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I have the following:

$f: A \rightarrow B \\ X, X' \subseteq A \\ Y,Y' \subseteq B$

I have to prove that:

$$f^{-1}(Y \cap Y') = f^{-1}(Y) \cap f^{-1}(Y')$$

What I have done is the following.
We have to prove actually two propositions:
$\Rightarrow ) \quad f^{-1}(Y \cap Y') \subseteq f^{-1}(Y) \cap f^{-1}(Y') \\ \Leftarrow ) \quad f^{-1}(Y) \cap f^{-1}(Y') \subseteq f^{-1}(Y \cap Y') $

I have tried to prove the first $(\Rightarrow)$:

what we have to demonstrate is
$\forall z \in f^{-1}(Y \cap Y') \Rightarrow z \in f^{-1}(Y) \cap f^{-1}(Y')$
hence, I consider an element $z$
$z \in f^{-1}(Y \cap Y') \Rightarrow z = f^{-1}(y) \mbox{ with } y \in Y \cap Y' \mbox{ i.e. } y \in Y \mbox{ and } y \in Y'$
since $z = f^{-1}(y) \mbox{ with } y \in Y$ it means that $z = f^{-1}(y) \in f^{-1}(Y)$
and in simiar way,
since $z = f^{-1}(y) \mbox{ with } y \in Y'$ it means that $z = f^{-1}(y) \in f^{-1}(Y')$
therefore
$z \in f^{-1}(Y) \cap f^{-1}(Y')$

I think that is correct,
but, I have some problems in proving the converse.

Please, can you give me any suggestions? Many thanks!

2 Answers 2

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Let $$z\in f^{-1}(Y)\cap f^{-1}(Y')\Rightarrow z\in f^{-1}(Y)\text{ and }z\in f^{-1}(Y')$$ that is, by definition, $f(z)\in Y$ and $f(z)\in Y'$, thus $$f(z)\in Y\cap Y'\Rightarrow z\in f^{-1}(Y\cap Y')$$

Notice that with this approach, you can reverse this argument to show the other implication.

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What you have done so far is correct.

In case of second implication,

Let $z \in f^{-1}(Y) \cap f^{-1}(Y')$. This implies that $z$ is in inverse image of both $Y$ and $Y'$. Hence $f(z) \in Y \cap Y' \Rightarrow z \in f^{-1}(Y \cap Y').$