Kernel of the infinitesimal orbit map

For every $g,h\in G$, ${\mathrm{\Phi }}_x(gh)=g\cdot {\mathrm{\Phi }}_x(h)$. Left translation by $g$ on $G$ and action by $g$ on $M$ are diffeomorphisms, so differentiating shows that $d({\mathrm{\Phi }}_x{)}_g$ has the same rank as $d({\mathrm{\Phi }}_x{)}_e$. Thus ${\mathrm{\Phi }}_x$ has constant rank.

The fibre ${\mathrm{\Phi }}_x^{-1}(x)$ is $G_x$ by definition. Apply the local normal form [F3] at $e$: the tangent space of this fibre is the kernel of $d({\mathrm{\Phi }}_x{)}_e$. Because [F2] gives the fibre its embedded structure, $\ker d({\mathrm{\Phi }}_x{)}_e=T_e{G}_x={\mathfrak{g}}_x$.

By [F1], the infinitesimal map is $-d({\mathrm{\Phi }}_x{)}_e$. Multiplication by $-1$ does not change kernel or image, so step 2.1 proves $\ker (X\mapsto X_M(x))={\mathfrak{g}}_x$ and identifies its image with $\mathrm{im}d({\mathrm{\Phi }}_x{)}_e$.

The orbit map is constant precisely on left cosets of $G_x$, so [F5] gives a bijection ${\bar{\mathrm{\Phi }}}_x:G\mathrm{/}{G}_x\to G\cdot x$. It is smooth because the submersion charts in [F4] provide local smooth sections $s$ of $q$ and ${\bar{\mathrm{\Phi }}}_x={\mathrm{\Phi }}_x\circ s$ locally. At $e{G}_x$, its differential is the map induced by $d({\mathrm{\Phi }}_x{)}_e$ on $\mathfrak{g}\mathrm{/}{\mathfrak{g}}_x$; steps 2.1 and 3.1 make it injective with image $\mathrm{im}d({\mathrm{\Phi }}_x{)}_e$. Equivariance translates this calculation to every coset, so ${\bar{\mathrm{\Phi }}}_x$ is an injective immersion and its image carries the canonical immersed-orbit structure.

Under that structure, step 4.1 gives $T_x(G\cdot x)=\mathrm{im}d({\mathrm{\Phi }}_x{)}_e=\{X_M(x):X\in \mathfrak{g}\}$. The stabilizer is nonempty and may be all of $G$; then the orbit tangent and quotient are zero. A trivial stabilizer gives kernel zero. Disconnected groups and noneffective actions are allowed. There is no metric, boundary, endpoint, or biconditional. ${\mathrm{A}\mathrm{C}}_{\omega }$ is used through [F2] and [F4], and the pointwise linear algebra adds no choice.