Suppose \(X\) is a topological space . The identity map \(\id\colon X\to X\) is continuous function .
Proof
Let \(U\subseteq X\) be open. Then \(\id^{-1}(U)=U\) is again open, and therefore \(\id\) is continuous.
\[
\DeclareMathOperator{\id}{id}
\]