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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4http://cuhkmath.wordpress.com/2011/09/06/closed-graph-theorem-implies-open-mapping-theorem/ (I haven't checked this...). – David Mitra Dec 10 '11 at 14:51
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1You might be interested in this thread – t.b. May 02 '12 at 17:40
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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.
t.b.
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I think its because we give the quotient topology to $\bar{X}$. The quotient topology says that a set $\bar{U}$ in $\bar{X}$ is open iff the pre-image under the quotient map is open, i.e. $\pi^{-1}(\bar{U})$ is open. And thus almost by definition you get that $\pi$ is open. – eatfood Mar 17 '20 at 08:01
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In general, the quotient map may not always be open, but it is true when we are quotienting by a closed subspace. See https://math.stackexchange.com/a/4330993/732515 for the proof. – S.L. May 21 '23 at 23:26