Corollary 5

— Let E and F be two Fréchet spaces and u a continuous linear mapping from E into F. The following conditions are equivalent : a) u is an isomorphism from E onto F. b) u is an isomorphism from E onto F for the weakened topologies. c) $^t u$ is an isomorphism from $F'$ onto $E'$ for the weak topologies. d) $^t u$ is an isomorphism from $F'$ onto $E'$ for the strong topologies. e) $^t u$ is an isomorphism from $F'_c$ onto $E'_c$. Since an isomorphism is none other than a strict bijective morphism, the equivalence of a) and b) follows from the equivalence of conditions a) and b) of th. 1. It is clear that a) implies each of the conditions c), d) and e). Conversely, suppose that one of the conditions c), d) or e) is satisfied. It follows from th. 1 and its cor. 3 that u is a strict morphism from E into F, and $^t u$ is evidently bijective. Let N (resp. I) be the kernel (resp. the image) of u. Since $^t u$ is bijective, we have $\operatorname{Im} ^t u = E'$ and $\operatorname{Ker} ^t u = \{ 0 \}$, and so $N^\circ = E'$ and $I^\circ = \{ 0 \}$ by prop. 2 of IV, p. 27. But N (resp. I) is a closed vector subspace of E (resp. F), and the theorem of bipolars (II, p. 44) implies that $N = \{ 0 \}$ and $I = F$, hence u is bijective. We have therefore proved that u is an isomorphism.