Image of Set under Mapping is Set iff Restriction is Set

Theorem
Let $V$ be a basic universe

Let $f: V \to V$ be a mapping.

Let $x$ be a set.

Let $f \sqbrk x$ denote the image of $x$ under $f$.

Let $f {\restriction} x$ denote the restriction of $f$ to $x$.

Then $f \sqbrk x$ is a set $f {\restriction} x$ is a set.

Proof
Let $f {\restriction} x$ be a set.

Then its image is also a set.

But then:
 * $\Img {f {\restriction} x} = f \sqbrk x$

Conversely let $f \sqbrk x$ be a set.

From Cartesian Product of Sets is Set:
 * $x \times f \sqbrk x$ is a set.

From Restriction of Mapping is Subclass of Cartesian Product:
 * $f {\restriction} A \subseteq A \times f \sqbrk A$

The result follows from Subclass of Set is Set.