Puzzling statement in elderly source work

I'm studying N. Jacobson's "Lectures in Advanced Algebra" (originally published in 1951, apparently a classic), and have encountered this puzzling paragraph:

:"This type of factorization of mappings ... is particularly useful when the set of inverse images $\alpha^{-1} \left({a'}\right)$ coincides with $\overline S$; for, in this case, the mapping $\overline a$ is 1-1. Thus if $\overline a \overline \alpha = \overline b \overline \alpha$, then $a \alpha = b \alpha$ and $a \sim b$. Hence $\overline a = \overline b$. Thus we obtain here a factorization $\alpha = \nu \overline \alpha$ where $\overline \alpha$ is 1-1 onto $T$ and $\nu$ is the natural mapping."

Note that in the above, Jacobson uses:
* $\alpha$ for a general mapping from $S$ to $T$;
* $a'$ for the Image of a representative element $a$ of $S$ under $\alpha$;
* $\overline S$ for the quotient set defined by the equivalence induced by $\alpha$;
* $\nu$ for the quotient mapping from $S \to \overline S$;
* $\overline a$ and $\overline b$ for representative elements of $\overline S$;
* $\overline \alpha$ for the renaming mapping $\overline S \to T$.

The above, then, is a fairly terse proof of the Quotient Theorem for Surjections: the fact that a surjection can be factored into the canonical surjection and the renaming mapping, the latter of which is a bijection.

The puzzling statement is the one: "when the set of inverse images $\alpha^{-1} \left({a'}\right)$ coincides with $\overline S$".

Surely, as $\alpha$ is a mapping, the quotient set of $S$ induced by $\alpha$ always forms a partition of $S$, and so the "set of inverse images MATH]\alpha^{-1} \left({a'}\right)[/tex] will always coincide with $\overline S$?

What am I missing here?