I am currently working on associating categories and quivers.
a) Let $C$ be an arbitraty category. I need to show that one can associate with $C$ a quiver $Q_C$ by sending an object $X$ in $C$ to a vertex and a morphism $f:X\longrightarrow Y$ to a directed arrow between vertices $X$ and $Y$. This seems so obvious to me that I don’t really know how to prove it explicitly. Can somebody help me here?
b) I was wondering whether the converse is also true, whether it is possible to associate a category with any quiver. I know that a quiver is a directed graph, ie. a collection of vertices joined by directed arrows. I think the answer is No, because if we take a quiver where two vertices $X$ and $Y$ are not joined by an arrow, this would mean that we don’t have a morphism between those objects. Or because if we take a quiver where there is no arrow from a vertex $X$ to itself, this would mean we don’t have an identity morphism for this object. Is that the right argumentation or am I completely wrong?
We can define a functor from the category of small categories to the category of quivers, which takes a category (viewed as a quiver with two operations – composition and identity – subject to equations expressing associativity and unit laws) to its underlying quiver, which is the operation you describe. The definition is obvious, because we're just forgetting some extra structure defined on top of the quiver (this sort of functor is called forgetful).
There is also a functor in the opposite direction, which takes a quiver, and freely adds composites for any pair of edges sharing a source and target, and adds an identity edge for each vertex. So, while a quiver can not necessarily be equipped with the structure of a category, we can always "complete" it to a category. This functor is related to the one we described above: it is left adjoint to the forgetful functor from quivers to small categories. This demonstrates that the construction of a category from a quiver is a free construction. (Intuitively, this says that the two operations are inverse in an appropriate categorical sense.)