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.