prove $f^{-1}(C \cap D) = f^{-1}(C) \cap f^{-1}(D)$ I need help with this proof:
$f: X\rightarrow Y$
$C,D\subseteq Y$
$f^{-1}(C \cap D) = f^{-1}(C) \cap f^{-1}(D)$
Thanks.
 A: To show set equality,$$f^{-1}(C \cap D) = f^{-1}(C) \cap f^{-1}(D)$$ we typically proceed by showing that both the following inclusions hold:
$$f^{-1}(C \cap D) \quad \subseteq \quad f^{-1}(C) \cap f^{-1}(D)\tag{1}$$
$$f^{-1}(C) \cap f^{-1}(D) \quad \subseteq \quad f^{-1}(C \cap D)\tag{2}$$
For each of $(1), (2),$ we can use "element chasing": In general, to show $A \subseteq B$, it suffices to show that $a \in A \implies a \in B$.
For $(2)$, to show $$f^{-1}(C) \cap f^{-1}(D) \subseteq f^{-1}(C \cap D)\tag{2}$$
We start by assuming $x \in f^{-1}(C) \cap f^{-1}(D)$. Then, by the definition of set intersection, we have that $x \in f^{-1}(C)$ AND $x \in f^{-1}(D)$. This means $x \in C$ AND $x \in D$, which means, by definition, that  $x \in C\cap D$. Now, $x\in C\cap D$ implies that $x \in f^{-1}(C\cap D).$ Hence, we've shown that $$f^{-1}(C) \cap f^{-1}(D) \subseteq f^{-1}(C \cap D)\tag{2}$$
Use the very same strategy for approaching the first inclusion $(1)$. When you show that inclusion $(1)$ also holds, then you can assert what you set out to prove: $$f^{-1}(C \cap D) = f^{-1}(C) \cap f^{-1}(D)$$
A: To solve these kinds of questions, you want to show that $$f^{-1}(C\cap D) \subseteq f^{-1}(C) \cap f^{-1}(D)$$ and $$f^{-1}(C\cap D) \supseteq f^{-1}(C) \cap f^{-1}(D).$$
I will show $\subseteq$ here; you should try the other direction.
Suppose $x \in f^{-1} (C \cap D)$. This means $f(x) \in (C\cap D)$, which further implies $f(x) \in C$ AND $f(x) \in D$. Thus, $x \in f^{-1}(C)$ AND $x \in f^{-1}(D)$.
A: Just pick an element in $f^{-1}( C \cap D) $ and verify the first inclusion. The same for the second
Maybe this definition will help
$ f^{-1}( C) := \{ x \in X \text { such that } f (x) \in C  \} $
Just apply it
