# What do you call $C$ if $[D,C] = D^\vee \otimes C$ for all $D$?

This is different from $C$ being dualizable ($[C,D] = C^\vee \otimes D$). (EDIT: It turns out to be the same -- see Mike Shulman's answer!) But for example, if $C$ is a locally free sheaf of finite rank on a scheme/locally ringed space $X$, then $C$ has this property in the category of quasicoherent sheaves, or in the category of $\mathcal O_X$-modules -- just check locally.

Here I'm working in a symmetric monoidal closed category $\mathcal C$ with monoidal product $\otimes$, unit $I$, internal hom $[-,-]$, and $E^\vee$ denotes the dual $E^\vee = [E,I]$.

I'm really tempted to call such an object $C$ "locally free (of finite rank)", because the condition says that you can understand maps $D \to C$ as long as you internally (i.e. locally) understand maps $D \to I$, i.e. maps into the canonical "free" object.

Perhaps I should say "locally Cauchy-free" instead of "locally free (of finite rank)", since presumably the significance of "locally being able to take finite sums of $I$" is that finite sums of $I$ are $\mathcal C$-enriched absolute colimits (a.k.a. $\mathcal C$-enriched Cauchy colimits) in the additive context. But I don't understand what's going on well enough to firmly draw this connection.

• It still implies finite rank, does not it? – მამუკა ჯიბლაძე Feb 25 '18 at 4:56
• Moreover the other condition does not imply dualizability in general: for modules over a ring that one is equivalent to being finitely generated projective – მამუკა ჯიბლაძე Feb 25 '18 at 4:59
• @მამუკაჯიბლაძე I edited to include the finite rank condition shortly before your comment, but I missed a reference -- sorry for the confusion! By taking $C=D$ in either condition, one obtains a unit map $\eta: I \to C \otimes C^\vee$ which I'm pretty sure satisfies the triangle identities. So either condition implies dualizability. The condition $[C,D] = C^\vee \otimes D$ definitely holds if $C$ is dualizable, so it follows that that condition is equivalent to dualizability. I agree that for modules over a commutative ring, finitely generated projective=locally free of finite rank=dualizable. – Tim Campion Feb 25 '18 at 6:05
• @მამუკაჯიბლაძე So actually, I'm not sure I understand your second comment. Which condition doesn't imply dualizability? – Tim Campion Feb 25 '18 at 6:10
• Seems like we use different notions of dualizability. For me an object $C$ is dualizable if there is $X$ with $C\otimes X$ and $X\otimes C$ both isomorphic to $I$. I now realize that this is wrong, such objects are called invertible rather than dualizable. – მამუკა ჯიბლაძე Feb 25 '18 at 6:33

First of all, a nitpick: the condition "$[D,C] = D^\vee\otimes C$" should be stated more precisely as "the canonical map $D^\vee\otimes C \to [D,C]$ is an isomorphism". Now as you mentioned in a comment, it's well-known that the $\forall C$ version of this condition is equivalent to dualizability of $D$, and indeed the $C=D$ case is already equivalent to dualizability.
However, at least in a symmetric monoidal category, it seems to me that the $\forall D$ version is also equivalent to dualizability of $C$. As noted, taking $D=C$ it implies dualizability of $C$. But conversely, using the fact that if $C$ is dualizable then so is $C^\vee$, we have (in the imprecise version)
$$[D,C] = [D,[C^\vee,I]] = [C^\vee,[D,I]] = [C^\vee,D^\vee] = D^\vee \otimes C.$$
The connection to your comment about Cauchy-ness is that if $C$ is dualizable, then "copowers by $C$" are also a Cauchy colimit, coinciding with powers by $C^\vee$.