I was thinking a little more about my previous quesion about whether there is a natural section of canonical representatives of the quotient map from the space $\mathcal{L}^2(\mathbb{R})$ of square-integrable functions to the Hilbert space $L^2(\mathbb{R})$ (in which we mod out the kernel of the $L^2$ norm). Another possibility occurs to me, which is manifestly basis-independent.
The Fourier transform is a unitary operator on the Hilbert space $L^2(\mathbb{R})$. The Fourier inversion theorem holds almost everywhere for functions $f \in \mathcal{L}^2$; that is, $\mathcal{F}^{-1}(\mathcal{F}f)(x)$ and $\mathcal{F}(\mathcal{F}^{-1}f)(x)$ both agree with $f(x)$ almost everywhere. Since modding out by the kernel eliminates the "almost everywhere" subtlety, we have that $\mathcal{F} \mathcal{F}^{-1}$ and $\mathcal{F}^{-1}\mathcal{F}$ both equal the identity on the Hilbert space $L^2(\mathbb{R})$.
I think, although I'm not positive, that the transform $\mathcal{F} \mathcal{F}^{-1}$ and/or $\mathcal{F}^{-1}\mathcal{F}$ actually implements the quotient map, in the sense that these transformations take all the functions in each equivalence class in $L^2(\mathbb{R})$ to a single canonical representative. In this case, a natural section of the quotient map would be the set of functions $f \in \mathcal{L}^2$ for which the Fourier inversion theorem holds everywhere, rather than almost everywhere. Is this correct?
You don't necessarily get convergence everywhere for $\mathcal{F}^{-1}\mathcal{F}f$. So your choice of $f$ would not be defined everywhere using this methodology. It is true that $\mathcal{F}f$ and $\mathcal{F}g$ are identical if $f=g$ a.e. because these are defined through limits of integrals. So that means you're generally not going to be able to find an $f$ in the equivalence class for which $\mathcal{F}^{-1}\mathcal{F}f$ converges everywhere.