I'm checking a result that seems to require that the divergence operator $\nabla\cdot$ be surjective from something like the Sobolev space $\textbf{H}^2(\Omega) \cap \textbf{H}_0^1(\Omega)$ onto $H_0^1(\Omega)$. The only way I found that I could try and prove this would be to show its dual, the gradient operator $-\nabla$ is isomorphic from $H^{-1}(\Omega)$ onto $(\textbf{H}^2(\Omega) \cap \textbf{H}_0^1(\Omega))'$, which I'm also not sure is possible to show without modding out by constants.
I found this proof showing that $\nabla \cdot$ maps $\textbf{H}_0^1(\Omega)$ onto $L^2_{avg} = \{v \in L^2(\Omega): \int_\Omega u = 0\}$, and a closer result by Amrouche and Girault showing that $\nabla \cdot$ maps $$\textbf{H}_0^2(\Omega)/V \text{ onto } H_0^1(\Omega) \cap L^2_{avg}(\Omega),$$ where $V$ is the closure of $\{u \in C_0^\infty(\Omega):\nabla \cdot u = 0\}$ in $H^2(\Omega)$.
Is it possible for $\nabla \cdot$ to be surjective onto $H_0^1(\Omega)$ from anything? Is the best range that $\nabla \cdot$ can map onto something intersect $L^2_{avg}(\Omega)$ if I want the domain and range to vanish on the boundary?
After some more searching, I'm not completely convinced, but I do think we need the mean zero average condition and to mod out by constants as a consequence of De Rham's theorem:
Let $\Omega$ be a bounded Lipschitz-continuous domain of $\mathbb{R}^n$ and $\mathbb{Z} \ni m \geq 0$. Then there exists a $p \in H^{-(m+1)}(\Omega)$ such that $f = \nabla p$, if and only if the distribution $f \in H^{-m}(\Omega)$ satisfies $$\langle f, \phi \rangle = 0, \quad \forall \, \phi \in \{u \in H_0^2(\Omega): \nabla \cdot u = 0\}.$$ Further, on a connected domain $\Omega$, $p$ is unique up to a constant.