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Homemonomorphisms of category of sets

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# monomorphisms of category of sets

###### Theorem 1.

Every monomorphism in the category of sets is an injection.

###### Proof.

Assume $f\colon A\to B$ is a monomorphism. Then, by definition of monomorphism, given any two maps $g,h\colon C\to A$, if $f\circ g=f\circ h$, then $g=h$. Suppose $x$ and $y$ are two elements of $A$ such that $f(x)=f(y)$. Let $C$ be a set with one element, let $g$ be the map which sends this one element to $x$ and let $h$ be the map which sends this one element to $y$. Because $f(x)=f(y)$, we have $f\circ g=f\circ h$. Since $f$ is a monomorphism, $g=h$, so $x=y$. This implies that $f$ is injective. ∎

###### Theorem 2.

Every injection is a split monomorphism.

###### Proof.

Assume $f\colon A\to B$ is injection. If $A$ is empty, the result is trivial, so we assume that $A$ is not empty; let $z$ be an element of $A$. Set

$g=\{(f(x),x)\mid x\in A\}\cup\{(x,z)\mid x\in B\land(\forall y\in A)x\neq f(y)\}$ |

We claim that $g$ is a function from $B$ into $A$. Suppose that $x$ is an element of $B$. If $x\neq f(y)$ for any $y\in A$, then we have exactly one element of $g$ with $x$ as the first element, namely $(x,z)$. If $x=f(y)$ for some $y\in A$, then we the pair $(x,y)$ with $x$ as first element; were there another pair with $x$ as first element, then we would have $(f(x_{1}),x_{1})=(f(x_{2}),x_{2})$ but, as $f$ is an injection, $f(x_{1})=f(x_{2})$ would imply $x_{1}=x_{2}$, so this would not be a distinct pair. Hence $g$ is a function. Furthermore, by construction $g\circ f(x)=x$ for all $x\in A$, so $f$ is a split monomorphism. ∎

## Mathematics Subject Classification

18-00*no label found*

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