Inverse of Identity Mapping

Theorem

Let $S$ be a set.

Let $I_S: S \to S$ be the identity mapping on $S$.

Then the inverse of $I_S$ is itself:

$\paren {I_S}^{-1} = I_S$

Proof

From the nature of the identity mapping, we have:

$I_S \circ I_S = I_S$

from which it follows by definition that $I_S$ is the inverse of itself.

$\blacksquare$

Sources

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 3.5$. Identity mappings
• 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Composition of Functions
• 1996: H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability ... (previous) ... (next): Appendix $\text{A}.7$: Inverses: Proposition $\text{A}.7.5 \ (1)$