It is well known that pentagon+triangle identity of type (a1b) implies "all diagrams commute" [monoidal category](http://ncatlab.org/nlab/show/monoidal+category), in particular triangle identities of type (1ab) and (ab1). My question is that whether pentagon+triangle identity of type (1ab), or (ab1), implies triangle identity of type (a1b)?



A question on triangle identities
The answer is: No for monoidal categories, but Yes for symmetric monoidal categories. Actually, you can get away with a lot less for the Yes. Intuitively, all you need is for the unit to commute with arbitrary elements, i.e. $\rho\_x = \lambda\_x \circ \tau: x\otimes 1 \rightarrow x$. 


The No is a result of the observation that if you only have the triangle identity of type (1ab) then you only have diagrams involving $\lambda$ and $\alpha$, which means you have no hope of getting the diagrams involving $\rho$ to commute. For the readers who don't want to dig through the nLab article, a diagram of type (1ab) means:


$\begin{array}{ccc} 1\otimes (a\otimes b) & \rightarrow & (1\otimes a)\otimes b\\\
\\\
\lambda\_{a\otimes b}\downarrow & & \downarrow \lambda\_{a} \otimes 1\_b\\\
\\\
a\otimes b & = & a\otimes b\\\
\end{array}$


The associative pentagon only deals with $\alpha$, and without some kind of twist map there is no way to relate $\alpha$ and $\lambda$ with $\rho$. Thus, I don't think the diagram of type (a1b) will commute, since the left vertical map in that diagram will be $\rho\_a \otimes 1\_b:(a\otimes 1)\otimes b \rightarrow a\otimes b$


Now, to get the Yes we require the following to commute:


$\begin{array}{ccc} x\otimes 1 & \rightarrow^\tau & 1\otimes x\\\
\\\
\lambda\_{x}\downarrow & & \downarrow \rho\_{x}\\\
\\\
x & = & x\\\
\end{array}$


From this it is clear how to relate $\rho$ and $\tau$. I verified that (a1b) holds by drawing a diagram with $a\otimes b$ in the center and with the outside a bunch of triangles which all commuted except one (and therefore the last must commute). Here's a list of the vertices of the triangles I used. I'll leave the diagram chase to you (mostly because it would be a nightmare to tex in this environment): $a(1b), a(b1), (ab)1, 1(ab), (1a)b, (a1)b$. There could very well be an easier proof.


