**Background and definitions.**

Let $k$ denote a field complete with respect to a non-trivial non-archimedean norm. Let $R$ be the integers in $k$, and say $\pi\in R$ with $0<|\pi|<1$ ($\pi$ doesn't have to be a uniformiser and the maximal ideal of $R$ doesn't even have to be principal, but I don't know the answer to my question even if $k={\mathbb{Q}_p}$ and $\pi=p$). Let $V$ be a $k$-vector space, which will be infinite-dimensional in practice. Let me say that a sub-$R$-module $L$ of $V$ is a *lattice* if it has the following two properties:

(i) $kL=V$ (i.e. $L$ contains a basis for $V$), and

(ii) $L$ contains no line, i.e. if $v\in V$ and $kv\subseteq L$ then $v=0$.

[I'm aware that other people might use the world "lattice" to mean something else.]

Given $V$ and a lattice $L$ in $V$, one can put a topology on $V$; a basis of open sets is $v+\pi^nL$ for $v\in V$ and $n\in\mathbf{Z}$. We can even complete $V$ with respect to $L$; the completion is the projective limit of $V/\pi^n L$. The completion $\widehat{V}$ is also a topological vector space and there's a natural continuous and injective map $V\to\widehat{V}$ with dense image (injectivity follows from (ii) ).

**The question.**

Notation as above. Say $V$ is a $k$-vector space and we choose two lattices $L_1$ and $L_2$ in $V$, with $L_1\subseteq L_2$. Let $V_1$ denote $V$ with the topology induced from $L_1$ and let $\widehat{V}_1$ denote its completion. Similarly we define $V_2$ and $\widehat{V}_2$.

Because $L_1\subseteq L_2$, $L_2$ is open with respect to the $L_1$ topology, and the map $V_1\to V_2$ (identity on the underlying sets) is bijective and continuous. On the other hand there's certainly no reason for the induced map on completions $\widehat{V}_1\to\widehat{V}_2$ to be bijective.

Is it always injective though?

**Example.**

Here's an example. If $V=k[T]$ the polynomial ring, and $L_1=R[T]$, $L_2=R[T/\pi]$ then the completion corresponds to restriction of analytic functions from a disc to a smaller disc, and this map is far from bijective. It is injective though -- an analytic function on a big disc is determined by its values on a small disc. More generally if the $L_i$ are free $R$-modules one can get quite a concrete handle on the completions. But I would imagine that in general a lattice can be quite pathological as an $R$-module and I don't have some good examples to hand in order to test out my question in these cases.