Let $X$ be a path-connected smooth manifold, it is known that: $$H^1(X):=H^1_{dR}(X)=\mathrm{Hom} (\pi_1(X),\mathbb R).$$ Explicitly, a closed one-form $\alpha$ gives a function on $\pi_1(X)$ by $[\gamma]\mapsto \int_\gamma \alpha$; and this only depends on the de Rham class $[\alpha]$ and the homotopy class of $[\gamma]$ due to Stokes' formula. So it is well-defined.

Now, let's consider the fundamental groupoid $\Pi_1(X)$ (objects are points of $X$ and morphisms are the homotopy classes of paths). We also take a closed one-form $\alpha$, then it seems that we can also get a map in a similar manner: $$ \mathrm{Mor}_{\Pi_1(X)}(p,q) \to \mathbb R, \qquad [\gamma]\mapsto \int_\gamma \alpha $$ where $p,q\in X$ and $\gamma$ is a path between $p$ and $q$. Although another path $\gamma'$ in the same homotopy class $[\gamma]$ will again yields the same value, the difference is that this is not invariant if we replace $\alpha$ by some $\alpha +d \eta$. Namely, we have to stay at $Z^1(X)$ and cannot go to $H^1(X)$. But in view of $H^1(X)=\mathrm{Hom} (\pi_1(X),\mathbb R)$, the collection of $\mathbb R$-valued homomorphisms on the morphism spaces of the fundamental groupoid $\Pi_1(X)$ should be still related to $Z^1(X)$ in some sense, I guess.

In short, as $\Pi_1(X)$ is often viewed as a generalization of $\pi_1(X)$, it should be interesting to ask in analogy: what should be a reasonable `generalization' of $H^1_{dR}(X)=\mathrm{Hom} (\pi_1(X),\mathbb R)$?

Picard groupoidis a strictly symmetric monoidal groupoid in which each object and morphism is strictly invertible. The category of Picard groupoids is equivalent to the category of 2-step chain complexes $C=(C_1\to C_0)$: the associated Picard groupoid $\mathcal{G}(C)$ has object set $C_0$ and morphism set $C_0\times C_1$. You can form the free Picard groupoid $\mathcal{F}(G)$ on a groupoid $G$. The claim is is then that $\mathcal{F}(\Pi_1(X))=\mathcal{G}(C_1(X)/B_1(X)\to C_0(X))$, where these are singular chains. $\endgroup$ – Charles Rezk Jul 18 '19 at 17:38