A given onto function f : A $\rightarrow$ B, where |A| $\neq$ 0 and |B| $\neq$ 0
An equivalence relation $\sim$ is defined on set A as :
$a_{1} \sim a_{2}$ if $f(a_{1}) = f(a_{2})$, where $a_{1}$, $a_{2}$ $\in$ A.
Let's try to understand this question by using an example.
Let, A = {1, 2, 3, 4, - - - - - - , 18, 19, 20} & B = {0, 1, 2, 3}
And, $\sim$ be Division Modulo 4.
NOTE : Division Modulo 4 is an Equivalence Relation.
If we try to observe Division Modulo 4 with respect to set A then, we will get such type of function mapping :
From above mapping :
$[1]_{R}$ = {1, 5, 9, 13, 17} = $[5]_{R} = [9]_{R} = [13]_{R} = [17]_{R}$
$[2]_{R}$ = {2, 6, 10, 14, 18} = $[6]_{R} = [10]_{R} = [14]_{R} = [18]_{R}$
$[3]_{R}$ = {3, 7, 11, 15, 19} = $[7]_{R} = [11]_{R} = [15]_{R} = [19]_{R}$
$[4]_{R}$ = {4, 8, 12, 16, 20} = $[5]_{R} = [9]_{R} = [13]_{R} = [17]_{R}$
IMPORTANT CONCLUSION : Here, f is mapping elements from A to the remainders that we will get
when divided by 4.
Inshort we can say elements are mapped to their remainders.
Now, $\varepsilon$ be the set of all Equivalence class under $\sim$. So,
$\varepsilon$ = { {1, 5, 9, 13, 17}, {2, 6, 10, 14, 18}, {3, 7, 11, 15, 19}, {4, 8, 12, 16, 20} }.
A new mapping F : $\varepsilon$ $\rightarrow$ B is defined as :
F([x]) = f(x), for all equivalence classes [x] in $\varepsilon$.
Let's try to observe the mapping of F, we will get this type of function mapping :
IMPORTANT COLNCLUSION : Here, Set of all equivalence class are mapped to their
remainders in B.
Let's try to observe the options one by one.
Option A : F is NOT well-defined.
This is completely FALSE statement. F is well-defined.
Option B : F is an onto (or surjective) function.
This is TRUE statement as all elemets of B are covered. So, F is onto function.
Option C : F is a one-to-one (or injective) function.
This is TRUE statement as we can see every element in $\varepsilon$ is mapped to only one element in B.
So, F is one-to-one function.
Option D : F is a bijective function.
This is TRUE statement a F is both one-to-one & onto. Hence F is an Bijective Function.
Correct Answer : B, C, D.