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Are there standard approaches to show that a non CCC category is distributive?

By the way, I know that a CCC category is distributive and I suppose that what makes the things work is that there is an exponential. But I have tried (not too hard) to prove the existence of the $(A\oplus B)\times C \rightarrow (A\times C) \oplus (B\times C)$ in a CCC category using exponentials, but I didn't succeed....any idea or pointer to an article ?

Thanks

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    $\begingroup$ In a CCC, $(-) \times C$ is a left adjoint and thus preserves colimits (including coproducts). $\endgroup$
    – varkor
    Commented Dec 14, 2020 at 22:27
  • $\begingroup$ @Varkor Thanks varkor but can you elaborate a little bit, I'm just trying to learn CT by myself and your answer is a bit short for my level of knowledge. Thanks $\endgroup$
    – Gianfranco
    Commented Dec 15, 2020 at 9:37
  • $\begingroup$ @Gianfranco Left adjoint functors preserve colimits, and right adjoint functors preserve limits. Coproducts are a special kind of colimit, so left adjoints will preserve coproducts since they preserve colimits. In a CCC, the functor $(-)\times C$ that sends an object $X$ to $X\times C$ and an arrow $f:X\to Y$ to $f\times1_C$ is a left adjoint, so in particular it preserves colimits meaning that the image of a colimit will be the colimit of an image. $\endgroup$
    – Alec Rhea
    Commented Dec 15, 2020 at 11:10

2 Answers 2

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Here is the explicit execution of "left adjoints preserve colimits":

\begin{align*} \mathrm{Hom}((A + B) \times C, D) &\cong \mathrm{Hom}(A + B, D^C) \\ &\cong \mathrm{Hom}(A, D^C) \times \mathrm{Hom}(B, D^C) \\ &\cong \mathrm{Hom}(A \times C, D) \times \mathrm{Hom}(B \times C, D) \\ &\cong \mathrm{Hom}((A \times C) + (B \times C), D) \end{align*}

Therefore, by the (covariant) Yoneda lemma, $(A + B) \times C$ is isomorphic to $(A \times C) + (B \times C)$.

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The distributivity condition is equivalent to asking that the functor $(-) \times C$ preserves coproducts for all $C \in \mathscr C$. This is easy to verify abstractly for any cartesian-closed category since $(-) \times C \dashv (-)^C$ and left adjoints preserve coproducts (indeed, any colimit), as Andrej Bauer proves in their answer.

Assuming you have a concrete characterisation of the products and coproducts in your category $\mathscr C$, you may simply unwind the definitions on both sides of the equation and check the two objects are isomorphic. This is because it suffices to show that any natural isomorphism exists between $(A + B) \times C$ and $A \times C + B \times C$ to imply distributivity.

For simpler heuristics, you can check that $\mathscr C$ satisfies the properties of any distributive category, e.g. the initial object must be strict.

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  • $\begingroup$ Thanks a lot to all of you for your answers, gives me enough tools to go forward. Thanks again. $\endgroup$
    – Gianfranco
    Commented Dec 15, 2020 at 16:38

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