To show that:

$f^{-1}(S \cap T) = f^{-1}(S) \cap f^{-1}(T),$

we will prove this set equality by showing that both sides are subsets of each other.

Proof:

Step 1: Show that $f^{-1}(S \cap T) \subseteq f^{-1}(S) \cap f^{-1}(T)$.

Let $x \in f^{-1}(S \cap T)$. By the definition of the inverse image, this means that:

$f(x) \in S \cap T.$

By the definition of the intersection of sets, this implies that:

$f(x) \in S \quad \text{and} \quad f(x) \in T.$

Therefore, $x \in f^{-1}(S)$ and $x \in f^{-1}(T)$, which means:

$x \in f^{-1}(S) \cap f^{-1}(T).$

Thus, we have shown that:

$f^{-1}(S \cap T) \subseteq f^{-1}(S) \cap f^{-1}(T).$

Step 2: Show that $f^{-1}(S) \cap f^{-1}(T) \subseteq f^{-1}(S \cap T)$.

Let $x \in f^{-1}(S) \cap f^{-1}(T)$. This means that:

$x \in f^{-1}(S) \quad \text{and} \quad x \in f^{-1}(T),$

which implies that:

$f(x) \in S \quad \text{and} \quad f(x) \in T.$

By the definition of the intersection, this means that:

$f(x) \in S \cap T,$

so $x \in f^{-1}(S \cap T)$.

Thus, we have shown that:

$f^{-1}(S) \cap f^{-1}(T) \subseteq f^{-1}(S \cap T).$

Conclusion:

Since we have shown both $f^{-1}(S \cap T) \subseteq f^{-1}(S) \cap f^{-1}(T)$ and $f^{-1}(S) \cap f^{-1}(T) \subseteq f^{-1}(S \cap T)$, it follows that:

$f^{-1}(S \cap T) = f^{-1}(S) \cap f^{-1}(T).$

This completes the proof.

Would you like further details on any part of the explanation or additional questions to explore? Here are some related questions:

1. What is the definition of a function's inverse image?
2. How does the inverse image behave with respect to union of sets?
3. Can we apply the same concept to more than two sets?
4. How would the result change if $f$ were not a function, but a relation?
5. How does the behavior of the inverse image compare to that of the image under $f$?

Tip: When dealing with inverse images, remember that the preimage of an intersection always equals the intersection of the preimages, but this doesn't necessarily hold for images of intersections.