Question

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

Answer 1

Answer on Question #75633 – Math – Discrete Mathematics

Consider the function $f$.

Let's denote the set $\{a, b, c\}$ as set $A$ and the set $\{1, 2, 3\}$ as set $B$. Then function $f$ is a mapping $f \colon A \to B$, $f(a) = 2$, $f(b) = 3$, $f(c) = 1$.

a) $f$ is an injective function ($f$ is an "into" function) because for $\forall m, n \in A: m \neq n \text{ } f(m) \neq f(n)$.

b) $f$ is a surjective mapping ($f$ is an "onto" function), because $f(A) = B$.

c) $f$ is an injective mapping and $f$ is an surjective mapping, therefore $f$ is a bijective mapping. Then there is an inverse mapping $f^{-1} \colon B \to A$, which can be defined as follows:

$$\forall y \in B \exists ! x \in A: f (x) = y. \text{ Then } x = f ^ {- 1} (y).$$

As a result, $a = f^{-1}(2), b = f^{-1}(3), c = f^{-1}(1)$.

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