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GATE CSE 2023 Q49: Map F([x]) = f(x) on equivalence classes of an onto f

Question

Let f:A→Bf : A \to B be an onto (or surjective) function, where AA and BB are nonempty sets. Define an equivalence relation ∼\sim on the set AA as

a1∼a2 if f(a1)=f(a2),a_1 \sim a_2 \text{ if } f(a_1) = f(a_2),

where a1,a2∈Aa_1, a_2 \in A. Let E={[x]:x∈A}\mathcal{E} = \{[x] : x \in A\} be the set of all the equivalence classes under ∼\sim. Define a new mapping F:E→BF : \mathcal{E} \to B as

F([x])=f(x), for all the equivalence classes [x] in E.F([x]) = f(x), \text{ for all the equivalence classes } [x] \text{ in } \mathcal{E}.

Which of the following statements is/are TRUE?

Try it: multiple select: one or more correct options

Select one or more options
Select all correct options. No negative marking, no partial marks.
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Answer: B, C, D (official GATE 2023 final answer key)

Concept used: The kernel of a function: ff induces a bijection from A/∼A/{\sim} onto its image.

First step: Check well-definedness: if [x]=[y][x] = [y] then f(x)=f(y)f(x) = f(y) by the definition of ∼\sim.

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Source: official GATE 2023 CS question paper and final answer key, organised by IIT Kanpur. Question paper · Answer key

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