Proposition: Nonconstant channels need not compose nonconstantly

Composability of channels does not imply transitivity of operational sensitivity.

Proof. Let $X\in \{0,1\}$, let $N$ be an independent fair bit, and consider

Both channels are nonconstant, but $LK(X)=X\oplus N$ has the same fair-bit law for either value of $X$. An intervention on the intermediate pair can change the terminal output although changing $X$ cannot change its law.