Each of the scalar random variables, $ Y $, $ X $, $ U $, and $ V $, is continuous and possibly has $ \mathbb{R} $ as its support. The random variable, $Z$, could be vector valued, but continuous.

I have the following \begin{align}\nonumber Y = \beta X + U\\\nonumber X = \pi Z + V, \end{align} where $ Z \perp (U,V) $ ( $ \perp $ denotes independence). But $ U\not\perp X$. In the second equation, given $ Z $, there is one-to-one mapping between $ X $ and $ V $.

Given the set-up, I need to know whether the following conditional independence holds: $ U\perp X\mid V $.

My question is: since, given $ Z $, there is one-to-one mapping between $ X $ and $ V $, is it true that $ \sigma(X, Z) = \sigma(V, Z) $, where $ \sigma(X, Z) $ is the sigma algebra generated by $ (X,Z) $ and $ \sigma(V, Z) $ that generated by $ (V,Z) $?

If $ \sigma(X, Z) = \sigma(V, Z) $, then I could write $ U\mid X,Z \sim U\mid V,Z $. But since $ Z \perp (U,V) $, I could write $ U\mid V,Z \sim U\mid V $. From this could we deduce that $ U\perp X\mid V $.