Math 811, Ping Li

Exercise 6B: Prove that in the category SET S there is no zero object. First, I give some symbols.

(a) One point set in category SET S is denoted by ∗.

(b) For M, N ∈ obj(SET S), the element’s number of C(M, N) is denoted by|C(M, N)|. Proof: There are three conditions.

1, For any M ∈ obj(SET S), M ≠ ϕ and M ≠ ∗, we can select an object N ∈ obj(SET S), satisfied N ≠ ϕ and N ≠ ∗, then

|C(M, N)| > 1 and |C(N, M)| > 1

So M isn’t the zero object.

2, Every one point set∗ in category SET S is not the zero object. Because for any object N ∈ obj(SET S), N ≠ ϕ and N ≠ ∗, there is only one morphism in C(N, ∗), and f ∈ C(N, ∗) is given by f(y) = ∗, ∀y ∈ N, ie

|C(N, ∗)| = 1

But |C(∗, N)| = the element’s number of N > 1

So every one point set is not the zero object.

3, The empty set ϕ in category SET S isn’t the zero object. Because, for any object N (N ≠ ϕ) in category SET S , from the definition, there is only one morphism from ϕ to N, i.e.,

|C(ϕ, N)| = 1

But C(N, ϕ) is empty

|C(N, ϕ)| = 0

So ϕ isn’t the zero object.

Then it follow the result: there is no zero object on the category SET S.