# A (simple) example on dual categories

I have two questions on a simple example of dual categories:

In the category of Groups and Group homomorphism I consider the following diagramm $\mathbb{Z}\stackrel{q}{\to}\frac{\mathbb{Z}}{2\mathbb{Z}}\to 0$. The map $q$ is the quotient map and obviously surjective, so this diagramm describes a "real" situation.

My first question is how do I (formally) get to the dual category? I read that I just reverse arrows, but is there a proper functor? At least I want to Change my categories so I need a map/functor to do so.

If I just reverse arrows I get to $\mathbb{Z}\stackrel{q}{\leftarrow}\frac{\mathbb{Z}}{2\mathbb{Z}}\leftarrow 0$. This cant be a useful expression from my point of view. The dual category has the same objects (true, the objects are Groups) with reversed morphisms (seems not to be true, because q is no longer a Group homomorphism but something weird.) This yields:

Second question: Can you explain how this reversed diagram fits in the dual category?

Third question: If I assume the reversed diagramm is right. I wonder how I can work with this. I expected if I can show a morphism is injective in a category, it is an epimorphism in the dual category. But as for $q$ in the reversed diagramm I can show nothing, because it does not make sense as a function, althouh in the dual category (the original diagram) $q$ is clearly a monomorphism.

• Make a distinction between $q$ as group homomorphism and as morphism in the category Grp. As morphism of the opposite category it is not expected to correspond with a group homomorphism. A look here might be useful. – drhab Sep 23 '15 at 13:40
• Yes. And looking at it as an element of $\mathcal C^{op}(a,b)$ it is a morphism having $a$ as domain and $b$ as codomain. Try to focus on that context and somehow "forget" that it is a group homomorphism. – drhab Sep 23 '15 at 14:03
• Ok, so indeed $q$ is a group homomorphism, i.e. it is a morphism, because the morphism-set of $Gr$ consists of group homs. Hence it is an element of $Gr(\mathbb{Z},\mathbb{Z}_2)$, the hom-set in your notation (cf. link). So $q$ with reversed domains is an element of $Gr^{op}(\mathbb{Z},Z_2)$, because it IS an group hom and its domain is $Z_2$ and its codomain is $\mathbb{Z}$. Is that right? – user221151 Sep 23 '15 at 14:08
• I would say: $q$ is an element of $Grp^{op}(\mathbb Z_2,\mathbb Z)$ because it is and element of $Grp(\mathbb Z,\mathbb Z_2)$. These homsets have the same elements. But the functions $dom$ and $cod$ on these sets are interchanged. If we look at $q$ as element of $Grp(\mathbb Z,\mathbb Z_2)$ then it can be recognized as a grouphomomorphism. Essentially nothing changes if we look at $q$ as element of $Grp^{op}(\mathbb Z_2,\mathbb Z)$. It is still that group homomorphism. But as arrow it has $\mathbb Z_2$ as domain and $\mathbb Z$ as codomain. This unlike the group homomorphism. – drhab Sep 23 '15 at 14:16
• Of $q$ as element of the opposite category you can only say that it is monic. This purely based on the fact that it is epi in the original category. You cannot call it "injective". I have to go now. Later I will have a second look and I hope that there will be more reactions on your question. Cheers. – drhab Sep 23 '15 at 14:32

## 1 Answer

The purpose of dualizing is not to "make sense" in the sense that the morphisms in the opposite of a familiar category will be as meaningful as the morphisms in the original category (although sometimes such an interpretation can be found.) The purpose, instead, is to essentially double the efficiency of the development of category theory. This is the principle of duality: for every statement about a general category which you've proved, you get a "dual" statement by reversing all the arrows, which is also true since opposite categories exist.

For a somewhat trivial example, take the fact that "In any category if a morphism $f$ admits a splitting on the left, i.e. $g$ such that $gf$ is an identity, then $f$ is a monomorphism." It's also true that "In any category if a morphism $f$ admits a splitting on the right, i.e. $g$ such that $fg$ is an identity, then $f$ is an epimorphism." But you don't have to prove these statements separately, because if $f$ satisfies the hypotheses of the latter statement in $\mathcal{C}$, then it satisfies the hypotheses of the former statement in $\mathcal{C}^{op}$ so that it's a mono in $\mathcal{C}^{op}$, equivalently, an epi in $\mathcal{C}$.

The above was mostly intended to clarify your third question. For your first, no, there is no functor between $\mathcal{C}$ and $\mathcal{C}^{op}$ most of the time.