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  • $\begingroup$ This seems very close (at least in spirit) to what I had in mind. What I do not understand, though, is the domain of definition of $I$. If $\mathcal M$ is the class of smooth manifolds with boundary, it seems that the domain of $I$ is the disjoint union of $\Omega^1(M) \times \mathcal C _M$, where $M \in \mathcal M$ and $\mathcal C _M$ is the set of smooth curves $: [0,1] \to M$. Weird, though... $\endgroup$ Commented Jul 16, 2020 at 13:52
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    $\begingroup$ You're right that the domain of $I$ is this funny disjoint union, but perhaps this isn't such a strange domain to get. We probably want to think of $I$ really as a collection of functions $I_M\colon\Omega^1(M)\times\mathcal C_M\to\mathbb R$, one for each $M\in\mathcal M$, and we're saying that this whole collection is uniquely determined by some compatibility condition with respect to smooth maps, plus a normalisation condition when $M=[0,1]$. $\endgroup$ Commented Jul 16, 2020 at 14:53
  • $\begingroup$ (ctd) These sorts of structures are actually quite common: if we view $I_M$ as a map $\mathcal C_M\to\mathrm{Hom}(\Omega^1(M),\mathbb R)$, then this "compatibility with smooth maps" condition is just saying that these $I_M$ are the components of a natural transformation (in the sense of category theory). And natural transformations crop up all over the place. $\endgroup$ Commented Jul 16, 2020 at 14:53
  • $\begingroup$ I think that I can come up with version better suited to my needs, that does away with the disjoint union: property (1) should hold for smooth maps $f : M \to M$ (yes, not $N$), and property (2) should be replaced by the condition that $I(\mathrm d \varphi, c) = \varphi(c(1)) - \varphi(c(0))$ for every smooth $\varphi : M \to \mathbb R$. This means that we may work with a single, fixed manifold $M$. Also, one may take it now to be boundaryless (which is what I need). (The boundary was needed by the case $M = [0,1]$ which becomes irrelevant in this approach.) $\endgroup$ Commented Jul 16, 2020 at 15:20
  • $\begingroup$ Glad to hear it! Though I would be a little surprised if those two conditions uniquely determine $I$ in general. Your second condition only pins down line integrals of exact $1$-forms; it's unclear how this helps calculate integrals of inexact forms (non-closed forms seem especially difficult). $\endgroup$ Commented Jul 16, 2020 at 15:40