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.