The flux of a single-component field through a graph face is a base integral of its trace

By [F2] the inner product of $R{e}_k(p)$ with any vector $\nu$ is $R(p){\nu }_k$, so by [F1] the flux integrand of $R{e}_k$ through the patch $(D_j,{\phi }_j)$ is $w\mapsto R({\phi }_j(w))({\phi }_{j,u}\times {\phi }_{j,v}{)}_k(w)$ on $D_j$.

Suppose $j\in {\mathrm{\Sigma }}^{+}$ and put $h(w):=R({\sigma }_k^{-1}(w,{\gamma }_2(w)))$ for $w\in D$; by [F3] the point ${\sigma }_k^{-1}(w,{\gamma }_2(w))$ lies in $E$ and $h$ is continuous on $D$, being $R$ composed with a continuous map. By [F4] the image of ${\phi }_j$ lies in the graph of ${\gamma }_2$, so for $w\in D_j$ the point ${\phi }_j(w)$ has ${\pi }_k$-projection ${\psi }_j(w)\in D$ and $k$th coordinate ${\gamma }_2({\psi }_j(w))$; hence ${\phi }_j(w)={\sigma }_k^{-1}({\psi }_j(w),{\gamma }_2({\psi }_j(w)))$ and $R({\phi }_j(w))=h({\psi }_j(w))$. For $j\in {\mathrm{\Sigma }}^{-}$ the same computation with ${\gamma }_1$ in place of ${\gamma }_2$ defines a continuous $h$ on $D$ with $R({\phi }_j(w))=h({\psi }_j(w))$.

By [L1] the factor $({\phi }_{j,u}\times {\phi }_{j,v}{)}_k$ in step 1.1 is $\det D{\psi }_j$, so the flux integrand is $w\mapsto R({\phi }_j(w))\det D{\psi }_j(w)$. The map ${\psi }_j$ is $C^1$ on an open neighbourhood of $D_j$ by [F5], since ${\phi }_j$ is and ${\pi }_k$ is linear.

The map ${\psi }_j$ is injective on $D_j^{\circ }$. Indeed let $a,b\in D_j^{\circ }$ with ${\psi }_j(a)={\psi }_j(b)$. By step 1.2 both ${\phi }_j(a)$ and ${\phi }_j(b)$ are determined by their common ${\pi }_k$-projection through the same graph function, so ${\phi }_j(a)={\phi }_j(b)$; by [F5] no point of $D_j^{\circ }$ shares its image with a distinct point of $D_j$, so $a=b$.

By [F4] and step 2.1, $\det D{\psi }_j$ is positive on $D_j^{\circ }$ when $j\in {\mathrm{\Sigma }}^{+}$ and negative there when $j\in {\mathrm{\Sigma }}^{-}$; in either case it is nonvanishing on $D_j^{\circ }$.

By [F5] the parameter region $D_j$ is compact and Jordan measurable, so steps 2.2 and 3.1 put the data $({\psi }_j,D_j)$ under the hypotheses of [L2]. Hence $V_j={\psi }_j[D_j^{\circ }]$ is bounded, open and Jordan measurable, it is contained in $D$ by step 1.2, and with the continuous $h$ of step 1.2 ${\int }_{D_j}h({\psi }_j(w))\mid \det D{\psi }_j(w)\mid dw={\int }_{V_j}h\mathrm{.}$ Both sides exist, the left by [L5] on the compact Jordan $D_j$ and the right as part of [L2], with integrals read as in [F6].

On $D_j^{\circ }$ the two functions $\mid \det D{\psi }_j\mid$ and ${\epsilon }_j\det D{\psi }_j$ coincide, where ${\epsilon }_j=+1$ for $j\in {\mathrm{\Sigma }}^{+}$ and ${\epsilon }_j=-1$ for $j\in {\mathrm{\Sigma }}^{-}$, by step 3.1. They can differ only on $\mathrm{\partial }{D}_j$, which has content zero by [F5] and [L4]; both are continuous on the compact Jordan $D_j$, hence bounded and integrable by [L5], so multiplying each by the bounded continuous $h\circ {\psi }_j$ and applying [L3] on $D_j$ gives ${\int }_{D_j}h({\psi }_j)\mid \det D{\psi }_j\mid ={\epsilon }_j{\int }_{D_j}h({\psi }_j)\det D{\psi }_j\mathrm{.}$

Let $j\in {\mathrm{\Sigma }}^{+}$, so ${\epsilon }_j=+1$. Combining steps 1.1, 1.2 and 2.1 the flux integral is ${\int }_{D_j}h({\psi }_j)\det D{\psi }_j$, which by step 5.1 equals ${\int }_{D_j}h({\psi }_j)\mid \det D{\psi }_j\mid$ and by step 4.1 equals ${\int }_{V_j}h={\int }_{V_j}R({\sigma }_k^{-1}(w,{\gamma }_2(w)))dw$. That is the first asserted identity.

Let $j\in {\mathrm{\Sigma }}^{-}$, so ${\epsilon }_j=-1$ and $h(w)=R({\sigma }_k^{-1}(w,{\gamma }_1(w)))$. Steps 1.1, 1.2 and 2.1 again make the flux integral ${\int }_{D_j}h({\psi }_j)\det D{\psi }_j$, and step 5.1 now reads ${\int }_{D_j}h({\psi }_j)\mid \det D{\psi }_j\mid =-{\int }_{D_j}h({\psi }_j)\det D{\psi }_j$, so the flux integral is $-{\int }_{D_j}h({\psi }_j)\mid \det D{\psi }_j\mid$, which by step 4.1 is $-{\int }_{V_j}R({\sigma }_k^{-1}(w,{\gamma }_1(w)))dw$. This is the second asserted identity, and the sign comes from that replacement of the absolute determinant and from nothing else.