Inclusion Mapping is Injection
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Theorem
Let $S, T$ be sets such that $S$ is a subset of $T$.
Then the inclusion mapping $i_S: S \to T$ defined as:
- $\forall x \in S: \map {i_S} x = x$
is an injection.
For this reason the inclusion mapping can be known as the canonical injection of $S$ to $T$.
Proof
Suppose $\map {i_S} {s_1} = \map {i_S} {s_2}$.
\(\ds \map {i_S} {s_1}\) | \(=\) | \(\ds s_1\) | Definition of Inclusion Mapping | |||||||||||
\(\ds \map {i_S} {s_2}\) | \(=\) | \(\ds s_2\) | Definition of Inclusion Mapping | |||||||||||
\(\ds \map {i_S} {s_1}\) | \(=\) | \(\ds \map {i_S} {s_2}\) | by definition | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds s_1\) | \(=\) | \(\ds s_2\) | from above |
Thus $i_S$ is an injection by definition.
$\blacksquare$
Sources
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 8$: Functions
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Example $5.4$
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.3$: Mappings