Trying to show the surjectivity of a map. I am trying to solve the following exercise.I'm stuck in between,please help.

Let $\Omega$ be a set with $n$ elements.Let $\mathcal A$ be the collection of all Boolean algebras of subsets of $\Omega$.A partition of $\Omega$ is a collection of nonempty disjoint subsets $A_1,A_2,...,A_k$ such that $\Omega = \cup_{j=1}^{k} A_j$.Let $\mathcal P$ be the collection of all possible partitions of $\Omega$.Given a subset $B$ of $\Omega$ we denote $B^1=B$ and $B^{-1} =B^c$.Show that the map $\Phi: \mathcal A \to \mathcal P$ given by $$\Phi(\sigma)= \{\sigma_{\epsilon} \mid \sigma_{\epsilon}= \cap_ { B \in \sigma} B^{(\epsilon_B)},\epsilon \in \{-1,1\}^{\vert  \sigma \vert }\}$$ is a bijection.

I have shown that the $\Phi$ is injective.I think for showing surjective we have to take the sigma algebra generated by that partition for the preimage.Any hints/ideas to do this formally?
 A: Assume $P=\{A_1,\dots, A_n\}$ is a partition. Consider then the Boolean algebra of the sets $A_1,\dots, A_n$. It looks like 
$$ D:=\sigma(A_1,\dots, A_n) = \{ \cup_{j\in J} A_j \mid J \subseteq \{1,\dots n\} \} $$
(see $σ$-algebra generated by countable disjoint sets).
It is then easy to see that $\phi(D)  \supseteq P $.  
Further note that for any $J  \subseteq \{1,\dots n\}$ you have $$ (\cup_{j\in J} A_j )^C = \cup_{j\in J^C} A_j  .$$ Therefore, an element $A \in \phi(D)$ is of the form $$ A = \cup_{i\in I} A_i$$ for some set $I \subseteq \{1,\dots,n\}$. Further for each $i=1,\dots,n$ you have either
$$ A = A \cap A_i  \quad or \quad A = A \cap A_i^c $$ because $A_i \in D$. This shows that $|I|>1$ is not possible, because then you would find $i,j\in I$ with $i\neq j$ and $$ A \cap A_i = A = A \cap A_j$$ which is not possible because of $A_i \cap A_j = \emptyset $. This shows $\phi(D)  \subseteq P $.

By the way: In your definition of $\phi$ you need to add that $\sigma_\epsilon \neq \emptyset$.
