Restriction of Mapping is Mapping

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

Let $$X \subseteq S$$.

Let $$f \restriction_X$$ be the restriction of $f$ to $X$.

Then $$f \restriction_X: X \to T$$ is a mapping:
 * whose domain is $$X$$;
 * whose preimage is $$X$$.

Proof
As $$f: S \to T$$ is a mapping, we have that:
 * $$\forall x \in S: \left({x, y_1}\right) \in f \and \left({x, y_2}\right) \in f \implies y_1 = y_2$$

From the definition of a subset, $$x \in X \implies x \in S$$, and so:
 * $$\forall x \in X: \left({x, y_1}\right) \in f \restriction_X \and \left({x, y_2}\right) \in f \restriction_X \implies y_1 = y_2$$

Also from the definition of a mapping:
 * $$\forall x \in S: \exists y \in T: \left({x, y}\right) \in f$$

Again from the definition of a subset, $$x \in X \implies x \in S$$, and so:
 * $$\forall x \in X: \exists y \in T: \left({x, y}\right) \in f \restriction_X$$

So $$f \restriction_X: X \to T$$ is a mapping.

The fact that the domain of $$f \restriction_X$$ is $$X$$ follows from the definition of domain.

The fact that the preimage of $$f \restriction_X$$ is also $$X$$ follows from Preimage of Mapping equals Domain.