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These two theorems are equivalent but I can not figure out how to deduce the open mapping from the closed graph. Can anyone give a hint or some reference?

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4… (I haven't checked this...). – David Mitra Dec 10 '11 at 14:51
You might be interested in this thread – t.b. May 2 '12 at 17:40

1 Answer 1

up vote 18 down vote accepted

The proof linked to by David is the standard one, but the write-up strikes me as somewhat clumsy.

Here's my take on this argument:

Suppose $T: X \to Y$ is continuous and onto.

  1. The map $\pi: X \to \bar{X} = X/ \operatorname{Ker}{T}$ is open, onto and continuous. The map $T$ factors over $\pi$ via a continuous linear bijection $\bar{T}: \bar{X} \to Y$ by definition of the quotient topology.

  2. By continuity of $\bar{T}$ the graph of $\bar{T}$ is closed. Switching coordinates $(\bar{x},y) \mapsto (y,\bar{x})$ is a homeomorphism $\bar{X} \times Y \to Y \times \bar{X}$ and it maps the graph of $\bar{T}$ to the graph of $\bar{T}^{-1}$, so the graph of the linear map $\bar{T}^{-1}$ is closed, too. By the closed graph theorem $\bar{T}^{-1}$ is continuous.

  3. For all open $U \subset X$ the set $T(U) \subset Y$ is open because it is the pre-image of the open set $\pi(U)$ under the continuous map $\bar{T}^{-1}$, hence $T$ is open.

Note: Step 2. is a proof of the inverse mapping theorem from the closed graph theorem.

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