How to prove image and preimage properties (ZFC) How to prove these properties (under ZFC) about images and preimages of a mapping $f : X\to Y$ :
$$ f\Big[\bigcup_{i\in I} A_i\Big] = \bigcup_{i\in I} f[A_i] $$
$$ f^{-1}\Big[\bigcup_{i\in I} B_i\Big] = \bigcup_{i\in I} f^{-1}[B_i] $$
$$ f^{-1}\Big[\bigcap_{i\in I} B_i\Big] = \bigcap_{i\in I} f^{-1}[B_i] $$
with the families $(A_i)_{i\in I} \in \mathfrak P(X)^I$ and $(B_i)_{i\in I} \in \mathfrak P(Y)^I$. I don't know how to proceed, to prove rigorously these properties (i.e. with predicate logic, quantification, set theory axioms...), because I've seen some unclear proof...
 A: A typical strategy to show that two sets $A$ and $B$ are equal is double inclusion: $A \subset B$ and $B \subset A$. Let us illustrate it on the first example.


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*Let $y$ belong to $f(\cup A_i)$. Thus, there is some $x$ in $\cup A_i$ such that $y=f(x)$. Also, there is some $i_0$ in $I$ such that $x$ belongs to $A_{i_0}$. Therefore, $y$ belongs to $f(A_{i_0})$, which is a subset of $\cup f(A_{i})$. This is true for all $y$ in $f(\cup A_i)$. Hence, $$f\Big(\bigcup_{i\in I} A_i\Big)\subset \bigcup_{i\in I} f(A_{i})\, .$$

*Inversely, let $y$ belong to $\cup f(A_i)$. There is some $i_0$ in $I$ such that $y \in f(A_{i_0})$. Thus, we can find $x$ in $A_{i_0}$ which satisfies $y=f(x)$. Since $A_{i_0}\subset \cup A_{i}$, $x$ belongs also to $\cup A_i$, and thus, $y$ belongs to $f(\cup A_i)$. This is true for all $y$ in $\cup f(A_i)$. Hence, $$ \bigcup_{i\in I} f(A_{i}) \subset f\Big(\bigcup_{i\in I} A_i\Big)\, .$$
Finally, we have shown $$f\Big(\bigcup_{i\in I} A_i\Big) = \bigcup_{i\in I} f(A_{i}) \, .$$ The other identities can be obtained in a similar manner.
