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Consider a category whose only isomorphisms are identities. The only examples that I can think of, are categories which are freely generated by a graph. Are all such categories free in this sense?

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These are called gaunt categories. –  Zhen Lin Dec 11 '12 at 13:24

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Consider the dense linear order $\mathbb{Q}$ as a category. Since it is a partially-ordered set, every isomorphism is the identity, and moreover, every endomorphism is the identity. However, it is also not the free category on any graph because every non-identity morphism can be factored in a non-trivial way. (The free category on an acyclic graph must have "irreducible" morphisms that cannot be factored further, and the free category on a non-acyclic graph would have non-trivial endomorphisms.)

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These are called skeletal categories and there are a lot of examples of them. In fact, if you believe in the axiom of choice, every small category has a skeleton, i.e. a skeletal subcategory that it is equivalent to via the inclusion functor.

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Is it correct that all categories freely generated by finite graphs are skeletal? Many thanks. –  Hooman Dec 11 '12 at 13:20
That's not correct. In a skeletal category there can be non-trivial automorphisms. What is true is that all isomorphisms are automorphisms. –  Zhen Lin Dec 11 '12 at 13:24
@ZhenLin: ah, yes you are right. I will leave this answer here just for the record. –  Adeel Dec 11 '12 at 13:29

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