Is incompatibility required for contextuality? A preprint just came out claiming that incompatibilty is not requried to demonstrate generalized contextuality. My question isn't about generalized contextuality---which I don't quite understand---but rather about "textbook" contextuality of the Kochen-Specker variety. The involvement of incompatible observables seems to be a ubiquitous ingredient for contextuality, but it has never been clear to me as to why it's needed.
EDIT:
I should de-emphasize my reference to the KS theorem since incompatibility is required for it by definition. My question is instead about the need for incompatibility to observe contextuality in general. I define the latter with Abramsky's definition as measurement statistics which are "locally consistent but globally inconsistent". In that same paper, he states that

the  key  ingedient  of  quantum  mechanics  which  enables  the
possibility  of  contextual phenomena, while still allowing a
consistent description of our actual empirical observations,is the
presence of incompatible observables.

Although that seems to be the case in almost all examples of contextuality, I can't find any justification for why incompatibility is a key ingredient. There could very well be a set of globally incompatible observations which can all be performed together. If not, why not?
 A: First, let's talk about what Abramsky means when he says that measurements are "locally consistent". Suppose we have a set of three measurements $M_{1},M_{2},M_{3}$ every which can take only values $\pm 1$, and our contexts are given by $\{ \, \{M_{1},M_{2} \}, ,\ \{M_{2},M_{3}\} \, \}$, that is, only $M_{1}$ and $M_{3}$ are not compatible. If we set the behavior of our system(the context's probabilities) to be the extreme distributions $p_{M_{1},M_{2}}(+1,+1) = 1$ and $p_{M_{2},M_{3}}(-1,-1) = 1$ we get
different probabilities for $M_{2}$ when we marginalize $p_{M_{1},M_{2}}$ and $p_{M_{2},M_{3}}$. We say that this is a disturbing scenario and this is what Abramsky would call "locally inconsistent". It happens that many theories, in particular classical probability theory and quantum mechanics, are non-disturbing. That is, if the measurement $M$ is in the context $C_{1}$ as well as in context $C_{2}$, we have
$$
Prob(M = m | C_{1}) = Prob(M = m |C_{2})
$$
Where $Prob(M=m|C)$ is just the probability of finding $M=m$ when we marginalize though the context $C$. Or, in general, if $U \subseteq C_{1} \cap C_{2}$, $p_{U}^{C_{1}} = p_{U}^{C_{2}}$ where $p_{U}^{C}$ is the probability distribution for finding a specific set of value for the measurements in $U$ obtained by marginalization in $C$.
If we accept only non-disturbing models and we suppose that every measurement is compatible, we will have only one big maximal context $C$ such that every measurement is in it. Therefore, of course, we can define a global distribution of probabilities that is simply $p_{C}$ and the distribution of every set of measurements is defined $p_{U} = p_{U}^{C}$. We have no contextuality effects!
This is enough to show that in quantum mechanics the answer is yes, we do need incompatibility for contextuality effects to arise. If, on the other hand, we allow for disturbing models, we are left with the question: if I have a set of compatible measurements $C$ and $U \subseteq C$, can I say that $p_{U}^{C} = p_{U}$? The answer is yes.
Proof:
Let's assume that our context is finite $C = \{ M_{1},..., M_{N} \}$ and that, without loss of generality, $U = \{ M_{1},...,M_{u}\}$ with $u\leq N$. Since $C$ is a context, it has a refinement $M$ (that is, a measurement) associated with it along with functions $f_{1},f_{2},...,f_{N}$ such that
$$
Prob(M_{1} = m_{1}, ... , M_{N} = m_{N}|C) = \sum_{f_{1}(m) = m_{1}   \, and \, \cdots \, and f_{N}(m)=m_{N}} Prob(M = m) 
$$
and, as well,
$$
Prob(M_{1} = m_{1}, ... , M_{u} = m_{u}|C) = \sum_{f_{1}(m) = m_{1}   \, and \, \cdots \, and f_{u}(m)=m_{u}} Prob(M = m) 
$$
what we wish to prove is that
$$
Prob(M_{1} = m_{1}, ... , M_{u} = m_{u}|C) = \sum_{m_{u+1},\cdots,m_{N}} Prob(M_{1} = m_{1}, ... , M_{N} = m_{N}|C)
$$
Well,
\begin{align*}
\sum_{m_{u+1},\cdots,m_{N}} Prob(M_{1} = m_{1}, ... , M_{N} = m_{N}|C) &=  \sum_{m_{u+1},\cdots,m_{N}} \left( \sum_{f_{1}(m) = m_{1}   \, and \, \cdots \, and f_{N}(m)=m_{N}} Prob(M = m) \right)  \\
& \text{since we are adding every value of $m_{u+1},\cdots,m_{N}$} \\
&= \sum_{f_{1}(m) = m_{1}   \, and \, \cdots \, and f_{u}(m)=m_{u}} Prob(M = m) \\
&= Prob(M_{1} = m_{1}, ... , M_{u} = m_{u}|C)
\end{align*}

The example of disturbing scenario and the formalism and notation used to define compatibility are taken from the book

Amaral, B., & Cunha, T. M. (2018). On Graph Approaches to Contextuality and their Role in Quantum Theory (SpringerBriefs in Mathematics) (1st ed. 2018 ed.). Springer. https://doi.org/10.1007/978-3-319-93827-1'

