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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.

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    $\begingroup$ 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). $\endgroup$
    – Martin Brandenburg
    Mar 27, 2013 at 18:48
  • $\begingroup$ @MartinBrandenburg What does the notation $A^B$ mean? $\endgroup$
    – Mateen Ulhaq
    May 22, 2018 at 4:38
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    $\begingroup$ @Mateen: it means the set of functions from $B$ to $A$. $\endgroup$
    – Qiaochu Yuan
    Nov 3, 2018 at 20:13

2 Answers 2

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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.)

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  • $\begingroup$ Is the function with empty domain and non-empty codomain unique? $\endgroup$
    – N.S.
    Sep 29, 2021 at 11:19
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I think you already answered your own question: Yes, this is because a mapping (in $\mathcal{Set}$) are defined to be total.

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