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1:48 PM
@SaalHardali IIRC in a stable setting the equation $Hom(X,Z)=DX⊗Z$ is true for all X iff Z is dualizable (this is of course different than the unstable setting, like R-modules!)
This was worked out in some MO answer, but I cannot remember which one...
 
2:08 PM
@DenisNardin I too vaguely recall something of the sort, but didn't find it either and then I became skeptical.
 
2:30 PM
@SaalHardali So, let's see if I can whip up a proof. If $Z$ is dualizable we have $[T,DX⊗Z]=[T⊗DZ⊗X,1]=[T⊗X,Z]=[T,F(X,Z)]$ for every $T$, so $F(X,Z)=DX⊗Z$
The viceversa should follow from taking the unit map $1→F(X,X)=DX⊗X$ and show it satisfies the triangular identities
 
 
3 hours later…
5:24 PM
@DenisNardin I'm cometely convinced. So much so that my initial example is in fact wrong because the same proof works in any symmetric monoidal category. A counter example to my assertion is Hom(Q/Z,Q/Z) \tensor Q != Hom( Q/Z, Q/Z \tensor Q)
Also I have a terrible Déjà Vu of asking the exact same question and coming to this exact realization before.
Here's simple proof for me to never forget this:
$Hom(X,Z)=Hom(X \otimes D(Z), 1)=Hom(D(Z),D(X))=D^2(Z) \otimes D(X)= Z \otimes D(X) = D(X) \otimes Z$
 
6:16 PM
(What I meant is to write a mnemonic for your proof Denis of course)
 

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