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I know some famous examples of large categories, where the set of objects is a proper class. How about a collection of arrows? Please give me examples if you can.

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Consider the category with a single object $*$ and an arrow $\alpha\colon *\to *$ for every ordinal $\alpha$. Define $\alpha\circ \beta = \max(\alpha,\beta)$, so that $0$ is the identity and associativity obviously holds. Since there are a proper class of ordinals, $\text{Hom}(*,*)$ is a proper class.

The point here is that a category with one object is a monoid, so to get a category with $\text{Hom}(*,*)$ large, we just need a large (proper class sized) monoid, and the ordinals do nicely.

For an even more trivial example, let $\mathbb{X}$ be any proper class, and consider the category with two objects $1$ and $2$, $\mathrm{Hom}(1,1) = {\text{id}_1}$, $\mathrm{Hom}(2,2) = {\text{id}_2}$, and $\mathrm{Hom}(1,2) = \mathbb{X}$. There is a unique way to define composition to make this into a category.

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    thanks for you answer. i am undergraduate student and new in Categories. i will study your answer2017-01-27
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    so it is true if i say: a category is not locally small iff the collection of arrows is proper class?2017-01-27
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    The definition is: A category is locally small iff for all objects $A$ and $B$, $\text{Hom}(A,B)$ is a set. So negating this: A category is not locally small iff there exist objects $A$ and $B$ such that $\text{Hom}(A,B)$ is a proper class.2017-01-27
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    What you said isn't quite right, because a locally small category may have a proper class of objects, so the collection of *all* arrows may be a proper class (there are already a proper class of identity arrows). It's just that the collection of all arrows between $A$ and $B$ is a set, for any fixed $A$ and $B$.2017-01-27
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All of the standard large categories have a proper class of arrows. Every category has at least as many arrows as objects, since every object has an identity arrow. If you don't want a proper class of objects, then you're stuck with relatively unnatural examples, as in the other answer.