Renaming Mapping is Well-Defined

Theorem
Let $f: S \to T$ be a mapping.

Let $r: S / \mathcal R_f \to \operatorname{Im} \left({f}\right)$ be the renaming mapping, defined as:


 * $r: S / \mathcal R_f \to \operatorname{Im} \left({f}\right): r \left({\left[\!\left[{x}\right]\!\right]_{\mathcal R_f}}\right) = f \left({x}\right)$

where:
 * $\mathcal R_f$ is the equivalence induced by the mapping $f$
 * $S / \mathcal R_f$ is the quotient set of $S$ determined by $\mathcal R_f$
 * $\left[\!\left[{x}\right]\!\right]_{\mathcal R_f}$ is the equivalence class of $x$ under $\mathcal R_f$.

The renaming mapping is always well-defined.

Proof
We have that $\mathcal R_f$ is an equivalence.

To determine whether $r$ is well-defined, we have to determine whether $r: S / \mathcal R_f \to \operatorname{Im} \left({f}\right)$ actually defines a mapping at all.

Consider a typical element $\left[\!\left[{x}\right]\!\right]_{\mathcal R_f}$ of $S / \mathcal R_f$.

Suppose we were to choose another name for the class $\left[\!\left[{x}\right]\!\right]_{\mathcal R_f}$.

Assume that $\left[\!\left[{x}\right]\!\right]_{\mathcal R_f}$ is not a singleton.

For example, let us choose $y \in \left[\!\left[{x}\right]\!\right]_{\mathcal R_f}, y \ne x$ such that:
 * $\left[\!\left[{x}\right]\!\right]_{\mathcal R_f} = \left[\!\left[{y}\right]\!\right]_{\mathcal R_f}$, then $r \left({\left[\!\left[{x}\right]\!\right]_{\mathcal R_f}}\right) = r \left({\left[\!\left[{y}\right]\!\right]_{\mathcal R_f}}\right)$.

From the definition, we have:


 * $y \in \left[\!\left[{x}\right]\!\right]_{\mathcal R_f} \implies f \left({y}\right) = f \left({x}\right) = r \left({\left[\!\left[{x}\right]\!\right]_{\mathcal R_f}}\right) = r \left({\left[\!\left[{y}\right]\!\right]_{\mathcal R_f}}\right)$

Thus $r$ is well-defined.