hcju/stacksqa huggingface dataset:
Set $X^* = X \amalg \{\infty\}$. We declare a subset $V$ of $X^*$ to be open if either $V \subset X$ is open in $X$, or $\infty \in V$ and $U = V \cap X$ is an open of $X$ such that $X \setminus U$ is quasi-compact. We omit the verification that this defines a topology. It is clear that $X \to X^*$ identifies $X$ with an open subspace of $X$. \medskip\noindent Since $X$ is locally quasi-compact, every point $x \in X$ has a quasi-compact neighbourhood $x \in E \subset X$. Then $E$ is closed (Lemma \ref{lemma-closed-in-quasi-compact}) and $V = (X \setminus E) \amalg \{\infty\}$ is an open neighbourhood of $\infty$ disjoint from the interior of $E$. Thus $X^*$ is Hausdorff. \medskip\noindent Let $X^* = \bigcup V_i$ be an open covering. Then for some $i$, say $i_0$, we have $\infty \in V_{i_0}$. By construction $Z = X^* \setminus V_{i_0}$ is quasi-compact. Hence the covering $Z \subset \bigcup_{i \not = i_0} Z \cap V_i$ has a finite refinement which implies that the given covering of $X^*$ has a finite refinement. Thus $X^*$ is quasi-compact. \medskip\noindent The map $X \to X^*$ factors as $X \to \beta(X) \to X^*$ by the universal property of the Stone-{\v C}ech compactification. Let $\varphi : \beta(X) \to X^*$ be this factorization. Then $X \to \varphi^{-1}(X)$ is a section to $\varphi^{-1}(X) \to X$ hence has closed image (Lemma \ref{lemma-section-closed}). Since the image of $X \to \beta(X)$ is dense we conclude that $X = \varphi^{-1}(X)$.
By Lemma \ref{lemma-bijective-map} it suffices to show that $f$ is injective. Suppose that $x, x' \in X$ are distinct points with $y = f(x) = f(x')$. Choose disjoint open neighbourhoods $U, U' \subset X$ of $x, x'$. Observe that $f$ is closed (Lemma \ref{lemma-closed-map}) hence $T = f(X \setminus U)$ and $T' = f(X \setminus U')$ are closed in $Y$. Since $X$ is the union of $X \setminus U$ and $X \setminus U'$ we see that $Y = T \cup T'$. By Lemma \ref{lemma-image-open-technical} we see that $y$ is contained in the closure of $Y \setminus T$ and the closure of $Y \setminus T'$. On the other hand, by Lemma \ref{lemma-intersection-empty}, this intersection is empty. In this way we obtain the desired contradiction.
We will use without further mention that the quasi-compact subsets of $X$ are exactly the closed subsets (Lemma \ref{lemma-closed-in-compact}). Consider the collection $\mathcal{E}$ of all quasi-compact subsets $E \subset X$ with $f(E) = Y$ ordered by inclusion. We will use Zorn's lemma to show that $\mathcal{E}$ has a minimal element. To do this it suffices to show that given a totally ordered family $E_\lambda$ of elements of $\mathcal{E}$ the intersection $\bigcap E_\lambda$ is an element of $\mathcal{E}$. It is quasi-compact as it is closed. For every $y \in Y$ the sets $E_\lambda \cap f^{-1}(\{y\})$ are nonempty and closed, hence the intersection $\bigcap E_\lambda \cap f^{-1}(\{y\}) = \bigcap (E_\lambda \cap f^{-1}(\{y\}))$ is nonempty by Lemma \ref{lemma-intersection-closed-in-quasi-compact}. This finishes the proof.
Let $Y = X$ but endowed with the discrete topology. Let $X' = \beta(Y)$. The continuous map $Y \to X$ factors as $Y \to X' \to X$. This proves the first statement of the lemma by Example \ref{example-stone-Cech-discrete}. \medskip\noindent By Lemma \ref{lemma-find-compact-subset} we can find a quasi-compact subset $E \subset X'$ surjecting onto $X$ such that no proper closed subset of $E$ surjects onto $X$. Because $X'$ is extremally disconnected there exists a continuous map $f : X' \to E$ over $X$ (Proposition \ref{proposition-projective-in-category-hausdorff-qc}). Composing $f$ with the map $E \to X'$ gives a continuous selfmap $f|_E : E \to E$. Observe that $f|_E$ has to be surjective as otherwise the image would be a proper closed subset surjecting onto $X$. Hence $f|_E$ has to be $\text{id}_E$ as otherwise Lemma \ref{lemma-rainwater} shows that $E$ isn't minimal. Thus the $\text{id}_E$ factors through the extremally disconnected space $X'$. A formal, categorical argument (using the characterization of Proposition \ref{proposition-projective-in-category-hausdorff-qc}) shows that $E$ is extremally disconnected. \medskip\noindent To prove uniqueness, suppose we have a second $X'' \to X$ minimal cover. By the lifting property proven in Proposition \ref{proposition-projective-in-category-hausdorff-qc} we can find a continuous map $g : X' \to X''$ over $X$. Observe that $g$ is a closed map (Lemma \ref{lemma-closed-map}). Hence $g(X') \subset X''$ is a closed subset surjecting onto $X$ and we conclude $g(X') = X''$ by minimality of $X''$. On the other hand, if $E \subset X'$ is a proper closed subset, then $g(E) \not = X''$ as $E$ does not map onto $X$ by minimality of $X'$. By Lemma \ref{lemma-isomorphism} we see that $g$ is an isomorphism.
By definition of constructible subsets, we can write each $T_i$ as a finite union of $U \cap V^c$ with $U, V \subset X$ retrocompact open. Hence we may assume that $T_i = U_i \cap V_i^c$ with $U_i, V_i \subset X$ retrocompact open. Let $S$ be the finite set of closed subsets of $X$ consisting of $\emptyset, X, U_i^c, V_i^c$ and finite intersections of these. Write $S = \{Z_s\}$. If $s \in S$, then $Z_s$ is constructible (Lemma \ref{lemma-constructible}). Moreover, $Z_s \cap Z_{s'} \in S$ for all $s, s' \in S$. Define a partial ordering on $S$ by inclusion. Then set $Y_s = Z_s \setminus \bigcup_{s' < s} Z_{s'}$ to get the desired stratification.
Let $X = \coprod X_i$ be a finite partition of $X$. Let $Z$ be an irreducible component of $X$. Since $X = \bigcup \overline{X_i}$ with finite index set, there is an $i$ such that $Z \subset \overline{X_i}$. Since $X_i$ is locally closed this implies that $Z \cap X_i$ contains an open of $Z$. Thus $Z \cap X_i$ contains an open $U$ of $X$ (Lemma \ref{lemma-Noetherian}). Write $X_i = U \amalg X_i^1 \amalg X_i^2$ with $X_i^1 = (X_i \setminus U) \cap \overline{U}$ and $X_i^2 = (X_i \setminus U) \cap \overline{U}^c$. For $i' \not = i$ we set $X_{i'}^1 = X_{i'} \cap \overline{U}$ and $X_{i'}^2 = X_{i'} \cap \overline{U}^c$. Then $$ X \setminus U = \coprod X^k_l $$ is a partition such that $\overline{U} \setminus U = \bigcup X_l^1$. Note that $X \setminus U$ is closed and strictly smaller than $X$. By Noetherian induction we can refine this partition by a finite good stratification $X \setminus U = \coprod_{\alpha \in A} T_\alpha$. Then $X = U \amalg \coprod_{\alpha \in A} T_\alpha$ is a finite good stratification of $X$ refining the partition we started with.
This follows from the discussion above and Categories, Lemma \ref{categories-lemma-colimits-coproducts-coequalizers}. Another proof of existence of colimits is sketched in Categories, Remark \ref{categories-remark-how-to-use-it}. It follows from the above that the forgetful functor commutes with colimits. Another way to see this is to use Categories, Lemma \ref{categories-lemma-adjoint-exact} and use that the forgetful functor has a right adjoint, namely the functor which assigns to a set the corresponding chaotic (or indiscrete) topological space.
It is enough to prove the existence and commutation for products and equalizers, see Categories, Lemma \ref{categories-lemma-limits-products-equalizers}. Let $G_i$, $i \in I$ be a collection of topological groups. Take the usual product $G = \prod G_i$ with the product topology. Since $G \times G = \prod (G_i \times G_i)$ as a topological space (because products commutes with products in any category), we see that multiplication on $G$ is continuous. Similarly for the inverse map. Let $a, b : G \to H$ be two homomorphisms of topological groups. Then as the equalizer we can simply take the equalizer of $a$ and $b$ as maps of topological spaces, which is the same thing as the equalizer as maps of groups endowed with the induced topology.
We have the corresponding result for topological spaces, see Lemma \ref{lemma-profinite}. Combined with Lemma \ref{lemma-topological-group-limits} we see that it suffices to prove that (1) implies (3). \medskip\noindent We first prove that every neighbourhood $E$ of the neutral element $e$ contains an open subgroup. Namely, since $G$ is the cofiltered limit of finite discrete topological spaces (Lemma \ref{lemma-profinite}), we can choose a continuous map $f : G \to T$ to a finite discrete space $T$ such that $f^{-1}(f(\{e\})) \subset E$. Consider $$ H = \{g \in G \mid f(gg') = f(g')\text{ for all }g' \in G\} $$ This is a subgroup of $G$ and contained in $E$. Thus it suffices to show that $H$ is open. Pick $t \in T$ and set $W = f^{-1}(\{t\})$. Observe that $W \subset G$ is open and closed, in particular quasi-compact. For each $w \in W$ there exist open neighbourhoods $e \in U_w \subset G$ and $w \in U'_w \subset W$ such that $U_wU'_w \subset W$. By quasi-compactness we can find $w_1, \ldots, w_n$ such that $W = \bigcup U'_{w_i}$. Then $U_t = U_{w_1} \cap \ldots \cap U_{w_n}$ is an open neighbourhood of $e$ such that $f(gw) = t$ for all $w \in W$. Since $T$ is finite we see that $\bigcap_{t \in T} U_t \subset H$ is an open neighbourhood of $e$. Since $H \subset G$ is a subgroup it follows that $H$ is open. \medskip\noindent Suppose that $H \subset G$ is an open subgroup. Since $G$ is quasi-compact we see that the index of $H$ in $G$ is finite. Say $G = Hg_1 \cup \ldots \cup Hg_n$. Then $N = \bigcap_{i = 1, \ldots, n} g_iHg_i^{-1}$ is an open normal subgroup contained in $H$. Since $N$ also has finite index we see that $G \to G/N$ is a surjection to a finite discrete topological group. \medskip\noindent Consider the map $$ G \longrightarrow \lim_{N \subset G\text{ open and normal}} G/N $$ We claim that this map is an isomorphism of topological groups. This finishes the proof of the lemma as the limit on the right is cofiltered (the intersection of two open normal subgroups is open and normal). The map is continuous as each $G \to G/N$ is continuous. The map is injective as $G$ is Hausdorff and every neighbourhood of $e$ contains an $N$ by the arguments above. The map is surjective by Lemma \ref{lemma-intersection-closed-in-quasi-compact}. By Lemma \ref{lemma-bijective-map} the map is a homeomorphism.
We will use the argument of Categories, Remark \ref{categories-remark-how-to-use-it} to prove existence of colimits. Namely, suppose that $\mathcal{I} \to \textit{Top}$, $i \mapsto G_i$ is a functor into the category $\textit{TopGroup}$ of topological groups. Then we can consider $$ F : \textit{TopGroup} \longrightarrow \textit{Sets},\quad H \longmapsto \lim_\mathcal{I} \Mor_{\textit{TopGroup}}(G_i, H) $$ This functor commutes with limits. Moreover, given any topological group $H$ and an element $(\varphi_i : G_i \to H)$ of $F(H)$, there is a subgroup $H' \subset H$ of cardinality at most $|\coprod G_i|$ (coproduct in the category of groups, i.e., the free product on the $G_i$) such that the morphisms $\varphi_i$ map into $H'$. Namely, we can take the induced topology on the subgroup generated by the images of the $\varphi_i$. Thus it is clear that the hypotheses of Categories, Lemma \ref{categories-lemma-a-version-of-brown} are satisfied and we find a topological group $G$ representing the functor $F$, which precisely means that $G$ is the colimit of the diagram $i \mapsto G_i$. \medskip\noindent To see the statement on commutation with the forgetful functor to groups we will use Categories, Lemma \ref{categories-lemma-adjoint-exact}. Indeed, the forgetful functor has a right adjoint, namely the functor which assigns to a group the corresponding chaotic (or indiscrete) topological group.
It is enough to prove the existence and commutation for products and equalizers, see Categories, Lemma \ref{categories-lemma-limits-products-equalizers}. Let $R_i$, $i \in I$ be a collection of topological rings. Take the usual product $R = \prod R_i$ with the product topology. Since $R \times R = \prod (R_i \times R_i)$ as a topological space (because products commutes with products in any category), we see that addition and multiplication on $R$ are continuous. Let $a, b : R \to R'$ be two homomorphisms of topological rings. Then as the equalizer we can simply take the equalizer of $a$ and $b$ as maps of topological spaces, which is the same thing as the equalizer as maps of rings endowed with the induced topology.
The exact same argument as used in the proof of Lemma \ref{lemma-topological-group-colimits} shows existence of colimits. To see the statement on commutation with the forgetful functor to rings we will use Categories, Lemma \ref{categories-lemma-adjoint-exact}. Indeed, the forgetful functor has a right adjoint, namely the functor which assigns to a ring the corresponding chaotic (or indiscrete) topological ring.
It is enough to prove the existence and commutation for products and equalizers, see Categories, Lemma \ref{categories-lemma-limits-products-equalizers}. Let $M_i$, $i \in I$ be a collection of topological modules over $R$. Take the usual product $M = \prod M_i$ with the product topology. Since $M \times M = \prod (M_i \times M_i)$ as a topological space (because products commutes with products in any category), we see that addition on $M$ is continuous. Similarly for multiplication $R \times M \to M$. Let $a, b : M \to M'$ be two homomorphisms of topological modules over $R$. Then as the equalizer we can simply take the equalizer of $a$ and $b$ as maps of topological spaces, which is the same thing as the equalizer as maps of modules endowed with the induced topology.
The exact same argument as used in the proof of Lemma \ref{lemma-topological-group-colimits} shows existence of colimits. To see the statement on commutation with the forgetful functor to $R$-modules we will use Categories, Lemma \ref{categories-lemma-adjoint-exact}. Indeed, the forgetful functor has a right adjoint, namely the functor which assigns to a module the corresponding chaotic (or indiscrete) topological module.
It is clear that (3) implies (2). On the other hand, if (2) holds and $X \to Z$ and $Y \to Z$ are as in (3), then (2) assures there is a section to the projection $X \times_Z Y \to X$ which implies a suitable dotted arrow exists (details omitted). Thus (3) is equivalent to (2). \medskip\noindent Assume $X$ is extremally disconnected and let $f : Y \to X$ be as in (2). By Lemma \ref{lemma-find-compact-subset} there exists a quasi-compact subset $E \subset Y$ such that $f(E) = X$ but $f(E') \not = X$ for all proper closed subsets $E' \subset E$. By Lemma \ref{lemma-isomorphism} we find that $f|_E : E \to X$ is a homeomorphism, the inverse of which gives the desired section. \medskip\noindent Assume (2). Let $U \subset X$ be open with complement $Z$. Consider the continuous surjection $f : \overline{U} \amalg Z \to X$. Let $\sigma$ be a section. Then $\overline{U} = \sigma^{-1}(\overline{U})$ is open. Thus $X$ is extremally disconnected.
Part (1) is immediate from the axioms of a site. Part (2) follows by the construction of fibre products in $\text{SR}(\mathcal{C})$ in the proof of Lemma \ref{lemma-coprod-prod-SR} and the requirement that the morphisms in a covering of $\mathcal{C}$ are representable. Part (3) follows by thinking of $A \times K \to B \times L$ as the composition $A \times K \to B \times K \to B \times L$ and hence a composition of basechanges of coverings. The final statement follows because $\text{SR}(\mathcal{C}, X) = \text{SR}(\mathcal{C}/X)$.
We will use the fact that if $\{U_i \to U\}_{i \in I}$ is a covering of the site $\mathcal{C}$, then the morphism $$ \amalg_{i \in I} h_{U_i} \to h_U $$ becomes surjective after sheafification, see Sites, Lemma \ref{sites-lemma-covering-surjective-after-sheafification}. Thus the first assertion follows immediately. \medskip\noindent For the second assertion, note that according to Simplicial, Example \ref{simplicial-example-cosk0} the simplicial object $\text{cosk}_0 \text{sk}_0 K$ has terms $K_0 \times \ldots \times K_0$. Thus according to the definition of a hypercovering we see that $(d^1_0, d^1_1) : K_1 \to K_0 \times K_0$ is a covering. Hence (2) follows from the claim above and the fact that $F$ transforms products into fibred products over $h_X$. \medskip\noindent For the third, we claim that $\text{cosk}_n \text{sk}_n F(K) = F(\text{cosk}_n \text{sk}_n K)$ for $n \geq 1$. To prove this, denote temporarily $F'$ the functor $\text{SR}(\mathcal{C}, X) \to \textit{PSh}(\mathcal{C})/h_X$. By Lemma \ref{lemma-coprod-prod-SR} the functor $F'$ commutes with finite limits. By our description of the $\text{cosk}_n$ functor in Simplicial, Section \ref{simplicial-section-skeleton} we see that $\text{cosk}_n \text{sk}_n F'(K) = F'(\text{cosk}_n \text{sk}_n K)$. Recall that the category used in the description of $(\text{cosk}_n U)_m$ in Simplicial, Lemma \ref{simplicial-lemma-existence-cosk} is the category $(\Delta/[m])^{opp}_{\leq n}$. It is an amusing exercise to show that $(\Delta/[m])_{\leq n}$ is a connected category (see Categories, Definition \ref{categories-definition-category-connected}) as soon as $n \geq 1$. Hence, Categories, Lemma \ref{categories-lemma-connected-limit-over-X} shows that $\text{cosk}_n \text{sk}_n F'(K) = \text{cosk}_n \text{sk}_n F(K)$. Whence the claim. Property (2) follows from this, because now we see that the morphism in (2) is the result of applying the functor $F$ to a covering as in Definition \ref{definition-covering-SR}, and the result follows from the first fact mentioned in this proof.
Let $\mathcal{G}' \subset \mathcal{G}$ be the image of the morphism $\mathcal{F} \to \mathcal{G}$. Let $U \in \Ob(\mathcal{C})$. Set $A = \mathcal{F}(U)$ and $B = \mathcal{G}'(U)$. Then the simplicial set $K(U)$ is equal to the simplicial set with $n$-simplices given by $$ A \times_B A \times_B \ldots \times_B A\ (n + 1 \text{ factors)}. $$ By Simplicial, Lemma \ref{simplicial-lemma-cosk-minus-one-equivalence} the morphism $K(U) \to B$ is a trivial Kan fibration. Thus it is a homotopy equivalence (Simplicial, Lemma \ref{simplicial-lemma-cosk-minus-one-equivalence}). Hence applying the functor ``free abelian group on'' to this we deduce that $$ \mathbf{Z}_K(U) \longrightarrow \mathbf{Z}_B $$ is a homotopy equivalence. Note that $s(\mathbf{Z}_B)$ is the complex $$ \ldots \to \bigoplus\nolimits_{b \in B}\mathbf{Z} \xrightarrow{0} \bigoplus\nolimits_{b \in B}\mathbf{Z} \xrightarrow{1} \bigoplus\nolimits_{b \in B}\mathbf{Z} \xrightarrow{0} \bigoplus\nolimits_{b \in B}\mathbf{Z} \to 0 $$ see Simplicial, Lemma \ref{simplicial-lemma-homology-eilenberg-maclane}. Thus we see that $H_i(s(\mathbf{Z}_K(U))) = 0$ for $i > 0$, and $H_0(s(\mathbf{Z}_K(U))) = \bigoplus_{b \in B}\mathbf{Z} = \bigoplus_{s \in \mathcal{G}'(U)} \mathbf{Z}$. These identifications are compatible with restriction maps. \medskip\noindent We conclude that $H_i(s(\mathbf{Z}_K)) = 0$ for $i > 0$ and $H_0(s(\mathbf{Z}_K)) = \mathbf{Z}_{\mathcal{G}'}$, where here we compute homology groups in $\textit{PAb}(\mathcal{C})$. Since sheafification is an exact functor we deduce the result of the lemma. Namely, the exactness implies that $H_0(s(\mathbf{Z}_K))^\# = H_0(s(\mathbf{Z}_K^\#))$, and similarly for other indices.
This proof is exactly the same as the proof of Lemma \ref{lemma-compare-cosk0} above. Namely, we first let $K_n' \subset K_n$ be the sub presheaf which is the image of the map $L_n \to K_n$. Assumption (2) means that the sheafification of $K_n'$ is equal to the sheafification of $K_n$. Moreover, since $L_i = K_i$ for all $i < n$ we see that get an $n$-truncated simplicial presheaf $U$ by taking $U_0 = L_0 = K_0, \ldots, U_{n - 1} = L_{n - 1} = K_{n - 1}, U_n = K'_n$. Denote $K' = \text{cosk}_n U$, a simplicial presheaf. Because we can construct $K'_m$ as a finite limit, and since sheafification is exact, we see that $(K'_m)^\# = K_m$. In other words, $(K')^\# = K^\#$. We conclude, by exactness of sheafification once more, that $H_i(K) = H_i(K')$. Thus it suffices to prove the lemma for the morphism $L \to K'$, in other words, we may assume that $L_n \to K_n$ is a surjective morphism of {\it presheaves}! \medskip\noindent In this case, for any object $U$ of $\mathcal{C}$ we see that the morphism of simplicial sets $$ L(U) \longrightarrow K(U) $$ satisfies all the assumptions of Simplicial, Lemma \ref{simplicial-lemma-section}. Hence it is a trivial Kan fibration. In particular it is a homotopy equivalence (Simplicial, Lemma \ref{simplicial-lemma-trivial-kan-homotopy}). Thus $$ \mathbf{Z}_L(U) \longrightarrow \mathbf{Z}_K(U) $$ is a homotopy equivalence too. This for all $U$. The result follows.
Denote $K^n = \text{cosk}_n \text{sk}_n K$ for $n \geq 1$. Define $K^0$ as the simplicial object with terms $(K^0)_n$ equal to the $(n + 1)$-fold fibred product $K_0 \times_\mathcal{G} \ldots \times_\mathcal{G} K_0$, see Simplicial, Example \ref{simplicial-example-fibre-products-simplicial-object}. We have morphisms $$ K \longrightarrow \ldots \to K^n \to K^{n - 1} \to \ldots \to K^1 \to K^0. $$ The morphisms $K \to K^i$, $K^j \to K^i$ for $j \geq i \geq 1$ come from the universal properties of the $\text{cosk}_n$ functors. The morphism $K^1 \to K^0$ is the canonical morphism from Simplicial, Remark \ref{simplicial-remark-augmentation}. We also recall that $K^0 \to \text{cosk}_1 \text{sk}_1 K^0$ is an isomorphism, see Simplicial, Lemma \ref{simplicial-lemma-cosk-minus-one}. \medskip\noindent By Lemma \ref{lemma-compare-cosk0} we see that $H_i(K^0) = 0$ for $i > 0$ and $H_0(K^0) = \mathbf{Z}_\mathcal{G}^\#$. \medskip\noindent Pick $n \geq 1$. Consider the morphism $K^n \to K^{n - 1}$. It is an isomorphism on terms of degree $< n$. Note that $K^n \to \text{cosk}_n \text{sk}_n K^n$ and $K^{n - 1} \to \text{cosk}_n \text{sk}_n K^{n - 1}$ are isomorphisms. Note that $(K^n)_n = K_n$ and that $(K^{n - 1})_n = (\text{cosk}_{n - 1} \text{sk}_{n - 1} K)_n$. Hence by assumption, we have that $(K^n)_n \to (K^{n - 1})_n$ is a morphism of presheaves which becomes surjective after sheafification. By Lemma \ref{lemma-acyclicity} we conclude that $H_i(K^n) = H_i(K^{n - 1})$. Combined with the above this proves the lemma.
Observe that for any object $Z = \{U_i \to X\}$ of $\text{SR}(\mathcal{C}, X)$ and any abelian sheaf $\mathcal{F}$ on $\mathcal{C}$ we have \begin{eqnarray*} \mathcal{F}(Z) & = & \prod \mathcal{F}(U_i) \\ & = & \prod \Mor_{\textit{PSh}(\mathcal{C})}(h_{U_i}, \mathcal{F})\\ & = & \Mor_{\textit{PSh}(\mathcal{C})}(F(Z), \mathcal{F})\\ & = & \Mor_{\textit{PAb}(\mathcal{C})}(\mathbf{Z}_{F(Z)}, \mathcal{F}) \\ & = & \Mor_{\textit{Ab}(\mathcal{C})}(\mathbf{Z}_{F(Z)}^\#, \mathcal{F}) \end{eqnarray*} Thus we see, for any simplicial object $K$ of $\text{SR}(\mathcal{C}, X)$ that we have \begin{equation} \label{equation-identify-cech} s(\mathcal{F}(K)) = \Hom_{\textit{Ab}(\mathcal{C})}(s(\mathbf{Z}_{F(K)}^\#), \mathcal{F}) \end{equation} see Definition \ref{definition-homology} for notation. The complex of sheaves $s(\mathbf{Z}_{F(K)}^\#)$ is quasi-isomorphic to $\mathbf{Z}_X^\#$ if $K$ is a hypercovering, see Lemma \ref{lemma-hypercovering-acyclic}. We conclude that if $\mathcal{I}$ is an injective abelian sheaf, and $K$ a hypercovering, then the complex $s(\mathcal{I}(K))$ is acyclic except possibly in degree $0$. In other words, we have $$ \check{H}^i(K, \mathcal{I}) = 0 $$ for $i > 0$. Combined with Lemma \ref{lemma-h0-cech} the lemma is proved.