Tangent and cotangent bundles extend over a boundary

If $n=0$, every smooth germ is constant, so [L2] makes every derivation zero and the empty coordinate family is a basis. Assume $n\ge 1$. For a boundary germ, define ${\mathrm{\partial }}_i{\mathrm{\mid }}_p$ by differentiating any smooth Euclidean extension in the $i$th coordinate. By [L1] this is well defined; linearity and the Euclidean product rule make it a derivation.

Let $v$ be any derivation and let $F$ extend a representative of a boundary germ $f$ near the coordinate point $a$. By [L2], write $F(x)-F(a)={\sum }_i(x^i-a^i)g_i(x)$ with $g_i(a)={\mathrm{\partial }}_{i}F(a)$. Restricting to the half-space and applying $v$, using [L2] and the Leibniz rule, gives $v(f)={\sum }_{i}v(x^i){\mathrm{\partial }}_i{\mathrm{\mid }}_p(f)$. Thus the coordinate derivations span. Applying a linear relation among them to each coordinate germ proves independence, so $\dim T_{p}M=n$.

By [L4], differentiating a boundary-chart transition and its inverse gives mutually inverse matrices; [L1] makes these derivatives extension independent, and their entries vary smoothly. They are the tangent transition maps. By [L3], the dual inverse matrices are the cotangent transition maps. Hence the usual tangent and cotangent bundles extend smoothly over all of $M$, including the $n=0$ case with empty matrices.