Let us consider two spaces $\mathbb{X},\mathbb{Y}$. For simplicity we put $\mathbb{X} = \mathbb{Y} = \mathbb{R}$. On the product space $\mathbb{S} = \mathbb{X}\times \mathbb{Y}$ we consider a Markov process $S = (X,Y)$ with a transition kernel $T$ such that $ T(B|s) = \mathsf P\{S_1\in B|S_0 = s\}. $
For any fixed $y\in\mathbb Y$ we construct a Markov process $X^y$ on $\mathbb{X}$ such that its transition kernel $T^y$ is given by $ T^y(A|x) = T(A\times \mathbb Y|(x,y)). $
For a set $A\subset \mathbb X$ we are interested in the following probability $ u(s) = \mathsf P\{S_n\in A\times \mathbb Y\text{ for all }n\geq 0|S_0 = s\}. $
What I have is that for all $y\in \mathbb Y$ and $x\in\mathbb X$ holds $ \mathsf P\{X^y_n\in A\text{ for all }n\geq 0|X_0 = x\} = 0. $
Can I derive from here that $u(s)\equiv 0$ or if there is a counterexample?
It will be even helpful if $\mathbb{X},\mathbb{Y}$ are finite (I mean for the proof or a counterexample).