Let $\phi:G \rightarrow H$ be a morphism of (semi-)normed groups. Then $\phi$ is said to be strict if the natural isomorphism $G/\ker(\phi)\rightarrow \mathrm{Im}(\phi)$ is a homeomorphism.
That means that $\phi$ is open onto its image.
By Proposition 4 in Chapter 1.1.9 of "Non-archimedean analysis" by Bosch, Güntzer, Remmert we know that if $\phi:G\rightarrow H$ is a strict morphism of (semi-)normed groups then so is the completion $\hat{\phi}: \hat{G}\rightarrow \hat{H}$.
Now I was wondering if the converse is also true?
EDIT: The (semi)-norm on the groups are ultrametric functions.
NO. Consider a Banach space $(H,\|\cdot\|)$ and a strictly finer norm $\||\cdot\||$ on $H$ (this requires the axiom of choice which provides linear discontinuous functionals $\varphi:X\to \mathbb R$ so that we can take $\||x\||=\|x\|+|\varphi(x)|$). Let $\phi: G\to H$ be the identical map where $G=H$ is endowed with the finer norm $\||\cdot\||$. Then $\phi$ is not strict because the unit ball of $G$ is not open in $H$ but $\tilde \phi$ is strict because of the open mapping theorem (b.t.w., $\tilde\phi$ is not injective although so is $\phi$).