Forgetting free category, same-same but different?

Question

I know almost nothing about categories but I was reading about the universal property of free categories. If C(G) is the free category generated by the graph $G$ and U is the forgetful functor, what is the difference between the graph $G$ and $U(C(G))$? What would be an instance where they are different? For example if I take a graph $G$ as $a\overset{f}{\underset{g}{\leftrightarrows}}b\overset{h}{\rightarrow} c$ the arrows of $C(G)$ are the three identity arrows and $f, g, h, f\circ g\, g\circ f, h, h\circ g, f\circ g\circ f, g\circ f\circ g, ...$ and $U(C(G))$ would be the same as $G$ no?

Answer 1

The underlying category of a free category has freely added compositions.

For example the free category on the graph $G=x \rightarrow y \rightarrow z$ will have a composite arrow $x\rightarrow z$, i.e. we have $$UC(G) = \begin{array}{ccc} x &\rightarrow&y\\ &\searrow&\downarrow\\ &&z \end{array}.$$

If you call this new graph above $G'$ and consider $UC(G')$, what will happen? By applying $U$ you forgot about compositions, so the diagonal arrow $x\rightarrow z$ is not to be regarded as the composite $x \rightarrow y \rightarrow z$. Thus $UC(G')$ has a newly added composite and may be depicted as $$UC(G') = \begin{array}{ccc} x &\rightarrow&y\\ &\searrow\searrow&\downarrow\\ &&z \end{array}.$$

Even worse, if you have a finite graph $G$ with a nontrivial cycle, then the graph $UC(G)$ will have infinitely many arrows. Take for example $G=\underset{\circlearrowright f}x$, then you will get a graph $UC(G)$ with infinitely many cylces, one for each $n$-fold composite $f^{n} = f\circ ... \circ f$.

So in fact it is rarely the case, that $UC(G) \cong G$.