Abstract isometries can give a nonunitary comparison

For every $z,w\in \mathbb{C}$, $({\mathrm{\Omega }}_{-}(z),{\mathrm{\Omega }}_{-}(w))=z\bar{w}=(z,w)$ and $({\mathrm{\Omega }}_{+}(z),{\mathrm{\Omega }}_{+}(w))=z\bar{w}=(z,w)$ in the standard first-variable-linear inner products, so both maps are isometries. Their ranges are $\mathbb{C}\times \{0\}$ and $\{0\}\times \mathbb{C}$, respectively, and are distinct proper subspaces of ${\mathbb{C}}^2$.

The Hilbert adjoint is ${\mathrm{\Omega }}_{+}^{\ast }(x,y)=y$, since $({\mathrm{\Omega }}_{+}z,(x,y))=(z,y)$ for all $z,x,y$. Hence ${\mathrm{\Omega }}_{+}^{\ast }{\mathrm{\Omega }}_{-}(z)={\mathrm{\Omega }}_{+}^{\ast }(z,0)=0$ for every $z$. The zero map is not unitary because it does not preserve the norm of $1\in \mathbb{C}$ and is not surjective. This refutes the universal claim: isometry alone need not give equal ranges or a unitary comparison. This abstract Hilbert-space example does not construct a QFT model or establish non-completeness in any particular physical theory.