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I trying to understand the proof of Proposition 2.10.5 from the book: TENSOR CATEGORIES, by P. Etingof, S. Gelaki, D. Nikshych, and V. Ostrik, See http://www-math.mit.edu/~etingof/egnobookfinal.pdf.

Proposition 2.10.5. If $X\in \mathcal{C}$ has a left (respectively, right) dual object, then it is unique up to a unique isomorphism.

Something I did not understand is why the three small squares of the following diagram commute? enter image description here

where $e_1, c_1, e_2, c_2$ the corresponding evaluation and coevaluation morphisms

Eric Wofsey
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Dan
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1 Answers1

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Let's take the top left square. We can rewrite it as $$\require{AMScd} \begin{CD} X_1^*\otimes 1\otimes 1 @>{id\otimes id\otimes c_1}>> X_1^*\otimes 1\otimes (X\otimes X_1^*)\\ @V{id\otimes c_2\otimes id}VV @V{id\otimes c_2\otimes id}VV \\ X_1^*\otimes (X\otimes X_2^*)\otimes 1 @>{id\otimes id\otimes c_1}>> X_1^*\otimes(X\otimes X_2^*)\otimes(X\otimes X_1^*) \end{CD}$$ where $1$ is the unit object. Commutativity of this square is now immediate from the fact that $\otimes$ is a bifunctor.

The other two squares are similar: once you explicitly write out the unit objects rather than implicitly eliminating them as is done in the original diagram, you are just using the functoriality of $\otimes$.

Eric Wofsey
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  • Thank you Eric! This is very nice explanation. I think there is a typo: the right vertical arrow $id \otimes c_2 \otimes c_1$ should be $id \otimes c_2 \otimes id$. – Dan Aug 17 '15 at 06:54
  • Oops, yes, corrected now. – Eric Wofsey Aug 17 '15 at 07:41
  • Could you tell me why the triangle in the upper right corner commutes by axiom (2.43) applied to X_2 ^* (that what the authors say)? – Dan Aug 17 '15 at 08:45
  • The upper right triangle is just (2.43) tensored with $id:X_1^\to X_1^$ on both sides. If it helps, the diagonal map can be labelled as $id\otimes id\otimes id$ rather than just $id$. – Eric Wofsey Aug 17 '15 at 08:54
  • Perfect! . Thank you Eric! – Dan Aug 17 '15 at 08:58