# Mapping is Extension iff Composite with Inclusion

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## Theorem

Let $S$ and $T$ be sets.

Let $A \subseteq S$.

Let $f: S \to T$ and $g: A \to T$ be mappings.

Then $f$ is an extension of $g$ if and only if:

- $f = g \circ i_A$

where $i_A$ is the inclusion mapping on $A$.

This can be illustrated using a commutative diagram as follows:

- $\begin {xy} \[email protected] + [email protected] + 1em { A \ar[r]^*{i_A} \[email protected]{-->}[rd]_*{f = g \circ i_A} & S \ar[d]^*{g} \\ & T } \end {xy}$

## Proof

### Necessary Condition

Let $f: S \to T$ be an extension of $g: A \to T$.

Then by definition:

\(\, \displaystyle \forall x \in A: \, \) | \(\displaystyle \map f x\) | \(=\) | \(\displaystyle \map g x\) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map g {\map {i_A} x}\) | Definition of Inclusion Mapping | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map {\paren {g \circ i_A} } x\) | Definition of Composition of Mappings |

That is:

- $f = g \circ i_A$

$\Box$

### Sufficient Condition

Let $f = g \circ i_A$ where $i_A$ is the inclusion mapping on $A$.

Then:

\(\, \displaystyle \forall x \in A: \, \) | \(\displaystyle \map f x\) | \(=\) | \(\displaystyle \map {\paren {g \circ i_A} } x\) | Definition of $f$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map g {\map {i_A} x}\) | Definition of Composition of Mappings | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map g x\) | Definition of Inclusion Mapping |

So, by definition, $f$ is an extension of $g$.

$\blacksquare$

## Sources

- 1975: Bert Mendelson:
*Introduction to Topology*(3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 9$: Inverse Functions, Extensions, and Restrictions