Preimages All Exist iff Surjection/Corollary

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

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


 * $\forall B \subseteq T, B \ne \O: f^{-1} \sqbrk B \ne \O$


 * $f$ is a surjection
 * $f$ is a surjection

where $f^{-1} \sqbrk B$ 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:
 * $\map {f^{-1} } t \ne \O$

As $t \in B$ it follows from Preimage of Subset is Subset of Preimage that:
 * $f^{-1} \sqbrk B \ne \O$

$B$ is arbitrary, so:
 * $\forall B \subseteq T, B \ne \O: f^{-1} \sqbrk B \ne \O$

Sufficient Condition
Suppose that:
 * $\forall B \subseteq T, B \ne \O: f^{-1} \sqbrk B \ne \O$

$f$ is not a surjection.

Then by definition:
 * $\exists t \in T: \neg \paren {\exists s \in S: \map f s = t}$

That is:
 * $\exists \set t \subseteq T: f^{-1} \sqbrk {\set t} = \O$

which contradicts the hypothesis.

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

Hence the result.