de Rham cohomology as locally constant integrals modulo the trivially constant ones

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On page 3 of Bott & Tu's Differential Forms in algebraic topology, the authors state

...we can think of [a differential 1-form] $\theta$ as a function on paths $\gamma$: $$\gamma \mapsto \int_\gamma \theta.$$

It then suggests itself to seek those $\theta$ which give rise to locally constant functions of $\gamma$, i.e., for which the integral $\int_\gamma \theta$ is left unaltered under small variations of $\gamma$---but keeping the endpoints fixed! (Otherwise, only the zero 1-form would be locally constant.) Stokes' theorem teaches us that these line integrals are characterized by the differential equations... $$d\theta=0.$$ On the other hand...the gradients are trivially locally constant.

One is here irresistibly led to the definition of $H^1(M)$ as the vector space of locally constant line integrals modulo the trivially constant ones. Similarly, the higher cohomology groups $H^k(M)$ are defined by simply replacing line integrals with their higher-dimensional analogues, the k-volume integrals.

I interpret their statement about "small variations" in $\gamma$ fixing the endpoints as referring to (smooth?) homotopy invariance rel endpoints of the integral. Is this correct?

I am also confused by why the authors state that by Stokes Theorem, closed forms are locally constant line integrals. I can't see how Stokes theorem can be used unless the path happens to be a boundary loop.

Can someone clarify for me why closed forms in general are "locally constant integrals", or point me to a reference that proves this? I've tried browsing ahead in Bott & Tu but haven't been able to find an explanation for this bit.

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In his motivation, what you are calling "locally constant path-integrals" are intended to mean locally constant with respect to ("proper") variations of the paths (he clarifies this).

That said, let $\gamma_s$ be a variation of $\gamma$ such that $\gamma_0=\gamma$. Visualizing this as a map $H: I \times I \to M$ with $H(\cdot,0)=\gamma_0(\cdot)$, we know that $$\int_{I \times I} d(H^*\theta)=\int_{\partial (I \times I)}H^*\theta$$ by Stokes. The right side is $$\int_{\gamma_0} \theta-\int_{\gamma_1} \theta,$$ whereas the left side is $\int\limits_{I \times I}H^*(d\theta)=0.$