The smooth structure on G/H is independent of the local complement

Define ${\mathrm{\Psi }}_i:{\mathfrak{m}}_i\times H\to G$ by ${\mathrm{\Psi }}_i(X,h)=\exp (X)h$. By [F1], its differential at $(0,e)$ is $(X,Y)\mapsto X+Y$, which is an isomorphism because $\mathfrak{g}={\mathfrak{m}}_i\oplus \mathfrak{h}$. The inverse function theorem in [F1] therefore supplies product neighborhoods $W_i\times V_i$ on which ${\mathrm{\Psi }}_i$ is a diffeomorphism; after the standard continuity shrinking, $S_i=\exp (W_i)$ meets each represented coset once. The corresponding quotient coordinate map sends $\exp (X)H$ to $X$.

Consider a point in the overlap of the two quotient chart domains. After translating both constructions to that point and shrinking, every representative from $S_1$ lies in the product neighborhood for ${\mathrm{\Psi }}_2$. If $X\in W_1$, write ${\mathrm{\Psi }}_2^{-1}(\exp X)=(Y(X),h(X))\mathrm{.}$ Both components are smooth, and $\exp (X)H=\exp (Y(X))H$; therefore the transition from the ${\mathfrak{m}}_1$-coordinate to the ${\mathfrak{m}}_2$-coordinate is precisely $X\mapsto Y(X)$, the first component of ${\mathrm{\Psi }}_2^{-1}\circ \exp {\mathrm{\mid }}_{W_1}$. It is smooth.

Interchanging $1$ and $2$ produces the smooth inverse transition. Translations are diffeomorphisms, so the same calculation handles all translated charts. Thus the two atlases are smoothly compatible and generate the same maximal atlas. Zero-dimensional complements, $H=G$, and $H=\{e\}$ cause no exception. Choice is inherited only through [A1].