Proposition 9

Let $\mathrm{E}$ and $\mathrm{F}$ be two sets, $\mathfrak{a}$ and $\mathfrak{b}$ their cardinals, $f$ a surjection of $\mathrm{E}$ onto $\mathrm{F}$ such that the sets $\overset{-1}{f}(y)$, where $y \in \mathrm{F}$, all have the same cardinal $\mathfrak{c}$. Then $\mathfrak{a} = \mathfrak{b}\mathfrak{c}$.

For the family $(\overset{-1}{f}(y))_{y \in \mathrm{F}}$ is a partition of $\mathrm{E}$, each element of which partition is a set of cardinal $\mathfrak{c}$; hence the result (§ 3, no. 4, Proposition 6, Corollary 2).