# Restriction of Mapping to Image is Surjection

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## Theorem

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

Let $g: S \to \Img f$ be the restriction of $f$ to $S \times \Img f$.

Then $g$ is a surjective restriction of $f$.

## Proof

From Image is Subset of Codomain:

- $\Img g \subseteq T$

Furthermore, by definition of image, we have:

- $\forall s \in S: g \paren s \in \Img g$

Therefore, $g$ may be viewed as a mapping $g: S \to \Img g$.

Thus, by definition, $g$ is a surjection.

$\blacksquare$

## Comment

Thus, for any mapping $f: S \to T$ which is not surjective, by restricting its codomain to its image, it can be considered as a surjection.

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 23$: Restriction of a Mapping