The Identity Function

# The Identity Function

Definition: Let $A$ be any set. The Identity Function on $A$ is the function $i : A \to A$ defined for all $x \in A$ by $i(x) = x$. |

*An alternative notation for the identity function on $A$ is "$id_A$".*

If $A = \mathbb{R}$, then the identity function $i : \mathbb{R} \to \mathbb{R}$ is the function defined for all $x \in \mathbb{R}$ by $i(x) = x$. The graph of $i$ is given below:

If we instead consider a finite set, say $B = \{ 1, 2, 3, 4, 5 \}$ then the identity function $i : B \to B$ is the function given by $i(1) = 1$, $i(2) = 2$, $i(3) = 3$, $i(4) = 4$, and $i(5) = 5$.

We will now prove some rather trivial observations regarding the identity function.

Theorem 1: Let $A$ be any set and let $i : A \to A$ be the identity function on $A$. Then $i$ is bijective. |

**Proof:**To show that $i$ is bijective we must show that $i$ is both injective and surjective.

- Suppose that $i(x) = i(y)$. Then $x = y$, so $i$ is injective.

- Now let $y \in A$. Then if we let $x = y$ we have that $i(x) = i(y) = y$. Therefore, for all $y \in A$ there exists an $x = y \in A$ such that $i(x) = y$, and so $i$ is surjective.

- Since $i$ is both injective and surjective we have that by definition that $i$ is bijective. $\blacksquare$

Theorem 2: Let $A$ be any set and let $i : A \to A$ be the identity function on $A$. Then for any function $f : A \to A$, we have that $f \circ i = f$ and $i \circ f = f$. |

**Proof:**Consider the function $f \circ i : A \to A$. For all $x \in A$ we have that:

\begin{align} \quad (f \circ i)(x) = f(i(x)) = f(x) \end{align}

- Therefore $f \circ i = f$. Now consider the function $i \circ f : A \to A$. For all $x \in A$ we have that:

\begin{align} \quad (i \circ f)(x) = i(f(x)) = f(x) \end{align}

- Therefore $i \circ f = f$. $\blacksquare$

Theorem 3: Let $A$ be any set and let $i : A \to A$ be the identity function on $A$. Then $i^{-1} : A \to A$ exists and $i^{-1} = i$. |

**Proof:**From Theorem 1 we know that $i$ is bijective and so $i^{-1}$ exists. We must have that $i^{-1}(i(x)) = x$ and $i(i^{-1}(x)) = x$. From the second equation we see that $i^{-1}(x) = x$ which also satisfies the first equation.

- Therefore $i^{-1} = i$. $\blacksquare$.