Image of a partition of a set through a bijective function is also a partition

Question

Let $A$ be a set and $\mathcal{P}$ a partition of $A$ (a partition of a set $A$ is a set of non-empty subsets of $A$ such that every element $x \in A$ is in exactly one of these subsets), and $f:A \to A$ a bijective function. Then, the set $B = \{ f(P) : P \in \mathcal{P} \}$ is also a partition of $A$. Is this true or am I wrong?

Intuitively, I think is true because if $f$ is bijective, then each element of $A$ has a unique correspondent in $A.$ Then, each subset $B⊂A$ has a unique correspondent subset in $A.$ Then, if two or more disjoint subsets of $A$ have unique correspondents in $A,$ because subsets are disjoint, and $f$ is bijective, the correspondents are disjoint. So if we have a partition, image of partition through a bijective function is also a partition. Is this ok?

@AnneBauval, intuitively, I think is true because if $f$ is bijective, then each element of $A$ has a unique correspondent in $A$. Then, each subset $B \subset A$ has a unique correspondent subset in $A$. Then, if two or more disjoint subsets of A have unique correspondents in A, because subsets are disjoint, and $f$ is bijective, the correspondents are disjoint. So if we have a partition, image of partition through a bijective function is also a partition. Is this ok? — MathLearner, Mar 11 '23 at 14:00
Yes it is ok intuitively (next time, include such thoughts in your post). To formalize it, denote by $f(B)={f(x)\mid x\in B}$ the "correspondent" of $B$ (unique by definition), and prove that $\cup_{i\in I}f(B_i)=f(\cup_{i\in I}B_i)$ and (using the bijectivity of $f$) that $f(B)\cap f(C)=f(B\cap C).$ — Anne Bauval, Mar 11 '23 at 14:11
@AnneBauval, thanks! So, from this will result that if I have a function $g : \mathcal{P} \to B$ (where $ \mathcal{P}$ and $B$ are the partitions from my question), $g(X) = f(X), X\in\mathcal{P}$, then $g$ is a bijection? Is a true result? Or am I wrong? — MathLearner, Mar 11 '23 at 14:21
This is a new question (should be another post) but let's go: $g$ is surjective by definition of $B.$ It is injective, and even $G:P(A)\to P(A),X\mapsto f(X)$ is bijective: $G^{-1}(Y)=f^{-1}(Y).$ See Showing if $f$ is an injection then $A=f^{-1}(f(A))$ and if $f$ is onto $Y$ then $B=f(f^{-1}(B))$. — Anne Bauval, Mar 11 '23 at 14:29
Oh. Ok. Now I understand. So the function $g$ is really bijective. Thanks you very much for all help! — MathLearner, Mar 11 '23 at 14:41

Image of a partition of a set through a bijective function is also a partition

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