Remark

Let $f$ be a mapping of A into B. For each subset X of A we have (no. 3) $\mathrm{X} \subset \overset{-1}{f}\langle f\langle \mathrm{X}\rangle\rangle$. Furthermore, for each subset Y of B we have $f\langle \overset{-1}{f}\langle \mathrm{Y}\rangle\rangle \subset \mathrm{Y}$, for the relation $y \in f\langle \overset{-1}{f}\langle \mathrm{Y}\rangle\rangle$ is equivalent to

and therefore implies the relation $(\exists z)(z \in \mathrm{Y} \text{ and } y = z)$ and consequently also the relation $y \in \mathrm{Y}$.

If $f$ is surjective, we have $f\langle \overset{-1}{f}\langle \mathrm{Y}\rangle\rangle = \mathrm{Y}$ for every subset Y of B, for the relation $y \in \mathrm{Y} \subset \mathrm{B}$ implies by hypothesis the relation $(\exists x)(y = f(x))$ and therefore also $(\exists x)(y \in \mathrm{Y} \text{ and } y = f(x))$; but "$y \in \mathrm{Y}$ and $y = f(x)$" implies $(\exists z)(z \in \mathrm{Y} \text{ and } z = f(x))$, and our assertion follows.

If $f$ is injective, we have $\overset{-1}{f}\langle f\langle X\rangle\rangle = X$ for every subset X of A. For the relation $x \in \overset{-1}{f}\langle f\langle X\rangle\rangle$ is equivalent to $f(x) \in f\langle X\rangle$, hence to

but the hypothesis means that $f(z) = f(x)$ implies $z = x$, hence $x \in \overset{-1}{f}\langle f\langle X\rangle\rangle$ implies $x \in X$.