hcju/stacksqa huggingface dataset:
Let us prove the canonical map $\colim_{a : j \to i} f_a^{-1}\mathcal{G}(f_a^{-1}(U_i)) \to p_i^{-1}\mathcal{G}(p_i^{-1}(U_i))$ is injective. Let $s, s'$ be sections of $f_a^{-1}\mathcal{G}$ over $f_a^{-1}(U_i)$ for some $a : j \to i$. For $b : k \to j$ let $Z_k \subset f_{a \circ b}^{-1}(U_i)$ be the closed subset of points $x$ such that the image of $s$ and $s'$ in the stalk $(f_{a \circ b}^{-1}\mathcal{G})_x$ are different. If $Z_k$ is nonempty for all $b : k \to j$, then by Topology, Lemma \ref{topology-lemma-inverse-limit-spectral-spaces-nonempty} we see that $\lim_{b : k \to j} Z_k$ is nonempty too. Then for $x \in \lim_{b : k \to j} Z_k \subset X$ (observe that $\mathcal{I}/j \to \mathcal{I}$ is initial) we see that the image of $s$ and $s'$ in the stalk of $p_i^{-1}\mathcal{G}$ at $x$ are different too since $(p_i^{-1}\mathcal{G})_x = (f_{b \circ a}^{-1}\mathcal{G})_{p_k(x)}$ for all $b : k \to j$ as above. Thus if the images of $s$ and $s'$ in $p_i^{-1}\mathcal{G}(p_i^{-1}(U_i))$ are the same, then $Z_k$ is empty for some $b : k \to j$. This proves injectivity. \medskip\noindent Surjectivity. Let $s$ be a section of $p_i^{-1}\mathcal{G}$ over $p_i^{-1}(U_i)$. By Topology, Lemma \ref{topology-lemma-directed-inverse-limit-spectral-spaces} the set $p_i^{-1}(U_i)$ is a quasi-compact open of the spectral space $X$. By construction of the pullback sheaf, we can find an open covering $p_i^{-1}(U_i) = \bigcup_{l \in L} W_l$, opens $V_{l, i} \subset X_i$, sections $s_{l, i} \in \mathcal{G}(V_{l, i})$ such that $p_i(W_l) \subset V_{l, i}$ and $p_i^{-1}s_{l, i}|_{W_l} = s|_{W_l}$. Because $X$ and $X_i$ are spectral and $p_i^{-1}(U_i)$ is quasi-compact open, we may assume $L$ is finite and $W_l$ and $V_{l, i}$ quasi-compact open for all $l$. Then we can apply Topology, Lemma \ref{topology-lemma-descend-opens} to find $a : j \to i$ and open covering $f_a^{-1}(U_i) = \bigcup_{l \in L} W_{l, j}$ by quasi-compact opens whose pullback to $X$ is the covering $p_i^{-1}(U_i) = \bigcup_{l \in L} W_l$ and such that moreover $W_{l, j} \subset f_a^{-1}(V_{l, i})$. Write $s_{l, j}$ the restriction of the pullback of $s_{l, i}$ by $f_a$ to $W_{l, j}$. Then we see that $s_{l, j}$ and $s_{l', j}$ restrict to elements of $(f_a^{-1}\mathcal{G})(W_{l, j} \cap W_{l', j})$ which pullback to the same element $(p_i^{-1}\mathcal{G})(W_l \cap W_{l'})$, namely, the restriction of $s$. Hence by injectivity, we can find $b : k \to j$ such that the sections $f_b^{-1}s_{l, j}$ glue to a section over $f_{a \circ b}^{-1}(U_i)$ as desired.
Recall that $p_i^{-1}(U_i)$ is a quasi-compact open of the spectral space $X$, see Topology, Lemma \ref{topology-lemma-directed-inverse-limit-spectral-spaces}. Hence Lemma \ref{lemma-directed-colimits-sections} applies and we have $$ \mathcal{F}(p_i^{-1}(U_i)) = \colim_{a : j \to i} p_j^{-1}\mathcal{F}_j(p_i^{-1}(U_i)). $$ A formal argument shows that $$ \colim_{a : j \to i} \mathcal{F}_j(f_a^{-1}(U_i)) = \colim_{a : j \to i} \colim_{b : k \to j} f_b^{-1}\mathcal{F}_j(f_{a \circ b}^{-1}(U_i)) $$ Thus it suffices to show that $$ p_j^{-1}\mathcal{F}_j(p_i^{-1}(U_i)) = \colim_{b : k \to j} f_b^{-1}\mathcal{F}_j(f_{a \circ b}^{-1}(U_i)) $$ This is Lemma \ref{lemma-compute-pullback-to-limit} applied to $\mathcal{F}_j$ and the quasi-compact open $f_a^{-1}(U_i)$.
First note that the map $\mathcal{F}(U) \to \prod\nolimits_{x \in U} \mathcal{F}_x$ is injective by the uniqueness in the sheaf condition of Definition \ref{definition-sheaf-basis}. Let $(s_x)$ be any element on the right hand side which satisfies $(*)$. Clearly this means we can find a covering $U = \bigcup U_i$, $U_i \in \mathcal{B}$ such that $(s_x)_{x \in U_i}$ comes from certain $\sigma_i \in \mathcal{F}(U_i)$. For every $y \in U_i \cap U_j$ the sections $\sigma_i$ and $\sigma_j$ agree in the stalk $\mathcal{F}_y$. Hence there exists an element $V_{ijy} \in \mathcal{B}$, $y \in V_{ijy}$ such that $\sigma_i|_{V_{ijy}} = \sigma_j|_{V_{ijy}}$. Thus the sheaf condition $(**)$ of Definition \ref{definition-sheaf-basis} applies to the system of $\sigma_i$ and we obtain a section $s \in \mathcal{F}(U)$ with the desired property.
We first construct a presheaf $\mathcal{F}^{ext}$ with the desired property. Namely, for an arbitrary open $U \subset X$ we define $\mathcal{F}^{ext}(U)$ as the set of elements $(s_x)_{x \in U}$ such that $(*)$ of Lemma \ref{lemma-condition-star-sections} holds. It is clear that there are restriction mappings that turn $\mathcal{F}^{ext}$ into a presheaf of sets. Also, by Lemma \ref{lemma-condition-star-sections} we see that $\mathcal{F}(U) = \mathcal{F}^{ext}(U)$ whenever $U$ is an element of the basis $\mathcal{B}$. To see $\mathcal{F}^{ext}$ is a sheaf one may argue as in the proof of Lemma \ref{lemma-sheafification-sheaf}.
By the conditions imposed on the pair $(\mathcal{C}, F)$ it suffices to come up with a presheaf $\mathcal{F}^{ext}$ which does the correct thing on the level of underlying presheaves of sets. Thus our first task is to construct a suitable object $\mathcal{F}^{ext}(U)$ for all open $U \subset X$. We could do this by imitating Lemma \ref{lemma-diagram-fibre-product} in the setting of presheaves on $\mathcal{B}$. However, a slightly different method (but basically equivalent) is the following: Define it as the directed colimit $$ \mathcal{F}^{ext}(U) := \colim_\mathcal{U} FIB(\mathcal{U}) $$ over all coverings $\mathcal{U} : U = \bigcup_{i\in I} U_i$ by $U_i \in \mathcal{B}$ of the fibre product $$ \xymatrix{ FIB(\mathcal{U}) \ar[r] \ar[d] & \prod\nolimits_{x\in U} \mathcal{F}_x \ar[d] \\ \prod\nolimits_{i\in I} \mathcal{F}(U_i) \ar[r] & \prod\nolimits_{i \in I} \prod\nolimits_{x\in U_i} \mathcal{F}_x } $$ By the usual arguments, see Lemma \ref{lemma-image-contained-in} and Example \ref{example-application-lemma-image-contained-in} it suffices to show that this construction on underlying sets is the same as the definition using $(**)$ above. Details left to the reader.
It suffices to construct the multiplication map on the level of presheaves of sets. Perhaps the easiest way to see this is to prove directly that if $(f_x)_{x \in U}$, $f_x \in \mathcal{O}_x$ and $(m_x)_{x \in U}$, $m_x \in \mathcal{F}_x$ satisfy $(*)$, then the element $(f_xm_x)_{x \in U}$ also satisfies $(*)$. Then we get the desired result, because in the proof of Lemma \ref{lemma-extend-off-basis} we construct the extension in terms of families of elements of stalks satisfying $(*)$.
This is trivial because the collection of maps amounts to a morphism between the restrictions of $\mathcal{G}$ and $f_*\mathcal{F}$ to $\mathcal{B}$. By Lemma \ref{lemma-restrict-basis-equivalence-structures} this is the same as giving a morphism from $\mathcal{G}$ to $f_*\mathcal{F}$, which by Lemma \ref{lemma-f-map} is the same as an $f$-map. See also Lemma \ref{lemma-f-map-sets-algebraic-structures} and the discussion preceding it for how to deal with the case of sheaves of algebraic structures.
Let us first proves this for sheaves of sets. Fix $V \subset Y$ open. Pick $s \in \mathcal{G}(V)$. We are going to construct an element $\varphi_V(s) \in \mathcal{F}(f^{-1}V)$. We can define a value $\varphi(s)_x$ in the stalk $\mathcal{F}_x$ for every $x \in f^{-1}V$ by picking a $U \in \mathcal{B}_X$ with $x \in U \subset f^{-1}V$ and setting $\varphi(s)_x$ equal to the equivalence class of $(U, \varphi_V^U(s))$ in the stalk. Clearly, the family $(\varphi(s)_x)_{x \in f^{-1}V}$ satisfies condition $(*)$ because the maps $\varphi_V^U$ for varying $U$ are compatible with restrictions in the sheaf $\mathcal{F}$. Thus, by the proof of Lemma \ref{lemma-extend-off-basis} we see that $(\varphi(s)_x)_{x \in f^{-1}V}$ corresponds to a unique element $\varphi_V(s)$ of $\mathcal{F}(f^{-1}V)$. Thus we have defined a set map $\varphi_V : \mathcal{G}(V) \to \mathcal{F}(f^{-1}V)$. The compatibility between $\varphi_V$ and $\varphi_V^U$ follows from Lemma \ref{lemma-condition-star-sections}. \medskip\noindent We leave it to the reader to show that the construction of $\varphi_V$ is compatible with restriction mappings as we vary $v \in \mathcal{B}_Y$. Thus we may apply Lemma \ref{lemma-f-map-basis-below-structures} above to ``glue'' them to the desired $f$-map. \medskip\noindent Finally, we note that the map of sheaves of sets so constructed satisfies the property that the map on stalks $$ \mathcal{G}_{f(x)} \longrightarrow \mathcal{F}_x $$ is the colimit of the system of maps $\varphi_V^U$ as $V \in \mathcal{B}_Y$ varies over those elements that contain $f(x)$ and $U \in \mathcal{B}_X$ varies over those elements that contain $x$. In particular, if $\mathcal{G}$ and $\mathcal{F}$ are the underlying sheaves of sets of sheaves of algebraic structures, then we see that the maps on stalks is a morphism of algebraic structures. Hence we conclude that the associated map of sheaves of underlying sets $f^{-1}\mathcal{G} \to \mathcal{F}$ satisfies the assumptions of Lemma \ref{lemma-f-map-sets-algebraic-structures}. We conclude that $f^{-1}\mathcal{G} \to \mathcal{F}$ is a morphism of sheaves with values in $\mathcal{C}$. And by adjointness this means that $\varphi$ is an $f$-map of sheaves of algebraic structures.
To map $j_{p!}\mathcal{F}$ into $\mathcal{G}$ it is enough to map $\mathcal{F}(V) \to \mathcal{G}(V)$ whenever $V \subset U$ compatibly with restriction mappings. And by Lemma \ref{lemma-j-pullback} the same description holds for maps $\mathcal{F} \to \mathcal{G}|_U$. The adjointness of $j_!$ and restriction follows from this and the properties of sheafification. The identification of stalks is obvious from the definition of the extension by the empty set and the definition of a stalk. Statements (4) and (5) follow by computing the value of the sheaf on any open of $U$.
If $x \not \in Z$, then there exist arbitrarily small open neighbourhoods $U$ of $x$ which do not meet $Z$. Because $\mathcal{F}$ is a sheaf we have $\mathcal{F}(i^{-1}(U)) = \{*\}$ for any such $U$, see Remark \ref{remark-confusion}. This proves the first case. The second case comes from the fact that for $z \in Z$ any open neighbourhood of $z$ is of the form $Z \cap U$ for some open $U$ of $X$. For the statement that $i^{-1}i_* = \text{id}$ consider the canonical map $i^{-1}i_*\mathcal{F} \to \mathcal{F}$. This is an isomorphism on stalks (see above) and hence an isomorphism. \medskip\noindent For sheaves of abelian groups, and sheaves of algebraic structures you argue in the same manner.
First proof. In this proof we give a formula for the set of sections of $\mathcal{F}$ over an open $W \subset X$. Namely, we define $$ \mathcal{F}(W) = \{ (s_i)_{i \in I} \mid s_i \in \mathcal{F}_i(W \cap U_i), \varphi_{ij}(s_i|_{W \cap U_i \cap U_j}) = s_j|_{W \cap U_i \cap U_j} \}. $$ Restriction mappings for $W' \subset W$ are defined by the restricting each of the $s_i$ to $W' \cap U_i$. The sheaf condition for $\mathcal{F}$ follows immediately from the sheaf condition for each of the $\mathcal{F}_i$. \medskip\noindent We still have to prove that $\mathcal{F}|_{U_i}$ maps isomorphically to $\mathcal{F}_i$. Let $W \subset U_i$. In this case the condition in the definition of $\mathcal{F}(W)$ implies that $s_j = \varphi_{ij}(s_i|_{W \cap U_j})$. And the commutativity of the diagrams in the definition of a glueing data assures that we may start with {\it any} section $s \in \mathcal{F}_i(W)$ and obtain a compatible collection by setting $s_i = s$ and $s_j = \varphi_{ij}(s_i|_{W \cap U_j})$. \medskip\noindent Second proof (sketch). Let $\mathcal{B}$ be the set of opens $U \subset X$ such that $U \subset U_i$ for some $i \in I$. Then $\mathcal{B}$ is a base for the topology on $X$. For $U \in \mathcal{B}$ we pick $i \in I$ with $U \subset U_i$ and we set $\mathcal{F}(U) = \mathcal{F}_i(U)$. Using the isomorphisms $\varphi_{ij}$ we see that this prescription is ``independent of the choice of $i$''. Using the restriction mappings of $\mathcal{F}_i$ we find that $\mathcal{F}$ is a sheaf on $\mathcal{B}$. Finally, use Lemma \ref{lemma-extend-off-basis} to extend $\mathcal{F}$ to a unique sheaf $\mathcal{F}$ on $X$.
It is clear that (b) implies (a). We now check that (a) implies (c). If $f$ is  not universally injective, we can find $M \in \text{Mod}_R$ such that the map $1_M \otimes f: M \to M \otimes_R S$ has nontrivial kernel $N$. The natural projection $M \to M/N$ is not an isomorphism, but its image in  $DD_{S/R}$ is an isomorphism. Hence $f^*$ is not fully faithful. \medskip\noindent We finally check that (c) implies (b). By Lemma \ref{lemma-descent-lemma}, for $(M, \theta) \in DD_{S/R}$, the natural map $f^* f_*(M,\theta) \to M$ is an isomorphism of $S$-modules. On  the other hand, for $M_0 \in \text{Mod}_R$, we may tensor (\ref{equation-equalizer-S}) with $M_0$ over $R$ to obtain an  equalizer sequence, so $M_0 \to f_* f^* M$ is an isomorphism. Consequently, $f_*$ and $f^*$ are  quasi-inverse functors, proving the claim.
To prove (a), choose a finite set $\{n_i\}$ of generators of $M \otimes_R S$ in $\text{Mod}_S$. Write each $n_i$ as $\sum_j m_{ij} \otimes s_{ij}$ with $m_{ij} \in M$ and $s_{ij} \in S$. Let $F$ be the finite free $R$-module with basis $e_{ij}$ and let $F \to M$ be the $R$-module map sending $e_{ij}$ to $m_{ij}$. Then $F \otimes_R S\to M \otimes_R S$ is surjective, so $\Coker(F \to M) \otimes_R S$ is zero and hence $\Coker(F \to M)$ is zero. This proves (a). \medskip\noindent To see (b) assume $M \otimes_R S$ is finitely presented. Then $M$ is finitely generated by (a). Choose a surjection $R^{\oplus n} \to M$ with kernel $K$. Then $K \otimes_R S \to S^{\oplus r} \to M \otimes_R S \to 0$ is exact. By Algebra, Lemma \ref{algebra-lemma-extension} the kernel of $S^{\oplus r} \to M \otimes_R S$ is a finite $S$-module. Thus we can find finitely many elements $k_1, \ldots, k_t \in K$ such that the images of $k_i \otimes 1$ in $S^{\oplus r}$ generate the kernel of $S^{\oplus r} \to M \otimes_R S$. Let $K' \subset K$ be the submodule generated by $k_1, \ldots, k_t$. Then $M' = R^{\oplus r}/K'$ is a finitely presented $R$-module with a morphism $M' \to M$ such that $M' \otimes_R S \to M \otimes_R S$ is an isomorphism. Thus $M' \cong M$ as desired. \medskip\noindent To prove (c), let $0 \to M' \to M'' \to M \to 0$ be a short exact sequence in  $\text{Mod}_R$. Since $\bullet \otimes_R S$ is a right exact functor, $M'' \otimes_R S \to M \otimes_R S$ is surjective. So by Lemma \ref{lemma-C-is-faithful} the map $C(M \otimes_R S) \to C(M'' \otimes_R S)$ is injective. If $M \otimes_R S$ is flat, then Lemma \ref{lemma-flat-to-injective} shows $C(M \otimes_R S)$ is an injective object of $\text{Mod}_S$, so the injection $C(M \otimes_R S) \to C(M'' \otimes_R S)$ is split in $\text{Mod}_S$ and hence also in $\text{Mod}_R$. Since $C(M \otimes_R S) \to C(M)$ is a split surjection by  Lemma \ref{lemma-split-surjection}, it follows that  $C(M) \to C(M'')$ is a split injection in $\text{Mod}_R$. That is, the sequence $$ 0 \to C(M) \to C(M'') \to C(M') \to 0 $$ is split exact.  For $N \in \text{Mod}_R$, by (\ref{equation-adjunction}) we see that  $$ 0 \to C(M \otimes_R N) \to C(M'' \otimes_R N) \to C(M' \otimes_R N) \to 0 $$ is split exact. By Lemma \ref{lemma-C-is-faithful},  $$ 0 \to M' \otimes_R N \to M'' \otimes_R N \to M \otimes_R N \to 0 $$ is exact. This implies $M$ is flat over $R$. Namely, taking $M'$ a free module surjecting onto $M$ we conclude that $\text{Tor}_1^R(M, N) = 0$ for all modules $N$ and we can use Algebra, Lemma \ref{algebra-lemma-characterize-flat}. This proves (c). \medskip\noindent To deduce (d) from (c), note that if $N \in \text{Mod}_R$ and $M \otimes_R N$  is zero, then $M \otimes_R S \otimes_S (N \otimes_R S) \cong (M \otimes_R N) \otimes_R  S$ is zero, so $N \otimes_R S$ is zero and hence $N$ is zero. \medskip\noindent To deduce (e) at this point, it suffices to recall that $M$ is finitely  generated and projective if and only if it is finitely presented and flat. See Algebra, Lemma \ref{algebra-lemma-finite-projective}.
To prove (a), choose a finite set $\{x_i\}$ of generators of $A \otimes_R S$ over $S$. Write each $x_i$ as $\sum_j y_{ij} \otimes s_{ij}$ with $y_{ij} \in A$ and $s_{ij} \in S$. Let $F$ be the polynomial $R$-algebra on variables $e_{ij}$ and let $F \to M$ be the $R$-algebra map sending $e_{ij}$ to $y_{ij}$. Then $F \otimes_R S\to A \otimes_R S$ is surjective, so $\Coker(F \to A) \otimes_R S$ is zero and hence $\Coker(F \to A)$ is zero. This proves (a). \medskip\noindent To see (b) assume $A \otimes_R S$ is a finitely presented $S$-algebra. Then $A$ is finite type over $R$ by (a). Choose a surjection $R[x_1, \ldots, x_n] \to A$ with kernel $I$. Then $I \otimes_R S \to S[x_1, \ldots, x_n] \to A \otimes_R S \to 0$ is exact. By Algebra, Lemma \ref{algebra-lemma-finite-presentation-independent} the kernel of $S[x_1, \ldots, x_n] \to A \otimes_R S$ is a finitely generated ideal. Thus we can find finitely many elements $y_1, \ldots, y_t \in I$ such that the images of $y_i \otimes 1$ in $S[x_1, \ldots, x_n]$ generate the kernel of $S[x_1, \ldots, x_n] \to A \otimes_R S$. Let $I' \subset I$ be the ideal generated by $y_1, \ldots, y_t$. Then $A' = R[x_1, \ldots, x_n]/I'$ is a finitely presented $R$-algebra with a morphism $A' \to A$ such that $A' \otimes_R S \to A \otimes_R S$ is an isomorphism. Thus $A' \cong A$ as desired. \medskip\noindent To prove (c), recall that $A$ is formally unramified over $R$ if and only if the module of relative differentials $\Omega_{A/R}$ vanishes, see Algebra, Lemma \ref{algebra-lemma-characterize-formally-unramified} or \cite[Proposition~17.2.1]{EGA4}. Since $\Omega_{(A \otimes_R S)/S} = \Omega_{A/R} \otimes_R S$, the vanishing descends by Theorem \ref{theorem-descent}. \medskip\noindent To deduce (d) from the previous cases, recall that $A$ is unramified over $R$ if and only if $A$ is formally unramified and of finite type over $R$, see Algebra, Lemma \ref{algebra-lemma-formally-unramified-unramified}. \medskip\noindent To prove (e), recall that by Algebra, Lemma \ref{algebra-lemma-etale-flat-unramified-finite-presentation} or \cite[Th\'eor\`eme~17.6.1]{EGA4} the algebra $A$ is \'etale over $R$ if and only if $A$ is flat, unramified, and of finite presentation over $R$.
Here are the first few maps where $\varphi(n \otimes 1) = \sum \alpha_i \otimes x_i$ $$ \begin{matrix} \delta^1_0 & : & N & \to & A \otimes N & n & \mapsto & 1 \otimes n \\ \delta^1_1 & : & N & \to & A \otimes N & n & \mapsto & \sum \alpha_i \otimes x_i\\ \sigma^0_0 & : & A \otimes N & \to & N & a_0 \otimes n & \mapsto & a_0n \\ \delta^2_0 & : & A \otimes N & \to & A \otimes A \otimes N & a_0 \otimes n & \mapsto & 1 \otimes a_0 \otimes n \\ \delta^2_1 & : & A \otimes N & \to & A \otimes A \otimes N & a_0 \otimes n & \mapsto & a_0 \otimes 1 \otimes n \\ \delta^2_2 & : & A \otimes N & \to & A \otimes A \otimes N & a_0 \otimes n & \mapsto & \sum a_0 \otimes \alpha_i \otimes x_i \\ \sigma^1_0 & : & A \otimes A \otimes N & \to & A \otimes N & a_0 \otimes a_1 \otimes n & \mapsto & a_0a_1 \otimes n \\ \sigma^1_1 & : & A \otimes A \otimes N & \to & A \otimes N & a_0 \otimes a_1 \otimes n & \mapsto & a_0 \otimes a_1n \end{matrix} $$ with notation as in Simplicial, Section \ref{simplicial-section-cosimplicial-object}. We first verify the two properties $\sigma^0_0 \circ \delta^1_0 = \text{id}$ and $\sigma^0_0 \circ \delta^1_1 = \text{id}$. The first one, $\sigma^0_0 \circ \delta^1_0 = \text{id}$, is clear from the explicit description of the morphisms above. To prove the second relation we have to use the cocycle condition (because it does not holds for an arbitrary isomorphism $\varphi : N \otimes_R A \to A \otimes_R N$). Write $p = \sigma^0_0 \circ \delta^1_1 : N \to N$. By the description of the maps above we deduce that $p$ is also equal to $$ p = \varphi \otimes \text{id} : N = (N \otimes_R A) \otimes_{(A \otimes_R A)} A \longrightarrow (A \otimes_R N) \otimes_{(A \otimes_R A)} A = N $$ Since $\varphi$ is an isomorphism we see that $p$ is an isomorphism. Write $\varphi(n \otimes 1) = \sum \alpha_i \otimes x_i$ for certain $\alpha_i \in A$ and $x_i \in N$. Then $p(n) = \sum \alpha_ix_i$. Next, write $\varphi(x_i \otimes 1) = \sum \alpha_{ij} \otimes y_j$ for certain $\alpha_{ij} \in A$ and $y_j \in N$. Then the cocycle condition says that $$ \sum \alpha_i \otimes \alpha_{ij} \otimes y_j = \sum \alpha_i \otimes 1 \otimes x_i. $$ This means that $p(n) = \sum \alpha_ix_i = \sum \alpha_i\alpha_{ij}y_j = \sum \alpha_i p(x_i) = p(p(n))$. Thus $p$ is a projector, and since it is an isomorphism it is the identity. \medskip\noindent To prove fully that $N_\bullet$ is a cosimplicial module we have to check all 5 types of relations of Simplicial, Remark \ref{simplicial-remark-relations-cosimplicial}. The relations on composing $\sigma$'s are obvious. The relations on composing $\delta$'s come down to the cocycle condition for $\varphi$. In exactly the same way as above one checks the relations $\sigma_j \circ \delta_j = \sigma_j \circ \delta_{j + 1} = \text{id}$. Finally, the other relations on compositions of $\delta$'s and $\sigma$'s hold for any $\varphi$ whatsoever.
By Simplicial, Lemma \ref{simplicial-lemma-push-outs-simplicial-object-w-section} the map $R \to (A/R)_\bullet$ is a homotopy equivalence of cosimplicial $R$-algebras (here $R$ denotes the constant cosimplicial $R$-algebra). Hence $M \to (A/R)_\bullet \otimes_R M$ is a homotopy equivalence in the category of cosimplicial $R$-modules, because $\otimes_R M$ is a functor from the category of $R$-algebras to the category of $R$-modules, see Simplicial, Lemma \ref{simplicial-lemma-functorial-homotopy}. This implies that the induced map of associated complexes is a homotopy equivalence, see Simplicial, Lemma \ref{simplicial-lemma-homotopy-s-Q}. Since the complex associated to the constant cosimplicial $R$-module $M$ is the complex $$ \xymatrix{ M \ar[r]^0 & M \ar[r]^1 & M \ar[r]^0 & M \ar[r]^1 & M \ldots } $$ we win (since the extended version simply puts an extra $M$ at the beginning).
If $(N, \varphi)$ is effective, then we may write $N = A \otimes_R M$ with $\varphi = can$. It follows that $H^0(s(N_\bullet)) = M$ by Lemmas \ref{lemma-canonical-descent-datum-cosimplicial} and \ref{lemma-ff-exact}. Conversely, suppose the map of the lemma is an isomorphism. In this case set $M = H^0(s(N_\bullet))$. This is an $R$-submodule of $N$, namely $M = \{n \in N \mid 1 \otimes n = \varphi(n \otimes 1)\}$. The only thing to check is that via the isomorphism $A \otimes_R M \to N$ the canonical descent data agrees with $\varphi$. We omit the verification.
Let $(N, \varphi)$ be a descent datum for $R \to A$. Set $N' = R' \otimes_R N = A' \otimes_A N$, and denote $\varphi' = \text{id}_{R'} \otimes \varphi$ the base change of the descent datum $\varphi$. Then $(N', \varphi')$ is a descent datum for $R' \to A'$ and $H^0(s(N'_\bullet)) = R' \otimes_R H^0(s(N_\bullet))$. Moreover, the map $A' \otimes_{R'} H^0(s(N'_\bullet)) \to N'$ is identified with the base change of the $A$-module map $A \otimes_R H^0(s(N)) \to N$ via the faithfully flat map $A \to A'$. Hence we conclude by Lemma \ref{lemma-recognize-effective}.
By (\ref{equation-adjunction}), for any $P \in \text{Mod}_R$ we have a  commutative diagram $$ \xymatrix@C=9pc{ \Hom_R( P, C(N)) \ar[r]_{\Hom_R(P,C(f))} \ar[d]^{\cong} & \Hom_R(P,C(M)) \ar[d]^{\cong} \\ C(P \otimes_R N ) \ar[r]^{C(1_{P} \otimes f)} & C(P \otimes_R M ). } $$ If $f$ is universally injective, then $1_{C(M)} \otimes f: C(M) \otimes_R M \to  C(M) \otimes_R N$ is injective, so both rows in the above diagram are surjective for $P = C(M)$. We may thus  lift $1_{C(M)} \in \Hom_R(C(M), C(M))$ to some $g \in \Hom_R(C(N), C(M))$ splitting  $C(f)$. Conversely, if $C(f)$ is a split surjection, then  both rows in the above diagram are surjective, so by Lemma \ref{lemma-C-is-faithful}, $1_{P} \otimes f$ is injective.