1
$\begingroup$

In Selinger, P. A survey of graphical languages for monoidal categories (New Structures for Physics, Springer, 2011, 813, 289-233), it is stated that:

Lemma 4.17 ([23, Prop. 7.2]). A braided monoidal category is autonomous if and only if it is right autonomous.

Proof. If $\eta : I \rightarrow B \otimes A$ and $\epsilon : A \otimes B \rightarrow I$ form an exact pairing, then so do $c^{-1}_{A,B} \circ \eta : I \rightarrow A \otimes B$ and $\epsilon \circ c_{B,A} : B \otimes A \rightarrow I$. Therefore any right dual of $A$ is also a left dual of $A$.

(The reference [23] is A. Joyal and R. Street. Braided tensor categories. Advances in Mathematics, 102:20–78, 1993.)

I don't understand why $c^{-1}_{A,B} \circ \eta$ and $\epsilon \circ c_{B,A}$ form an exact pairing. We need to prove that $((\epsilon \circ c_{B,A}) \otimes 1_B) \circ (1_B \otimes (c^{-1}_{A,B} \circ \eta)) = 1_B$ and $(1_A \otimes (\epsilon \circ c_{B,A})) \circ ((c^{-1}_{A,B} \circ \eta) \otimes 1_A) = 1_A$, but I can't see what axiom can be applied here.

In the graphical language, this is equivalent to:

diagrammatic version of the equation

Any hint is welcome.

$\endgroup$
0
$\begingroup$

The question has been answered on mathoverflow.net. The equality I forgot to use was that $c_{I,A} = 1_A$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.