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] + 2mu@ + 1em { A \ar[r]^*{i_A} \ar@{-->}[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:

\(\ds \forall x \in A: \, \) \(\ds \map f x\) \(=\) \(\ds \map g x\)
\(\ds \) \(=\) \(\ds \map g {\map {i_A} x}\) Definition of Inclusion Mapping
\(\ds \) \(=\) \(\ds \map {\paren {g \circ i_A} } x\) 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:

\(\ds \forall x \in A: \, \) \(\ds \map f x\) \(=\) \(\ds \map {\paren {g \circ i_A} } x\) Definition of $f$
\(\ds \) \(=\) \(\ds \map g {\map {i_A} x}\) Definition of Composition of Mappings
\(\ds \) \(=\) \(\ds \map g x\) Definition of Inclusion Mapping

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

$\blacksquare$


Sources

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