# Injection iff Monomorphism in Category of Sets

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## Theorem

Let $\mathbf{Set}$ be the category of sets.

Let $f: X \to Y$ be a morphism in $\mathbf{Set}$, i.e. a mapping.

Then $f$ is an injection if and only if it is a monomorphism.

## Proof

### Necessary Condition

Suppose that $f$ is injective.

Suppose further that we have mappings $g, h: Z \to X$ such that $g \ne h$.

Then necessarily there exists some $z \in Z$ such that $g \left({z}\right) \ne h \left({z}\right)$ by Equality of Mappings.

As $f$ is injective, it follows that:

$\map f {\map g z} \ne \map f {\map h z}$

which, again by Equality of Mappings, means that $f \circ g \ne f \circ h$.

Hence $f$ is monic, by the Rule of Transposition.

$\Box$

### Sufficient Condition

Suppose that $f: X \rightarrowtail Y$ is a monomorphism.

By definition of injection, it will suffice to show that:

$x \ne x' \implies \map f x \ne \map f {x'}$

To this end, consider a singleton $\set a$, and define:

$\bar x: \set a \to X, \map {\bar x} a := x$
$\bar x': \set a \to X, \map {\bar x'} a := x'$

In particular, $\bar x \ne \bar x'$, and so, $f$ being monic, we deduce:

$f \circ \bar x \ne f \circ \bar x'$

It follows that it must be that:

 $\ds \map f x$ $=$ $\ds \map f {\map {\bar x} a}$ $\ds$ $\ne$ $\ds \map f {\map {\bar x'} a}$ Equality of Mappings $\ds$ $=$ $\ds \map f {x'}$

Hence $f$ is injective.

$\blacksquare$