The ultraweak Poisson trial operator has trivial kernel

Compactly supported vector tests on each cell give $\sigma +\mathrm{\nabla }u=0$ distributionally; compactly supported scalar tests give $div\sigma =0$. Thus $u{\mathrm{\mid }}_K\in H^1(K)$, $\sigma {\mathrm{\mid }}_K\in H(div;K)$, and both identities hold cellwise.

With $\sigma =-\mathrm{\nabla }u$, the first local equation and [F3] give $⟨\widehat{u}-u{\mathrm{\mid }}_{\mathrm{\partial }K},{\gamma }_n\tau ⟩=0$ for every $\tau \in H(div;K)$. Surjectivity [F4] makes the scalar trace classes equal. The second local equation gives $⟨{\widehat{\sigma }}_n-{\gamma }_n\sigma ,v⟩=0$ for every $v\in H^1(K)$, so those normal-trace functionals are equal as well.

Take a global H1_0 lift $u_0$ of $\widehat{u}$. On each cell, $u-u_0$ has zero scalar trace, hence belongs to $H_0^1(K)$; zero-extending compactly supported approximants and summing over the finite mesh shows $u-u_0\in H_0^1(\mathrm{\Omega })$, so $u\in H_0^1(\mathrm{\Omega })$. Take a global H(div) lift ${\sigma }_0$ of ${\widehat{\sigma }}_n$. The cell field $\sigma -{\sigma }_0$ has zero weak normal trace on every cell. Summing its local Green identities against restrictions of $C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ shows its piecewise $L^2$ divergence is its global distributional divergence; thus it glues to H(div), and $\sigma \in H(div;\mathrm{\Omega })$. Now $\sigma =-\mathrm{\nabla }u$ and $div\sigma =0$; the weak Green identity tested with $u\in H_0^1$ gives $\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2^2=0$. By [F6], $u=0$, then $\sigma$ and both trace components vanish.