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Definition: Write SmoothOrientedManifold for the category of smooth oriented real manifolds.

Definition: Write ChainComplex for the category of chain complexes, and write $\otimes$ for the tensor product of chain complexes.

Definition: Write $c_{\mathbb{R}}$ for the functor from SmoothOrientedManifold to ChainComplex which sends each object to the constant complex of $\mathbb{R}$ concentrated in degree $0$, and each smooth function to an identity map of chain complexes.

Definition: Write DR : SmoothOrientedManifold $\rightarrow$ ChainComplex for the DeRham complex functor, which on objects is the complex of Kähler differentials of $C^{\infty}(X,\mathbb{R})$.

Definition: Write S : SmoothOrientedManifold $\rightarrow$ ChainComplex for the singular complex functor.

Definition: Write $\otimes$ for the tensor product of chain complexes.

Lemma: There is a natural transformation p${}_{S}$ : S(-) $\otimes$ S(-) $\rightarrow$ S(- $\times$ -) which is an inclusion on objects.

Lemma: There is a natural transformation p${}_{DR}$ : DR(-) $\otimes$ DR(-) $\rightarrow$ DR(- $\times$ -) which is an inclusion on objects.

Stokes' Theorem: there is a unique natural transformation $\int : S \otimes DR \rightarrow c_{\mathbb{R}}$ such that:

(A) For each $X$ and each $Y$, $$\int_{X \times Y} \circ \left( p_{S} \otimes p_{DR} \right) \circ 1 \otimes \tau \otimes 1 = \left( \int_{X} \circ p_{S} \right) \otimes \left( \int_{Y} \circ p_{S} \right)$$

(B) $\int \text{Id}_{I} \otimes \chi_{1} = 1$, where $\chi_{I} : I \rightarrow \mathbb{R}$ is the constant function sending all elements to 1 and Id${}_{I}$ is the identity function on the unit interval.

My question is, is this true, and also, what is the easiest way to prove it?

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  • $\begingroup$ Might have a look here $\endgroup$ Commented May 26 at 6:43
  • $\begingroup$ @TimCampion wow this is great. Here Stoke's theorem follows from the preservation of the differential by the natural transformation in the theorem. $\endgroup$
    – Ron Z.
    Commented May 26 at 10:02
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    $\begingroup$ the name is Stokes, not Stoke $\endgroup$ Commented May 26 at 11:05

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