) (i) Let f be the function from {a, b, c} to {1, 2, 3} such that f(a) = 2, f(b) = 3 and f(c) = 1. Is f invertible, and if it is, what is it’s inverse?

Suppose A and B are nonempty sets and f : A → B is a function.

If a function is one-to-one and onto, then it is invertible.

Here , A={a,b,c} and B={1,2,3}.

Given that -> f(a) = 2 , f(b) = 3 , f(c) = 1

Now here, f maps every element of A to a unique element of B, so f is one-to-one.

Also every element in B has a pre-image in A , so f is onto.

Thus, by definition f is invertible.

A function f-1 : B → A is called an inverse function for f if it satisfies the following condition:

For every x ∈ A and y ∈ B, f(x) = y if and only if f-1(y) = x.

So, here $f(a)=2\implies f$-1$(2)=a \\$

$f(b)=3 \implies f$-1(3) = b

$f(c)=1 \implies f$-1(1)$\;=c$