Example 5 Let $A=\{1,2, \cdots, 2012\}$,
$$
B=\{1,2, \cdots, 19\},
$$

and let $S$ be the set of all subsets of $A$. Find the number of mappings $f: S \rightarrow B$ such that
$$
f\left(A_{1} \cap A_{2}\right)=\min \left\{f\left(A_{1}\right), f\left(A_{2}\right)\right\}
$$

for any $A_{1}, A_{2} \in S$. ${ }^{[2]}$
【Analysis】According to the conditions, equation (1) can be extended to any number of sets, that is,
$$
\begin{array}{l}
f\left(A_{1} \cap A_{2} \cap \cdots \cap A_{k}\right) \\
\quad=\min \left\{f\left(A_{1}\right), f\left(A_{2}\right), \cdots, f\left(A_{k}\right)\right\} .
\end{array}
$$

Since sets with fewer elements can be written as the intersection of sets with more elements, starting from the extreme case, once the correspondence for sets with more elements is determined, the correspondence for sets with fewer elements is naturally determined.
Solution The number of functions $f$ that satisfy the conditions is
$$
1^{2012}+2^{2012}+\cdots+19^{2012} \text {. }
$$

For any finite integer sets $U, V$, $f(U \cap V)$ is the minimum of $f(U)$ and $f(V)$.
Let $f(A)=n$. Then for all $X \in S$,
$$
f(X) \leqslant n \text {. }
$$

If $X \in S$ and $|X|=2011$, then there are 2012 such $X$. Thus, there are $n^{2012}$ different ways to determine the value of $f(X)$.

Next, we prove that the values of $f$ on other sets are uniquely determined.

First, we use mathematical induction on $k$ to prove that for all $a_{1}, a_{2}, \cdots, a_{k} \in A$,
$$
\begin{array}{l}
f\left(A \backslash\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}\right) \\
=\min \left\{f\left(A \backslash\left\{a_{1}\right\}\right), \cdots, f\left(A \backslash\left\{a_{k}\right\}\right)\right\} .
\end{array}
$$

When $k=1$, equation (2) is trivial.
When $k=2$,
$$
\begin{array}{l}
f\left(A \backslash\left\{a_{1}, a_{2}\right\}\right)=f\left(\left(A \backslash\left\{a_{1}\right\}\right) \cap\left(A \backslash\left\{a_{2}\right\}\right)\right) \\
=\min \left\{f\left(A \backslash\left\{a_{1}\right\}\right), f\left(A \backslash\left\{a_{2}\right\}\right)\right\} .
\end{array}
$$

Assume that the conclusion holds for $k \geqslant 2$.
By the induction hypothesis,
$$
\begin{array}{l}
f\left(A \backslash\left\{a_{1}, a_{2}, \cdots, a_{k+1}\right\}\right) \\
=f\left(\left(A \backslash\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}\right) \cap\left(A \backslash\left\{a_{k+1}\right\}\right)\right) \\
=\min \left\{f\left(A \backslash\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}\right), f\left(A \backslash\left\{a_{k+1}\right\}\right)\right\} \\
=\min \left\{f\left(A \backslash\left\{a_{1}\right\}\right), f\left(A \backslash\left\{a_{2}\right\}\right), \cdots, f\left(A \backslash\left\{a_{k+1}\right\}\right)\right\},
\end{array}
$$

so the conclusion also holds for $k+1$.
If equation (2) holds, then for all $A_{1}, A_{2} \in S$,
$$
f\left(A_{1} \cap A_{2}\right)=\min \left\{f\left(A_{1}\right), f\left(A_{2}\right)\right\} .
$$

Since $1 \leqslant n \leqslant 19$, the number of functions that satisfy the conditions is
$$
1^{2012}+2^{2012}+\cdots+19^{2012} \text {. }
$$