For any category $C$, there is a terminal functor $C\to \ast$ to the singleton category. With this, a monoidal category $(C,\otimes,I)$ has natural candidates to make it a category internal to $\textbf{Cat}$ whose

  • object of objects is $\ast$
  • object of morphisms is $C$
  • source and target maps are the unique functor $C\to\ast$
  • identity-assigning morphism is the unit $I:\ast\to C$
  • composition is the product $\otimes:C\times C\to C$

In the last line, we used that $C\times_\ast C\cong C\times C$ because $\ast$ is a terminal object. Is this indeed a category in Cat?

  • 1
    $\begingroup$ I don't really understand what you mean by "cartesian category", is it a category with products? If yes, it doesn't make sense to say that there is a morphism $C\to *$ if $*$ is an object of the category $C$... $\endgroup$
    – Arnaud D.
    Apr 12, 2020 at 13:31
  • $\begingroup$ The question was edited for clarification. I meant cartesian monoidal category i.e. monoidal category with the category-theoretical product as the tensor product and the terminal object as the unit object. $\endgroup$ Apr 12, 2020 at 14:35
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    $\begingroup$ A double category (i.e. an internal category in $\mathbf{Cat}$) whose category of objects has just one object and its identity morphism is just a strict monoidal category. Of course, a cartesian monoidal category is not in general a strict monoidal category. $\endgroup$ Apr 12, 2020 at 14:45
  • $\begingroup$ There are a few points where you should be careful about your notation: Note that $C$ is a category, while $*$ is an object of $C$, so as Arnaud D. says, a morphism from one to the other does not make sense. However, if you take $*$ to also denote the terminal category, the remainding of the constructions you do is well-defined. I will write an answer as soon as I can figure out if the necessary axioms are fulfilled... $\endgroup$ Apr 12, 2020 at 17:47
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    $\begingroup$ Note that there's nothing gained from the Cartesianness of your monoidal structure here, once you clarify your confusion about the difference between the terminal category and the terminal object of $C$. $\endgroup$ Apr 12, 2020 at 18:04

1 Answer 1


I will refer to https://ncatlab.org/nlab/show/internal+category for the axioms of an internal category, as I don't want to format that many diagrams :)

As there is only one functor from any category into the terminal category by definition, the diagrams for the laws of specifying the source and target of the identities are trivially fulfilled, as are the laws specifying the source and target of a composition.

However, and this is a critical point, the associativity and the unit law are NOT fulfilled. The reason for that, to spell it out, is that $ X \times * \neq X \neq * \times X$ and $X \times (Y \times Z) \neq (X \times Y) \times Y$ in general, but they are only canonically isomorphic. Maybe there is some notion of internal category in higher category theory that allows for this additional freedom (i.e. that these diagrams need only commute up to coherent isomorphisms), but at least with the standard definition your statement seems to not be true.

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    $\begingroup$ It's a pseudo-category object in the 2-category of categories. $\endgroup$
    – Berci
    Apr 12, 2020 at 18:54
  • $\begingroup$ I figure the result is true for strict monoidal categories, where the inequalities you highlighted actually are equalities? $\endgroup$ Apr 14, 2020 at 12:59
  • $\begingroup$ Yes, in that case it should hold. $\endgroup$ Apr 15, 2020 at 15:30

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