Let $X,Y$ be two continuous random variables with density functions $f_X,f_Y$. Suppose $f_{X|Y}(x|y)=\delta(x-F(y))$, where $F$ is some function of $y$. Does this mean that $f_{Y|X}(y|x) = \delta(y-G(x))$ where $G$ is some function of $x$?
I tried to approach this using Bayes' theorem
$$ f_{Y|X}(y|x) = \frac{f_{X|Y}(x|y)f_Y(y)}{f_X(x)} = \frac{\delta(x-F(y))f_Y(y)}{f_X(x)} = \frac{\delta(x-F(y))f_Y(y)}{f_X(F(y))} $$ where I use the property of $\delta$ such that $\delta(x-x_0)f(x) = f(x_0)\delta(x-x_0)$. I am not so sure how to continue from here. I am also not able to think of a counterexample.