**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.



injective implies completion injective?
Let us take $V=Q\_p[T]$ and $L\_2 = \oplus\_{i \geq 0} Z\_p \cdot T^i$ and $L\_1 = Z\_p \cdot p \oplus (\oplus\_{i \geq 0} Z\_p (T^i + p T^{i+1}))$. The element $x = 1 \cdot p - p \cdot (1+pT) + p^2 \cdot (T+pT^2) - \cdots $ belongs to the $p$-adic completion of $L\_1$ and is nonzero in it, but its image in the $p$-adic completion of $L\_2$ is zero. This should give you an example where your map $\hat{V}\_1 \to \hat{V}\_2$ is not injective.


