Suppose $(x,y,z)$ is (mutually) independent. Let $a=f(x,z)$ and $b=g(y,z)$. Is $x \perp y|(a,b)$, where the notation means $x$ is independent of $y$ conditional on $(a,b)$?
At first I thought that this is true by the following argument:
$(x,y,z)$ independent $\Rightarrow$ $(x \perp y)|z$ $\Rightarrow$ $(x \perp y)|(\tilde f(z),\tilde g(z))$ (for functions $ \widetilde f$ and $\widetilde g$) $\Rightarrow$ $(x \perp y)|(f(x,z),g(y,z))$ $\Rightarrow$ $(x \perp y)| (a, b)$.
I know the first and second $\Rightarrow$ are correct but I'm not so sure about the third $\Rightarrow$. My reasoning for the third $\Rightarrow$ is that the functions include $x$ and $y$, which are independent by assumption, won't affect the independence but I'm skeptical since I couldn't prove the statement by showing that $f_{x,y|a,b}=f_{x|a,b}f_{y|a,b} $.
edit: My claim is false. @antkam provides a counterexample. I just wanted to add my intuition for why in general $x \not\perp y|(a,b)$. Consider the causal diagram $$x\rightarrow a=f(x,z)\leftarrow z\rightarrow b=g(y,z) \leftarrow y.$$ Now suppose that $a$ and $b$ are fixed. Then for any value of $z$, we must have $x$ and $y$ compensating for the fact that $a$ and $b$ are fixed (so they can't just be independently chosen).
I'm not sure what you mean by "$f, g$ are invertible in $z$", but here's a counter-example. If this doesn't work, lemme know why not, and I will change my answer.
Flip $3$ independent fair coins and let $x, y, z\in \{0,1\}$ indicate whether each coin shows Head.
$f(x,z) = x+z, ~~~g(y,z) = y+z$
Conditioned on $(a,b) = (1,1)$, we have two possible sample points $(x,y,z) = (1,1,0)$ or $(0,0,1)$ and so $x \not\perp y \mid (a,b)$