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 $$r \left({\left[\!\left[{x}\right]\!\right]_{\mathcal{R}_f}}\right)$$ actually defines a mapping at all.

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.