Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to find a monic in category OSet defined as

"sets with unary operation, $(A,x)$, where $x:A\rightarrow A$, and morphism preserving that operation, that is a morphism from $(A,x)$ to $(B,y)$ is $f:A\rightarrow B$ with $f\circ x=y\circ f$"

I am doing it exactly how we prove injective functions are monic in category Set, i.e, I prove any injective function $\space$ $m:(A,x)\rightarrow (B,y)$ in OSet is monic by proving for each parallel pair of arrows $f,g:(S,z)\rightarrow (A,x)$, we have$\space$ $m\circ f=m\circ g\implies f=g$.

My confusion is this that do I need to do anything more with it as this time $m$ is not a function but 'operation preserving function'.

Thank you.

share|improve this question

1 Answer 1

Basically, for a given morphism $m:(A,x)\rightarrow (B,y)$ in $\mathbf{OSet}$ you want to prove

$m$ is monic $\iff$ the underlying map $m:A\rightarrow B$ is injective.

The proof of the $\Leftarrow$ direction works as you described above, just like for $\mathbf{Set}$. But for the proof of the $\Rightarrow$ direction you need to do a little bit more.

What makes the proof work in the case of ordinary sets is the isomorphism $A\cong \mathbf{Set}(1,A)$ (where $1$ is some one-element set), which allows you to view an element $a\in A$ as a map $a:1\rightarrow A$ so that e.g. $m(a) = m(a')$ ''means'' the same as $m\circ a= m\circ a'$.

So you need to identify a suitable object $(N,s)$ in $\mathbf{OSet}$ such that there is an isomorphism $A\cong \mathbf{OSet}((N,s),(A,x)))$. A one-element set (with the identity map) will not work here because maps in $\mathbf{OSet}((1,id),(A,x))$ only correspond to fixpoints of $x$. But perhaps you can guess by my notation a suitable $(N,s)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.