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I'm having trouble understanding the following statement:

"In a rigid category, duals are unique up to unique isomorphism."

It seems to me that this isomorphism is not unique.

Let me try to give a counterexample: Let $(X,Y,\epsilon: X \otimes Y \to I,\eta: I: Y \otimes X)$ be a dual pair (satisfying the snake identities).

Now, let $f:Y \to Y$ be any isomorphism other than the identity. We define $$h := \left(f^{-1} \otimes 1_X\right) \circ \eta$$ $$e := \epsilon \circ (1_X \otimes f)$$ Claim: $(X, Y, e, h)$ is a dual pair. Let's prove the snake identities (in a strict monoidal category): $$(e \otimes 1_X)\circ(1_X \otimes h) = \left((\epsilon \circ \left(1_X \otimes f\right)) \otimes 1_X) \circ \left(1_X \otimes \left(\left(f^{-1} \otimes 1_X\right) \circ \eta\right)\right)\right)\\ =(\epsilon \otimes 1_X) \circ (1_X \otimes f \otimes 1_X) \circ \left(1_X \otimes f^{-1} \otimes 1_X\right) \circ (1_X \otimes \eta)\\ =(\epsilon \otimes 1_X) \circ (1_X \otimes \eta) = 1_X$$ Similarly: $$(1_Y \otimes e) \circ (h\otimes 1_Y) = (1_Y \otimes (\epsilon \circ (1_X \otimes f))) \circ \left(\left(\left(f^{-1} \otimes 1_X\right) \circ \eta\right)\otimes 1_Y\right)\\ = (1_Y \otimes (\epsilon \circ (1_X \otimes f))) \circ \left(\left(\left(f^{-1} \otimes 1_X\right) \circ \eta\right)\otimes 1_Y\right)\\ = (1_Y \otimes \epsilon) \circ (1_Y \otimes 1_X \otimes f) \circ \left(f^{-1} \otimes 1_X \otimes 1_Y\right) \circ (\eta \otimes 1_Y)\\ = f^{-1} \circ (1_Y \otimes \epsilon) \circ (1_Y \otimes 1_X \otimes f) \circ (\eta \otimes 1_Y)\\ = f^{-1} \circ (1_Y \otimes \epsilon) \circ (\eta \otimes 1_Y) \circ f\\ = f^{-1} \circ f = 1_Y $$ I'm sorry for the lengthy formulae, it is an easy exercise when done in graphical calculus.

So if $Y$ has more than one automorphism (which will frequently be the case), we can define any number of different dualities. It seems that duals may be unique up to isomorphism, but this isomorphism itself is not unique. Am I wrong?

Turion
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    When we say "unique up to unique isomorphism", it must be understood in the right sense: namely, that there is a unique isomorphism of structured objects. For a much easier example, see here. – Zhen Lin Jul 22 '14 at 14:37
  • @ZhenLin, yes, that's why I prove the snake identities for the new dualities. Or is there any more structure that the new duality must be compatible with? – Turion Jul 22 '14 at 14:39
  • That's not the point. A dual for $X$ is a triple $(Y, \epsilon, \eta)$ such that etc.; and for any two duals, there is a unique isomorphism connecting them that is compatible with all the data. That in no way says anything about the automorphisms of $Y$ as a bare object. – Zhen Lin Jul 22 '14 at 14:47
  • @ZhenLin, what does "compatible with all that data" mean, other than "define the new $\epsilon$ and $\eta$ by composition" and "the new dual satisfies the snake identities"? As I see it now, my $(Y, e, h)$ is another dual and $f$ is a unique iso compatible with the data. – Turion Jul 22 '14 at 14:51
  • Oh, maybe I understand what you're getting at: While $Y$ and $Y$ are the same objects, $(Y,\epsilon,\eta)$ and $(Y,e,h)$ are different duals. So the whole dual data is unique, but not the underlying object? – Turion Jul 22 '14 at 14:53
  • Yes. The same applies to limits, colimits, adjoints, etc. – Zhen Lin Jul 22 '14 at 14:56
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    @ZhenLin, great, thanks! Do you want to make this into an answer? (If you don't have the time I can do that as well) – Turion Jul 22 '14 at 15:02

1 Answers1

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Thanks to Zhen Lin to bring me on the right track. The misunderstanding is the following:

The dual object ($Y$ in my example) is unique up to isomorphism, but this isomorphism is not unique. However, a dual is more than merely the object, it's the triple $(Y,\epsilon,\eta)$. This one is indeed unique up to unique isomorphism, and the calculation in the question shows exactly this.

So the same dual object $Y$ can be part of different duals.

Turion
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