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?



Sufficent condition for strict morphism of normed vector spaces
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.


