Proof. We adapt the proof of [23] and show that \(\boldsymbol{BS}\) has arbitrary coproducts and coequalizers.

Coproducts: Let \((X_i,\mathop{\mathrm{\mathcal{O}}}_{X_i})_{i \in I}\) be a family of band spaces. Their coproduct \((S,\mathop{\mathrm{\mathcal{O}}}_S)\) consists of the topological space \(S \mathrel{\vcenter{:}}= \coprod_{i \in I} X_i\) together with the sheaf of bands \(\mathop{\mathrm{\mathcal{O}}}_S\), defined as follows: for every open subset \(U = \coprod_{i \in I} U_i\), where \(U_i\) is an open subset of \(X_i\), we define \(\mathop{\mathrm{\mathcal{O}}}_S(U)\) as the product, in the category of bands, of the bands \(\mathop{\mathrm{\mathcal{O}}}_{X_i}(U_i)\). Moreover, for every \(x \in X_i \subseteq S\), we have an isomorphism \[\mathop{\mathrm{\mathcal{O}}}_{S,x} \cong \mathop{\mathrm{\mathcal{O}}}_{X_i,x}.\] This shows that \((S,\mathop{\mathrm{\mathcal{O}}}_S)\), together with the natural morphisms \(\iota_i: (X_i,\mathop{\mathrm{\mathcal{O}}}_{X_i}) \to (S,\mathop{\mathrm{\mathcal{O}}}_S)\), is indeed the coproduct in \(\boldsymbol{BS}\) of the family \((X_i,\mathop{\mathrm{\mathcal{O}}}_{X_i})_{i \in I}\).

Coequalizers: Let \((f,f^\#)\) and \((g,g^\#)\) be two morphisms in \(\boldsymbol{BS}\) from \((X,\mathop{\mathrm{\mathcal{O}}}_X)\) to \((Y,\mathop{\mathrm{\mathcal{O}}}_Y)\). Define the band space \((Z,\mathop{\mathrm{\mathcal{O}}}_Z)\) as follows:

• \(Z\) is the coequalizer, in the category of topological spaces, of the continuous maps \(f\) and \(g\); that is, \(Z\) is the quotient space \(Y/\sim\), where \(\sim\) is the smallest equivalence relation such that \(f(x) \sim g(x)\) for all \(x \in X\).

• Let \(p \colon Y \to Z\) be the canonical projection, and let \(W\) be an open subset of \(Z\). Observe that if \(V = p^{-1}(W)\), then \(f^{-1}(V) = g^{-1}(V)\). Denote this common set by \(U\). We define \(\mathop{\mathrm{\mathcal{O}}}_Z\) by \[\mathop{\mathrm{\mathcal{O}}}_Z(W) \mathrel{\vcenter{:}}= \left\{ s \in \mathop{\mathrm{\mathcal{O}}}_Y(V)\;\middle|\; f^\#_U(s) = g^\#_U(s) \in \mathop{\mathrm{\mathcal{O}}}_X(U) \right\}.\] For open subsets \(W' \subseteq W\) of \(Z\), the restriction map \(\mathop{\mathrm{\mathcal{O}}}_Z(W) \to \mathop{\mathrm{\mathcal{O}}}_Z(W')\) is induced by the restriction maps of \(\mathop{\mathrm{\mathcal{O}}}_Y\), and the morphism of presheaves \(p^\# \colon \mathop{\mathrm{\mathcal{O}}}_Z \to p_*\mathop{\mathrm{\mathcal{O}}}_Y\) is defined, for every open subset \(W \subseteq Z\), as the inclusion \(\mathop{\mathrm{\mathcal{O}}}_Z(W) \hookrightarrow \mathop{\mathrm{\mathcal{O}}}_Y(p^{-1}(W))\). The sheaf axioms for \(\mathop{\mathrm{\mathcal{O}}}_Z\) follow from those for \(\mathop{\mathrm{\mathcal{O}}}_Y\).

Finally, we prove that for every \(y \in Y\), the morphism \[\begin{align} p^\#_{y}: \mathop{\mathrm{\mathcal{O}}}_{Z,p(y)} &\longrightarrow \mathop{\mathrm{\mathcal{O}}}_{Y,y}\\ [(s,W)] &\longmapsto [(s,p^{-1}(W))], \end{align}\] where \((s,W)\) is a pair of an open subset \(W \subseteq Z\) and a section \(s \in \mathop{\mathrm{\mathcal{O}}}_Z(W)\), is a local morphism of bands. Indeed, suppose that there exists \([(t, U)] \in \mathop{\mathrm{\mathcal{O}}}_{Y,y}\) such that \[[(s,p^{-1}(W))]\cdot [(t, U)] = [(s|_{p^{-1}(W) \cap U} \cdot t|_{p^{-1}(W) \cap U}, p^{-1}(W) \cap U)] = [(1, p^{-1}(W) \cap U)].\] Then, there exists an open subset \(W' \subseteq Z\) such that \(y \in p^{-1}(W') \subseteq p^{-1}(W) \cap U\) and \[\label{identityasdf} s|_{p^{-1}(W')} \cdot t|_{p^{-1}(W')} = 1.\tag{5}\] Since \(s \in \mathop{\mathrm{\mathcal{O}}}_Z(W)\), we have \[f^\#(s|_{p^{-1}(W')}) = g^\#(s|_{p^{-1}(W')}).\] Applying \(f^\#\) and \(g^\#\) to the identity 5 and using cancellation, we obtain \[f^\#(t|_{p^{-1}(W')}) = g^\#(t|_{p^{-1}(W')}).\] Hence \(t|_{p^{-1}(W')} \in \mathop{\mathrm{\mathcal{O}}}_Z(W')\), which shows that \([(t|_{p^{-1}(W')}, p^{-1}(W'))]\) defines an element of \(\mathop{\mathrm{\mathcal{O}}}_{Z,p(y)}\) which is the inverse of \([(s,W)]\). Therefore, \(p^\#_y\) is a local morphism of bands. ◻

Proof. The first part follows from [22]. Let \(X\) be a band scheme, \(\{U_i \to X\}_{i \in I}\) an affine open covering and \(U_{i,j} \mathrel{\vcenter{:}}= U_i \cap U_j\). Then, the following isomorphism holds. \[\varinjlim\left(\coprod_{(i,j)\in I^2} U_{i,j} \rightrightarrows \coprod_{i\in I} U_i \right) \overset{\sim}{\longrightarrow} X.\] ◻

Proof. Since a morphism of \(B\)-modules is completely determined by the images of the underlying elements, every injective (resp.surjective) morphism is necessarily a monomorphism (resp.epimorphism).

Conversely, suppose that \(f\) is a monomorphism and that \(f(n)=f(n')\) for some \(n,n' \in N\). Consider the maps \(g,h \colon B \to N\) defined by \(g(b)=b\cdot n\) and \(h(b)=b\cdot n'\) for all \(b \in B\). Both maps are compatible with the action of \(B\). Moreover, if \(\sum_i b_i \in N_B\), then by definition both \(\sum_i b_i \cdot n\) and \(\sum_i b_i \cdot n'\) belong to \(I_N\). Hence \(g\) and \(h\) are well-defined morphisms of \(B\)-modules satisfying \(f \circ g = f \circ h\). Since \(f\) is a monomorphism, we conclude that \(g=h\), and therefore \(n=n'\). Thus \(f\) is injective.

Finally, suppose that \(f\) is an epimorphism, and let \(\pi \colon M \to M/\mathrm{im}(f)\) be the canonical projection. By construction, we have \(\pi \circ f = 0 \circ f\). Since \(f\) is an epimorphism, this implies that \(\pi = 0\). Consequently, \(M/\mathrm{im}(f)=0\), and hence \(f\) is surjective. ◻

Proof. Let \(\sim\) be the equivalence relation defining \(M / \mathop{\mathrm{Im}}(f)\), and let \(\phi : M \to P\) be a morphism of \(B\)-modules such that \(\phi \circ f = 0\).

If \(m \sim m'\), then \(\phi(m) = \phi(m')\), so the map \[\tilde{\phi} : M / \mathop{\mathrm{Im}}(f) \longrightarrow P, \qquad \tilde{\phi}([m]) = \phi(m),\] is well defined. By the definition of the null set of \(M / \mathop{\mathrm{Im}}(f)\), we see that \(\tilde{\phi}\) is a morphism of \(B\)-modules. Moreover, \(\tilde{\phi}\) is the unique morphism of \(B\)-modules satisfying \(\tilde{\phi} \circ \pi = \phi\). ◻

Proof. By usual arguments, we see that \(\Phi\) is a well-defined bijective morphism of \(B\)-modules. It remaines for us to verify the correspondence between their null sets. By definition, \(I_{S^{-1}M}\) is generated by formal sums \(\sum_j m_j/1 \in (S^{-1}M)^+\) with \(\sum_j m_j \in I_M\). Since each \(m_j/1\) is the image of \(m_j \otimes 1\), we conclude, by the definition of null sets of tensor product, that \(\sum_j m_j/1\) is in the image of \(I_{M \otimes_B S^{-1}B }\) by \(\Phi^+\). ◻

Proof. \((1) \implies (2)\) is trivial. Suppose \((2)\) holds and let \(Q\) be a \(B\)-module such that there exists an isomorphism \[\varphi: \bigvee_{i \in I} B\cdot e_i \overset{\sim}{\longrightarrow} P \vee Q.\] Let \(J\) be the set of indexes \(j \in I\) such that \(\varphi(e_j) \in P\). We claim that the map \[\varphi_J: \bigvee_{j \in J} B\cdot e_j \longrightarrow P,\quad \varphi_J(e_j) \mathrel{\vcenter{:}}= \varphi(e_j) \text{ for every } j \in J\] is a well defined isomorphism of \(B\)-modules. Indeed, for every \(j \in J\) and \(b \in B\), we have \(\varphi(b\cdot e_j) = b\cdot \varphi(e_j) \in P\), hence \(\varphi_J\) is well define. Now, for every \(m \in P\), there exists an \(i \in I\) and an \(b \in N\) such that \(\varphi(b \cdot e_i) = m\). If, \(i \not\in J\), then we would have \(\varphi(b \cdot e_i) = b \cdot \varphi(e_i) \in Q\), which is a contradiction. So \(\varphi_J\) is surjective. The injectivity of \(\varphi_J\) follows from that of \(\varphi\).

Finally, we just need to verify the correspondence of the null sets. By the isomorphism \(\varphi\), and the definition of null set of \(P \vee Q\), we see that the null set of \(P\) is \[I_P = \left\{\sum_{l} b_l \cdot \varphi(e_j)\;\middle|\; \sum_{l} b_l \in N_B,\; j \in J \right\}.\] From this, we conclude that \(\varphi_I\) is an isomorphism. ◻

Proof. The equivalence \((2) \iff (3)\) is the Proposition 24, and the proof of \((2) \implies (1)\) is the same as the case of rings. We prove \((1) \implies (3)\). Suppose that \(P\) satisfies \((1)\) and consider the following epimorphism of \(F\)-modules: \[\pi: N\mathrel{\vcenter{:}}= \bigvee_{p \in P}F\cdot e_p \longrightarrow P,\quad \pi(e_p) \mathrel{\vcenter{:}}= p\text{ for every }p \in P.\] Then, by hypothesis, there exists a morphism \(\varphi : P \rightarrow N\) such that \(\pi \circ \varphi = \mathop{\mathrm{id}}_P\). Consider the following subset of \(P\): \[\begin{align} \tilde{P} &\mathrel{\vcenter{:}}= \left\{p \in P\;\middle|\; \text{There exists a } q \in P \text{ and } b \in F \text{ such that } \varphi(q) = b\cdot e_p \right\}\\ & \;= \left\{p \in P\;\middle|\; \text{There exists a } q \in P \text{ such that } \varphi(q) = e_p \right\}\\ & \;= \left\{p \in P\;\middle|\; \varphi(p) = e_p \right\}, \end{align}\] where the second equality follows from the properties of the idyll \(F\), and the last equality follows from \(\pi \circ \varphi = \mathop{\mathrm{id}}_P\). We claim that \[\tilde{\pi}: \tilde{N} \mathrel{\vcenter{:}}= \bigvee_{p \in \tilde{P}} F\cdot e_p \longrightarrow P,\quad \tilde{\pi}(e_p) \mathrel{\vcenter{:}}= p \text{ for every } p \in \tilde{P},\] is an isomorphism. Since \(\tilde{\pi}\) is the composition of the inclusion \(\bigvee_{p \in \tilde{P}} F\cdot e_p \hookrightarrow \bigvee_{p \in P}F\cdot e_p\) with \(\pi\), it is a morphism of \(F\)-modules.

Now, we prove the surjectivity. For every \(m \in P\), there exists a \(b \in F\) and \(q \in P\) such that \(\varphi(m) = b \cdot e_q\). By definition, \(q \in \tilde{P}\) and \(\tilde{\pi}(b\cdot e_q) = \pi(b\cdot e_q) = (\pi \circ \varphi)(m) = m\), hence \(\tilde{\pi}\) is surjective.

Next, we prove the injectivity. Suppose \(\tilde{\pi}(b\cdot e_p) = \tilde{\pi}(c\cdot e_q)\) for some \(b,c \in F\) and \(p,q \in \tilde{P}\). Using that \(\varphi(e_p) = e_p\), \(\varphi(e_q) = e_q\) and \(\pi\circ \varphi = \mathop{\mathrm{id}}_P\), we conclude that \(b\cdot e_p = c \cdot e_q\), hence \(\tilde{\pi}\) is injective.

From surjectiviry of \(\tilde{\pi}\) and the projectivity of \(P\), there exists a morphism \(\phi : P \rightarrow \tilde{N}\) such that \(\tilde{\pi} \circ \phi = \mathop{\mathrm{id}}_P\). This implies that \(\tilde{\pi}\) is an isomorphism. ◻

Proof. Let \(M\) be a finitely generated projective module over \(F\). By Proposition 26 and Corollary 1, \(M\) is isomorphic to a free module of the form \[\phi: M \overset{\sim}{\longrightarrow} \bigvee_{i = 1}^n F\cdot e_i\;,\] with a uniquely determined \(n\). Hence, closedness by coproduct follows directly. Now, suppose \(N\) is a strict submodule of \(M\), and consider the set \(I'\) defined as \[I' \mathrel{\vcenter{:}}= \left\{i \in I\;\middle|\; e_i \in N \right\}.\] Then, since \(F\) is an idyll, the restriction of \(\phi\) to \(N\) gives an isomorphism from \(N\) to the coproduct \(\bigvee_{i \in I'}F \cdot e_{i'}\). Moreover, the quotient \(M/N\) is isomorphic to the coproduct \[M/N \overset{\sim}{\longrightarrow} \bigvee_{i \in I\setminus I'} F\cdot e_i.\] Finally, since, by definition, kernel is a strict submodule and, by Lemma 2, cokernel is a quotient by a strict submodule, we conclude that \(\mathop{\mathrm{\mathbf{BMod}}}_F^{\mathrm{fin, proj}}\) is also closed by kernel and cokernel. ◻

Proof. Define the map \(\tilde{\varphi}\) as \[\tilde{\varphi} \left( \sum m_i + \left<I_M \right>\right) \mathrel{\vcenter{:}}= \sum f(m_i),\] for every \(\sum m_i + \left<I_M \right> \in M_{\mathop{\mathrm{\mathbb{Z}}}}^+\). Observe that, since we are considering \(P\) as an object of \(\mathop{\mathrm{\mathbf{BMod}}}_B\), its null set is \[I_P = \left\{ \sum p_i \in P^+ : \sum p_i = 0\text{ in } P \right\}\] Hence, by the definition of morphisms of \(B\)-modules, for every \(\sum m_i \in I_M\), we have \(\sum f(m_i) = 0\) in \(N\), proving the well-definess of \(\tilde{\varphi}\).

From the definition of \(\tilde{\varphi}\), it follows that it preserves both the sum and the action of \(B_{\mathop{\mathrm{\mathbb{Z}}}}^+\). ◻

Proof. Bilinearity of \(\otimes_B\) follows directly from the defining relations of \(M \otimes_B N\): \[(b \cdot m) \otimes n = b \cdot (m \otimes n), \qquad m \otimes (b \cdot n) = b \cdot (m \otimes n),\] and from the definition of its null set.

Let \(\varphi : M \times N \to P\) be a bilinear morphism of \(B\)-modules. The map \[\tilde{\varphi} : M \otimes_B N \longrightarrow P, \qquad \tilde{\varphi}(m \otimes n) \mathrel{\vcenter{:}}= \varphi(m,n),\] is well-defined by the defining relations of the tensor product, the fact that every element of \(M\otimes_BN\) is a pure tensor, and the bilinearity of \(\varphi\). Let \(\sum_i m_i \otimes n \in I_{M\otimes_B N}\) such that \(\sum_i m_i \in I_M\). Then, by the bilinearity of \(\varphi\), we have \(\sum_i \varphi(m_i,n) \in I_P\). Analogously, if \(\sum_i m \otimes n_i \in I_{M\otimes_B N}\) such that \(\sum_i n_i \in I_N\), we also have \(\sum_i \varphi(m,n_i) \in I_P\). This proves that \(\varphi\) is a morphism of \(B\)-modules and is uniquely determined by the condition \(\tilde{\varphi} \circ \otimes = \varphi\). ◻

Proof. Since we have a universal property (Theorem 29), we can use the same proof as for modules over commutative rings. ◻

Proof. First, we prove that \((C \otimes_B D)_{\mathrm{mod}}\) has a structure of \(B\)-algebra. For this, cosider the map \[\begin{align} \Phi: (C \otimes_B D)_{\mathrm{mod}} \times (C \otimes_B D)_{\mathrm{mod}} &\longrightarrow (C \otimes_B D)_{\mathrm{mod}}\\ (c \otimes d, c' \otimes d')\;\;\;\;\;\;\;\;\;\; &\longmapsto (cc' \otimes dd'). \end{align}\] This map is a well-defined bilinear morphism of \(B\)-modules. Indeed, it clearly preserves the action of \(B\) and for \(\sum c_i \otimes d \in I_{(C \otimes_B D)_{\mathrm{mod}}}\) with \(\sum_i c_i \in I_C\), we have \(\sum c_ic' \in I_C\), hence \(\sum c_ic' \otimes dd' \in I_{(C \otimes_B D)_{\mathrm{mod}}}\). Since \(I_{(C \otimes_B D)_{\mathrm{mod}}}\) is generated by the sums \(\sum c_i \otimes d\) and \(\sum c\otimes d_i\) with \(\sum c_i \in I_C\) and \(\sum d_i \in I_D\), we conclude that \(\Phi\) is a bilinar morphism.

This map makes the underlying set of \((C \otimes_B D)_{\mathrm{mod}}\) into a pointed monoid, with identity \(1 \otimes 1\) and zero \(0 \otimes 0\). Now, we verify that \(I_{(C \otimes_B D)_{\mathrm{mod}}}\) is a null set for this pointed monoid. Since we already have some properties from the null set as modules, we only need to check that it is closed by multiplication of elements of \((C \otimes_B D)_{\mathrm{mod}}\), which follows from the bilinarity proved above. Hence \((C \otimes_B D)_{\mathrm{mod}}\) is a band, and when we consider the morphism \(B \rightarrow (C \otimes_B D)_{\mathrm{mod}}\) sending \(b \in B\) to \(b \otimes 1\), we see that it is a \(B\)-algebra.

Finally, consider the map \[\begin{align} \Psi: (C \otimes_B D)_{\mathrm{mod}} &\longrightarrow (C \otimes_B D)_{\mathrm{alg}}\\ c \otimes b &\longmapsto c \otimes b. \end{align}\] Then, by the definitions of the underlying pointed monoids of both sides and their null sets, we conclude that \(\Psi\) is an isomorphism of \(B\)-algebras. ◻

Proof. Completeness and cocompleteness follows from Proposition 19. The tensor product \(\otimes_B\) makes \(\mathop{\mathrm{\mathbf{BMod}}}_B\) into a closed symmetric monoidal category by Lemma 5 and Corollary 2. ◻

Proof. We produce two functors which are inverse to each other. First, let \(A\) be a \(B\)-algebra, and regard \(A\) as an \(B\)-module. The multiplication \[\mu_A : A\times A \longrightarrow A,\qquad (a,a')\mapsto a\cdot_A a',\] is a bilinear morphism of \(B\)-modules. By Theorem 29, there is a unique morphism of \(B\)-modules \[\widetilde{\mu}_A : A\otimes_B A \longrightarrow A\] such that \(\widetilde{\mu}_A(a\otimes a')=\mu_A(a,a')\) for all \(a,a'\in A\). The identity \(1_A\) defines a morphism of \(B\)-modules \[\eta_A : B \longrightarrow A,\qquad b\mapsto b\cdot_A 1_A.\] This makes \((A,\tilde{\mu}_A,\eta_A)\) into a commutative monoid in the category \(\mathop{\mathrm{\mathbf{BMod}}}_B\).

Conversely, let \((M,\mu,\eta)\) be a commutative monoid in \(\mathop{\mathrm{\mathbf{BMod}}}_B\), i.e. \(M\) is a \(B\)-module, \(\mu:M\otimes_B M\to M\) and \(\eta:B\to M\) are morphisms in \(\mathop{\mathrm{\mathbf{BMod}}}_B\). Define a multiplication on the underlying \(B\)-module \(M\) by \[m\cdot_M m' \mathrel{\vcenter{:}}= \mu(m\otimes m')\qquad(m,m'\in M),\] and define the unit element \(1_M \mathrel{\vcenter{:}}= \eta(1_B)\). This makes \(M\) into a \(B\)-algebra.

The two functors are inverse to each other, proving the isomorphism of categories. ◻

Proof. Every isomorphism is flat, so we just prove the converse. Suppose \(f\) is flat. Assume, without loss of generality, that \(B = C\) as pointed monoids and \(f\) is the identity map. In this case, we have \(N_B \subseteq N_C\). We claim that \(N_B = N_C\). Suppose, by contradiction, that \(N_B \neq N_C\) and consider the formal sum \[\sigma \mathrel{\vcenter{:}}= x_1+ \cdots + x_n \in N_C\setminus N_B \quad (x_i \in C)\] with minimal length. In particular, \(x_i \neq 0\) for every \(i\). Now, consider the following formal sums in \(N_{B\otimes_B C}\): \[\begin{align} \tau&\mathrel{\vcenter{:}}= 1\otimes x_1 + \cdots + 1\otimes x_{n-1} + 1\otimes x_n &\in N_{B \otimes_B C}\\ \nu&\mathrel{\vcenter{:}}= (-1)^0 \otimes 1 + \cdots (-1)^{n-2}\otimes1 + r_n \otimes 1 &\in N_{B \otimes_B C}, \end{align}\] where \(r_n = -1\) if \(n\) is even and \(r_n = 0\) if \(n\) is odd, that is, the sum \((-1)^0 + \cdots + (-1)^{n-2} + r_n \in N_B\). Observe that, since \(B = C\) as sets, we have \[\tau = x_1 \otimes 1 + \cdots x_n \otimes 1.\] Now, since \(f\) is flat, we have an isomorphism of \(C\)-modules \[\Phi: \left(B \times B\right) \otimes_B C \overset{\sim}{\longrightarrow} \left(B \otimes_B C \right) \times \left(B \otimes_B C \right);\quad (m_1, m_2) \otimes c \mapsto (m_1 \otimes c, m_2 \otimes c).\] Observe that, by the definition of the null set of tensor product, the formal sum \[\begin{align} \mu = (x_1 \otimes 1, (-1)^0 \otimes 1) + (x_2 \otimes 1, &(-1)^1 \otimes 1) + \cdots\\ &\cdots + (x_{n-1} \otimes 1, (-1)^{n-2} \otimes 1) + (x_n \otimes 1, r_n\otimes 1), \end{align}\] belongs to \(N_{\left(B \otimes_B C \right) \times \left(B \otimes_B C \right)}\). Hence, its pre-image by \(\Phi^+\) \[\begin{align} (\Phi^+)^{-1}(\mu) = (x_1, (-1)^0)\otimes 1 + (x_2, (-1)^1)\otimes 1 + &(x_3, (-1)^2)\otimes 1 + \cdots\\ &\cdots + (x_{n-1}, (-1)^{n-2}) + (x_n, r_n)\otimes 1 \end{align}\] belongs to \(N_{\left(B \times B\right) \otimes_B C}\). Now, by the definition of null set of tensor product, \((\Phi^+)^{-1}(\mu)\) is a \(B\)-linear combination of formal sums of the following two forms: \[\begin{align} \sum_i (m_1^i,m_2^i) &\otimes 1, \text{ with } \sum_i (m_1^i,m_2^i) \in N_{B \times B}, \text{ or }\\ \sum_i(m_1, m_2) &\otimes c_i, \text{ with } \sum_i c_i \in N_C. \end{align}\] Suppose the formal sums of type \((2)\) appears as subsums of \(\sigma\). Since \(\sum_i c_im_1 \in N_C\), the minimality of \(\sigma\) implies that \(\sum_i c_im_1 \in N_B\). Therefore, this implies that \(\sigma \in N_B\), a contradiction. ◻

Proof. We need to check that \(- \otimes_B S^{-1}B\) preserves finite products and equalizers. Then, from Lemma 3, we conclude that \(\iota_S\) is flat.

Finite products: Let \((M_i)_{i = 1}^n\) be a family of \(B\)-modules and consider the following map \[\begin{align} \Phi: S^{-1}\left(\prod_{i =1}^n M_i\right) &\longrightarrow \prod_{i =1}^n S^{-1}M_i\\ (m_i)_{i = 1}^n /s &\longmapsto \left(m_i/s \right)_{i = 1}^n. \end{align}\] The proof that \(\Phi\) is well-defined and bijective follows by usual arguments. We check the correspondece of their null sets. By definition, we have the following equalities: \[\begin{align} I_{S^{-1}(\prod M_i)} &= \left< \iota_S^+(I_{\prod M_i}) \right>_{S^{-1}B}\\ &= \left<\sum_j \frac{(m_{i,j})_i}{1} \;\middle|\; \sum_j (m_{i,j})_i \in I_{\prod M_i} \right>_{S^{-1}B}\\ &= \left<\sum_j \frac{(m_{i,j})_i}{1} \;\middle|\; \sum_j m_{i,j} \in I_{M_i} \text{ for all } i \right>_{S^{-1}B}. \end{align}\] Hence, the image of \(I_{S^{-1}(\prod M_i)}\) by \(\Phi^+\) is \[\begin{align} \Phi^+(I_{S^{-1}(\prod M_i)}) &= \left<\sum_j \left(\frac{m_{i,j}}{1}\right)_i \;\middle|\; \sum_j m_{i,j} \in I_{M_i} \text{ for all } i \right>_{S^{-1}B}\\ &= \left<\sum_j \left(\frac{m_{i,j}}{1}\right)_i \;\middle|\; \sum_j \frac{m_{i,j}}{1} \in I_{S^{-1}M_i} \text{ for all } i \right>_{S^{-1}B}\\ & = I_{\prod S^{-1}M_i}. \end{align}\] This proves that \(\Phi\) is an isomorphism.

Equalizers: Let \(f\) and \(g\) be morphisms of \(B\)-modules from \(M\) to \(N\). Then, we have the following bijection that preserves the action of \(B\). \[\begin{align} \Psi: S^{-1}\mathop{\mathrm{eq}}(f,g) &\longrightarrow \mathop{\mathrm{eq}}(S^{-1}f, S^{-1}g)\\ x/s &\longmapsto x/s \end{align}\] We check the correspondence of their null sets. Observe that for every \(m/s \in \mathop{\mathrm{eq}}(S^{-1}f, S^{-1}g)\), there exists \(m' \in \mathop{\mathrm{eq}}(f,g)\) and \(s' \in S\) such that \(m/s = m'/s'\) in \(S^{-1}M\). With this in mind, we have the following equalities: \[\begin{align} I_{\mathop{\mathrm{eq}}(S^{-1}f, S^{-1}g)} &= \left\{ \sum_i \frac{m_i}{s_i} \in I_{S^{-1}M}\;\middle|\; \frac{m_i}{s_i} \in \mathop{\mathrm{eq}}(S^{-1}f, S^{-1}g) \right\}\\ &= \left\{ \sum_i \frac{m'_i}{s'} \in I_{S^{-1}M}\;\middle|\; m'_i \in \mathop{\mathrm{eq}}(f,g) \right\}\\ &= \left< \sum_i \frac{m'_i}{1} \in I_{S^{-1}M} \;\middle|\; m'_i \in \mathop{\mathrm{eq}}(f,g) \right>_{S^{-1}B}\\ &= I_{S^{-1}\mathop{\mathrm{eq}}(f,g)}. \end{align}\] This proves that \(\Psi\) is an isomorphism. ◻

Proof of Proposition 35. We follow the idea of the proof of [5]. By Lemma 8, every localization map is flat, so it remains to prove the converse. Let \(f: B \to C\) be a flat morphism of bands and define \(S \mathrel{\vcenter{:}}= f^{-1}(C^\times)\). By [22], there exists a unique morphism \(f_S: S^{-1}B \to C\) making the diagram \[\begin{tikzcd} B \arrow[rr, "f"] \arrow[rd, swap, "\iota_S"] & & C \\ & S^{-1}B \arrow[ru, swap, "f_S"] \end{tikzcd}\] commute. We claim that \(f_S\) is an isomorphism.

Since every element of a tensor product of band modules is a pure tensor, the proof that \(f_S\) is surjective is exactly the same as in [5]. Hence, we focus on the proof of injectivity. Let \(a/s\) and \(b/t\) be elements of \(S^{-1}B\) such that \(f_S(a/s) = f_S(b/t)\). By the definition of \(f_S\), this means that \[\label{sonma} f(at) = f(bs).\tag{6}\]

For each \(\alpha \in B\), let \(g_\alpha : B \to B\) be the morphism of \(B\)-modules defined by \(g_\alpha(x) = \alpha x\) for all \(x \in B\). Since \(f\) is flat, we have the following isomorphism of \(C\)-modules: \[\begin{align} \mathop{\mathrm{eq}}(g_{at}, g_{bs}) \otimes_B C &\overset{\sim}{\longrightarrow} \mathop{\mathrm{eq}}(g_{at} \otimes_B \mathop{\mathrm{id}}_C, g_{bs} \otimes_B \mathop{\mathrm{id}}_C),\\ x \otimes c &\longmapsto x \otimes c. \end{align}\] Observing that, by 6 , \(g_{at} \otimes_B \mathop{\mathrm{id}}_C = g_{bs} \otimes_B \mathop{\mathrm{id}}_C\), we obtain the following isomorphism of \(C\)-modules: \[\begin{align} \mathop{\mathrm{eq}}(g_{at}, g_{bs}) \otimes_B C &\overset{\sim}{\longrightarrow} C,\\ x \otimes c &\longmapsto f(x)\, c. \end{align}\] Hence, there exists \(l \in \mathop{\mathrm{eq}}(g_{at}, g_{bs})\) such that \(f(l) \in C^\times\), that is, \(l \in S\). Therefore, we have \(l a t = l b s\) and \(a/s = b/t\) in \(S^{-1}B\).

Finally, since \(f_S\) is a flat bijective morphism of bands, Lemma 7 implies that \(f_S\) is an isomorphism. ◻

Proof. Apply [5] for both sides. ◻

Proof of Proposition 37. Suppose \(f\) satisfies \((ii)\). Let \(\mathop{\mathrm{\mathscr{D}}}= (D_i, g_{i,j})_{i,j \in \Lambda}\) be a directed system of \(B\)-algebras \(D_i\), with morphisms of \(B\)-algebras \(g_{i,j} : D_i \to D_j\) for every \(i \leq j\), and let \(\Psi_{\mathop{\mathrm{\mathscr{D}}}}\) be the canonical map 10 . First, we show that \(\Psi_{\mathop{\mathrm{\mathscr{D}}}}\) is injective. If \(\Psi_{\mathop{\mathrm{\mathscr{D}}}}([\varphi_i]) = \Psi_{\mathop{\mathrm{\mathscr{D}}}}([\varphi_j])\) for some \(\varphi_i \in \mathop{\mathrm{Hom}}_B(C,D_i)\) and \(\varphi_j \in \mathop{\mathrm{Hom}}_B(C,D_j)\), then, by the representation 8 of directed colimit of algebras, there exists \(k \geq i,j\) such that \[(g_{i,k} \circ \varphi_i \circ \xi)([T_l]) = (g_{j,k} \circ \varphi_j \circ \xi) ([T_l])\text{ for every } l \in I.\] Now, for every \(c = b\prod_{l \in I} T_l^{e_l} \in B[T_l]_{l \in I}\sslash \left<S\right>\), we have \[(g_{i,k} \circ \varphi_i)(\xi(c)) = b\prod_{l \in I}(g_{i,k} \circ \varphi_i \circ \xi)([T_l])^{e_l} = b\prod_{l \in I}(g_{j,k} \circ \varphi_j \circ \xi)([T_l])^{e_l} = (g_{j,k} \circ \varphi_j)(\xi(c)).\] Since \(\xi\) is surjective, this proves that, in \(\varinjlim_{i \in \Lambda} \mathop{\mathrm{Hom}}_B(C,D_i)\), we have \[[\varphi_i]=[g_{i,k} \circ \varphi_i] = [g_{j,k} \circ \varphi_j] = [\varphi_j].\]

Next, we show that \(\Psi_{\mathop{\mathrm{\mathscr{D}}}}\) is surjective. Let \(\varphi : C \rightarrow \varinjlim \mathop{\mathrm{\mathscr{D}}}\) be a morphism of \(B\)-algebras. Then, since \(I\) and \(S\) are finite sets, there exist \(k \in \Lambda\) and a list \((y_l)_{l \in I}\) in \(D_k\) such that \(\iota_k(y_l) = (\varphi \circ \xi)[x_l]\) for every \(l \in I\), and \((\varphi^+\circ \xi^+)(S) \subseteq N_{D_k}\). Hence, by [22], there exists a morphism \(\psi : B[x_l]_{l \in I}\sslash \left<S\right> \to D_k\) such that \(\iota_k \circ (\psi \circ \xi^{-1}) = \varphi\), that is, \(\Psi_{\mathop{\mathrm{\mathscr{D}}}}([\psi \circ \xi^{-1}]) = \varphi\). Therefore, \(\Psi_{\mathop{\mathrm{\mathscr{D}}}}\) is a bijection.

Conversely, suppose that \(f\) is of finite presentation. We know, by [22], that there exists an isomorphism of \(B\)-algebras of the form \[B[x_i]_{i \in I}\sslash \left<S \right> \overset{\sim}{\longrightarrow} C.\] Consider an isomorphism as above with \(I\) minimal and for this minimal \(I\) let \(S\) be minimal. We claim that both \(I\) and \(S\) are finite sets. Let \(\mathop{\mathrm{\mathscr{D}}}= (D_{J,T})_{J,T}\) be the directed system of \(B\)-algebras constructed as in Remark 36 for \(B[x_i]_{i \in I}\sslash \left<S\right>\). Since \(\Phi_{\mathop{\mathrm{\mathscr{D}}}}\) in bijective, the identity map \(\mathop{\mathrm{id}}_C\) is the image of the class of some morphism of \(B\)-algebras \(\varphi_{J,T} : C \rightarrow D_{J,T}\). That is, we have \[\label{eq:321} \mathop{\mathrm{id}}_C = \iota_{J,T} \circ \varphi_{J,T},\tag{11}\] where \(\iota_{J,T} : D_{J,T} \to \varinjlim \mathop{\mathrm{\mathscr{D}}}\cong C\) is the canonical map. First, observe that \(J = I\). Indeed, if there were an \(n \in I \setminus J\), then, since \(\iota_{J,T}\) is surjective, there is an element \(a\prod_{i \in J}x_i^{e_i}\) such that \(x_n - a\prod_{i \in J}x_i^{e_i} \in \left<S\right>\), which by Lemma 9 contradicts the minimality of \(I\). Next, notice that 11 holds for every finite subset \(T' \subseteq S\) containing \(T\), hence we may choose \(T\) containing, for every \(l \in I\), the sum of the form \(x_l - y_l\), where \(y_l \in B[x_i]_{i \in I}\) is such that \([y_l] = \varphi([x_l])\). We claim that \(\iota_{J,T}\) is an isomorphism. To prove this, observe that by the hypothesis on \(\left<T\right>\) above, we have \(\varphi([x_l]) = [x_l]\) for every \(l \in I\). Hence, \(\varphi\) and \(\iota_{I,T}\) are inverse morphisms to each other. Therefore, \(\iota_{J,T}\) is an ismorphism. ◻

Proof. Let \(f: B \to C\) be a finite localization map. Then, there exists a finitely generated multiplicative subset \(S = \left<s_1, \cdots, s_n \right>\) of \(B\) and an isomorphism \(\phi: S^{-1}B \to C\) such that \(\phi \circ \iota_S = f\). Since the following map \[\varphi: S^{-1}B \overset{\sim}{\longrightarrow} B[x_1, \cdots, x_n]\sslash \left<s_1x_1 - 1, \cdots, s_nx_n - 1 \right>,\] is an isomorphism, and every localization map is a flat epimorphism by Lemma 8, we conclude that \(f\) is a flat epimorphism of finite presentation.

Conversely, suppose that \(f: B \to C\) is a flat epimorphism of finite presentation. Since \(f\) is of finite presentation, we may assume that \(C = B[x_1, \cdots, x_n]\sslash \left<T\right>\), where \(T\) is a finite set. Now, since \(f\) is flat, there exist a multiplicative subset \(S \subseteq B\) and an isomorphism \(\phi: S^{-1}B \to C\) making the following diagram commutative: \[\begin{tikzcd} B \arrow[r, "f"] \arrow[d, "\iota_S"] & B[x_1, \cdots, x_n]\sslash \left<T \right>\\ S^{-1}B \arrow[ru, "\phi"]. \end{tikzcd}\] By the proof of Proposition 35, we know that \[\begin{align} S &= f^{-1}\left\{\left( B[x_1, \cdots, x_n]\sslash \left<T\right> \right)^\times\right\}\\ &= \left< f^{-1}([x_i]) \;\middle|\; [x_i] \in \left(B[x_1, \cdots, x_n]\sslash \left<T\right>\right)^\times \right>. \end{align}\] Therefore, \(S\) is finitely generated and \(f\) is a finite localization map. ◻

Proof. For every \(i \in I\), let \(f_i: B \to B_i\) be the map on \(\mathop{\mathrm{Comm}}(\mathop{\mathrm{\mathbf{BMod}}}_{\mathop{\mathrm{\mathbb{F}}}_1^\pm})\) corresponding to \(\varphi_i\). Let \(J \subseteq I\) be a finite subset such that the functor \(\prod_{j \in J} (- \otimes_B B_j)\) is conservative. By definition of Zariski covering, each \(f_j\) is a flat epimorphisms of finite presentation, hence, by Corollary 3, we have \(B_j \cong B[h_j^{-1}]\) for some \(h_j \in B\) for every \(j \in J\). Suppose that \(h_j \not\in B^\times\) for every \(j \in J\), and consider the \(B\)-module \(M\mathrel{\vcenter{:}}= B/(B\setminus B^\times)\) where \(B\setminus B^\times\) is the \(B\)-submodule of non-invertible elements of \(B\). Then, for every \(j \in J\), we have an isomorphism \(M \otimes_B B_j \cong 0\) since \[[b] \otimes \frac{c}{h_j} = [b] \otimes \frac{ch_j}{h_j^2} = [h_j]\otimes \frac{bc}{h_j^2} = 0.\] This implies that \(M \to 0\) is an isomorphism, hence \(M = 0\) and \(B^\times = \emptyset\), which is not possible since \(B^\times\) always contains \(1\). Therefore, there exists a \(j \in J\) such that \(h_j \in B^\times\), which implies that the morphism \(f_j: B \to B_j\) is an isomorphism. ◻

Proof. Follows from [22]. ◻

Proof. Same proof as [9]. ◻

Proof. The proof is verbatim the same as that of [9], except by the following modifications:

• [9] is replaced by Proposition 10.

• [9] is replaced by Proposition 14.

• [9] is replaced by Proposition 13.

• [9] is replaced by Proposition 45.

• [9] is replaced by Proposition 10.

 ◻

Proof. Let \(B\) be a band satisfying \((H1)\) to \((H4)\). For every \(x, y \in B\), define the hyperoperation as follows: \[x \boxplus y \mathrel{\vcenter{:}}= \left\{ z \in B \;\middle|\; x + y + (-z) \in N_B \right\} \subseteq B.\] Condition \((H1)\) ensures that \(x \boxplus y \neq \emptyset\). The commutativity \(x \boxplus y = y \boxplus x\) holds in general, while \((H4)\) implies associativity. Indeed, using \((H4)\) we have, for every \(x, y\) and \(z\) in \(B\), the following equalities. \[\begin{align} (x \boxplus y) \boxplus z &= \left\{ \mu \in B\;\middle|\; ^\exists a \in B \text{ such that } x + y + (-a) \in N_B \text{ and } a + z + (-\mu) \in N_B \right\}\\ &= \left\{\mu \in B \;\middle|\; x + y + z + (-\mu) \in N_B\right\}\\ &= \left\{ \mu \in B\;\middle|\; ^\exists b \in B \text{ such that } y + z + (-b) \in N_B \text{ and } x + b + (-\mu) \in N_B \right\}\\ &= x \boxplus( y \boxplus z). \end{align}\] By \((H2)\), we see that \(0\) is the unique element of \(B\) such that \[0_B \boxplus x = x \boxplus 0_B = \{x\},\] for all \(x \in B\). The axioms of a band also imply the existence and uniqueness of additive inverses, as well as the reversibility property \[x \in y \boxplus z \;\Longleftrightarrow\; z \in x \boxplus (-y).\] Let \(x,y\) and \(z\) be elements of \(B\), then, by \((H3)\), we have \[\begin{align} (x \boxplus y) z &= \left\{ \mu z \in B \;\middle|\; x + y + (-\mu) \in N_B \right\}\\ &\overset{(H3)}{=} \left\{ \mu z \in B \;\middle|\; xz + yz + (-\mu z) \in N_B \right\}\\ &= xz \boxplus yz \end{align}\] Hence, \(B\) is a hyperring.

Conversely, if \(R\) is a hyperring, then the associated band \(R^{\mathrm{band}}\) satisfies \((H1)\) to \((H4)\). ◻

Proof. Same idea as Proposition 40. ◻

Proof. Let \(I\) be a hyperideal of \(R\). Then it is clearly an \(m\)-ideal of \(R^{\mathrm{band}}\). We claim that the condition \(I \boxplus I \subseteq I\) is equivalent to condition (3) in the definition of \(k\)-ideals (Definition 5).

Suppose first that \(I \boxplus I \subseteq I\). Let \(a, b \in I\) and let \(y \in R^{\mathrm{band}}\) be such that \(a + b + y \in N_{R^{\mathrm{band}}}\). By definition of the hyperaddition, this implies that \(-y \in a \boxplus b \subseteq I \boxplus I\). Hence \(-y \in I\), and therefore \(y \in I\). This shows that \(I\) is a \(k\)-ideal.

Conversely, suppose that \(I\) is a \(k\)-ideal of \(R^{\mathrm{band}}\). Let \(a, b \in I\) and let \(z \in a \boxplus b\). By definition, this means that \(a + b + (-z) \in N_{R^{\mathrm{band}}}\). Since \(I\) is a \(k\)-ideal, it follows that \(z \in I\). Hence \(a \boxplus b \subseteq I\), and therefore \(I \boxplus I \subseteq I\). ◻

Proof. This follows from the definition of the hyperaddition on \(A/G\) as given in Example 1. ◻

Proof. Follows from Proposition 43. ◻

Proof. We claim that the following map is an isomorphism of bands. \[\begin{align} \Phi : \mathop{\mathrm{\mathcal{O}}}^k_{X,\mathfrak{p}} &\longrightarrow B_\mathfrak{p}\\ [s] &\longmapsto s(\mathfrak{p}). \end{align}\] where \([s]\) is the equivalence class of a section \(s \in \mathop{\mathrm{\mathcal{O}}}^k_X(D(f))\). By the definition of null sets of \(\mathop{\mathrm{\mathcal{O}}}_X^k(D(f))\) and the representation of filtered colimits of bands in Section 5.3, we have that \(\Phi\) is a well-defined morphism of bands.

We show that \(\Phi\) is injective. Let \([s], [t] \in \mathop{\mathrm{\mathcal{O}}}^k_{X,\mathfrak{p}}\) be such that \(s(\mathfrak{p}) = t(\mathfrak{p})\). Then, by property \((\star_2)\) in the definition of \(\mathop{\mathrm{\mathcal{O}}}^k_X\), there exist a open neighborhood \(V\) of \(\mathfrak{p}\) and elements \(a,f, b\) and \(g\) of \(B\) such that \(s(\mathfrak{q}) = a/f\) and \(t(\mathfrak{q}) = b/g\) for all \(\mathfrak{q} \in V\). In particular, \(a/f = b/g\) in \(B_\mathfrak{p}\) implying the existence of an \(h \not\in \mathfrak{p}\) such that \(hag = hbf\) in \(B\). Hence, \(V \cap D(h)\) is an open neighborhood of \(\mathfrak{p}\) such that \(s|_{V \cap D(h)} = t|_{V \cap D(h)}\), and \([s] = [t]\) in \(\mathop{\mathrm{\mathcal{O}}}^k_{X,\mathfrak{p}}\).

We show that \(\Phi\) is surjective. For every \(a/f \in B_\mathfrak{p}\), consider the open neighborhood \(D^k(f)\) of \(\mathfrak{p}\) and the section \(s\) on \(D^k(f)\) defined by \(s(\mathfrak{q}) = a/f\) for every \(\mathfrak{q} \in D^k(f)\). Then, \(\Phi([s]) = a/f\).

Finally, we verify the correspondece between their null sets. Let \(\sum_{i=1}^n [s_i]\) be such that \(\sum_{i=1}^n s_i(\mathfrak{p}) \in N_{B_\mathfrak{p}}\). Let \(U\) be an open neighborhood of \(\mathfrak{p}\) such that \(s_i\) is defined on (after a restriction if necessary) \(U\) for every \(i\). By definition, we heve \(\sum_i s_i \in N_{\mathop{\mathrm{\mathcal{O}}}^k_X(U)}\), hence by the representation of \(N_{\mathop{\mathrm{\mathcal{O}}}^k_{X,\mathfrak{p}}}\) as in Section 5.3, we conclude that \(\sum_i [s_i] \in N_{\mathop{\mathrm{\mathcal{O}}}^k_{X,\mathfrak{p}}}\). This proves that \(\Phi\) is an isomorphism. ◻

Proof. By [22], we have \(\langle S \rangle_k = \bigcup_{n=0}^{\infty} \langle S \rangle_k^{(n)}\), where \(\langle S \rangle_k^{(0)} \mathrel{\vcenter{:}}= S\) and, for every \(n \ge 1\), \[\label{recurrence} \langle S \rangle_k^{(n)} \mathrel{\vcenter{:}}= \left\{ a \in B \;\middle|\; a-\sum_i b_i s_i \in N_B \text{ for some } b_i \in B \text{ and } s_i \in \langle S \rangle_k^{(n-1)} \right\}.\tag{12}\]

Let \(l \ge 0\) be such that \(f \in \langle S \rangle_k^{(l)}\). If \(l=0\), then \(f \in S\), and there is nothing to prove. Thus, assume that \(l \ge 1\). By 12 , there exists a finite subset \(F^{l-1} \subseteq \langle S \rangle_k^{(l-1)}\) such that \(f \in \langle F^{l-1} \rangle_k\). If \(l-1=0\), we may take \(F=F^{l-1}\). Otherwise, for each \(g \in F^{l-1}\), applying 12 again yields a finite subset \(F_g^{\,l-2} \subseteq \langle S \rangle_k^{(l-2)}\) such that \(g \in \langle F_g^{\,l-2} \rangle_k\). Since \(F^{l-1}\) is finite, the union \[F^{l-2} \mathrel{\vcenter{:}}= \bigcup_{g \in F^{l-1}} F_g^{\,l-2}\] is finite, and \(h \in \langle F^{l-2} \rangle_k\) for every \(h \in F^{l-1}\).

Repeating this construction, we eventually obtain a finite subset \(F^{0} \subseteq \langle S \rangle_k^{(0)} = S\). A straightforward descending induction on \(i\) shows that \[f \in \langle F^{0} \rangle_k.\] Setting \(F=F^{0}\) completes the proof. ◻

Proof. By [22], we have \(\left<h_1, \cdots, h_n \right>_k = \bigcup_{n = 0}^\infty \left<h_1, \cdots, h_n \right>_k^{(n)}\). We proceed by induction on \(n\). Suppose that for every \(s \in \left<h_1, \cdots, h_n \right>_k^{(n-1)}\) and every \(j\), there exists \(\beta_{s,j} \in B\) such that \(s a_j = \beta_{s,j} h_j\). Let \(f \in \left<h_1, \cdots, h_n \right>_k^{(n)}\). By definition, there exist \(s_i \in \left<h_1, \cdots, h_n \right>_k^{(n-1)}\) and \(b_i \in B\) such that \(f - \sum_i b_i s_i \in N_B\). Taking the product with \(a_j\), we obtain \[f a_j - \sum_i b_i (s_i a_j) = f a_j - \left(\sum_i b_i \beta_{s_i,j}\right) h_j.\] That is, \(f a_j \in \left< h_j \right>_k\). By property \((D2)\), there exists \(\beta_j \in B\) such that \(f a_j = \beta_j h_j\). ◻

Proof of Theorem 47. We follow the proof of [20], and indicate only the necessary modifications.

Consider the following well-defined morphism of bands: \[\Phi \colon R_f \longrightarrow \mathop{\mathrm{\mathcal{O}}}^k_X(D^k(f)),\quad \frac{a}{f^n} \longmapsto s,\quad \text{where } s(\mathfrak{p}) = \frac{a}{f^n} \text{ in } R_{\mathfrak{p}}.\]

The injectivity of \(\Phi\) follows as in loc. cit., the only difference being the use of property \((D1)\) to show that the implication \[h f^m a - h f^n b \in N_D \;\Longrightarrow\; f^m a = f^n b \in N_D\] holds.

We now prove that \(\Phi\) is surjective. Let \(s \in \mathop{\mathrm{\mathcal{O}}}^k_X(D^k(f))\). Using [22] and property \((D2)\), one shows, as in the classical case, that \(D^k(f)\) admits a covering by basic open subsets \(\{D^k(h_i)\}_{i \in I}\) such that for each \(i \in I\), there exists \(a_i \in B\) satisfying \[s(\mathfrak{q}) = \frac{a_i}{h_i} \quad \text{for all } \mathfrak{q} \in D^k(h_i).\] Let \(I\) be the \(k\)-ideal generated by the elements \(h_i\). Since \(D^k(f) = \bigcup_{i \in I} D^k(h_i)\), we obtain \[V^k((f)) = \bigcap_{i \in I} V^k((h_i)) = V^k(I).\] By [22], it follows that \(f^n \in I\) for some \(n\). By Lemma 11, there exists a finite subset \(J \subseteq I\) such that \[\label{1432} f^n \in \langle h_j \mid j \in J \rangle_k.\tag{13}\] Hence \(D^k(f)\) is covered by the finite family \(\{D^k(h_j)\}_{j \in J}\). Write this family as \(D^k(h_1),\) \(\dots, D^k(h_n)\).

Using property \((D1)\), one shows, as in [20], that \[a_i h_j = a_j h_i \quad \text{for all } i,j.\] By 13 and Lemma 12, for each \(j\) there exists \(\beta_j \in B\) such that \[f^n a_j = \beta_j h_j.\] Applying again property \((D1)\), one deduces, as in [20], that \(\beta_i = \beta_j\) for all \(i,j\). Denoting this common value by \(\beta\), we obtain \[\Phi\!\left(\frac{\beta}{f^n}\right) = s,\] which proves that \(\Phi\) is surjective.

Finally, we verify the compatibility of the null sets. Let \(\sum_i \frac{a_i}{f^n} \in (B_f)^+\) (after taking a common denominator) be such that \(\sum_i s_i \in N_{\mathop{\mathrm{\mathcal{O}}}^k_X(D^k(f))}\), where \(\Phi\!\left(\frac{a_i}{f^n}\right) = s_i\) for every \(i\). By definition, for every \(\mathfrak{p} \in D^k(f)\), we have \[\sum_i s_i(\mathfrak{p}) = \sum_i \frac{a_i}{f^n} \in N_{B_{\mathfrak{p}}} = \left\langle \sum_j \frac{b_j}{1} \;\middle|\; \sum_j b_j \in N_B \right\rangle_{B_\mathfrak{p}}.\] Hence, there exist \(b_i \in B\) and \(h \notin \mathfrak{p}\) such that \(a_i/f^n = b_i/h\) for every \(i\), and \(\sum_i b_i \in N_B\). By property \((D1)\), we have \(a_i h = b_i f^n\) for every \(i\). Taking the sum, we obtain \[h \cdot \left(\sum_i a_i\right) = f^n \cdot \left(\sum_i b_i\right) \in N_B.\] Again, by property \((D1)\), we have \(\sum_i a_i \in N_B\), and therefore \[\sum_i \frac{a_i}{f^n} \in N_{B_f}.\] This shows that \(\Phi\) is an isomorphism of bands. ◻

Proof. Let \(f : N \hookrightarrow M\) be a monomorphism, and let \(\pi : M \to M/\mathop{\mathrm{Im}}(f)\) be the canonical projection. As sets, we have \(N = \mathop{\mathrm{eq}}(\pi, 0).\) It remains to verify that the band–pre-additions agree.

Since \(\mathop{\mathrm{Im}}(f)\) is a sub-\(B\)-module of \(M\), the equalizer \(\mathop{\mathrm{eq}}(\pi,0)\) is the sub-\(B\)-module whose underlying pointed set is \(\{m \in M \mid \pi(m) = \star\} = \mathop{\mathrm{Im}}(f)\), hence identified with \(N\). Therefore the question is whether \(I_N = I_{\mathop{\mathrm{eq}}(\pi,0)}\).

If \(f\) is strict, then for any sum \(\sum N_i \in N^{+}\), we have \[\sum n_i \in I_N \quad\Longleftrightarrow\quad \sum f(n_i) \in I_M,\] which exactly means \(I_N = I_{\mathop{\mathrm{eq}}(\pi,0)}\). Hence \(f\) is the kernel of \(\pi\), so \(f\) is normal.

Conversely, if \(f\) is the kernel of some morphism, then by definition it is the equalizer \(\mathop{\mathrm{eq}}(\pi,0)\), so the pre-additions agree and \(f\) is strict. ◻

Proof. First observe that the projection \(\pi : M \to M/\mathop{\mathrm{Im}}(f)\) satisfies all the conditions in the statement.

Conversely, suppose \(f : N \to M\) is an epimorphism satisfying the two conditions in the statement. We show that \(f\) is the cokernel of the strict monomorphism \[f^{-1}(\star) \hookrightarrow N.\]

Consider the map \[\phi : N / f^{-1}(\star) \to M, \qquad \phi([n]) = f(n).\] This map is well defined because if \(n \sim n'\), then either \(n = n'\) or both lie in \(f^{-1}(\star)\), and in both cases we have \(f(n) = f(n')\). Moreover, \(\phi\) preserves the \(B\)-action.

To show that \(\phi\) is strict, let \(\sum [n_i]\) be a sum in \(I_{N/f^{-1}(\star)}\). Then there exist \(c_j \in f^{-1}(\star)\) such that \[\sum n_i + \sum c_j \in I_N.\] Since \(f\) is strict, this is equivalent to \[\sum f(n_i) \in I_M,\] which means exactly that \(\sum \phi([n_i]) \in I_M\). Thus \(\phi\) is strict.

We now show that \(\phi\) is bijective. Surjectivity follows from the surjectivity of \(f\). For injectivity, suppose \(\phi([n]) = \phi([n'])\), i.e.\(f(n) = f(n')\). By hypothesis on the fibers, either \(n = n'\) or both lie in \(f^{-1}(\star)\). In both cases we obtain \([n] = [n']\). Hence \(\phi\) is injective.

Therefore \(\phi\) is an isomorphism of \(B\)-modules, satisfying \(\phi \circ \pi = f\). This shows that \(f\) is the cokernel of the monomorphism \(f^{-1}(\star) \to N\), and hence \(f\) is normal. ◻

Proof. First, we prove that \(\overline{f}\) is well defined. Indeed, if \(m \sim m'\), then \(m = m'\) or \(m, m' \in \ker(f)\), and both of the cases imply that \(f(m) = f(m')\).

It is easy to see that \(\overline{f}\) preserves the action of \(B\). Hence, to prove that \(\overline{f}\) is a morphism of \(B\)-modules consider a sum \(\sum [m_i] \in I_{M/\ker(f)}\). By the definition of quotient, we have \(\sum m_i' \in I_M\) for some \(m_i' \sim m_i\) for every \(i\). Since \(f\) is a morphism of \(B\)-modules we conclude that \[\sum f(m_i) = \sum f(m_i') \in I_N \cap \mathop{\mathrm{Im}}(f)^+ = I_{\mathop{\mathrm{Im}}(f)}.\]

\(\overline{f}\) is clearly surjective. To prove the injectivity, let \(\overline{f}([m]) = \overline{f}([m'])\). Then, \(f(m) + f(-m') \in I_N\). Since \(f\) is strict, there exists \(c_i \in \ker(f)\) such that \(m - m' + \sum c_i \in I_M\). Taking the projection to \(M/\ker(f)\) and by the uniqueness of \(-[m]\), we conclude that \([m] = [m']\).

Finally, we prove that \(\overline{f}\) is strict. Let \(\sum [m_i] \in (M/\ker(f))^+\)be such that \(\sum f(m_i) \in I_{\mathop{\mathrm{Im}}(f)} \subseteq I_N\). By the strictness of \(f\), there exist \(c_j \in \ker(f)\) such that \(\sum m_i + \sum c_j \in I_M\). Taking the projection to \(M/\ker(f)\) we get that \(\sum [m_i] \in I_{M/\ker(f)}\). ◻

Proof. First, we prove that \(i'\) is well defined. Denote by \(\sim_P\) the equivalence relation in \(N\) defined by the sub \(B\)-module \(P\) (see Definition 35). If \(a \sim_P a'\) in \(N\), then \(i(a) \sim_P i(a')\) in \(M\), hence \(i'([a]) = [i(a)]\) is well defined. Since \(i\) is injective, and we are taking the quotient of both \(M\) and \(N\) by \(P\), \(i'\) is also injective. Now, let \(\sum [n_i] \in (N/P)^+\). Then, it belongs to \(I_{N/P}\) if and only if \(\sum n_i' \in I_N\) for some \(n_i' \sim n_i\). Since \(i\) is a strict monomorphism, this is equivalent to \(\sum i(n_i') \in I_M\), which in turn is equivalent to \(\sum i'([n_i]) = \sum [i(n_i')] \in I_{M/P}\). This concludes that \(i'\) is a strict monomorphism. The commutativity of the diagram is immediate by the definition of \(i'\).

We prove that the diagram is coCartesian. Let \(Q\) be a \(B\)-module with two morphisms \(\alpha : N/P \rightarrow Q\) and \(\beta: M/P \rightarrow Q\) such that \(\alpha \circ j = \beta \circ i\). \[\begin{tikzcd} N \arrow[r, hook, "i"] \arrow[d, two heads, "j"] & M \arrow[d, two heads, "j'"] \arrow[rdd, bend left, "\beta"]& \\ N/P \arrow[r, hook, "i'"] \arrow[rrd, bend right, "\alpha"] & M/P \arrow[rd, dotted, "\gamma"] & \\ & & D \end{tikzcd}\] Define the map \(\gamma : M/P \rightarrow D\) as \(\gamma([m]) = \beta(m)\). This is well defined since \(m \sim m'\) implies \(m = m'\) or \(m, m' \in P\). The former case is obvious. Suppose the latter cases occur, that is, both \(m,m' \in P\). Then, since \(P \subseteq N\), \(\beta(m) = (\beta \circ i)(m) = (\alpha \circ j)(m) = \alpha([m]) = \star_D\). In the same way we get \(\beta(m') = \star_D\), hence \(\beta(m) = \beta(m')\).

The morphism \(\gamma\) clearly preserves the action of \(B\). Now, let \(\sum [m_i] \in I_{M/P}\). This implies that \(\sum m_i' \in I_M\) for some \(m'_i \sim m_i\), from which we get that \(\sum \gamma([m_i]) =\sum \beta(m_i') \in I_D\). Hence \(\gamma\) is a morphism of \(B\)-modules.

Finally, we prove that \(\gamma \circ i' = \alpha\): for every \(n \in N\), we have \[(\gamma \circ i')([n]) = (\gamma \circ i' \circ j)(n) = (\gamma \circ j' \circ i)(n) = (\beta \circ i)(n) = (\alpha \circ j)(n) = \alpha([n]).\]

We prove that the diagram is Cartesian. Let \(R\) be a \(B\)-module with two morphisms \(\alpha: R \rightarrow N/P\) and \(\beta : R \rightarrow M\) such that \(i' \circ \alpha = j' \circ \beta\). \[\begin{tikzcd} R \arrow[rrd, bend left, "\beta"] \arrow[rdd, bend right, "\alpha"] \arrow[rd, dotted, "\gamma"] & & \\ & N \arrow[r, hook, "i"] \arrow[d, two heads, "j"] & M \arrow[d, two heads, "j'"] \\ & N/P \arrow[r, hook, "i'"] & M/P & & \end{tikzcd}\] We claim that \(\mathop{\mathrm{Im}}(\beta) \subseteq \mathop{\mathrm{Im}}(i)\). Indeed, for every \(r \in R\), \((j' \circ \beta)(r) = (i' \circ \alpha)(r)\). Since \(j\) is surjective, there exists an \(n \in N\) such that \(j(n) = \alpha(r)\), hence \(j'(\beta(r)) = (i' \circ j)(n) = j'(i(n))\) which implies that \(\beta(r) \sim i(n)\). This means that either \(\beta(r) = i(n)\) or \(\beta(r), i(n) \in P\), and since \(P \subseteq N\), both cases implie that \(\beta(r) \in \mathop{\mathrm{Im}}(i)\). Hence, \(\beta\) induces a morphism of \(B\)-modules \(\gamma: R \to N\) such that \(i \circ \gamma = \beta\).

Finally, we prove that \(j \circ \gamma = \alpha\): For every \(r \in R\), we have \[(i' \circ j \circ \gamma)(r) = (j' \circ i \circ \gamma)(r) = (j' \circ \beta)(r) = (i' \circ \alpha)(r),\] and by the injectivity of \(i'\), we conclude that \((j \circ \gamma)(r) = \alpha(r)\). ◻

Proof. Define \(N\) as \((j')^{-1}(\mathop{\mathrm{Im}}(i')) \in \mathscr{E}\). Then, \(N\) is a \(B\)-submodule of \(M\) and the inclusion morphism \(i: N \hookrightarrow M\) is a strict monomorphism.

Observe that by the definition of \(N\) we can define a map \(j : N \rightarrow N'\) as \(j(n) = (i')^{-1}(j'(i(n)))\) for every \(n \in N\). We claim that \(j\) is a strict epimorphism of \(B\)-modules. Indeed, by definition, it is surjective and it clearly preserves the action of \(B\). Let \(\sum n_i \in N^+\). By the strictness of \(i, j'\) and \(i'\), we get that \[\sum n_i \in I_N \iff \sum i(n_i) \in I_M \iff \sum (j'\circ i)(n_i) \in I_{M'} \iff \sum j(n_i) \in I_N',\] hence, \(j\) is a strict epimorphism.

Finally, we prove that the square 15 is isomorphic to a square as in 14 which we know is bi-cartesian. Define \(P\) as the \(B\)-submodule \(i^{-1}(\ker(j'))\) of \(N\). Since \(i\) is injective, we know that \(P \cong \ker(j')\) and by the Proposition 51, we get that \(N/P \cong N'\) and \(M/P \cong M'\) and the following diagram commutes \[\begin{tikzcd}[row sep=scriptsize, column sep=scriptsize] N \arrow[dd, two heads, "j"] \arrow[rr, hook, "i"] \arrow[dr, equal] & & M \arrow[dd, two heads,near start, "j'"] \arrow[rd, equal] & \\ & N \arrow[rr, crossing over, hook, near start, "i"] & & M \arrow[dd, two heads, "\pi'"] \\ N' \arrow[rr, hook, near end, "i'"] \arrow[dr, "\sim"] & & M' \arrow[rd, "\sim"] & \\ & N/P \arrow[from=uu, two heads, crossing over, near start, "\pi"] \arrow[rr, hook] & & M/P \end{tikzcd}\] This proves that the diagram (15 ) is bi-cartesian. ◻

Proof. Let \(P \coloneq \ker(j)\), \(M' \coloneq M/P\), and \(j' : M \rightarrow M/P\) be the canonical projection. By Theorem 51 we have an isomorphism \(\phi: N' \overset{\sim}{\rightarrow} N/P\). Put \(i' \mathrel{\vcenter{:}}= \psi \circ \phi\). By Lemma 15, there exists a morphism \(\psi : N/P \hookrightarrow M/P\) making the following diagram commutate \[\begin{tikzcd} M \arrow[r, hook, "i"] \arrow[d, two heads, "j"] & M \arrow[dd, two heads, "j'"] \\ N' \arrow[d, "\phi"] \arrow[dr, "i'"] & \\ N/P \arrow[r, hook, "\psi"] & M/P\; (= M'). \end{tikzcd}\] ◻

Proof. Observe that every admissible short exact sequence of \(\mathop{\mathrm{\mathbf{BMod}}}_F^{\mathrm{fin,proj}}\) is of the following form \[\begin{tikzcd} \bigvee_{i = 1}^nF\cdot e_i \arrow[r, hook] & \bigvee_{i = 1}^{n + m}F\cdot e_i \arrow[r, two heads] & \bigvee_{i = 1}^mF\cdot e_i. \end{tikzcd}\] Hence for every \(M = \bigvee_{i = 1}^n F\cdot e_i\), we have \[[M] = n[F],\] and we define \(\dim_F(M) \mathrel{\vcenter{:}}= n\). Since \(\mathop{\mathrm{\mathbf{BMod}}}_F^{\mathrm{fin,proj}}\) admits coproducts and split admissible sequences, the standard description of the \(K_0\)-group of such a proto-exact category yields an isomorphism of groups \[K_0(\mathop{\mathrm{\mathbf{BMod}}}_F^{\mathrm{fin,proj}}) \longrightarrow \mathop{\mathrm{\mathbb{Z}}},\quad [M] \longmapsto \dim_F(M).\] ◻