Consider the C-shaped region $C\subset \mathbb{R}^2$ shown below, which has arms I, II $\subset C$ given by I $= I_0\times I_1$ and II $= I_0\times I_2$ for some intervals $I_0, I_1, I_2\subset\mathbb{R}$, with $I_1$ and $I_2$ disjoint.
Let further $r : C\rightarrow \mathbb{R}^2$ be a $C^1$-diffeomorphism which acts trivially on $C$ in the sense that
$$\tag{1} r(x,y) = (\alpha_1(x), y) \ \ \forall\, (x,y)\in \text{I}, \qquad \text{and} \qquad r(x,y) = (\alpha_2(x), y) \ \ \forall\, (x,y)\in \text{II} $$
for some $\alpha_i : I_0\rightarrow \mathbb{R}$ strictly monotone, and $\left.r\right|_{C\setminus{(\text{I}\cup\text{II})}} = \left.\mathrm{id}\right|_{C\setminus{(\text{I}\cup\text{II})}}$.
Suppose now that $X = (X^1_t, X^2_t)_{t\in[0,1]}$ is a continuous-time stochastic process in $\mathbb{R}^2$ that evolves inside the C-shaped region shown above, and let the process
$$\tag{2} Y=(Y^1_t, Y^2_t)_{t\in[0,1]}:=r(X)\equiv(r(X^1_t, X^2_t))_{t\in[0,1]}$$
be the image of $X$ under $r$. The components $X^1$ and $X^2$ of $X$ are assumed to be statistically independent
My question: Assume that $\alpha_1\neq \alpha_2$ with $\alpha_1(U) > \alpha_2(U)$ for some (nonempty) open $U\subset I_0$ and $\mathbb{P}(X_{t_j}\in U\times I_j)>0$, $j=1,2$, for some $t_1, t_2\in [0,1]$. Does this imply that the processes $Y^1$ and $Y^2$ must be statistically dependent?
Remark: It seems plausible that $Y^1$ and $Y^2$ must be dependent since (by the fact that $\alpha_1$ and $\alpha_2$ are different) the evolution of $Y^2(=X^2)$ influences the trajectories of $Y^1$ via the arm I, II that $Y=(Y^1,Y^2)$ lands in. Yet it seems to me that the given assumptions might be too weak to actually prove the independence of $Y^1$ and $Y^2$.
$\rhd$ Here are some ideas. First we show that $Y_1$ and $Y_2$ can in fact be independant as long as $\alpha_1(I_0) \cap \alpha_2(I_0) \neq \emptyset$. Then we look at some information we can get on $X_1$ which can be helpful
$\rhd$ First, let $\gamma : \mathbb{R} \to \mathbb{R}$ be a continuous function such that $\gamma(\mathbb{R}) = \alpha_1(I_0) \cap \alpha_2(I_0)$ (think of gamma as a path which travel along the common absisses of the images of $\alpha_1$ and $\alpha_2$). One can consider the process where: with probability $\dfrac{1}{2}$, we have $(X^1_t,X^2_t)=(\alpha_1^{-1}(\gamma(t)),1) \in I$ ; and with probability $\dfrac{1}{2}$, we have $(X^1_t,X^2_t)=(\alpha_2^{-1}(\gamma(t)),-1) \in II$.
Then $(Y_t^1,Y_t^2) = (\gamma(t), Z)$ where $Z$ follows the uniform distribution on $\{-1,1\}$ independently of $Y_1$ (as $Y_1$ is not random at all anymore !).
Thus, as long as $\alpha_1(I_0) \cap \alpha_2(I_0) \neq \emptyset$, we can build a stochastic process satisfying your assumptions such that $Y_1$ and $Y_2$ are independent.
$\rhd$ Let's still find some general information that may be useful if you manage to get more assumption on $X_1$. Using your notations, we have $Y^2_t= X^2_t$, thus we ask whether $Y^1_t$ and $X^2_t$ are independent. We have,as $X_1$ and $X_2$ are independant, $$P(Y^1_t \in \alpha_2(U) \mid X^2_t \in I_2) = P(X^1_t \in U \mid X^2_t \in I_2) = P(X^1_t \in U) $$ and $$P(Y^1_t \in \alpha_2(U) \mid X^2_t \in I_1) = P(X^1_t \in \alpha_1^{-1}(\alpha_2(U)) \mid X^2_t \in I_1) = P(X^1_t \in \alpha_1^{-1}(\alpha_2(U))).$$ If $Y^1_t$ and $X^2_t$ are independant, then both of these probabilities must be equal to $P(Y^1_t \in \alpha_2(U))$, and therefore $$P(X^1_t \in \alpha_1^{-1}(\alpha_2(U))) = P(X^1_t \in U).$$
By your hypothesis, $\alpha_1^{-1}(\alpha_2(U))$ and $U$ are disjoint sets. If you have some more assumptions that may contradict this, this will provide an answer.