Proof. Since f is continuous, its graph is closed. Hence its inverse also has closed graph. Now apply the Closed Graph Theorem -- e.g., in the form of 27.28.c.

Lemma. Let X be a topological vector space, and let K be a linear subspace. Let Q = X/K be the quotient, and equip it with the quotient topology. Let p be the quotient map from X onto Q. Then: p is an open map; i.e., whenever G is an open subset of X, then p(G) is an open subset of Q. Also: If K is closed, then Q is Hausdorff. If K is closed and X is an F-space, then Q is an F-space, when equipped with the quotient F-norm given by ||q|| = inf{||x|| : x is a member of q}.

Sketch of some parts of the proof. To prove p is open, refer to 15.31.f: The p-saturation of an open set G is equal to G+K. That's the union of the open sets G+k (for k in K), so it is open. For the F-space results, the completeness of Q can be proved most easily, not by using Cauchy sequences, but by using the characterization of completeness in terms of absolutely convergent series mentioned in 22.20.

Open Mapping Theorem. Let X and Y be F-spaces. Suppose f is a continuous linear map from X onto Y. Then f is an open map.

Outline of proof. Let K be the kernel of f, and define Q and p as in the preceding lemma. The map f factors naturally through p -- we have a map g from Q to Y defined by g(p(x)) = f(x). Verify that g is a continuous linear bijection from Q onto Y; hence it is an isomorphism of topologies; hence it is an open map. Also p is an open map, by the preceding lemma. Hence the composition f is an open map.