Question

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

• A. $F$ is NOT well-defined.
• B. $F$ is an onto (surjective) function.
• C. $F$ is a one-to-one (injective) function.
• D. $F$ is a bijective function.

Hint:

Quotienting $A$ by the relation $a_1\sim a_2 \Leftrightarrow f(a_1)=f(a_2)$ collapses each fiber of $f$ into a single class. The induced map $F$ sends each class (fiber) to its common value in $B$, giving a natural bijection $\mathcal{E}\cong B$ whenever $f$ is surjective.

Answer 1

The Correct Option is B

Step 1: Well-definedness.
Suppose $[x]=[y]$ in $\mathcal{E}$, i.e., $x\sim y \Rightarrow f(x)=f(y)$.
Then $F([x])=f(x)=f(y)=F([y])$. Hence $F$ does not depend on the representative and is well-defined.
Therefore (A) is False.
Step 2: Surjectivity of $F$.
Given $f$ is surjective: for every $b\in B$, there exists $a\in A$ with $f(a)=b$.
Let $[a]\in\mathcal{E}$. Then $F([a])=f(a)=b$. Thus every $b$ has a preimage under $F$.
Hence $F$ is surjective \Rightarrow (B) is True.
Step 3: Injectivity of $F$.
Assume $F([x])=F([y])$. Then $f(x)=f(y)$, which means $x\sim y$ by definition.
Therefore $[x]=[y]$ in $\mathcal{E}$, proving $F$ is injective. Thus (C) is True.
Step 4: Bijectivity.
Since $F$ is both injective and surjective, it is bijective.
Therefore (D) is True.
\[ \boxed{\text{True statements: (B), (C), and (D).}} \]