# Why is an empty set not a terminal object in categories $\mathsf{Top}$ and $\mathsf{Sets}$?

From Awodey: In any category $\mathsf{C}$,

an object $0$ is initial if for any object $C$ there is a unique morphism $0 \to C$,

an object $1$ is terminal if for any object $C$ there is a unique morphism $C \to 1$.

First I was puzzled by the fact that we can do mappings from a zero set to any other set in a category $\mathsf{Sets}$. But then after I accepted it conceptually, I do not understand why there is no unique morphism to an empty set from every object in a category $\mathsf{Sets}$. Is this because we cannot map a non-empty set to an empty set (since our mapping is total) although we can map an empty set to a non-empty set?

Please, note, I am familiar with the proposition saying: Initial (terminal) objects are unique up to isomorphism.

• It is also easy to remember using $|A^B|=|A|^{|B|}$, which holds for all sets $A,B$, also when $A$ is empty (which shows that $\emptyset$ is not terminal) or when $B$ is empty (which shows that $\emptyset$ is initial). – Martin Brandenburg Mar 27 '13 at 18:48
• @MartinBrandenburg What does the notation $A^B$ mean? – Mateen Ulhaq May 22 '18 at 4:38
• @Mateen: it means the set of functions from $B$ to $A$. – Qiaochu Yuan Nov 3 '18 at 20:13

## 2 Answers

Is this because we cannot map a non-empty set to an empty set (since our mapping is total) although we can map an empty set to a non-empty set?

Yep. A function has to take values on inputs. A function with non-empty domain but empty codomain can't take any values. (By contrast, a function with empty domain but non-empty codomain doesn't need to take values because it has no inputs.)

I think you already answered your own question: Yes, this is because a mapping (in $\mathcal{Set}$) are defined to be total.