Let $\gamma\colon H^1(\Omega) \to H^{\frac 12}(\partial\Omega)$ be the linear trace map which has a right continuous inverse $\xi\colon H^{\frac 12}(\partial\Omega) \to H^1(\Omega)$.
Is the image of $\xi$ dense in $H^1(\Omega)$? i.e., for every $u \in H^1(\Omega)$, do there exist $w_n \in H^{\frac 12}(\partial\Omega)$ such that $\xi(w_n) \to u$ in $H^1(\Omega)$?
I wish $\xi$ was surjective but is this the best I can do?
First, $\xi$ cannot be surjective. Consider two distinct functions in $H^1(\Omega)$ with the same boundary values; they cannot both be in the range of $\xi$. This is a general property of quotient maps (remember that $H^{1/2}(\partial\Omega)=H^1(\Omega)/H^1_0(\Omega)$).
But $\xi$ cannot even have a dense image. Suppose $u\in H^1(\Omega)$ is such that there is a sequence $(w_n)\subset H^{1/2}(\partial \Omega)$ so that $\xi(w_n)\to u$. This implies that $w_n=\gamma(\xi(w_n))\to\gamma(u)$ since $\gamma$ is continuous. It then follows from the continuity of $\xi$ that $\xi(w_n)\to \xi(\gamma(u))$ and by uniqueness of limits $\xi(\gamma(u))=u$. Therefore $u$ is in the range of $\xi$. Since $\xi$ is not surjective, this cannot hold for all $u\in H^1(\Omega)$.
If you give up continuity, I suspect you could make $\xi$ have dense range. But I don't know if this could be done explicitly (without relying on Hamel bases or such).
