Proving a commutative diagram involving the associator and two 2-cells in a bicategory Let $f \; : \; A \to A$, $g \; : \; A \to B$ and $h \; : \; B \to B$ be 1-cells, and $\sigma \; : \; g \cdot f \; \to \; g$ and $\theta \; : \; g \; \to \; h \cdot g$ be 2-cells in a bicategory.
I believe that the following equation holds in any bicategory:
$$
\theta \circ \sigma = \text{id}_h \cdot \sigma \; \circ \; \alpha_{h, g, f} \; \circ \; \theta \cdot \text{id}_f
$$
(where $\alpha$ is the associator.)
Am I right?
If yes, what is an elementary proof of it?
If no, what is a counterexample? What is the closest statement that is true? What is its elementary proof?
Edit: This is wrong as it is as pointed out by @Berci in a comment below. What I am looking for is general conditions that allow σ and θ to be swapped (possibly with additional unitors and associators).
 A: If you view a monoidal category $\mathcal V$ as a one-object bicategory, your question reduces to asking for general conditions for the following to hold: you have $U,V,W\in\mathcal V$, with morphisms $\sigma:V\otimes U\to V$ and $\theta:V\to W\otimes V$ such that
$$
\theta\circ\sigma = (W\otimes\sigma)\circ\alpha_{UVW}\circ(\theta\otimes U) \tag{1}
$$
Consider $\mathcal V = \mathbf{Set}$ with its Cartesian monoidal structure (which is even symmetric!), then the above equation is saying for all $(v,u)\in V\times U$ that $\theta(\sigma(v,u))=(\theta_1(v),\sigma(\theta_2(v),u))$, where $\theta=(\theta_1,\theta_2)$.
In other words, $(1)$ holds in $\mathbf{Set}$ iff $\theta_1(\sigma(v,u))=\theta_1(v)$ and $\theta_2(\sigma(v,u))=\sigma(\theta_2(v),u)$. Similar to Berci's comment, even if $U=W=*$ are singletons (so that $\sigma,\theta:V\to V$), equation $(1)$ reduces to $\theta\circ\sigma=\sigma\circ\theta$. What are general conditions for functions over a set to commute? This isn't even necessarily true if $\theta,\sigma$ are isomorphisms.
This is a long-winded way of saying that even general conditions for your desired property to hold are going to be hard to come by. An unsatisfactory sufficient condition for your property to hold is that the bicategory is locally thin, but this by definition just means that parallel $2$-cells are necessarily equal.
That being said, your statement strikes me as perhaps somehow related to the exchange law, since for example this tells us that $2$-cells $\operatorname{id}_x\Rightarrow\operatorname{id}_x$ will necessarily commute by the Eckmann-Hilton argument. However, this only directly applies when $A=B$ and $g=f=h=\operatorname{id}_A$ and can immediately fail once you deviate from this. (In the context of a monoidal category, these conditions are saying that $U,V,W$ are all taken to be the tensor unit).
In fact, even under these restrictions, this property only holds up to unitors.
