# Category of graphs.

Hello, I'm writting something about Malcev categories and monadicity. The fact is that I need to know if Graph is or not complete (have all finite limits). It seems easy but I would like a real answer (not my feelings saying that it is) and I don't find that information anywhere.

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The category of quivers is complete and cocomplete because it is the category of presheaves of sets over the category with two objects and two nonidentity arrows A=>O. However, complete is stronger than having finite limits. It means that it has all limits. What do you mean by the category of graphs? Can our graphs have loops or multiple arrows? –  Harry Gindi Jan 17 '10 at 18:59

First let me just mention that complete usually means "has all small limits". If you want to say "has all finite limits", you could use the term finitely complete or just has finite limits.

You did not explain which particular category of graphs you are talking about, but luckily almost any reasonable choice will have an affirmative answer: yes, the category of graphs has finite limits. I can tell you why if you tell me what objects and morphisms you have in mind.

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Wow, you are fast! I need exactly to be lex (left exact) the category of directed graphs. That's all and I guess it is already answered. Thank you all. –  Doctor Gibarian Jan 17 '10 at 19:24

Consider the category G that has exactly two distinct objects (call them a and b) and two distinct arrows a->b. A graph is precisely a functor G -> Set. This means that the category of graphs is a presheaf category. The rest follows.

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It has all finite products, and I think (depending on your defintion of Graph), it is easy to show that it has equalizers, and thus all finite limits.

If you are talking about directed graphs, then this is the topos of presheaves of sets over the category which consists of a pair of parallel arrows, and thus has a slew of wonderful properties, finite limits included.

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Here's another one: The category of graphs is a topological category over SET an therefore complete and cocomplete.

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