Consider the set of all functions $f\colon \{0, 1, \dots, 2014\} \to \{0, 1, \dots, 2014\}$ such that $f(f(i)) = i$, for all $0 \le i \le 2014$. Consider the following statements: $P.$ For each such function it must be the case that for every $i$, $f(i) = i$. $Q.$ For each such function it must be the case that for some $i$, $f(i) = i$. $R.$ Each such function must be onto. Which one of the following is CORRECT?

Let $S = \{0, 1, \dots, 2014\}$. The number of elements in $S$ is $|S| = 2015$, which is odd.

1. Statement $R$: Since $f(f(i)) = i$, $f$ is its own inverse ($f = f^{-1}$). Thus, $f$ is a bijection, which means $f$ is both one-to-one (injective) and onto (surjective). Therefore, statement $R$ is true.

2. Statement $Q$: Since $f$ is an involution on a finite set, its cycle decomposition consists only of fixed points (1-cycles) and transpositions (2-cycles). Let $k$ be the number of 2-cycles and $m$ be the number of fixed points (elements where $f(i) = i$). Then:

   $$2k + m = 2015$$

   Since $2k$ is even and $2015$ is odd, $m$ must be an odd integer, which means $m \ge 1$. Hence, there must exist at least one $i$ such that $f(i) = i$. Therefore, statement $Q$ is true.

3. Statement $P$: $P$ claims that all elements must be fixed points ($m = 2015$). We can construct a valid function $f$ with $f(0) = 1$, $f(1) = 0$, and $f(i) = i$ for all $i \ge 2$. In this case, $f(f(i)) = i$ holds for all $i$, but $f(0)

e 0$. Thus, statement $P$ is false.

Hence, Only $Q$ and $R$ are true, which corresponds to option (B).