Mumbai University > Computer Engineering > Sem 3 > Discrete Structures

Marks: 4 Marks

Year: Dec 2013

f is one to one and onto hence $$a_1, a_2 € A \\ f(a_1)=f(a_2)-\gt a_1=a_2 \\ b € B a € A$$

s.t. b =f(a)

To prove that $f^{-1}$ is one one onto.

Let $f (a_1) = b_1 \ \ \ and \ \ \ f (a_2) = b_2$

$$f^{-1}(b_1) = a_1 and f^{-1}(b_2) = a_2$$

Now $f^{-1}(b_1) = f^{-1}(b_2) \\ a_1= a_2 \\ f (a_1)=f( a_2) \text{[as f is well defined]} \\ b_1= b_2$

$f^{-1}$ is one to one function.

Again b € B a € A, s.t f (a) = b

$$f^{-1} (b) = a$$

Hence a $€ A b € B$

$s.t f^{-1} (b) =a$

Hence f^(-1) is onto mapping.

Therefore $f^{-1}$ is one to one onto mapping from B to A.