Mapping to Singleton is Surjection

Theorem

Let $S$ be a non-empty set.

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

Let $T$ be a singleton.

Then $f$ is a surjection.

Proof

Let $T = \left\{{t}\right\}$.

For $f$ to be a surjection, all we need to do is show:

$\forall y \in T: \exists x \in S: f \left({x}\right) = y$.

As $S \ne \varnothing$, $\exists s \in S$.

As $f: S \to T$ is a mapping, it follows that $f \left({s}\right) \in T$.

So as $f \left({s}\right) \in T$ it follows that $t = f \left({s}\right)$.

As $T = \left\{{t}\right\}$, it follows that $\forall y \in T: \exists x \in S: y = f \left({x}\right)$.

Hence the result.

$\blacksquare$