0$ for all $z \in X$, i.e. $k=0$. This is a contradiction, and the proof is complete.

*Proof of Proposition \[P: uneq=cc\].* If there is a unitary equivalence $(\ga, u)$ then it is clear that ${{\mathfrak{L}}}(X)$ and ${{\mathfrak{L}}}(Y)$ are $*$-isomorphic. Hence $X$ and $Y$ are completely isometric as operator bimodules.

Conversely suppose there is completely isometric right module map $(\ga, u)$; we need to show that $u$ is adjointable. As in [@BleLeM04 Remark 3.6.1] we get the completely isometric isomorphism $$j \colon {{\mathcal{A}}}(X) \to {{\mathcal{A}}}(Y)
:
\begin{bmatrix} a & 0 \\ x & 0 \end{bmatrix} \mapsto \begin{bmatrix} \ga(a) & 0 \\ u(x) & 0 \end{bmatrix}.$$ By Lemma \[L: cenv linking\] then the linking algebras are $*$-isomorphic by some $\Phi$. Let $v \colon Y \to X$ such that $$\begin{bmatrix} 0 & 0 \\ v(y) & 0 \end{bmatrix}
=
\Phi^{-1}(\begin{bmatrix} 0 & 0 \\ y & 0 \end{bmatrix}).$$ Then we obtain $$\begin{aligned}
\ga({\left\langlex, v(y)\right\rangle}_X)
& =
\Phi({\left\langlex, v(y)\right\rangle}_X)
=
\Phi(x)^* y
=
{\left\langleu(x), y\right\rangle}_Y\end{aligned}$$ where we omit the zero entries of the matrices. Then $u^* = v$ and the proof is complete.

We remark that Proposition \[P: uneq=cc\] does not hold for bounded bimodule maps. Dor-On [@Dor15] illustrates this by examining a 