Lemma 2

Let $\mathfrak{h}$ be a splitting Cartan subalgebra of $\mathfrak{g}$, and $s\in$ Aut$_0(\mathfrak{g},\mathfrak{h})$. Assume that the restriction of $s$ to $\sum_{\alpha\in R}\mathfrak{g}^{\alpha}$ does not have 1 as an eigenvalue.

Then $\varepsilon (s) = 1$.

By extension of $k$, we are reduced to the case where $s\in$ Aut$_e(\mathfrak{g},\mathfrak{h})$. The dimension of the nilspace of $s-1$ is at least dim $\mathfrak{h}$ (Chap. VII, §4, no. 4, Prop. 9). Hence $(s-1)|\mathfrak{h}$ is nilpotent. Since $s|\mathfrak{h}\in A(R^{\vee}),s|\mathfrak{h}$ is of finite order, and hence semi-simple (Chap. V, Appendix, Prop. 2). Consequently, $(s-1)|\mathfrak{h}= 0$, which proves that $\varepsilon (s) = 1$.