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If $(\mathcal C, \otimes)$ is a monoidal category, the pushout product turns the arrow-category ${\mathcal C}^{[1]}$ into a monoidal category.

Adding more and more properties to $\otimes$ results in analogous properties being enjoyed by $\widehat{\otimes}$: for example, if $\otimes$ is symmetric, then so is $\widehat\otimes$, and if $(\mathcal C, \otimes)$ is left/right closed then so is $(\mathcal C^{[1]}, \widehat\otimes)$. Now,

Assume that $(\mathcal C, \otimes)$ is compact closed. It is not difficult to show that if $f \in\mathcal C^{[1]}$ (say $f :A\to B$ to fix notation) is invertible, then $f^\dagger = (f^{-1})^* : A^*\to B^*$ serves as its dual. But is this necessary? In poor words, if all objects are $\otimes$-dualizable, then an arrow is $\widehat\otimes$-dualizable iff it is invertible?

(the best guess I've been given is that $0\to X$ is dualizable even for nonzero objects -a compact closed category has a zero object, so the notation-; this is reasonable and I expect it, nevertheless I get lost in diagrams when I try to prove that there are suitable unit and counit).

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    $\begingroup$ It's certainly not true that every compact closed category has a zero object -- consider a group regarded as a discrete monoidal category. $\endgroup$ Feb 20, 2018 at 19:02
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    $\begingroup$ To get the pushout product don't you need ⊗ to preserve some kind of colimits? (otherwise my feeling is that you get only a symmetric multicategory, a priori) $\endgroup$ Feb 20, 2018 at 19:14
  • $\begingroup$ @DenisNardin I think you need it in order to make it symmetric; I have to check the other comment too. $\endgroup$
    – fosco
    Feb 20, 2018 at 21:26
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    $\begingroup$ Yes, for the pushout-product to exist you of course need $C$ to have pushouts, and for it to be associative I'm near-certain you need $\otimes$ to preserve pushouts in each variable. But the latter is automatic in a compact closed category that has pushouts, because it is in particular closed. I don't know offhand an example of a compact closed category that has pushouts but not a zero object, but I also can't think of any reason a zero object should be implied. $\endgroup$ Feb 21, 2018 at 10:06
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    $\begingroup$ And the unit object of the pushout product is $\emptyset \to I$, and for this to in fact be a unit you neet $\otimes$ to preserve $\emptyset$ (the initial object) in each variable. $\endgroup$ Feb 21, 2018 at 10:08

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Assuming that $C$ has finite colimits preserved in each variable by $\otimes$ (as is necessary for the pushout product to exist and be associative and unital), the functor $X\mapsto (\emptyset\to X)$ embeds $C$ strong-monoidally in $C^{[1]}$. In particular, it preserves dualizable objects, so if $C$ is compact closed then all objects of the form $\emptyset\to X$ are dualizable in $C^{[1]}$.

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  • $\begingroup$ If $C$ is closed, $\otimes$ does preserve any existing colimits $\endgroup$ Feb 21, 2018 at 10:30
  • $\begingroup$ Hm, thanks, it was easy. :-) Are initial arrows and isos all the dualizables of $C^{[1]}$ then? $\endgroup$
    – fosco
    Feb 21, 2018 at 11:14

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