Consider a symmetric monoidal category, and assume that it is closed, i.e. internal Homs exist. Recall that an object $X$ is dualizable if the canonical map $X \otimes DX \to \operatorname{Hom}(X,X)$ (obtained as the adjoint of the map $X \otimes DX \otimes X \to X \otimes 1 \to X$) is an isomorphism.

Apparently it's a well-known fact that dualizability of $X$ implies that the map $Y \otimes DX \to \operatorname{Hom}(X,Y)$ (obtained similarly) is also an isomorphism, for all $Y$. I did not find a complete proof of this fact anywhere.

I have attempted it myself. My first attempt was to use the Yoneda Lemma, which failed. Later I managed to find a map in the opposite direction, namely a composition $\operatorname{Hom}(X,Y) \to \operatorname{Hom}(X,X) \otimes \operatorname{Hom}(X,Y) \to X \otimes DX \otimes \operatorname{Hom}(X,Y) \to DX \otimes Y$, where the first arrow is obtained from the unit of the adjunction, and the second from dualizability. I strongly suspect this provides an inverse, thus proving the result. I did not mention to show this. Any hints would be appreciated.


This is an awkward definition of dualizability for proving things with; among other things, with the correct definition you don't need to assume closure, and dualizable objects are obviously preserved by monoidal functors. With the standard definition involving unit-counit / evaluation-coevaluation maps, it's clear that we always have an adjunction

$$\text{Hom}(X \otimes Z, Y) \cong \text{Hom}(Z, X^{\ast} \otimes Y)$$

(I'm being a little sloppy about which things should go on the left vs. the right because we're in a symmetric monoidal category; if we weren't I'd have to be a lot more careful). On the other hand, by the definition of closure we have

$$\text{Hom}(X \otimes Z, Y) \cong \text{Hom}(Z, [X, Y])$$

so we're done by the Yoneda lemma.

It remains to prove that the definition you give is equivalent to the standard definition. The argument above implies that the standard definition implies your definition; to go from your definition to the standard definition requires first finding the evaluation and coevaluation maps. The evaluation map is the dual pairing between $X$ and $DX = [X, 1]$; the coevaluation map comes from the inclusion of the identity $1 \to [X, X] \cong X \otimes DX$. I don't know off the top of my head how to show the zigzag identities, though.

  • $\begingroup$ Thank you. Ultimately to prove the equivalence you still end up having to prove some set of identities though. I'll give it a try, perhaps just working straight from the coherence axioms. $\endgroup$ – I I Jun 6 '18 at 15:09

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