Let $\mathbf{X}$ be a Grothendieck topos and let $A$ be an abelian group in $\mathbf{X}$.
Verdier's Theorem allows one to describe $\mathrm{H}^n(\mathbf{X},A)$ in terms of hypercoverings, namely, as the colimit of $\mathrm{H}^n(U\_\bullet,A)$ where $U\_\bullet$ ranges over the category of hypercoverings (of the terminal object of $\mathbf{X}$).


One can further form the Cech cohomology $\check{\mathrm{H}}^n(\mathbf{X},A)$, which is the colomit of $\mathrm{H}^n(U\_\bullet,A)$ as $U\_\bullet$ ranges over hypercoverings with $U\_\bullet=\mathrm{cosk}\_0(U\_\bullet)$. Let me call them *Cech hypercoverings*. (These are the hypercoverings with $U\_n=U\_0^{\times (n+1)}$ and the standard simplicial structure.)


A well-known spectral sequence relates the derived-functor cohomology with the Cech cohomology, implying in particular that $\mathrm{H}^1(\mathbf{X},A)$ is canonically isomorphic to $\check{\mathrm{H}}^1(\mathbf{X},A)$. I am not absolutely certain, but it seems correct that under the description provided by Verdier's Theorem, the isomorphism $\check{\mathrm{H}}^1(\mathbf{X},A)\to \mathrm{H}^1(\mathbf{X},A)$ is the obvious one, namely, if a *Cech* cohomology class is represented by a $1$-cocycle in $Z^1(U\_\bullet,A)$, then its image in $\mathrm{H}^1(\mathbf{X},A)$ is the cohomology class represented by that $1$-cocycle.


The latter statement means that for a hypercovering $U\_\bullet$ (not necessarily Cech) and any $1$-cocycle $\alpha\in Z^1(U\_\bullet,A)$, one can find a *Cech* hypercovering $U'\_\bullet$ and $\alpha'\in Z^1(U'\_\bullet,A)$ representing the same cohomology class. 
**I am looking for a way to construct these $U'$ and $\alpha'$ directly.**
The main problem is that $U\_\bullet$ cannot be refined to a Cech hypercovering in general.


An alternative approach to the problem (which also applies to non-abelian $A$) is via the correspondence with $A$-torsors. It is a standard fact that $\check{H}^1(\mathbf{X},A)$ is in 1-1 correspondence with isomorphism classes of $A$-torsors. Explicitly, if $\alpha\in Z^1(U\_\bullet,A)$ with $U\_\bullet$ being a Cech hypercovering, then the $A$-torsor $P$ corresponding to $\alpha$ can be described by
$$
P(V)=\{a\in A(U\_0\times V)~:~ \alpha\_V \cdot d\_0^1a =d\_1^1 a\}
$$
where $d^1\_i:A(U\_0\times V)\to A(U\_1\times V)=A(U\_0\times U\_0\times V)$ is induced by $d^1\_i:U\_1\to U\_0$.
Suppose now that $U\_\bullet$ is an arbitrary hypercovering (not necessarily Cech). The question will be resolved if one can show that $P$ constructed above is still an $A$-torsor. The difficult thing to check is that $P\neq \emptyset$. (However, when $U\_\bullet$ is Cech, one can check that $\alpha\in P(U\_0)$.)



An explicit isomorphism between the 1st Cech cohomology and the 1st hypercohomology
Some hints in the literature led me to an answer, which I find a bit surprising:
One can take $U'\_\bullet=\mathrm{cosk}\_0(U\_\bullet)$ and the $1$-cocycle $\alpha\in Z^1(U\_\bullet,A)\subseteq A(U\_1)$ descends uniquely to a $1$-cocycle in $\alpha'\in Z^1(\mathrm{cosk}\_0(U\_\bullet),A)\subseteq A(U\_0\times U\_0)$ along $(d\_0,d\_1):U\_1\to U\_0\times U\_0$. In other words:


**Proposition:** For any hypercovering $U\_\bullet$, the canonical map $Z^1(\mathrm{cosk}\_0(U\_\bullet),A)\to Z^1(U\_\bullet,A)$ is an isomorphism.
Consequently, the map $\mathrm{H}^1(\mathrm{cosk}\_0(U\_\bullet),A)\to \mathrm{H}^1(U\_\bullet,A)$ is an isomorphism.


This is a nontrivial statement so let me sketch the ad-hoc proof I have. 


**Step 1:** We may assume $U\_\bullet=\mathrm{cosk}\_1(U\_\bullet)$. 


Indeed, since $U\_\bullet$ is a hypercovering, the map $U\_2\to \mathrm{cosk}\_1(U\_\bullet)\_2$ is a covering, and this easily implies that $Z^1(\mathrm{cosk}\_1(U\_\bullet),A)\to Z^1(U\_\bullet,A)$ is an isomorphism, so replace $U\_\bullet$ with its $1$-coskeleton. 


One consequence of this assumption is that $U\_2=\underline{\mathrm{Hom}}\_{\mathrm{Simp}}(\partial \Delta^2, U\_\bullet)$, where $\partial \Delta^2$ is the boundary of the $2$-simplex, realized as a constant simplicial object (i.e. sheaf) in $\mathbf{X}$, and $\underline{\mathrm{Hom}}$ denote internal $\mathrm{Hom}$ in $\mathbf{X}$.


**Step 2:** Let $\alpha\in Z^1(U\_\bullet,A)$. We claim that $a\in A(U\_1)$ descends along $(d\_0,d\_1):U\_1\to U\_0\times U\_0$ to some $\alpha'\in G(U\_0\times U\_0)$.


Let $V=U\_1\times\_{U\_0\times U\_0}U\_1$ and let $\pi\_1,\pi\_2:V\to U\_1$ denote the first and second projections. Let $S$ denote the simplicial object of $\mathbf{X}$ obtained by gluing two copies of $\Delta^1$ along their vertices. Then $V=\underline{\mathrm{Hom}}\_{\mathrm{Simp}}(S, U\_\bullet)$.
There is a simplicial map $\partial \Delta^2\to S$ which degenerate the edge $\{1,2\}$ into a vertex. This gives rise to a map $V\to U\_2$, which in turn gives rise a map $A(U\_2)\to A(V)$. Applying this map to the cocycle equation $d\_0^2\alpha-d\_1^2\alpha+d^2\_2\alpha=0$ in $A(U\_2)$ gives $0-\pi\_2^\*\alpha+\pi\_1^\*\alpha=0$ in $A(V)$, which means that $\alpha$ descends to $\alpha'\in G(U\_0\times U\_0)$. 


**Step 3:** We finally claim that $\alpha'$ lies in $Z^1(\mathrm{cosk}\_0(U\_\bullet),A)$, which proves the surjectivity of $Z^1(\mathrm{cosk}\_0(U\_\bullet),A)\to Z^1(U\_\bullet,A)$. The injectivity follows easily from the fact that $(d\_0,d\_1):U\_1\to\mathrm{cosk}\_0(U\_\bullet)\_1=U\_0\times U\_0$ is a covering.


To show the claim, it is enough to show that the canonical map $U\_2\to \mathrm{cosk}\_0(U\_\bullet)\_2=U\_0\times U\_0\times U\_0$ is a covering. Indeed, if this holds, then the fact that the $1$-cocycle equation holds for $\alpha$ in $A(U\_2)$ implies that it holds for $\alpha'$ in $A(U\_0\times U\_0\times U\_0)$. Proving that $U\_2\to U\_0\times U\_0\times U\_0$ is a covering amounts to showing that any $3$ "vertices" in $U\_0$ can be joined by a "triangle" in $U\_2$ locally. But this follows from $U\_2=\underline{\mathrm{Hom}}\_{\mathrm{Simp}}(\partial \Delta^2, U\_\bullet)$ and the fact that $(d\_0,d\_1):U\_1\to U\_0\times U\_0$ is a covering.


**I would value references for this proposition in the literature, if you know them.**


