Lemma 2

For every point $x\in X$, the mapping $\mu_x: E_x\times M\rightarrow X$ given by $(g, u)\mapsto g\cdot u$ is universally strict, and its image $E_x\cdot M$ is a neighbourhood of $x$ in X. In particular, $S\cdot M$ is a neighbourhood of M.

Under assumption (i), the mapping $\mu_x$ is open, since $E_x\times M$ is open in $G\times M$. Its image is therefore open in X.

Now suppose that assumption (ii) is satisfied. The mapping $\mu_x$ is proper (TG, I, p. 6, prop. 4, and p. 75, th. 1), hence universally strict (I, p. 20, corollary 11). Moreover, $(G-E_x)\cdot M$ is a closed subset of X which does not contain $x$, so that $E_x\cdot M$ is a neighbourhood of $x$.

The last assertion results from the fact that $E_x\subset S$ if $x\in M$.