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I revisited this popular tutorial about category theory after some time and I realised that right at the beginning there is a statement about what a category is:

Identities are omitted

In a category, if there is an arrow going from A to B and an arrow going from B to C then there must also be a direct arrow from A to C that is their composition.

Is that true in a standard category? All this time I was happy with my composition function

$$\circ : \operatorname{Hom}_\mathcal{C}(A,B)\times\operatorname{Hom}_\mathcal{C}(B,C)\to\operatorname{Hom}_\mathcal{C}(A,C)$$

To my knowledge this function never complained when the codomain was the empty set. Was I operating in something that is not a category? Or is the statement in the tutorial an oversimplification and I'm being too pedantic?

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  • $\begingroup$ +1 just by the picture. Very funny :D $\endgroup$
    – Dog_69
    May 16 '18 at 9:49
  • $\begingroup$ I don't take any credit @Dog_69 it's from the page in the link $\endgroup$
    – gurghet
    May 16 '18 at 9:50
  • $\begingroup$ OK. But still... It's worth. $\endgroup$
    – Dog_69
    May 16 '18 at 11:42
  • $\begingroup$ Maybe you were operating in $\mathsf{Rel}$-enriched categories without knowing ($\mathsf{Rel}$ is the category of sets and relations between them) ? $\endgroup$
    – Pece
    May 16 '18 at 12:27
  • $\begingroup$ No smartpants :) $\endgroup$
    – gurghet
    May 16 '18 at 12:38
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If the codomain is the empty set and the domain is not, then the composition map cannot exist as there is no map from a non empty set into the empty set. Hence the existence of such a composition map rules out this situation.

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    $\begingroup$ In other words, if $Hom(A,C)$ is empty the composition function proves that one of $Hom(A,B)$ or $Hom(B,C)$ must be empty $\endgroup$ May 16 '18 at 9:47
  • $\begingroup$ Thanks, sometimes my brain doesn't go that extra step. $\endgroup$
    – gurghet
    May 16 '18 at 9:49
  • $\begingroup$ @Chessanator This holds for all functions, not just composition; there is no function from a non-empty set to the empty set. Put another way, the set of all functions from non-empty sets to the empty set is the empty set. $\endgroup$
    – Alec Rhea
    Dec 11 '20 at 10:08

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