When is a natural map between completions injective? Let $X$ be a vector space equipped with a norm $p$ and a seminorm $q$. Denote the completion of $X$ with respect to $p$ with $X_p$ and with respect to $p+q$ by $X_{p+q}$. Then the induced map $\iota : X_{p+q} \to X_p$ is well-defined and continuous but not necessarily injective as can be seen in analogy to this answer to Kernel of the Extension of a Bounded Linear Operator by taking $X = l^2$,
\begin{equation}
p(x) = \sqrt{\sum_{n = 1}^\infty \left( \frac{x_n - x_{n+1}}{n} \right)^2}, \qquad
q(x) = \sqrt{\sum_{n = 1}^\infty \left( \frac{x_n}{n} \right)^2} \, .
\end{equation}
Then setting $x^m_n = m/(m+n)$ we have $\lim_{m \to \infty} p(x^m) = 0$, $x^m$ being Cauchy in $q$ and $\lim_{n \to \infty} q(x^m) = \sqrt{\pi^2/6}$ such that the equivalence class of $x^m$ in $X_{p+q}$ lies in the kernel of $\iota$ while being nonzero in $X_{p+q}$.
On the other hand we can take the Schwartz space $X = \mathcal{S}(\mathbb{R})$ and let $p = \lVert \cdot \rVert_{L^1}$ and $q = \lVert \cdot \rVert_{L^2}$.
Then $\iota$ is certainly injective because any sequence $f_n$ in $\mathcal{S}(\mathbb{R})$ with $\lim_{n \to \infty} \lVert f_n \rVert_{L^1} = 0$ has a subsequence converging to zero almost everywhere such that if $f_n$ is Cauchy in $L^2 (\mathbb{R})$, the corresponding limit $f$ in $L^2 (\mathbb{R})$ (which exists by completeness) has to be the zero function. Thus $\lim_{n \to \infty} \lVert f_n \rVert_{L^2} = 0$ as well proving that the equivalence class of $f_n$ is zero in $X_{p+q}$.
In the latter example we have much more structure: $X$ is a reflexive, complete and metrisable nuclear space and $p$ and $q$ are continuous norms on $X$.
But I suspect that these conditions are not sufficient.
What are some necessary and sufficient conditions for $\iota$ to be injective?
 A: A simple sufficient condition for two norms $p\le r$ on $X$ to induce an injective continuous linear map $i:X_r\to X_p$ between the completions is that the unit ball $B_r=\{x\in X: r(x)\le 1\}$ is $p$-closed.
Indeed, for $x\in X_r$ with $i(x)=0$ choose a sequence $x_n\in X$ with $x_n\to x$ in $X_r$. This sequence is $r$-Cauchy so that, for every $\varepsilon>0$, there is $n_\varepsilon\in \mathbb N$ with $x_n-x_m\in\varepsilon B_r$ for all $m,n\ge n_\varepsilon$. The continuity of $i$ and $i|_X=id_X$ yield $x_m=i(x_m)\to i(x)=0$ in $X_p$ and hence in $(X,p)$. For every $n\ge n_\varepsilon$, this implies $$x_n=\text{$p$-}\lim\limits_{m\to\infty} x_n-x_m+x_m\in \overline{\varepsilon B_r}^p=  \varepsilon B_r.$$ This shows that $x_n$ converges to $0$ in $(X,r)$ so that $x=0$.

You can relax the assumption to $\overline{B_r}^p \subseteq cB_r$ for some constant. Moreover, this principle can be generalized to general uniform spaces. In one form or another you find it in most text books about locally convex spaces, for instance in Grothendieck's Topological Vector Spaces.
A: The following theorem can be found in: Wendy Robertson, Completions of Topological Vector Spaces, Proc. London Math. Soc. Volume s3-8, Issue2, 242-257 (https://doi.org/10.1112/plms/s3-8.2.242)\
Suppose that $E$ is a Hausdorff topological vector space under
each of the two topologies $\xi$ and $\eta$, with $\xi$ finer than $\eta$, and that $i$ is the identity
mapping of $E$ under $\xi$ onto $E$ under $\eta$. Then the extension $j$ of $i$ is a (1-1)
mapping of $\hat{E_\xi}$ into $\hat{E_\eta}$  if and only if the filter condition holds:
if $\phi$ is a Cauchy filter base on $E$ and $i(\phi)$ is convergent
then $\phi$ is convergent.
