GATE CS 2023 Q39 — Sets, Relations and Functions

To determine the properties of the mapping $F: \mathcal{E} \rightarrow B$ defined by $F([x]) = f(x)$, let us analyze each property systematically:

1. Well-definedness of $F$:

   For $F$ to be well-defined, the output must not depend on the choice of the representative of the equivalence class. Let $[a_1] = [a_2]$ for some $a_1, a_2 \in A$.

   By definition of equivalence classes, $[a_1] = [a_2] \implies a_1 \sim a_2 \implies f(a_1) = f(a_2)$.

   Thus, $F([a_1]) = f(a_1) = f(a_2) = F([a_2])$. This proves that $F$ is well-defined. Therefore, statement A is FALSE.

2. Injectivity (One-to-One) of $F$:

   To check if $F$ is injective, let $F([a_1]) = F([a_2])$ for some $[a_1], [a_2] \in \mathcal{E}$.

   By definition of $F$, this means $f(a_1) = f(a_2)$.

   By definition of the equivalence relation $\sim$, $f(a_1) = f(a_2) \implies a_1 \sim a_2 \implies [a_1] = [a_2]$.

   Since $F([a_1]) = F([a_2]) \implies [a_1] = [a_2]$, $F$ is injective. Therefore, statement C is TRUE.

3. Surjectivity (Onto) of $F$:

   To check if $F$ is surjective, let $b \in B$. Since the original function $f: A \rightarrow B$ is onto, there exists some element $a \in A$ such that $f(a) = b$.

   For this element $a$, we have the equivalence class $[a] \in \mathcal{E}$. Applying $F$, we get:

   $$F([a]) = f(a) = b$$

   Since every element $b \in B$ has a pre-image $[a] \in \mathcal{E}$, $F$ is surjective. Therefore, statement B is TRUE.

4. Bijectivity of $F$:

   Since $F$ is both injective (one-to-one) and surjective (onto), it is a bijective function. Therefore, statement D is TRUE.