I have been trying to proove the following with no success:
Suppose you have $X$ a discrete random variable over $A$, and $f:A \rightarrow B$ a bijective function.
It seems to me that if you define $Y:=f(X)$ then the conditional probability of $Y$ knowing $X=x$ should be either $1$ or $0$:
$$p(y|x) = \begin{cases} 0 & y \neq f(x)\\ 1 & y = f(x) \end{cases}$$ $$x \in A, y \in B$$
I want to prove this formally, and don't know how to get started. It seems to me that the solution should follow from $X^{-1} x = Y^{-1} y$ --if $y = f(x)$-- but I do not know how to keep going from there.
$$ \mathbb P(f(X) = y | X = x) = \frac{\mathbb P(f(X) = y , X = x)}{\mathbb P(X = x)} = \frac{\mathbb P(f(x) = y , X = x)}{\mathbb P(X = x)} = \bigg\{ \begin{array}{cl} 1 & \text{if } f(x) = y \\ 0 & \text{otherwise.} \end{array}$$ The event $\{ f(x) = y \} $ is either $\Omega$ or $\emptyset$.