Why do we have $s.\chi_x=\chi_x$ implies that $s(x)=x u$ for some $u \in \mathfrak{C}^{*, 0}$ in an inverse semigroup?

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This is a detail in Theorem 4.4 on Page 9 of Xin Li's paper Left Regular Representations of Garside Categories I. C-star-Algebras and Groupoids.

In the proof he said that: $$s.\chi_x=\chi_x \text{ implies that } s(x)=x u \text{ for some } u \in \mathfrak{C}^{*, 0}$$

where $\mathfrak{C}$ is a left cancellative small category, $\mathfrak{C}^{*}$ denotes the set of invertible element in $\mathfrak{C}$ and $\mathfrak{C}^{*, 0} = \{u \in \mathfrak{C}^{*}, \mathfrak{t}(u) = \mathfrak{s}(u)\}$. Let $I_l$ be the inverse semigroup generated by the partial bijections arising from $\mathfrak{C}$. Let $J$ be the semilattice of idempotents in $I_l$. The multiplicative character $\chi_x: J \rightarrow \{0,1\}$ is defined by $\chi_x(e) = 1$ if $x\mathfrak{C} \leq e$, and $\chi_x(e) = 0$ otherwise. For any $s \in I_l$, we define $s.\chi_x(e) = \chi_x(s^{-1} e s)$. Details can be seen in the paper.

What I have no idea about is why this statement $s.\chi_x=\chi_x \text{ implies that } s(x)=x u \text{ for some } u \in \mathfrak{C}^{*, 0}$ holds? I am seeking an argument for this. I would appreciate it if anyone could give a convincing proof of this statement.