# Why is a braided left autonomous category also right autonomous?

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:

Any hint is welcome.

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