TITLE: Path Algebra for Categories
QUESTION [12 upvotes]: For a while I had been thinking that the path algebra of a quiver $Q$ over a commutative ring $R$ is the same as the "category ring" $R[P]$ (analogous to "group ring", "monoid ring", "semigroup ring", and the like), where $P$ consists of paths in $Q$, and multiplication in $P$ is composition (or zero when the domain doesn't match the codomain). 
However, as I have been writing something in more detail, I find that it's not always possible to find a corresponding quiver for some given category. For example, when $C$ is the thin category representing the partial order of $\mathbb R$ (objects are real numbers, and morphisms are pairs $(x, y)$ with $x \le y$), I cannot find the corresponding quiver.
My questions are

Am I just not aware of the quiver that will give rise to the category in question?
If it is really the case that there are no corresponding quivers for some categories, then this "category ring" is a more general object that the path algebra. Should it still be called the path algebra?
Why do people usually start with a quiver, then make it into a category to define the path algebra? Why not start with a category?

Edit:
Thank you for the answers below by Aaron and Julian. So the answer to my question number 1 is that I was not aware of the quiver algebra with relations. Now that I am, I have a follow-up question. Is the path algebra with relations the same as the category ring when $R$ is a field? (Why would one want to consider the quotient over a two-sided ideal that does not come from identifying paths anyway?)

REPLY [8 votes]: That's a trap I used to fall into as well.  This is the example that you should always always always keep in mind:
Let $G$ be a finite group, then construct a category with one object, and all arrows correspond to elements of $G$ with composition of arrows being group element multiplication, then the category algebra is a group algebra.  This is the reason why people, such as Peter Webb and Markus Linklemann, are interested in category algebras as they are generalisation of group algebras.
On the other hand, given a group, it is extremely difficult to construct the corresponding path algebra (basic algebra).
Note the "corresponding path algebra" always exists in the following sense (a result of Gabriel):
Let $A$ be a finite dimensional algebra with direct sum decomposition $ A= \bigoplus_{i=1}^n P_i^{\oplus m_i}$ (i.e. $P_i$ are indecomposable projective) and take $M=\bigoplus_{i=1}^n P_i$, then the path algebra $kQ/I$ corresponding to $A$ (constructed using Ext-quiver) is isomorphic to $End_A(M)$.
So to answer your question (1) and (2):  Given a path algebra $A=kQ/I$, you can construct a category which capture homological behaviour of $kQ/I$ is simply given by treating $Q$ as a category, and relations of compositions given by $I$.  Here each object (=vertex in $Q$) in the category has ONLY ONE invertible endomorphism corresponding to the primitive idempotent of that vertex.  In this way you can say category algebra is a generalisation of path algebra.
Concerning to your third question, I can't give a very clear answer, these are what my supervisor told me years ago, so I hope someone can give some supplementary comments:
There are various reason that we go from quiver to category.  Sometimes because we do not want to deal with the algebra directly.  (for example, V. Mazorchuk (and possibly Y. Drozd) is one of the people who uses this approach very often.)  A better reason is that "path algebra of an infinite quiver" is (was?) an unclear concept, as you don't have things like Gabriel's theorem/lemma as mentioned above, and you need to deal with certain issues arising from infinite dimensional algebras.  On the other hand, there is no obvious obstruction when we look at a category with infinite objects.
So why do we still favour path algebras?  The reason is most techniques that work on path algebras do not carry over to category algebras.  When you have a path algebra, most homological work becomes combinatorial.  Yet if you have a category algebra, it is still very difficult to see what the projective indecomposables are (remember the group algebra!).