At interior base points, the graph faces of an adapted presentation induce the outward unit normal

By [F3] the map ${\phi }_j$ is $C^1$ on an open neighbourhood of $D_j$, so ${\psi }_j={\pi }_k\circ {\phi }_j$ is $C^1$ there by [F7], and by [F2] and [L2] its Jacobian determinant at $c$ is nonzero, hence $D{\psi }_j(c)$ is invertible. So [L1] supplies open sets $P\ni c$ and $Q\ni w_0$ with ${\psi }_j{\mathrm{\mid }}_P:P\to Q$ bijective and with $C^1$ inverse $\lambda :Q\to P$; shrinking $P$ and $Q$, which stays possible because $D_j^{\circ }$ and $D^{\circ }$ are open and contain $c$ and $w_0$, we may take $P\subseteq D_j^{\circ }$ and $Q\subseteq D^{\circ }$.

Let $w\in Q$. By [F2] the point ${\phi }_j(\lambda (w))$ lies in the graph of $\gamma$ over $D$, and its ${\pi }_k$-projection is ${\psi }_j(\lambda (w))=w$, so ${\phi }_j(\lambda (w))={\sigma }_k^{-1}(w,\gamma (w))$. Reading the $k$th coordinate, $\gamma (w)=({\phi }_j(\lambda (w)){)}_k$ on $Q$, a composite of $C^1$ maps and therefore $C^1$ on $Q$ by [F7] and [L4]; by [L5] it is totally differentiable at $w_0$, and by [L3] and [F7] its total derivative there acts by $v\mapsto ⟨g,v⟩$ with $g:=\mathrm{\nabla }\gamma (w_0)$.

The map $\mathrm{\Phi }(w):={\sigma }_k^{-1}(w,\gamma (w))$ on $Q$ equals ${\phi }_j\circ \lambda$ by step 2.1, and its two parameter derivatives at $w_0$ are ${\tau }_i={\sigma }_k^{-1}(e_i,{\mathrm{\partial }}_i\gamma (w_0))$ for $i=0,1$. By [L4] the derivative $D\mathrm{\Phi }(w_0)=D{\phi }_j(c)\circ D\lambda (w_0)$ with $D\lambda (w_0)$ invertible, so $\mathrm{span}\{{\tau }_0,{\tau }_1\}$ and $\mathrm{span}\{{\phi }_{j,u}(c),{\phi }_{j,v}(c)\}$ are the same subspace, namely the tangent plane $T$ of [F4].

By [F3] and [F5] the vector $N$ is defined at $c$, has norm $1$ and is orthogonal to $T$. Write $N={\sigma }_k^{-1}(a,b)$ with $a\in {\mathbb{R}}^2$ and $b=N_k$; since ${\sigma }_k^{-1}$ merely permutes coordinates, [F7] gives $⟨{\sigma }_k^{-1}(a,b),{\sigma }_k^{-1}(a^{\mathrm{\prime }},b^{\mathrm{\prime }})⟩=⟨a,a^{\mathrm{\prime }}⟩+b{b}^{\mathrm{\prime }}$. Orthogonality to ${\tau }_i$ of step 3.1 therefore reads $a_i+b{\mathrm{\partial }}_i\gamma (w_0)=0$ for $i=0,1$, that is $a=-bg$. If $b$ were $0$ then $a=0$ and $N=0$, contradicting $\parallel N{\parallel }_2=1$; so $b\mathrm{\ne }0$, and by [F2] and [L2] the number $b$ has the sign of $\det D{\psi }_j(c)$, hence $b>0$ for $j\in {\mathrm{\Sigma }}^{+}$ and $b<0$ for $j\in {\mathrm{\Sigma }}^{-}$.

For real $t$ near $0$ the projection ${\pi }_k(p+tN)=w_0+ta$ lies in the open $Q$, so $u(t):=(p+tN{)}_k-\gamma ({\pi }_k(p+tN))=\gamma (w_0)+tb-\gamma (w_0+ta)$ is defined there, using $p_k=\gamma (w_0)$ from step 2.1. Then $u(0)=0$, and by step 2.1 and [L3] the function $u$ is differentiable at $0$ with $u^{\mathrm{\prime }}(0)=b-⟨g,a⟩=b+b\parallel g{\parallel }_2^2=b(1+\parallel g{\parallel }_2^2)$, using $a=-bg$ from step 4.1. Since $u(0)=0$, the difference quotient at $0$ is $u(t)\mathrm{/}t$, so [F8] and [L6] give ${\epsilon }_0>0$ such that $u(t)\mathrm{/}t$ has the sign of $b$ for every $t$ with $0<\mid t\mid <{\epsilon }_0$; hence $u(t)$ has the sign of $tb$ there.

Suppose $j\in {\mathrm{\Sigma }}^{+}$, so $\gamma ={\gamma }_2$ and $b>0$ by step 4.1. Shrink ${\epsilon }_0$ so that ${\pi }_k(p\pm tN)\in Q\subseteq D$ for $0<t<{\epsilon }_0$ and so that, ${\gamma }_1$ and ${\gamma }_2$ being continuous with ${\gamma }_1(w_0)<{\gamma }_2(w_0)$ by [F1], one also has ${\gamma }_1({\pi }_k(p-tN))<{\gamma }_2(w_0)-tb$ there. For $0<t<{\epsilon }_0$, step 5.1 gives $u(t)>0$, that is $(p+tN{)}_k>{\gamma }_2({\pi }_k(p+tN))$, so $p+tN\mathrm{\notin }E$ by [F1]; and $u(-t)<0$, that is $(p-tN{)}_k<{\gamma }_2({\pi }_k(p-tN))$, while $(p-tN{)}_k={\gamma }_2(w_0)-tb>{\gamma }_1({\pi }_k(p-tN))$ by the choice of ${\epsilon }_0$, so $p-tN\in E$ by [F1]. Hence $N$ is outward at $p$ by [F6].

Suppose instead $j\in {\mathrm{\Sigma }}^{-}$, so $\gamma ={\gamma }_1$ and $b<0$ by step 4.1. Shrink ${\epsilon }_0$ so that ${\pi }_k(p\pm tN)\in Q\subseteq D$ for $0<t<{\epsilon }_0$ and so that ${\gamma }_2({\pi }_k(p-tN))>{\gamma }_1(w_0)-tb$ there, which is possible since ${\gamma }_1(w_0)<{\gamma }_2(w_0)$ by [F1] and $-tb>0$ tends to $0$. For $0<t<{\epsilon }_0$, step 5.1 gives $u(t)<0$, that is $(p+tN{)}_k<{\gamma }_1({\pi }_k(p+tN))$, so $p+tN\mathrm{\notin }E$ by [F1]; and $u(-t)>0$, that is $(p-tN{)}_k>{\gamma }_1({\pi }_k(p-tN))$, while $(p-tN{)}_k={\gamma }_1(w_0)-tb<{\gamma }_2({\pi }_k(p-tN))$ by the choice of ${\epsilon }_0$, so $p-tN\in E$ by [F1]. Hence $N$ is outward at $p$ by [F6].

In both cases $p\in E$ while $p+tN\mathrm{\notin }E$ for arbitrarily small $t>0$, so $p$ is not interior to $E$ and therefore $p\in \mathrm{\partial }E$ by [F8]. Replacing $N$ by $-N$ exchanges the two conditions of [F6], which then fail, so $-N$ is not outward at $p$; since $\pm N$ are the only unit vectors orthogonal to the two-dimensional $T$, the vector $N$ is the outward unit normal to $T$ at $p$.