Proposition 1

Let E and F be locally convex spaces, $E_1$ a closed subspace of finite codimension of E and $u$ a continuous linear mapping of E into F. In order that $u$ be a strict morphism with closed image, it is necessary and sufficient that $u|E_1$ be one.

Suppose first that the linear mapping $u|E_1$ is injective. Then we have $E_1\cap$ Ker($u$) $=\{0\}$, so that Ker($u$) is of finite dimension. Let S be a vector subspace of E supplementary to $E_1+$ Ker($u$). The space E is a topological direct sum of $E_1$, Ker($u$) and S (EVT, I, p. 15, cor. 4 and prop. 3). If $u$ is a strict morphism with closed image, it defines by restriction an isomorphism of $E_1\oplus S$ onto a closed vector subspace of F, and fortiori an isomorphism of $E_1$ onto a closed vector subspace of F. Conversely, suppose that $u$ defines by restriction an isomorphism of $E_1$ onto $u(E_1)$ and that $u(E_1)$ is closed in F. We have $u(E) =u(E_1)\oplus u(S)$. It follows that $u(E)$ is a topological direct sum of $u(E_1)$ and $u$(S), and is closed in F (loc. cit.). Since $u(E_1)$ is closed and we have $u(E_1)\cap u(S) =\{0\}$, the space $u(S)$ is separated and $u$ defines by restriction an isomorphism of S onto $u(S)$ (EVT, I, p. 15, cor. de la prop. 3), hence also of $E_1\oplus S$ onto $u(E)$. This proves that $u$ is a strict morphism with closed image.

Passing to the general case. Put $N = E_1\cap$ Ker($u$) and $G = E/N$. The locally convex space $E_1/N$ identifies with a closed vector subspace of finite codimension $G_1$ of G. Let $v: G\rightarrow F$ be the continuous linear mapping deduced from $u$ by passing to the quotient. In order that $u$ (resp. $u|E_1$) be a strict morphism with closed image, it is necessary and sufficient that $v$ (resp. $v|G_1$) be one. This reduces us to the case already treated.