Preimages All Exist iff Surjection/Corollary

Theorem
Let $f: S \to T$ be a mapping.

Let $f^{-1}$ be the inverse of $f$.

Let $f^{-1} \left[{B}\right]$ denote the preimage of $B \subseteq T$.

Then:
 * $\forall B \subseteq T, B \ne \varnothing: f^{-1} \left[{B}\right] \ne \varnothing$


 * $f$ is a surjection

where $f^{-1} \left[{B}\right]$ denotes the preimage of $B \subseteq T$.

Necessary Condition
Let $f$ be a surjection.

Let $B \subseteq T$ such that $B \ne \varnothing$.

Then:
 * $\exists t \in T: t \in B$

From Preimages All Exist iff Surjection:
 * $f^{-1} \left({t}\right) \ne \varnothing$

As $t \in B$ it follows from Preimage of Subset is Subset of Preimage that:
 * $f^{-1} \left[{B}\right] \ne \varnothing$

$B$ is arbitrary, so:
 * $\forall B \subseteq T, B \ne \varnothing: f^{-1} \left[{B}\right] \ne \varnothing$

Sufficient Condition
Suppose that:
 * $\forall B \subseteq T, B \ne \varnothing: f^{-1} \left[{B}\right] \ne \varnothing$

Suppose $f$ is not a surjection.

Then by definition:
 * $\exists t \in T: \neg \left({\exists s \in S: f \left({s}\right) = t}\right)$

That is:
 * $\exists \left\{{t}\right\} \subseteq T: f^{-1} \left[{t}\right] = \varnothing$

which contradicts the hypothesis.

So by Proof by Contradiction, $f$ is a surjection.

Hence the result.