# Inclusion Mapping is Monomorphism

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## Theorem

Let $\struct {S, \circ}$ be an algebraic structure.

Let $\struct {T, \circ}$ be an algebraic substructure of $S$.

Let $\iota: T \to S$ be the inclusion mapping from $T$ to $S$.

Then $\iota$ is a monomorphism.

## Proof

We have that the inclusion mapping is an injection.

Now let $x, y \in T$:

\(\ds \map \iota {x \circ y}\) | \(=\) | \(\ds x \circ y\) | Definition of Inclusion Mapping | |||||||||||

\(\ds \) | \(=\) | \(\ds \map \iota x \circ \map \iota y\) | Definition of Inclusion Mapping |

demonstrating that $\iota$ has the morphism property.

So $\iota$ is a homomorphism which is also an injection.

Thus by definition, $\iota$ is a monomorphism.

$\blacksquare$