I need to show that $\sigma(P)$ is a bijective mapping from the set of partitions of a finite set $X$ to a set of $\sigma$-algebras on $X$. I managed to show surjectivity of this mapping, but I struggle with injectivity. I tried to prove it thorugh contradiction:
Let $P_1=\{A_i\}_{i=1}^n, P_2=\{B_i\}_{i=1}^m$ be partitions of $X$. Assume $P_1 \neq P_2 $ and $\sigma(P_1) = \sigma(P_2)$. Then there exists $C \in P_1 \backslash P_2$, and $C, C^c\in \sigma(P_1)=\sigma(P_2)$. I can use the fact that: $$\sigma(P_1) = \left\{\bigcup_{i\in I}A_i: I\subset\{1,...,n\}\right\},\quad \sigma(P_2) = \left\{\bigcup_{i\in I}B_i: I\subset\{1,...,m\}\right\}$$ And conclude that $$C = \bigcup_{i\in I_A}A_i = \bigcup_{i\in I_B}B_i ,\quad C^c = \bigcup_{i\in I_A'}A_i = \bigcup_{i\in I'_B}B_i$$ for some indexing sets $I_A, I_B, I'_A, I'_B$. My aim was to explicitly create a set that will be in $\sigma(P_1)\backslash \sigma(P_2)$ but I'm stuck.
You should use the fact that the $A_i$ and $B_i$ are disjoint: Let $A_i \in \sigma(P_1)=\sigma(P_2)$ and $A_i \notin P_2$. Then for (multiple) $B_j$'s we have $$A_i= \bigsqcup_j B_j.$$ Hence, for some $j$, $B_j \subsetneq A_i$. Since the $A_i$ are disjoint, $B_j \notin \sigma(P_1)$, a contradiction.