# Does the comma category of a set over category of graphs have initial and terminal objects?

As the question states. Let $$S$$ be a set and $$U : \textbf{Graph} \to \textbf{Set}$$ be the forgetful functor. Then does the comma category $$S \downarrow U$$ have initial or terminal objects?

I think the answer is no to both questions. Every construction of an initial/terminal object that I've tried has failed for some reason or another. Maybe I'm thinking about the question in the wrong way.

An object of $$S \downarrow U$$ is an $$S$$-pointed graph, so there definitely is an initial object: which $$S$$-pointed graph has a unique $$S$$-embedding into every graph?
Answer: $$S$$ as a discrete graph with the natural $$S$$-pointing.
If you allow loops, there also is a terminal object: which $$S$$-pointed graph admits a unique $$S$$-homomorphism from any graph?
Answer: $$S$$ as a complete graph with all edges, including all loops.
• The graph with set of vertices equal to $S$ and with no edges. (I invented the word, but compare with en.wikipedia.org/wiki/Discrete_category ) You may also want to call it the free graph on $S$: en.wikipedia.org/wiki/Free_object – punctured dusk Feb 28 at 12:21
• Let's say that we define the category of graphs so that the edge set must be reflexive i.e. every vertex has an edge to itself. I think the edge set of the initial object would then change to $\{(a, a) : a \in S \}$. Does that make sense? – real_father Feb 28 at 12:27