Corollary 3

Let E and F be two complete metrisable vector spaces over a non-discrete valued division ring. In order that a continuous linear mapping u of E in F should be a strict morphism, it is necessary and sufficient that $u(E)$ be closed in F.

The condition is necessary, because if $u$ is a strict morphism, the image $u(E)$, being isomorphic to the quotient $E/u^{-1}(0)$, is complete (I, p. 17) and therefore closed in F. The condition is sufficient, since, if $u(E)$ is closed in F, then $u(E)$ must be a complete metrisable vector space and thus by theorem 1 $u$ is a strict morphism of E on $u(E)$.