Let $f : A \rightarrow B$ be an onto (or surjective) function, where $A$ and $B$ are nonempty sets. Define an equivalence relation $\sim$ on the set $A$ as $$a_1 \sim a_2 \text{ if } f(a_1) = f(a_2) ,$$ where $a_1, a_2 \in A$. Let $\mathcal{E} = \{[x] : x \in A\}$ be the set of all the equivalence classes under $\sim$. Define a new mapping $F : \mathcal{E} \rightarrow B$ as $$F([x]) = f(x), \quad \text{for all the equivalence classes } [x] \text{ in } \mathcal{E}.$$ Which of the following statements is/are TRUE?

To analyze the mapping $F : \mathcal{E} \to B$ defined by $F([x]) = f(x)$:

1. Well-definedness: Let $[x] = [y]$ in $\mathcal{E}$. By definition of equivalence classes, $x \sim y \implies f(x) = f(y)$. Therefore, $F([x]) = F([y])$, so $F$ is well-defined. Thus, option (A) is false.

2. Injectivity (One-to-one): Suppose $F([x]) = F([y])$ for $[x], [y] \in \mathcal{E}$. Then $f(x) = f(y)$, which implies $x \sim y$, and hence $[x] = [y]$. Thus, $F$ is injective (one-to-one). Statement (C) is true.

3. Surjectivity (Onto): Since $f: A \to B$ is surjective, for every $b \in B$, there exists some $a \in A$ such that $f(a) = b$. Then $[a] \in \mathcal{E}$ satisfies $F([a]) = f(a) = b$. Therefore, $F$ is surjective (onto). Statement (B) is true.

4. Bijectivity: Since $F$ is both injective and surjective, $F$ is a bijective function. Statement (D) is true.

Hence, the correct statements are B, C, and D.