TITLE: Why is a braided left autonomous category also right autonomous?
QUESTION [8 upvotes]: 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.
Any hint is welcome.

REPLY [12 votes]: (Apologies for problems with the LaTeX; I’m on a dodgy internet connection and having trouble previewing to fix it.  Will attempt to fix it later when home.)
With any question like this — deriving an equation in some class of structures with a graphical language — my modus operandi is:

work it out in string diagrams;
translate back to algebraic language, if necessary.

This one comes out in string diagrams as:

(Note that my diagrams go bottom-to-top for morphisms, and left-to-right for tensoring.) Translating this gives:
\begin{align}
& \mathrel{\phantom{=}} (\epsilon \otimes B) \cdot (c_{B,A} \otimes B) \cdot (B \otimes c_{A,B}^{-1}) \cdot (B \otimes \eta) \\
\quad & = (\epsilon \otimes B) \cdot (A \otimes c^{-1}_{B,B}) \cdot (c^{-1}_{A,B} \otimes B) \cdot (B \otimes c_{B,A}) \cdot (c_{B,B} \otimes B) \cdot (B \otimes \eta) \\
\quad & = (\epsilon \otimes B) \cdot (A \otimes c^{-1}_{B,B}) \cdot (c^{-1}_{A,B} \otimes B) \cdot (\eta \otimes B) \\
\quad & = (B \otimes \epsilon) \cdot (\eta \otimes B) \\
\quad & = 1_B
\end{align}
Here the first step is by the “Yang-Baxter” hexagon equation for braids; the second uses the fact that $(B \otimes c_{B,A}) \cdot (c_{B,B} \otimes B) = c_{B,B \otimes B}$, plus the naturality of braiding against $\eta$; the third is similar to the second; and the last is the triangle equality for the original pair of duals.
The other triangle equality for the new pairing is similar.
(To be honest, I often use this sketch-then-algebraise approach even when I don’t know yet if the graphical language is formally valid in the setting — it can still be very helpful for quickly finding algebraic proofs, and if I’ve used some illegal manipulation, then I’ll find that out when I try to translate it.)