6.Let f: A-> B.Let f" be the inverse relation ,then show that

Question

a) f is afunction from Range(f) in to A if and only if fis 1-1
b) f*is afunction from B in to A if and only if fis 1-1 and onto

Answer 1

To show these statements, we start by clarifying some definitions and concepts.

Definitions:

1. Function: A relation \( f: A \to B \) is called a function if every element in \( A \) is related to exactly one element in \( B \).

2. Inverse Relation: The inverse relation \( f' \) (often denoted \( f^{-1} \)) of a relation \( f \) is defined such that if \( (a,b) \in f \), then \( (b,a) \in f' \).

3. One-to-One Function (Injective): A function \( f \) is one-to-one (1-1 or injective) if for all \( x_1, x_2 \in A\), if \( f(x_1) = f(x_2) \), then \( x_1 = x_2 \).

4. Onto Function (Surjective): A function \( f \) is onto (surjective) if the range of \( f \) is equal to \( B \), meaning every element in \( B \) is the image of at least one element in \( A \).

Now, we will prove the statements:

Part (a):

Statement: \( f \) is a function from \( \text{Range}(f) \) into \( A \) if and only if \( f \) is 1-1.

Proof:

1. (⇒) Assume \( f \) is a function from \( \text{Range}(f) \) into \( A \):

   • Let \( f(x_1) = f(x_2) \) for some \( x_1, x_2 \in A \).
   • Since \( f \) is a function from \( \text{Range}(f) \) into \( A \), for every output in the range, there is a unique corresponding input.
   • This means \( x_1 \) must equal \( x_2 \). Therefore, \( f \) is injective (1-1).

2. (⇐) Assume \( f \) is 1-1:

   • For any \( b \in \text{Range}(f) \), there exists \( a \in A \) such that \( f(a) = b \).
   • Since \( f \) is 1-1, there cannot be two different inputs \( a_1 \) and \( a_2 \in A \) that map to the same \( b \). Thus for each \( b \in \text{Range}(f) \), there is a unique \( a \), making \( f \) a function from \( \text{Range}(f) \) into \( A \).

Thus, we have shown that \( f \) is a function from \( \text{Range}(f) \) into \( A \) if and only if \( f \) is 1-1.

Part (b):

Statement: \( f' \) is a function from \( B \) into \( A \) if and only if \( f \) is 1-1 and onto.

Proof:

1. (⇒) Assume \( f' \) is a function from \( B \) into \( A \):

   • For every \( b \in B \), there is a unique \( a \in A \) such that \( f(a) = b \).
   • Since there must be a unique \( a \) for each \( b \), \( f \) must be 1-1; if \( f(a_1) = f(a_2) = b \), then \( a_1 \) must equal \( a_2 \).
   • Moreover, since \( f' \) is defined for every \( b \in B \), it means every element in \( B \) must be the image of some \( a \in A \) (i.e., \( f \) is onto).

2. (⇐) Assume \( f \) is 1-1 and onto:

   • Since \( f \) is onto, for every \( b \in B \), there exists at least one \( a \in A \) such that \( f(a) = b \).
   • Since \( f \) is 1-1, this mapping from \( b \) back to \( a \) must be unique for each \( b \).
   • Therefore, we can define the inverse relation \( f' \) where for every \( b \in B \), there is a unique associated \( a \in A \), meaning \( f' \) is a function from \( B \) into \( A \).

Thus, we have shown that \( f' \) is a function from \( B \) into \( A \) if and only if \( f \) is 1-1 and onto.

Overall, both parts of the statement are proven successfully.