Composite of Injections is Injection
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Theorem
A composite of injections is an injection.
That is:
- If $f$ and $g$ are injections, then so is $f \circ g$.
Proof
Let $f$ and $g$ be injections.
Then:
\(\ds \map {f \circ g} x\) | \(=\) | \(\ds \map {f \circ g} y\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map f {\map g x}\) | \(=\) | \(\ds \map f {\map g y}\) | Definition of Composition of Mappings | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map g x\) | \(=\) | \(\ds \map g y\) | as $f$ is injective | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds y\) | as $g$ is injective |
$\blacksquare$
Sources
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- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.10 \ (1)$
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