TITLE: What are the dualizable objects with respect to pushout-product?
QUESTION [6 upvotes]: 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).

REPLY [6 votes]: 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 $C^{[1]}$.