# How are non-bijective morphisms “reversed”?

I have the following general definitions:

• 1) The dual graph $G^\mathrm{op}$ of a graph $G$ is obtained by simply reversing all the arrows (e.g. by interchanging the roles of the source and target maps $\sigma,\tau:E_G\to V_G$).
• 2) The dual category $C^\mathrm{op}$ of a category $C$ is simply the dual graph of the category, so again, we just reverse the arrows.
• 3) A monomorphism is a morphism $f\in\mathrm{Mor}(C)$ such that for each pair or morphisms $g,h\in\mathrm{Mor}(C)$, the equality $f\circ g=f\circ h$ implies $g=h$.
• 4) An epimorphism
• 4a) is an epimorphism is a morphism $f$ such that for each pair or morphisms $g,h\in\mathrm{Mor}(C)$, the equality $g\circ f=h\circ f$ implies $g=h$.
• 4b) or, equivalently, is a morphism $f$ whose dual $f^\mathrm{op}$ is a monomorphism.

Please correct me if these standard definitions are not correctly stated.

Let's take a morphism $f:A\to B$ in the category $(\mathrm{Sets})$, and let's say that $f$ is surjective but not injective.

What is the dual morphism $f^\mathrm{op}$ of $f$? In other words, if I want to use definition 4b) to check if it's an epi, how can I show that its dual is a mono? Or, in other words, how are duals of non-bijective morphisms defined in $(\mathrm{Sets})$?

You've already given the definition: the category $\text{Set}$ can be thought of as a giant graph whose objects are sets and morphisms are functions, and taking the dual category just means we reverse the direction of every arrow. This is a purely formal procedure, and in particular even if $C$ has some interpretation in terms of sets and functions (that is, a functor $C \to \text{Set}$), that doesn't mean $C^{op}$ will.
So $f^{op}$ is just $f$ again, but with its source and target reversed, as a purely formal object. The fact that the dual of an epi is a mono is again a purely formal consequence of the fact that the two definitions (in terms of canceling) are dual. And bijectivity has nothing to do with it: even if $f$ happens to be a bijection, $f^{op}$ has nothing to do with $f^{-1}$. The two morphisms can't even be compared because they live in two different categories, namely $\text{Set}^{op}$ and $\text{Set}$.
Now it happens that $\text{Set}^{op}$ can be given an interpretation in terms of sets and functions, but it's not trivial to give and it doesn't come automatically from being an opposite category.
• Speculating on that final paragraph, I guess the most natural way to view $\mathbf{Set}^{op}$ as concrete over $\mathbf{Set}$ is to declare that the forgetful functor $\mathbf{Set}^{op} \rightarrow \mathbf{Set}$ is the contravariant powerset functor. Is this what you had in mind? – goblin Jan 29 '16 at 15:46
• @goblin: yes, but you can do better and even identify exactly what extra structure this equips the "underlying sets" with. One way to say it is that $\text{Set}^{op}$ is equivalent to the category of profinite Boolean rings. A more traditional way to say it is that it's equivalent to the category of complete atomic Boolean algebras. – Qiaochu Yuan Jan 29 '16 at 15:47