The constant-rank theorem for manifolds

Choose charts $(U_0,\alpha )$ at $p$ and $(V_0,\beta )$ at $F(p)$ as in [L3], and let $f:=\beta \circ F\circ {\alpha }^{-1}$. By [L3], the maps ${\alpha }^{-1}$ and $\beta$ are diffeomorphisms, so their differentials are linear isomorphisms. Therefore [L1] gives $Df(\alpha (x))=d{\beta }_{F(x)}\circ d{F}_x\circ d({\alpha }^{-1}{)}_{\alpha (x)}$ for every $x$ near $p$, and the rank of $Df(\alpha (x))$ equals the rank of $d{F}_x$. Using [F1], after shrinking $U_0$ the map $f$ has constant rank $r$ on an open Euclidean neighbourhood of $\alpha (p)$.

After permuting source and target coordinates, use [L2] to form the explicit source map $\mathrm{\Phi }$ from the first $r$ components of the smooth map $f$ and the remaining source coordinates. Thus $\mathrm{\Phi }$ is smooth. Its local inverse is unique and is $C^k$ for every finite $k$ by [L4], hence is smooth. Put $g:=f\circ {\mathrm{\Phi }}^{-1}$. By [L5], after shrinking to a product neighbourhood, $g(u,v)=(u,h(u))$; here $h$ is smooth because $g$ is smooth. The target shear $\tilde{\beta }(z,w):=(z,w-h(z))$ and its explicit inverse $(z,w)\mapsto (z,w+h(z))$ are smooth. Therefore $\tilde{\beta }\circ f\circ {\mathrm{\Phi }}^{-1}(u,v)=(u,0)$ near the distinguished point.

Compose the original source chart with $\mathrm{\Phi }$ and the target chart with the shear: $\phi :=\mathrm{\Phi }\circ \alpha$ and $\psi :=\tilde{\beta }\circ \beta$. By [L3] these are smooth charts, and their coordinate representative for $F$ is the normal form from step 2.1.

Therefore $F$ has the claimed local slice form around $p$. When $r=0$ or $r=\min \{m,n\}$, the empty blocks in the displayed formula are interpreted in the usual way.