Image of Subset is Image of Restriction

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:
 * $$f \left({X}\right) = \operatorname{Im} \left({f \restriction_X}\right)$$

where $$\operatorname{Im} \left({f}\right)$$ is the image of $$f$$, defined as:
 * $$\operatorname {Im} \left ({f}\right) = \left\{{t \in T: \exists s \in S: t = f \left({s}\right)}\right\}$$

Proof
Suppose $$y \in f \left({X}\right)$$.

Then by defintion of image of a subset:
 * $$\exists x \in X: f \left({x}\right) = y$$

or equivalently:
 * $$\exists x \in X: \left({x, y}\right) \in f$$

But then by definition of restriction of $f$ to $X$:
 * $$f \restriction_X = \left\{{\left({x, y}\right) \in f: x \in X}\right\}$$

Thus by definition of the image set of $$f \restriction_X$$:


 * $$\operatorname {Im} \left ({f \restriction_X}\right) = f \restriction_X \left ({X}\right) = \left\{{y \in T: \exists x \in X: \left({x, y}\right) \in f \restriction_X}\right\}$$

Hence it can be seen that $$y \in \operatorname {Im} \left ({f \restriction_X}\right)$$.

So $$f \left({X}\right) \subseteq \operatorname {Im} \left ({f \restriction_X}\right)$$.

Now suppose that $$y \in \operatorname {Im} \left ({f \restriction_X}\right)$$.

Then by definition of the image set of $$f \restriction_X$$:
 * $$\exists x \in S: \left({x, y}\right) \in f \restriction_X$$

By definition of restriction of $f$ to $X$, we have that $$x \in X$$.

Thus by defintion of image of a subset, $$y \in f \left({X}\right)$$.

So $$\operatorname {Im} \left ({f \restriction_X}\right) \subseteq f \left({X}\right)$$.

We have:
 * $$f \left({X}\right) \subseteq \operatorname {Im} \left ({f \restriction_X}\right)$$
 * $$\operatorname {Im} \left ({f \restriction_X}\right) \subseteq f \left({X}\right)$$

The result follows by Equality of Sets.