Stokes' theorem
Second step in the elementary proof (defining the pullback)

The previous step suggests we define the function P (u, v) = (F (ψ (u, v)) ⋅ ∂ ψ ∂ u (u, v)) e u + (F (ψ (u, v)) ⋅ ∂ ψ ∂ v (u, v)) e v {\displaystyle \mathbf {P} (u,v)=\left(\mathbf {F} ({\boldsymbol {\psi }}(u,v))\cdot {\frac {\partial {\boldsymbol {\psi }}}{\partial u}}(u,v)\right)\mathbf {e} _{u}+\left(\mathbf {F} ({\boldsymbol {\psi }}(u,v))\cdot {\frac {\partial {\boldsymbol {\psi }}}{\partial v}}(u,v)\right)\mathbf {e} _{v}}

Now, if the scalar value functions P u {\displaystyle P_{u}} and P v {\displaystyle P_{v}} are defined as follows, P u (u, v) = (F (ψ (u, v)) ⋅ ∂ ψ ∂ u (u, v)) {\displaystyle {P_{u}}(u,v)=\left(\mathbf {F} ({\boldsymbol {\psi }}(u,v))\cdot {\frac {\partial {\boldsymbol {\psi }}}{\partial u}}(u,v)\right)} P v (u, v) = (F (ψ (u, v)) ⋅ ∂ ψ ∂ v (u, v)) {\displaystyle {P_{v}}(u,v)=\left(\mathbf {F} ({\boldsymbol {\psi }}(u,v))\cdot {\frac {\partial {\boldsymbol {\psi }}}{\partial v}}(u,v)\right)} then, P (u, v) = P u (u, v) e u + P v (u, v) e v. {\displaystyle \mathbf {P} (u,v)={P_{u}}(u,v)\mathbf {e} _{u}+{P_{v}}(u,v)\mathbf {e} _{v}.}

This is the pullback of F along ψ, and, by the above, it satisfies ∮ ∂ Σ F (x) ⋅ d l = ∮ γ P (y) ⋅ d l = ∮ γ (P u (u, v) e u + P v (u, v) e v) ⋅ d l {\displaystyle \oint _{\partial \Sigma }{\mathbf {F} (\mathbf {x})\cdot \,\mathrm {d} \mathbf {l} }=\oint _{\gamma }{\mathbf {P} (\mathbf {y})\cdot \,\mathrm {d} \mathbf {l} }=\oint _{\gamma }{({P_{u}}(u,v)\mathbf {e} _{u}+{P_{v}}(u,v)\mathbf {e} _{v})\cdot \,\mathrm {d} \mathbf {l} }}

We have successfully reduced one side of Stokes' theorem to a 2-dimensional formula; we now turn to the other side.
punctuation:no_comma