In [Selinger, P. *A survey of graphical languages for monoidal categories* (New Structures for Physics, Springer, 2011, 813, 289-233)](http://www.mscs.dal.ca/~selinger/papers/graphical.pdf%E2%80%8E), 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.


Any hint is welcome.



Why is a braided left autonomous category also right autonomous?
Here's a hint: what you should probably use here is the method of string diagrams (due to Joyal and Street, but by now ubiquitous). In other words, draw a picture in terms of tangles; you will see two crossings, and you want to apply a series of Reidemeister II and III moves to make those crossings cancel out; these moves can be justified in any braided monoidal category. (It turns out that what you want to prove is an algebraic form of the Whitney trick applied to ordinary knots; see Kauffman's On Knots; look carefully on [page 172](http://books.google.com/books?id=BLvGkIY8YzwC&pg=PA171&lpg=PA171&dq=on+knots+kauffman+whitney+trick&source=bl&ots=m3bG8g1ezd&sig=W1sr2I3_VRkqGlXCTMw-McoRmAQ&hl=en&sa=X&ei=Qq5CU4_iNqbp2QWbx4HwAQ&ved=0CDkQ6AEwAg#v=onepage&q=on%20knots%20kauffman%20whitney%20trick&f=false).) 


If this hint is insufficient, I can come back and explain more (and I probably will if no one else answers; I'd have to sit down and LaTeX the commutative diagram though). The Reidemeister II move is essentially based on invertibility of the braiding $c$, and the Reidemeister III move on naturality of $c$ and the braiding axiom (a hexagonal coherence condition, but really a triangular coherence condition if you assume, as you may, that the associativity is strict). 


**Edit:** I have written out a complete derivation [here](http://ncatlab.org/toddtrimble/published/Whitney+trick). 


