Let $Z$ be an abelian von Neumann algebra, and let $A$ and $B$ be two von Neumann algebras that receive central maps $Z \to Z(A)$ and $Z \to Z(B)$.
We may then construct the <b>fiber product</b> of $A$ and $B$ over $Z$, denoted $A \times_Z B$.


The fiber product is defined as follows.
First consider $L^2A \boxtimes_Z L^2B$, namely the Connes fusion of $L^2A$ with $L^2B$ over $Z$.
Then define $A \times_Z B$ to be the algebra generated by left actions of $A$ and of $B$ on $L^2A \boxtimes_Z L^2B$.

> I would like there to be a canonical unitary isomorphism between $L^2(A \times_Z B)$ and $L^2A \boxtimes_Z L^2B$.<br> How do I do it?

When all the von Neumann algebras are separable, then I think that I know how to do it using disintegration theory.
But I'm unsure whether the same construction can be used in the absence of any separability assumptions. So I'd be very interested in a construction that doesn't rely on measure spaces and disintegration theory.

<hr>

Here's the argument when $A$, $B$, and $Z$ are separable.
In that case, we may write $Z=L^\infty(X, \mu)$ for $X$ some standard measure space, and use disintegration theory to write $A=\int^\oplus A_x d\mu(x)$ and $B=\int^\oplus B_x d\mu(x)$. The formula 
$$
\left(\int^\oplus_{x\in X} \sqrt{\phi_x} d\mu(x)\right)\boxtimes_\mu
\left(\int^\oplus_{x\in X} \sqrt{\psi_x} d\mu(x)\right)
\mapsto \int^\oplus_{x\in X} \sqrt{\phi_x\otimes \psi_x} d\mu(x)
$$
then provides the desired unitary
\begin{align}
&L^2A \boxtimes_Z L^2B =\\ &\left(\int^\oplus_{x\in X} L^2A_x d\mu(x)\right) \boxtimes_{L^\infty(X)} \left(\int^\oplus_{x\in X} L^2B_x d\mu(x)\right)\to \int^\oplus_{x\in X} L^2\big(A_x\bar\otimes B_x\big) d\mu(x) \\
&\hspace{12cm} = L^2\big(A \times_Z B\big).
\end{align}