I have the problem that I cannot solve: Under what conditions does $H(X∣f(Y))=H(X∣Y)$? I would like to draw a result about the relation between $p_X(\cdot | g(Y))$ and $p_X(\cdot | Y)$. Are they equal?
This is an exercise in the textbook. There is a solution, but I don't think it's correct (more precisely, it is not satisfactory enough). The provided solution is as below.
Suggested Solution (not satisfactory). If $H(X|g(Y )) = H(X|Y )$, then $H(X)−H(X|g(Y )) = H(X) − H(X|Y )$, i.e., $I(X; g(Y )) = I(X; Y )$. This is the condition for equality in the data processing inequality. From the derivation of the inequality, we have equality iff $X → g(Y ) → Y$ forms a Markov chain. Hence $H(X|g(Y )) = H(X|Y )$ iff $X → g(Y ) → Y$ . This condition includes many special cases, such as $g$ being one-to-one, and $X$ and $Y$ being independent. However, these two special cases do not exhaust all the possibilities.
The data processing inequality lemma already provides the complete answer. $I(X;Y)=I(X;g(Y))$ holds if and only if $$\tag{1}X\rightarrow Y \rightarrow g(Y)$$ and $$\tag{2} X\rightarrow g(Y) \rightarrow Y.$$
Now note that $$ \begin{align} p(x,g(y),y) &= p(y, g(y))p\left(x\mid y, g(y)\right)\\ &=p(y, g(y))p\left(x\mid y\right)\tag{3}, \end{align} $$ where the last equality holds since (1) implies that $g(Y)\rightarrow Y \rightarrow X$, i.e., $X$ is independent of $g(Y)$, when conditioned on $Y$. Also note that $$ \begin{align} p(x,g(y),y) &= p(y, g(y))p\left(x\mid y, g(y)\right)\\ &=p(y, g(y))p\left(x\mid g(y)\right) \tag{4}, \end{align} $$ where the last equality holds since (2) implies $Y\rightarrow g(Y) \rightarrow X$.
It follows from (3) and (4) that it must hold $$ p\left(x\mid g(y)\right) = p(x \mid y), $$ for all valid values of $x$ and $y$. This is the most general condition you seek, which captures the cases of $g$ being one to one and $Y$ independent of $X$.