The normal bundle of the diagonal is canonically the tangent bundle

In a chart of $M$, the product tangent coordinates identify the splitting [F2] with the two coordinate-vector blocks, and the difference map $\mathrm{\Phi }(v,w)=w-v$ has the constant matrix $[-II]$. Hence the splitting and $\mathrm{\Phi }$ are smooth in the product charts [F3], without requiring a general smooth-differential theorem. The diagonal inclusion has coordinate expression $x\mapsto (x,x)$, so its differential [F4] sends $v$ to $(v,v)$. Thus $\ker \mathrm{\Phi }=T{\mathrm{\Delta }}_M$. By [F5] and [F6] the quotient is a smooth bundle, and its induced map to $TM$ has smooth inverse $u\mapsto [(0,u)]$, as is also seen in these local trivializations [F7]. These formulas include $n=0$ and empty $M$.

Orientation. In an oriented basis $(e_1,\dots ,e_n)$ of $T_{x}M$ the product orientation of $T_{x}M\oplus T_{x}M$ is the class of $((e_1,0),\dots ,(e_n,0),(0,e_1),\dots ,(0,e_n))$ [F8]. The diagonal basis ${\delta }_i=(e_i,e_i)$ is obtained from it by the block matrix $\left(\begin{array}{cc}I& 0\\ I& I\end{array}\right)$ of determinant $+1$, while the lifts $(0,e_j)$ of the normal classes satisfy $\mathrm{\Phi }(0,e_j)=e_j$. Hence the ordered basis $({\delta }_1,\dots ,{\delta }_n,(0,e_1),\dots ,(0,e_n))$ is positive in the product orientation exactly when the $e_j$ are positive in $T_{x}M$: with the tangent-first convention [F9] the induced normal orientation is carried by $\mathrm{\Phi }$ to the tangent orientation transported from $M$, so the isomorphism is orientation-preserving. For $n=0$, write the supplied tangent orientation unit as $o=\pm 1$: the ambient product unit is $o^2=1$, and the tangent-first rule gives normal unit $o^{-1}=o$, preserved by the unique rank-zero isomorphism.