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.
 A: 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.
