I'd like to show: if $A : X \rightarrow Y$ is a bounded linear operator between Banach-spaces, then $\pi : X \rightarrow X / \mathrm{ker}(A)$ is a open map.

I found a proof, which I do not really like, namely by first embedding $X$ into $X \times (X / \ker (A))$ by $x \mapsto (x, \pi(x))$ - which is not too hard to be proven an open map - and then using the general fact from topology that any projection from a product-space is open. Then $\pi$ is the composition of the two maps that have been shown to be open.

However, I was hoping that there would be a more elegant way to prove this.

Thanks for any suggestions!


Notice that $\pi$ is surjective linear mapping. Since $\mathrm{ker}(A)$ is closed subspace of $X$ the quotient space $X/\mathrm{ker}(A)$ endowed with the quotient topology is Banach space. Now you can apply the Open Mapping theorem.

  • 1
    $\begingroup$ Yes, that works of course. But I actually wanted to use the fact to prove that the Closed Graph Theorem implies the Open Mapping Theorem. $\endgroup$ – Steven Jan 3 '16 at 9:37

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