Let $K$ be a non-archimedean field of char 0 and a morphism $f:V \rightarrow W$ of normed $K$-vector spaces given. The map $f$ is said to be strict if $V/\ker(f)$ with the quotient topology is homeomorph to $\mathrm{im}(f)$ with subspace topology.
I read without a proof that if $\mathrm{im}(f)$ is closed then $f$ is strict. But I cannot think of a proof.
Is this really true?
Counterexamples are trivially obtained by taking a stricly coarser norm on a Banach space, e.g., the $\ell_\infty$-norm on $\ell_2$. For $V=(\ell_2,\|\cdot\|_2)$, $W=(\ell_2,\|\cdot\|_\infty)$ and the identical map, the image is closed but the map isn't strict.
I don't know anything about normed spaces over non-archimedian fields but I guess that the situation is the same.
I think you need the assumption that $W$ is Banach, at least I don't see a way around that. If $W$ is Banach, then the reasoning is as follows. Let $\phi:V/\ker(f)\to \text{im}(f), [v]\mapsto f(v)$. Then $\phi$ is well-defined, continuous and invertible, say with inverse $\psi$. Moreover, since $V/\ker(f)$ is Hausdorff, the graph of $\psi$ is closed. If $\text{im}(f)$ is closed in $W$ (and $W$ is Banach), then $\text{im}(f)$ is Banach, and the Closed Mapping Theorem tells you that $\psi$ is continuous.
