How to deduce open mapping theorem from closed graph theorem? 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?
 A: 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.


*

*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.

*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.

*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.
