Let $E$ be a Banach space, $N\subset E$ a closed linear subspace and $q:E\rightarrow E/N$ the natural quotient map.
Question. If $X\subset E$ is a Borel-measurable linear subspace, is $q(X)\subset E/N$ also Borel-measurable (w.r.t. the quotient topology)?
- I am not sure whether the condition that $X$ is a linear subspace is really needed. Ideally it could be left away
- By the open mapping theorem, $q$ is an open map and thus the collection of sets $\mathcal{S}=\{A\subset E: q(A) \text{ is Borel measurable }\}$ contains all open sets. If $\mathcal{S}$ was a Dynkin system, then we would be done, but this does not seem to be the case (e.g. I don't see why $q(A^c)$ should be measurable, when $q(A)$ is).
- The example I have in mind is $E=C(M)$ for a closed manifold $M$. Then for $\alpha>\dim M/2$ one can show that the Sobolev space $X=H^\alpha(M)\subset E$ is measurable. I am interested whether this generalises to a smooth domain $\Omega\subset M$, i.e. after taking the quotient with respect to $N=\{u: u\vert_{M\backslash \Omega} = 0\}$.