What is the free category on the underlying graph of a category? Let $\mathcal{D}$ be a category.
Ittay Weiss wrote about Free($\mathcal{D}$) in chat with me.
He said Free($\mathcal{D}$) is the free category on the underlying graph of $\mathcal{D}$.
Is Free($\mathcal{D}$) different from $\mathcal{D}$?
I would like to know the exact definition of Free($\mathcal{D}$) and applications of the notion.
EDIT(Jan. 14, 2013)
Is Free($\mathcal{D}$) isomorphic or equivalent to $\mathcal{D}$?
Counterexamples?
 A: I didn't notice until now that this isn't a new question, but since Makoto asked for a fleshed-out example, I may as well post this.
The category on one object with only the identity morphism has as underlying graph a single vertex with a loop. The free category on the single vertex with a loop is the category on one object with countably many non-identity arrows $a,a\circ a,...$ i.e. as a monoid it is $\mathbb{N}$. The counit functor, of course, just sends all these arrows to the identity. The possibly-subtle point is that these non-identity arrows $a$ are generated by an arrow that was the identity before we applied the forgetful functor. But this is necessary because not every node in every graph has a distinguished loop to map to the identity in generating a free category. 
A: For a graph $G$ with object set $O$ let $C=FG$ be the category with the same objects and whose arrows from $a$ to $b$ are strings
$$a=a_1\stackrel{f_1}⟶a_2\stackrel{f_2}⟶...\stackrel{f_{n-1}}⟶a_n=b$$
of length $n-1$. We define the composition of two strings by juxtaposition, where the lengths are added. This is by its very nature associative. A string of length $0$ is thus the identity. Let's write such strings as $(a_1,f_1,f_2,...,f_{n-1},a_n)$. It can be decomposed as $(a_{n-1},f_{n-1},a_n)∘...∘(a_1,f_1,a_2)$. Let $η:G→UC$ be the graph map sending $f:a→b$ to $(a,f,b)$.
Now, consider any graph map $g:G→UB$ for a category $B$. Using the decomposition of a string of length $n$ into strings of length $1$, it is easy to see that there is a unique functor $g':C→B$ such that $Ug'∘η=g$. Hence $η$ is universal from $G$ to $U$. The object function $F$ can then be made into a free functor $\mathbf{Grph}\to\mathbf{Cat}$.
