Renaming Mapping is Well-Defined

Theorem
The renaming mapping $$r: S / \mathcal{R}_f \to \mathrm{Im} \left({f}\right)$$ is defined as:

$$r: S / \mathcal{R}_f \to \mathrm{Im} \left({f}\right): r \left({\left[\!\left[{x}\right]\!\right]_{\mathcal{R}_f}}\right) = f \left({x}\right)$$

where $$\mathcal{R}_f$$ is the equivalence induced by the mapping $$f$$.

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.

We need to show that if another name is chosen for the class $$\left[\!\left[{x}\right]\!\right]_{\mathcal{R}_f}$$, for example, choosing $$y \in \left[\!\left[{x}\right]\!\right]_{\mathcal{R}_f}, y \ne x$$ (assuming that $$\left[\!\left[{x}\right]\!\right]_{\mathcal{R}_f}$$ is not a singleton), 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:

$$y \in \left[\!\left[{x}\right]\!\right]_{\mathcal{R}_f} \Longrightarrow f \left({y}\right) = f \left({x}\right) = r \left({\left[\!\left[{x}\right]\!\right]_{\mathcal{R}_f}}\right) = r \left({\left[\!\left[{y}\right]\!\right]_{\mathcal{R}_f}}\right)$$

Thus $$r$$ is well-defined.