# Spelling out uniqueness in a rigid category

As has been discussed in other posts, the dual of an object in a rigid category is unique up to unique isomorphism. As highlighted here, this does not mean that, for any two duals $(X^*,\epsilon,\nu)$ and $(X^\wedge,\epsilon',\nu')$, there exists a unique isomorphism (as objects in the category) between $X^*$ and $X^{\wedge}$. Instead it means a unique isomorphism that respects the evaluation and coevaluation maps of each dual.

I am having trouble finding what respects actually means, and how it implies uniqueness. I would guess that, for $\phi:X^* \to X^\wedge$ saying that such an isomorphism respects the structures means $$\nu' = (\text{id} \otimes \phi)\circ \nu:1 \to X \otimes X^{\wedge}, ~~~~~ \epsilon = \epsilon' \circ (\phi \otimes \text{id}):X^{*} \otimes X \to 1.$$ However, I can't conclude uniqueness from this.

## 1 Answer

You can recover $\phi$ from $\nu'=(\mathrm{id}\otimes\phi)\circ \nu$ by applying $\epsilon$ and using the relation between $\epsilon$ and $\nu$. This forces $\phi$ to be unique.

In fact, you can just write down what $\phi$ has to be: $$(\epsilon\otimes\mathrm{id})\circ(\mathrm{id}\otimes\nu')$$ (Or rather $\phi$ is the map $X^*\rightarrow X^{\wedge}$ arising in the canonical way from this map $X^*\otimes 1\rightarrow 1\otimes X^{\wedge}$)

• Thanks for the answer Oscar. However, I'm still a little confused: Applying $\epsilon \otimes \text{id}_{X^\wedge}$ to $\text{id}_{X^*} \otimes \nu' = (\text{id}_{X^*} \otimes \text{id}_X \otimes \phi) \circ (\text{id}_{X^*} \otimes \nu)$ gives on the left hand side $(\epsilon \otimes \text{id}_{X^\wedge}) \circ (\text{id}_{X^*}\otimes \nu')$ as you have. (Other side in the next comment.) – Pavel Katzo Nov 13 '15 at 13:37
• Now on the right hand side it gives $(\epsilon \otimes \text{id}_{X^\wedge})\circ (\text{id}_{X^*} \otimes \text{id}_{X} \otimes \phi) \circ (\text{id}_{X^*} \otimes \nu)$. You say that this is equal to $\phi$. Does this follow from the fact that we can "commute" $(\epsilon \otimes \text{id}_{X^\wedge})$ and $(\text{id}^{\otimes 2} \otimes \phi)$ by bifunctoriallity, and then use the defining identity $(\epsilon \otimes \text{id}_{X^*}) \circ (\text{id}_{X^*} \otimes \nu) = \text{id}_{X^*}$ to get the desired expression for $\phi$? – Pavel Katzo Nov 13 '15 at 13:37
• Yeah, exactly what you said. – Oscar Cunningham Nov 13 '15 at 18:48