Prove a function is bijective Given any set _A_ , the _identity function on A_ is $I_A: A \rightarrow A$ defined by $I_A(x) = x$ for any $x ∈ A$. Show that the identity function is a bijection. **Attempt:** Suppose $\,f: A \rightarrow A$, then $f$ is **bijective** if it is both _injective_ and _surjective_. _Injective:_ Pick $x,y \in A$ such that $f(x) = f(x)$. So $I_A(x) = I_A(y)$, that is, $x = y$. Thus $I_A$ is injective by its definition. _Subjective:_ Next pick $y \in A$ such that $f(y) = x$. So $I_A(y) = x$. So, $A \subseteq I_A$, that is, A is in the range of $I_A$ and $y \in A$. Thus $I_A$ is subjectively by its definition that any element in the function must be in the range of the function. $$

**Hint** :

* The function is injective if $I_A(x)=I_A(y)\Rightarrow x=y$
* The function is surjective if you can find some $y$ such that $I_A (y)=x$