$

3) $\mathbb{C}_p : = \hat{\overline{\mathbb{Q}}_p}$

They are all the same cardinality, algebraically closed, and of characteristic 0. Therefore they are all isomorphic as fields. However: $\overline{\mathbb{Q}}_p \to \mathbb{C}_p$, since it is the topological completion, and is not surjective. Is there something wrong with this argument or not?

Edit (To make my question more clear): Is $\overline{\mathbb{Q}}_p \to \mathbb{C}_p$ surjective? Or can there exist embeddings of fields into isomorphic fields which are not surjective?

• @EinsNull now I see your edit. No, that map is certainly not surjective. For a concrete example of something not in the image, choose a sequence $\zeta_1, \zeta_2, \ldots$ where $\zeta_i$ is a primitive $p^i$th root of unity. Then $1 + \zeta_1p + \zeta_2p^2 + \ldots$ is in $\mathbb{C}_p$ but not $\overline{\mathbb{Q}}_p$. – hunter Feb 24 '16 at 10:03
• @EinsNull maybe an easier example to grasp: $\mathbb{Q}(x_1, x_2, \ldots)$ embeds into itself by the "Hilbert hotel map" $x_i \mapsto x_{i+1}$ – hunter Feb 24 '16 at 14:37