# Do empty category objects have self morphisms?

I understand that all category objects must have self morphisms, but allowing morphisms to or from empty sets breaks my intuition of what morphisms are (i.e. a relation between the origin and target objects).

If a category object that is the empty set has a self morphism, is it permitted to participate in other morphisms?

N.B. I’m just starting to learn category theory, so right now I only understand category objects as sets (or simpler objects like integers), which may be part of my problem.

In this setting, your question asks what the morphisms with domain $$\emptyset$$ are. The point is that the definition of a function $$f:A \to B$$ is "for each $$a \in A$$ there is a unique $$b \in B$$ such that $$f(a) = b$$" (In set theoretic language $$f:A \to B$$ is actually a subset of $$A \times B$$ such that for each $$a \in A$$, there is a unique $$b \in B$$ such that $$(a,b) \in f$$).
In particular, this definition means that for any set $$B$$ there is a unique function from $$\emptyset$$ to $$B$$; namely $$\emptyset$$ itself! (though in the category we distinguish between the object $$\emptyset$$ and each of the morphisms $$\emptyset$$ since the morphisms come with a domain and codomain) This may seem strange, but the definition requires it since there are no $$a \in \emptyset$$ so the needed condition vacuously holds for $$f = \emptyset$$. In particular, in Set for every object $$B$$ there is a unique morphism $$f:\emptyset \to B$$ (In technical language, this says that $$\emptyset$$ is an initial object in Set). So, yes, in Set, $$\emptyset$$ has an identity morphism $$\emptyset \to \emptyset$$ and also participates in morphisms to all other sets. This means your intuition about the category Set is incorrect.