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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.

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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.

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  • $\begingroup$ What do you mean by S as a discrete graph? $\endgroup$ – real_father Feb 28 at 12:20
  • $\begingroup$ 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 $\endgroup$ – punctured dusk Feb 28 at 12:21
  • $\begingroup$ 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? $\endgroup$ – real_father Feb 28 at 12:27
  • $\begingroup$ Yes, that's right. That would make more sense, if you view a graph as a category (so that it has identity maps). $\endgroup$ – punctured dusk Feb 28 at 12:28

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