Proof. This is because $\mathrm {Mod}^0(\Lambda )$ is abelian.

Proof. Note that $l\cdot s$ is nonzero for any $l\in \Lambda {\backslash }\{0\}$. We have an inclusion $\Lambda \cdot s\hookrightarrow V$. Since the left-hand side is a free $\Lambda$-module, the tensoring $(-)\otimes _\Lambda \mathbb {k}$ preserves the inclusion. Hence $s\otimes 1$ is nonzero in $V\otimes _\Lambda \mathbb {k}$.

Proof. This is a case of Lemma 2.7 by setting $V=\operatorname {Hom}_{\mathrm {Mod}^ {\mathbb {R}}(\Lambda _X)}(\tilde {\mathcal{V} }, \tilde {\mathcal{W} })$ and $s:=f$.

Proof. Take a lift $f={\bigoplus} _{c}f_c\in {\bigoplus} _{c}\operatorname {Hom}^c_{\mathrm {Mod}^ {\mathbb {R}}(\Lambda _X)}(\tilde {\mathcal{V} },\tilde {\mathcal{W} })$. Since $f_c$ is zero except for finite $c$, we can take $b$ to be a real number which is greater than or equal to the maximum of $c$ for which $f_c$ is nonzero. Then we set

(2.7)\begin{equation} f':=\bigoplus_{c}T^{b-c}f_c\in \operatorname{Hom}^b_{\mathrm{Mod}^{\mathbb{R}}(\Lambda_X)}(\tilde{\mathcal{V}},\tilde{\mathcal{W}}). \end{equation}

Since $T^{b-c}=1$ on $\operatorname {Hom}_{\mathrm {Mod}^{\mathfrak {I}}_{pre}(\Lambda _X)}( {\mathcal {V}},{\mathcal {W}})$, the element $f'$ represents $f$.

Proof. By multiplying some $T^a$ values we can assume that $b_1=b_2$. Since $f_1-f_2$ represents $0$ in $\operatorname {Hom}_{\mathrm {Mod}_{pre}^{ {\mathfrak {I}}}(\Lambda _X)}( {\mathcal {V}},{\mathcal {W}})$, there exists $b\in \mathbb{R} _{\geqslant 0}$ such that $T^b(f_1-f_2)=0$ by Lemma 2.8.

Proof. By Lemma 2.10, it suffices to prove the objects defined for $f'$ and $T^af'$ are isomorphic. We have morphisms $f'\colon \tilde {\mathcal{V} }\rightarrow \tilde {\mathcal{W} }\langle b\rangle$ and $T^af'\colon \tilde {\mathcal{V} }\rightarrow \tilde {\mathcal{W} }\langle a+b\rangle$ in $\mathrm {Mod}^0(\Lambda _X)$. Note that $\ker f'\hookrightarrow \ker T^af'$. Hence for any $\tilde {\mathcal{P} }\in \mathrm {Mod}^0(\Lambda _X)$, we have

(2.8)\begin{equation} \tilde{c}\colon \bigoplus_{b\in {\mathbb{R}}}\operatorname{Hom}^b_{\mathrm{Mod}^{\mathbb{R}}(\Lambda_X)}(\tilde{\mathcal{P}}, \ker f')\hookrightarrow\bigoplus_{b\in {\mathbb{R}}}\operatorname{Hom}^b_{\mathrm{Mod}^{\mathbb{R}}(\Lambda_X)}(\tilde{\mathcal{P}}, \ker T^af'), \end{equation}

which induces a comparison morphism

\[ c\colon \operatorname {Hom}_{\mathrm {Mod}^{\mathfrak {I}}_{pre}(\Lambda _X)}([\tilde {\mathcal{P} }], [\ker f'])\rightarrow \operatorname {Hom}_{\mathrm {Mod}^{\mathfrak {I}}_{pre}(\Lambda _X)}([\tilde {\mathcal{P} }], [\ker T^af']) \]

It suffices to show that $c$ is an isomorphism.

For any $g\in \operatorname {Hom}^b_{\mathrm {Mod}^ {\mathbb {R}}(\Lambda _X)}(\tilde {\mathcal{P} }, \ker T^af')$, consider $T^ag\in \operatorname {Hom}^{a+b}_{\mathrm {Mod}^ {\mathbb {R}}(\Lambda _X)}(\tilde {\mathcal{P} }, \ker T^af')$. Since $f'\langle a+b\rangle \circ T^a g=T^af'\langle b\rangle \circ g=0$, $T^ag$ factors through $\ker f'\langle a+b\rangle$ i.e.

\[ T^ag\in \operatorname {Hom}^{a+b}_{\mathrm {Mod}^ {\mathbb {R}}(\Lambda _X)}(\tilde {\mathcal{P} }, \ker f'). \]

Hence $T^ag$ is in the image of $\tilde {c}$. Since $g$ and $T^ag$ represents the same morphism in $\mathrm {Mod}^{\mathfrak {I}}_{pre}(\Lambda _X)$, we have the surjectivity of $c$.

On the other hand, let $g\in \operatorname {Hom}_{\mathrm {Mod}^{\mathfrak {I}}_{pre}(\Lambda _X)}([\tilde {\mathcal{P} }],[\ker f'])$ be zero in $\operatorname {Hom}_{\mathrm {Mod}^{\mathfrak {I}}_{pre}(\Lambda _X)}([\tilde {\mathcal{P} }], [\ker T^af'])$. For a lift $g'$ of $g$, we have $T^bg'=0$ for some $b\in \mathbb{R} _{\geqslant 0}$ by Lemma 2.8. Hence $g=0\in \operatorname {Hom}_{\mathrm {Mod}^{\mathfrak {I}}_{pre}(\Lambda _X)}([P], [\ker f'])$. This gives the injectivity of $c$.

Similar arguments prove the claims for $\mathrm {im}(f')$, $\operatorname {coker}(f')$, and $\mathrm {coim}(f')$.

Proof. Again, we will only prove for kernel and the others can be proved by similar arguments.

Let ${\mathcal {P}}\xrightarrow {g} {\mathcal {V}}\xrightarrow {f} {\mathcal {W}}\in \mathrm {Mod}^{\mathfrak {I}}_{pre}(\Lambda _X)$ satisfy $f\circ g=0$. We have lifts

(2.9)\begin{equation} \tilde{\mathcal{P}}\xrightarrow{\tilde{g}}\tilde{\mathcal{V}}\xrightarrow{\tilde{f}}\tilde{\mathcal{W}} \end{equation}

in $\mathrm {Mod}^0(\Lambda _X)$. By replacing $\tilde {\mathcal{W} }$ with $\tilde {\mathcal{W} }\langle c\rangle$ with sufficiently large $c$ and $\tilde {f}$ with $T^c\tilde {f}$, we can take so that $\tilde {f}\circ \tilde {g}=0$ by Lemma 2.8. Then there exists a morphism $\tilde {\mathcal{P} }\rightarrow \ker \tilde {f}$ by the universality of the kernel. The commutative diagram

(2.10)

descends to the commutative diagram

(2.11)

in $\mathrm {Mod}^{\mathfrak {I}}_{pre}(\Lambda _X)$, hence we only have to check the uniqueness of the morphism $h$.

Let $h'\colon {\mathcal {P}}\rightarrow [\ker f']$ be another morphism in $\mathrm {Mod}^{\mathfrak {I}}_{pre}(\Lambda _X)$ which fits into the following diagram.

(2.12)

We can lift $h'$ to $\tilde {h}'\colon \tilde {\mathcal{P} }\rightarrow \ker \tilde {f}\langle a\rangle$ for some $a\in \mathbb{R} _{\geqslant 0}$. Take $b\in {\mathbb {R}}_{\geqslant 0}$ so that $T^b\tilde {\iota }\langle a\rangle \circ \tilde {h}'=T^{a+b}\tilde {g}$ is satisfied. Then we again get a commutative diagram.

(2.13)

On the other hand, we have the following commutative diagram.

(2.14)

By the universality of $\ker \tilde {f}\langle a+b\rangle$, we have $T^{a+b}\tilde {h}=T^b\tilde {h}'$. Hence $h=h'$. This completes the proof.

Proof. This is obvious from Lemma 2.12.

Proof. Take a lift $\tilde {\mathcal{V} }'\xrightarrow {\tilde {f}'}\tilde {\mathcal{W} }\xrightarrow {\tilde {g}'}\tilde {\mathcal{X} }'$ such that $\tilde {g}'\circ \tilde {f}'=0$. Set $\tilde {\mathcal{V} }:=\ker \tilde {g}'$ and $\tilde {\mathcal{X} }:=\operatorname {im} \tilde {g}'$. Then we have an exact sequence

(2.17)\begin{equation} 0\rightarrow \tilde{\mathcal{V}}\xrightarrow{\tilde{f}} \tilde{\mathcal{W}}\xrightarrow{\tilde{g}} \tilde{\mathcal{X}}\rightarrow 0. \end{equation}

Here $\tilde {f}$ and $\tilde {g}$ are canonical morphisms. We have an associated morphism $\tilde {\mathcal{V} }'\rightarrow \tilde {\mathcal{V} }$. In $\mathrm {Mod}^{\mathfrak {I}}_{pre}(\Lambda _X)$, this associates a morphism $ {\mathcal {V}}\rightarrow [\tilde {\mathcal{V} }]=[\ker \tilde {g}']=\ker g$. By the exactness of the given sequence, we have $ {\mathcal {V}}\xrightarrow {\cong } [\tilde {\mathcal{V} }]$. Hence $\tilde {\mathcal{V} }$ is a lift of $ {\mathcal {V}}$. In a similar way, one can see that $\tilde {\mathcal{X} }$ is a lift of ${\mathcal {X}}$. This completes the proof.

Proof. Let $\{U_i\}_{i\in I}$ be a cover of $U$ in $\mathrm {Open}_{({\bar {Y}}, D_Y)}$. Let $J\subset I$ be as in the definition of the cover. Then $\{f^{-1}(U_i)\}$ is an open covering of $f^{-1}(U)$ in ${\bar {X}}$. Take $x\in {\bar {X}}$, then there exists a small neighborhood $V$ of $f(y)$ such that $V$ only intersects with a finite subset of $\{U_j\}_{j\in J}$. Then $f^{-1}(V)$ also only intersects with a finite subset of $\{f^{-1}(U_j)\}_{j\in J}$. Hence $\{f^{-1}(U_i)\}_{i\in I}$ is a cover of $f^{-1}(U)$ in $\mathrm {Open}_{({\bar {X}}, D_X)}$.

Proof. A short exact sequence in $\mathrm {Mod}^{\mathfrak {I}}_{pre}(\Lambda _U)$ can be lifted to a short exact sequence in $\mathrm {Mod}^0(\Lambda _U)$ by Corollary 2.13. Then we can restrict it to an exact sequence in $\mathrm {Mod}^0(\Lambda _V)$. By Lemma 2.12, this also gives an exact sequence in $\mathrm {Mod}^{\mathfrak {I}}_{pre}(\Lambda _V)$.

Proof. We will only consider the case of kernels. A similar argument holds for cokernels, images and coimages.

Let $f\colon {\mathcal {V}}\rightarrow {\mathcal {W}}$ be a morphism in $\mathrm {Mod}^{\mathfrak {I}}_{({\bar {X}}, D_X)}(U)$. Then there exists a covering $\{U_i\}$ of $U$ such that we have a descent data $f_i\colon {\mathcal {V}}_i\rightarrow {\mathcal {W}}_i$ in ${\mathrm {Mod}^{\mathfrak {I}}_{ps}}_{({\bar {X}}, D_X)}(U_i)$. If it is necessary, we can replace the covering with a finer covering so that each $f|_{U_i}$ is represented by a morphism $f_{i}\colon {\mathcal {V}}_i\rightarrow {\mathcal {W}}_i$ in $\mathrm {Mod}^{\mathfrak {I}}_{pre}(U_{i})$. On each intersection $U_{i}\cap U_{j}$, we have a further covering $\{U_{ijk}\}_k$ such that $(f_i-f_j)|_{U_{ijk}}=0$.

Then we have $\ker (f_{i})$ since $\mathrm {Mod}^{\mathfrak {I}}_{pre}(\Lambda _{U_{i}})$ is an abelian category. Since the restriction functors are exact (Lemma 3.7), we have $\ker (f_i|_{U_{ij}})|_{U_{ijk}}=\ker (f_i|_{U_{ijk}})=\ker (f_j|_{U_{ijk}})=\ker (f_j|_{U_{ij}})|_{U_{ijk}}$ in $\mathrm {Mod}^{\mathfrak {I}}_{pre}(\Lambda _{U_{ijk}})$. Hence we have $\ker (f_i)|_{U_{ij}}=\ker (f_i|_{U_{ij}})=\ker (f_j|_{U_{ij}})=\ker (f_j)|_{U_{ij}}$ in $\mathrm {Mod}^{\mathfrak {I}}_{{({\bar {X}}, D_X)}}(U_{ij})$. This further gives a descent data and glues up to an object ${\mathcal {K}}\in \mathrm {Mod}^{\mathfrak {I}}_{({\bar {X}}, D_X)}(U)$.

For a morphism $g\colon {\mathcal {X}}\rightarrow {\mathcal {V}}$ with $f\circ g=0$, by taking a sufficiently fine cover $\{U_i\}$, we can represent $f, g, {\mathcal {X}}, {\mathcal {V}}$ in $\mathrm {Mod}^{\mathfrak {I}}_{pre}(\Lambda _{U_i})$. Then one gets a unique factorizing morphism ${\mathcal {X}}|_{U_i}\rightarrow \ker f_i$. Again by taking a finer covering as in the previous part of the proof and the universality, the set of these factorizing morphisms gives a descent data and can be glued up into the unique factorizing ${\mathcal {X}}\rightarrow {\mathcal {K}}$. This shows ${\mathcal {K}}$ is $\ker f$.

Proof. Since kernels, cokernels, images, and coimages are defined locally, the assertion is obvious.

Proof. We set $D_U:=D_X\cap \bar {U}$. To show the claim, it is enough to prove $\operatorname {Hom}_{\mathrm {Mod}^ {\mathfrak {I}}_{pre}(\Lambda _U)}( {\mathcal {V}}, {\mathcal {W}})$ is a sheaf over the site $\mathrm {Open}_{(\bar {U},D_U)}$. Since $\bar {U}$ is compact, any cover in $\mathrm {Open}_{(\bar {U}, D_U)}$ has a finite subcover.

We first assume that there exists a finite cover $\{U_i\}$ of $U$ such that the restriction of $f\in \operatorname {Hom}_{\mathrm {Mod}^ {\mathfrak {I}}_{pre}(\Lambda _X)}( {\mathcal {V}}, {\mathcal {W}})$ to each open subset is zero. Let $\tilde {f}\in \operatorname {Hom}_{\mathrm {Mod}^0(\Lambda _X)}(\tilde {\mathcal{V} }, \tilde {\mathcal{W} })$ be a lift. Then the restriction of $f$ to each open subset $U_i$ is represented by $\tilde {f}|_{U_i}$. Since $f|_{U_i}=0$, there exists a $T^a$ such that $T^a\tilde {f}|_{U_i}=0$ by Lemma 2.8. Let $A$ be the maximum of those $a$ values. Then $T^A\tilde {f}=0$. Hence $f=0$.

Let $\{f_{i}\}\in \prod \operatorname {Hom}_{\mathrm {Mod}^ {\mathfrak {I}}_{pre}(\Lambda _{U_i})}( {\mathcal {V}}|_{U_i}, {\mathcal {W}}|_{U_i})$ satisfy the descent condition. Depending on $i$, we have a set of lifts $\tilde {f}_i\colon \tilde {\mathcal{V} }|_{U_i}\rightarrow \tilde {\mathcal{W} }|_{U_i}\langle a_i\rangle$ in $\mathrm {Mod}^0(\Lambda _{U_i})$. In our situation, we can take $a_i=a_j$ for any $i,j$, since the indexes are finite. Then we replace $\tilde {\mathcal{W} }$ with $\tilde {\mathcal{W} }\langle a_i\rangle$ by Lemma 2.12. On $U_i\cap U_j$, $\tilde {f}_i$ and $\tilde {f}_j$ may not coincide, but $f_i=[\tilde {f}_i]$ coincides with $f_j=[\tilde {f}_j]$. Hence there exists $T^{a_{ij}}$ such that $T^{a_{ij}}\tilde {f}_i=T^{a_{ij}}\tilde {f}_j$. By taking the maximum among $a_{ij}$, we replace $\tilde {f}_j$ with $T^a \tilde {f}_j$ then the set $\{T^a \tilde {f}_i\}$ satisfies the descent condition in $\mathrm {Mod}^0(\Lambda _U)$. Hence we get a glued morphism.

Proof. The embedding is given by the proof of Lemma 3.11.

Proof. This is a consequence of Lemma 3.10 and Corollary 2.13.

Proof. We will only prove the first equivalence. The second equivalence can be proved in a similar manner.

It is enough to show that the left-hand side satisfies the universality of the right-hand side. Let $ {\mathcal {V}}$ be an object which is over $[F]$. Locally, we have a lift $\tilde {\mathcal{V} }\rightarrow F$. By the universality, we get a morphism $\tilde {\mathcal{V} }\rightarrow \displaystyle {\lim _{\substack {\longleftarrow \\ {\mathfrak {U}}}}}F$, which induces a morphism $ {\mathcal {V}}\rightarrow [\displaystyle {\lim _{\substack {\longleftarrow \\ {\mathfrak {U}}}}}F]$ locally. The uniqueness of this morphism can be shown by a method similar to the proof of Lemma 2.12. The uniqueness glue up these local morphisms to obtain the desired result.

Proof. On $U_{ij}:=U_i\cap U_j$, we have the isomorphism $f\colon [\tilde {\mathcal{V} }_i|_{U_{ij}}]\rightarrow [\tilde {\mathcal{V} }_j|_{U_{ij}}]$ in ${\mathrm {Mod}^{\mathfrak {I}}_{ps}}_{({\bar {X}}, D_X)}(U_{ij})$. Then there exists an open covering $\{U_{ijk}\}$ of $U_{ij}$ where there exists a descent data $f_{ijk}\colon [\tilde {\mathcal{V} }_i|_{U_{ijk}}]\rightarrow [\tilde {\mathcal{V} }_j|_{U_{ijk}}]$ for the isomorphism $f\colon [\tilde {\mathcal{V} }_i|_{U_{ij}}] \rightarrow [\tilde {\mathcal{V} }_j|_{U_{ij}}]$.

We can take a lift $\tilde {f}_{ijk}\colon \tilde {\mathcal{V} }_i|_{U_{ijk}}\rightarrow \tilde {\mathcal{V} }_j|_{U_{ijk}}\langle a\rangle$ of this morphism and that of the inverse $\tilde {g}_{ijk}\colon \tilde {\mathcal{V} }_j|_{U_{ijk}}\rightarrow \tilde {\mathcal{V} }_i|_{U_{ijk}}\langle b\rangle$ for some $a, b$. The difference $\tilde {g}_{ijk}\langle a\rangle \circ \tilde {f}_{ijk}-T^{a+b}$ and $\tilde {f}_{ijk}\langle b\rangle \circ \tilde {g}_{ijk}-T^{a+b}$ becomes $0$ after multiplying $T^c$ for sufficiently big $c$ by Lemma 2.8.

For simplicity, let us assume that ${\mathcal {W}}$ has a global lift, i.e. ${\mathcal {W}}=[\tilde {\mathcal{W} }]$, although we can do the same for general ${\mathcal {W}}$. We have the following induced morphisms:

(3.5)\begin{equation} \begin{gathered} \mathop{{{\mathcal{H}}}om}\nolimits(\tilde{\mathcal{V}}_i|_{U_{ijk}}, \tilde{\mathcal{W}}|_{U_{ijk}})\xrightarrow{p(\tilde{g}_{ijk})} \mathop{{{\mathcal{H}}}om}\nolimits(\tilde{\mathcal{V}}_j|_{U_{ijk}}, \tilde{\mathcal{W}}|_{U_{ijk}})\langle -b\rangle, \\ \mathop{{{\mathcal{H}}}om}\nolimits(\tilde{\mathcal{V}}_j|_{U_{ijk}}, \tilde{\mathcal{W}}|_{U_{ijk}}), \xrightarrow{p(\tilde{f}_{ijk})} \mathop{{{\mathcal{H}}}om}\nolimits(\tilde{\mathcal{V}}_i|_{U_{ijk}}, \tilde{\mathcal{W}}|_{U_{ijk}})\langle -a\rangle, \end{gathered} \end{equation}

where $p(\tilde {f}_{ijk})$ and $p(\tilde {g}_{ijk})$ are precompositions of $f$ and $g$. Since $p(\tilde {f}_{ijk})\langle a\rangle \circ p(\tilde {g}_{ijk})-T^{a+b}\operatorname {id}$ and $p(\tilde {g}_{ijk})\circ p(\tilde {f}_{ijk})\langle b\rangle -T^{a+b}\operatorname {id}$ vanish by multiplying $T^c$ for big $c$, we can conclude $[\mathop {{{\mathcal {H}}}om}\nolimits (\tilde {\mathcal{V} }_i|_{U_{ijk}}, \tilde {\mathcal{W} }|_{U_{ijk}})]\cong [\mathop {{{\mathcal {H}}}om}\nolimits (\tilde {\mathcal{V} }_j|_{U_{ijk}}, \tilde {\mathcal{W} }|_{U_{ijk}})]$. A similar argument as was done in Proposition 3.9 gives a gluing of these isomorphisms to give a global object in $\mathrm {Mod}^{\mathfrak {I}}(\Lambda _X)$. The independence of the choice of local lifts is also clear.

Proof. This is obvious from the construction.

Proof. We can prove in a similar manner as in the proof of Lemma 3.16.

Proof. We have a canonical isomorphism

(3.11)\begin{equation} \mathop{{{\mathcal{H}}}om}\nolimits(\tilde{\mathcal{V}}_i\otimes \tilde{\mathcal{W}}_i, \tilde{\mathcal{X}}_i)\cong \mathop{{{\mathcal{H}}}om}\nolimits(\tilde{\mathcal{V}}_i, \mathop{{{\mathcal{H}}}om}\nolimits(\tilde{\mathcal{W}}_i,\tilde{\mathcal{X}}_i)). \end{equation}

By tensoring $\mathbb {k}$ and taking sheafification over $\mathrm {Open}_{(\overline {U_i}, \overline {U_i}\cap D_X)}$, we have an isomorphism

(3.12)\begin{equation} \operatorname{Hom}_{\mathrm{Mod}^{\mathfrak{I}}_{({\bar{X}}, D_X)}}({\mathcal{V}}\otimes {\mathcal{W}}, {\mathcal{X}})\cong \operatorname{Hom}_{\mathrm{Mod}^{\mathfrak{I}}_{({\bar{X}}, D_X)}}({\mathcal{V}},\mathop{{{\mathcal{H}}}om}\nolimits({\mathcal{W}}, {\mathcal{X}})) \end{equation}

as a sheaf over $\mathrm {Open}_{(\overline {U_i}, \overline {U_i}\cap D_X)}$. The isomorphisms over the $U_i$ are glued up and give a desired result.

Proof. From the above proposition, we have

(3.14)\begin{align} \operatorname{Hom}_{\mathrm{Mod}^{\mathfrak{I}}(\Lambda_{({\bar{X}}, D_X)})}({\mathcal{Y}}, \mathop{{{\mathcal{H}}}om}\nolimits({\mathcal{V}}\otimes {\mathcal{W}}, {\mathcal{X}}))&\cong \operatorname{Hom}_{\mathrm{Mod}^{\mathfrak{I}}(\Lambda_{({\bar{X}}, D_X)})}({\mathcal{Y}}\otimes {\mathcal{V}}\otimes {\mathcal{W}} ,{\mathcal{X}})\nonumber\\ &\cong \operatorname{Hom}_{\mathrm{Mod}^{\mathfrak{I}}(\Lambda_{({\bar{X}}, D_X)})}({\mathcal{Y}}\otimes {\mathcal{V}}, \mathop{{{\mathcal{H}}}om}\nolimits({\mathcal{W}},{\mathcal{X}}))\nonumber\\ &\cong \operatorname{Hom}_{\mathrm{Mod}^{\mathfrak{I}}(\Lambda_{({\bar{X}}, D_X)})}({\mathcal{Y}},\mathop{{{\mathcal{H}}}om}\nolimits({\mathcal{V}}, \mathop{{{\mathcal{H}}}om}\nolimits({\mathcal{W}},{\mathcal{X}}))). \end{align}

Yoneda's lemma implies the desired statement.

Proof. Let $\tilde {\mathcal{V} }'$ be another lift. Take a covering $\{U_i\}$ of $\mathrm {Open}_{({\bar {X}}, D_X)}$ such that we have lifts of the isomorphisms $\tilde {g}_i\colon \tilde {\mathcal{V} }|_{U_i}\rightarrow \tilde {\mathcal{V} }'\langle a_i\rangle |_{U_i}$ and $\tilde {h}_i\colon \tilde {\mathcal{V} }'|_{U_i}\rightarrow \tilde {\mathcal{V} }|_{U_i}\langle b\rangle$ over each $U_i$. Hence $\tilde {g}_i\langle b\rangle \circ \tilde {h}_i-T^{a+b}$ and $\tilde {h}_i\langle a\rangle \circ \tilde {g}_i-T^{a+b}$ vanish by large $T^c$. By pushing forward these equations, we have

(3.16)\begin{equation} 0=f_*(T^c(\tilde{g}\langle b\rangle\circ\tilde{h}-T^{a+b}\operatorname{id}_{\tilde{\mathcal{V}}}))=T^c(f_*\tilde{g}\langle b\rangle\circ f_*\tilde{h}-T^{a+b}\operatorname{id}_{f_*\tilde{\mathcal{V}}}). \end{equation}

Hence we have $[f_*\iota _{i*}\tilde {\mathcal{V} }|_{U_i}]\cong [f_*\iota _{i*}\tilde {\mathcal{V} }'|_{U_i}]$ where $\iota _i\colon U_i\rightarrow X$ is the inclusion map.

Let $ {\mathfrak {U}}$ be the Cech nerve of $\{U_i\}$ and $\iota _U$ for $U\in {\mathfrak {U}}$ the inclusion map. Since $f$ is tame, $\{f(U_i)\}$ is locally finite in ${\bar {Y}}$ i.e. there exists a covering of $Y$ in $\mathrm {Open}_{({\bar {Y}}, D_Y)}$ such that there are only finite $U_i$ in each open subset. Hence we have

(3.17)\begin{equation} \lim_{\substack{\longleftarrow \\ _{U\in {\mathfrak{U}}}}}[f_*\iota_{U*}\tilde{\mathcal{V}}_U]\cong \big[\lim_{\substack{\longleftarrow \\ _{U\in {\mathfrak{U}}}}}f_*\iota_{U*}\tilde{\mathcal{V}}_U\big]\cong [f_*\tilde{\mathcal{V}}] \end{equation}

by Lemma 3.14. Combining with the first part of the proof, we get an isomorphism $[f_*\tilde {\mathcal{V} }]\cong [f_*\tilde {\mathcal{V} }']$.

Proof. It can be proved by a similar argument as in Lemma 3.23.

Proof. It is enough to prove the statement over each $V_i$. There exists a finite covering $\{U_j\}$ of $f^{-1}(V_i)$ with lifts $\{\tilde {\mathcal{W} }_j\}$. Then $\mathop {{{\mathcal {H}}}om}\nolimits (f^{-1} {\mathcal {V}}, {\mathcal {W}})$ is represented by $\{\mathop {{{\mathcal {H}}}om}\nolimits (f^{-1}\tilde {\mathcal{V} }_i|_{U_j}, \tilde {\mathcal{W} }_j)\}$.

Let $ {\mathfrak {U}}$ be the Cech nerve of $\{U_j\}$. By the definition of the push forward, we have

(3.20)\begin{equation} f_*\mathop{{{\mathcal{H}}}om}\nolimits(f^{{-}1}{\mathcal{Y}}, {\mathcal{V}})|_{V_i}\simeq \lim_{\substack{\longleftarrow \\ U\in {\mathfrak{V}}}}([f_*i_{U*}\mathop{{{\mathcal{H}}}om}\nolimits((f^{{-}1}\tilde{\mathcal{Y}}_i)|_{U}, \tilde{\mathcal{V}}|_U)]). \end{equation}

Here $\tilde {\mathcal{V} }|_U$ means $\tilde {\mathcal{V} }_i|_U$ for some $U\subset U_i$. We also have

(3.21)\begin{align} f_*i_{U*}\mathop{{{\mathcal{H}}}om}\nolimits((f^{{-}1}\tilde{\mathcal{Y}}_i)|_{U}, \tilde{\mathcal{V}}|_U)&\simeq f_*\mathop{{{\mathcal{H}}}om}\nolimits(f^{{-}1}\tilde{\mathcal{Y}}_i, i_{U*}\tilde{\mathcal{V}}|_U)\nonumber\\ &\simeq \mathop{{{\mathcal{H}}}om}\nolimits(\tilde{\mathcal{Y}}_i, f_*\iota_{U*}\tilde{\mathcal{V}}|_U) \end{align}

for $U\in {\mathfrak {U}}$. Hence

(3.22)\begin{align} \lim_{\substack{\longleftarrow \\ {\mathfrak{U}}}}([f_*i_{U*}\mathop{{{\mathcal{H}}}om}\nolimits((f^{{-}1}\tilde{\mathcal{Y}}_i)|_{U}, \tilde{\mathcal{V}}|_U)])&\simeq \lim_{\substack{\longleftarrow \\ {\mathfrak{U}}}} \mathop{{{\mathcal{H}}}om}\nolimits([\tilde{\mathcal{Y}}_i], [f_*\iota_{U*}\tilde{\mathcal{V}}|_U])\nonumber\\ &\simeq \mathop{{{\mathcal{H}}}om}\nolimits([\tilde{\mathcal{Y}}_i], \lim_{\substack{\longleftarrow \\ {\mathfrak{U}}}} [f_*\iota_{U*}\tilde{\mathcal{W}}|_U])\nonumber\\ &\simeq \mathop{{{\mathcal{H}}}om}\nolimits([\tilde{\mathcal{Y}}_i], [f_*(\tilde{\mathcal{V}}|_{f^{{-}1}(V_i)})])\nonumber\\ &\simeq \mathop{{{\mathcal{H}}}om}\nolimits({\mathcal{Y}}, f_*{\mathcal{V}})|_{V_i}. \end{align}

This completes the proof.

Proof. Taking $\otimes \mathbb {k}$ and the global sections (as in the paragraph above Lemma 3.17) of both sides of Lemma 3.25, the right-hand side becomes $\operatorname {Hom}_{\mathrm {Mod}^{\mathfrak {I}}(\Lambda _{({\bar {X}}, D_X)})}( {\mathcal {V}}, f_*{\mathcal {W}})$ and the left-hand side becomes

(3.24)\begin{align} (f_*\mathop{{{\mathcal{H}}}om}\nolimits(f^{{-}1}{\mathcal{V}}, {\mathcal{W}}) \otimes \mathbb{k}) (Y)&\cong f_*(\mathop{{{\mathcal{H}}}om}\nolimits(f^{{-}1}{\mathcal{V}},{\mathcal{W}})\otimes \mathbb{k}) (Y)\nonumber\\ &\cong \operatorname{Hom}_{\mathrm{Mod}^{\mathfrak{I}}_{({\bar{X}}, D_X)}}(f^{{-}1}{\mathcal{V}}, {\mathcal{W}}). \end{align}

This completes the proof.

Proof. This can be proved in the same way as the proof of Lemma 3.23.

Proof. The first part is almost trivial. Let us prove the second part.

Let $0\rightarrow {\mathcal {V}}\xrightarrow {f} {\mathcal {W}}$ be an injection in the category $\mathrm {Mod}^{\mathfrak {I}}(\Lambda _{({\bar {X}}, D_X)})$ with a map $ {\mathcal {V}}\xrightarrow {g} [\prod _{x\in X}{\mathcal {F}}_x]$. Let us take a locally finite covering $\{U_i\}$ of $X$ with lifts $0\rightarrow \tilde {\mathcal{V} }_i\xrightarrow {\tilde {f}_i} \tilde {\mathcal{W} }_i$ and $\tilde {\mathcal{V} }_i\xrightarrow {\tilde {g}_i} \prod _{x\in U_i}{\mathcal {F}}_x\langle a_i\rangle $. We also get a lift $\tilde {\mathcal{W} }_i\xrightarrow {\tilde {h}_i}\prod _{x\in U}{\mathcal {F}}_x\langle a_i\rangle $.

For each $x\in V$, we choose $i_x$ from finite candidates of $i$ values satisfying $x\in U_i$. We set $\tilde {\mathcal{W} }_x:=(\tilde {\mathcal{W} }_{i_x})_x$. Over each $U_i$, the morphism $\tilde {h}_i$ gives an element of $({\bigoplus} _{a\in {\mathbb {R}}}\prod _{x\in U_i}\operatorname {Hom}^{a}(\tilde {\mathcal{W} }_{x}, {\mathcal {F}}_x))\otimes _\Lambda k\cong \operatorname {Hom}(W|_{U_i}, [\prod _{x\in V}{\mathcal {F}}_x])$ which is zero on $i_x\neq i$. Then they are trivially glued up to give a desired lift of $g$.

Proof. As usual, one can embed $\tilde {\mathcal{V} }_i$ to an injective object $\tilde {\mathcal{I} }_i$ which is a product of skyscraper sheaves.

Hence we have the inclusion $[\tilde {\mathcal{V} }_i]\hookrightarrow [\tilde {\mathcal{I} }_i]$. This induces the inclusion $ {\mathcal {V}}\hookrightarrow {\bigoplus} [\iota _{i*}\tilde {\mathcal{I} }_i]$, where the latter is a locally finite direct sum hence it exits. By Lemma 4.1, ${\bigoplus} [\iota _{i*}\tilde {\mathcal{I} }_i]$ is also an injective object. This completes the proof.

Proof. This follows from the fact that $[\cdot ]$ is an exact functor (Lemma 3.13) and Lemma 4.1.

Proof. Let $ {\mathcal {V}}\rightarrow {\mathcal {W}}\in \mathrm {Mod}^{\mathfrak {I}}(\Lambda _{({\bar {X}}, D_X)})$ be an injection. Let us take an open covering $\{U_i\}$ of $X$ with lifts $\{\tilde {\mathcal{V} }_i\}$, $\{\tilde {\mathcal{W} }_i\}$ and $\tilde {f}_i\colon \tilde {\mathcal{V} }_i\rightarrow \tilde {\mathcal{W} }_i$. Here one can take $\tilde {f}_i$ as an injection by Lemma 2.14. Then $ {\mathcal {V}}\otimes [\tilde {\mathcal{F} }]$ (respectively ${\mathcal {W}}\otimes [\tilde {\mathcal{F} }]$) is represented by $\tilde {\mathcal{V} }_i\otimes \tilde {\mathcal{F} }|_{U_i}\rightarrow \tilde {\mathcal{W} }_i\otimes \tilde {\mathcal{F} }|_{U_i}$, which is an injection. Then by Lemma 2.12, the morphism $f\otimes [\tilde {\mathcal{F} }]$ is also an injection. This completes the proof.

Proof. By the same construction as in [Reference Kashiwara and SchapiraKS90, Proposition 2.4.12], there exists a flat object $\tilde {\mathcal{F} }_i$ with a surjection $\tilde {\mathcal{F} }_i\rightarrow \tilde {\mathcal{V} }_i$. Let $\iota _i\colon U_i\hookrightarrow X$ be the open imbedding. Hence $[\iota _{i!}\tilde {\mathcal{F} }_i]$ is also a flat object by Lemma 4.4 and we have a surjection ${\bigoplus} _i[\iota _{i!}\tilde {\mathcal{F} }_i]\rightarrow {\mathcal {V}}$. This completes the proof.

Proof. Since $[\cdot ]$ is an exact functor, it suffices to show the statements for an object $\mathrm {Mod}^0(\Lambda _X)$ by a standard argument in homological algebra. Then $\tilde {\mathcal{V} }$ has an injective resolution $\tilde {\mathcal{I} }^\bullet$ such that $[\tilde {\mathcal{I} }^\bullet ]$ is an injective resolution of $[\tilde {\mathcal{V} }]$ by Corollary 4.3. For $F\in \{f_*, f_!, \mathop {{{\mathcal {H}}}om}\nolimits ([\tilde {\mathcal{W} }], -)\}$, we have

(4.1)\begin{equation} [{\mathbb{R}} F(\tilde{\mathcal{V}})]\simeq [F(\tilde{\mathcal{I}}^\bullet)]\simeq F[\tilde{\mathcal{I}}^\bullet]\simeq {\mathbb{R}} F([\tilde{\mathcal{V}}]). \end{equation}

This completes the proof.

Proof. It follows from the definition of $f^{-1}$ and its exactness on $\mathrm {Mod}^0(\Lambda _Y)$.

Proof. One can prove by the same argument as in Lemma 4.7 by using Corollary 4.6.

Proof. This can be proved by a standard argument. Let us take a flat resolution ${\mathcal {F}}$ of ${\mathcal {W}}$ and an injective resolution ${\mathcal {I}}$ of ${\mathcal {X}}$. Then ${\mathbb{R} \mathcal{H} om}({\mathcal {F}}, {\mathcal {I}})\simeq \mathop {{{\mathcal {H}}}om}\nolimits ({\mathcal {F}}, {\mathcal {I}})$ is again an injective object. Actually, we have

(4.3)\begin{equation} \mathop{{{\mathcal{H}}}om}\nolimits(-, \mathop{{{\mathcal{H}}}om}\nolimits({\mathcal{F}}, {\mathcal{I}})))\cong \mathop{{{\mathcal{H}}}om}\nolimits((-)\otimes {\mathcal{F}}, {\mathcal{I}}) \end{equation}

by Proposition 3.19. Then both sides of the equality in the statement is quasi-isomorphic to $\mathop {{{\mathcal {H}}}om}\nolimits ( {\mathcal {V}}\otimes {\mathcal {F}}, {\mathcal {I}})$. This completes the proof.

Proof. By Lemma 3.26 and the exactness of $f^{-1}$ imply that push-forward of an injective is again injective. Also, pull-back of a flat object is again flat. Let ${\mathcal {F}}$ be a flat resolution of ${\mathcal {Y}}$ and ${\mathcal {I}}$ be an injective resolution of $ {\mathcal {V}}$. By replacing with these resolutions, we can work with underived functors, then Lemma 3.25 completes the proof.

Proof. This is done by the same argument as in the proof of [Reference Kashiwara and SchapiraKS90, Lemma 3.1.3].

Proof. For $\tilde {\mathcal{W} }\in K^+(\mathrm {Mod}^0(\Lambda _X))\cong D^+(\mathrm {Mod}^0(\Lambda _X))$, we have

(4.9)\begin{equation} \operatorname{Hom}_{K^+(\mathrm{Mod}^0(\Lambda_X))}(f_!(\tilde{\mathcal{W}}\otimes_{{\mathbb{Z}}_Y}K), \tilde{\mathcal{V}})\cong \operatorname{Hom}_{K^+(\mathrm{Mod}^0(\Lambda_X))}(\tilde{\mathcal{W}}, f^!\tilde{\mathcal{V}}) \end{equation}

by the above lemma. Since $f_!\tilde {\mathcal{W} }\otimes K\simeq {\mathbb {R}} f_!\tilde {\mathcal{W} }$, we complete the proof of the first assertion. The second assertion can also be proved by the argument of the proof of [Reference Kashiwara and SchapiraKS90, Proposition 3.1.10].

Proof. Over $V_{ij}:=V_i\cap V_j$, we have $f_{ij}\colon [\tilde {\mathcal{Z} }_i]|_{V_{ij}}\xrightarrow {\cong } [\tilde {\mathcal{Z} }_j]|_{V_{ij}}$. We can lift this map to $\tilde {f}_{ij}\colon \tilde {\mathcal{Z} }_i|_{V_{ij}}\rightarrow \tilde {\mathcal{Z} }_j\langle a_{ij}\rangle |_{V_{ij}}$ (by taking a refined covering if necessary). We also have a lift of the inverse map $\tilde {f}_{ji}$. Then $\tilde {f}_{ij}\circ \tilde {f}_{ji}-T^{a_{ij}+a_{ji}}\operatorname {id}$ vanishes by multiplying some $T^a$. The map $\tilde {f}_{ij}$ induces a map $f_K^!\tilde {f}_{ij}\colon f^!_K\tilde {\mathcal{Z} }_i|_{V_{ij}}\rightarrow f^!_K\tilde {\mathcal{Z} }_j|_{V_{ij}}$. Then we also have $f_K^!\tilde {f}_{ji}$. Then $f_K^!\tilde {f}_{ij}\circ f_K^!\tilde {f}_{ji}-T^{a_{ij}+{a_{ji}}}\operatorname {id}=f_K^!(\tilde {f}_{ij}\tilde {f}_{ji}-T^{a_{ij}+a_{ji}}\operatorname {id})$ also vanishes by multiplying $T^a$. This completes the proof.

Proof. The exactness easily follows from the exactness of $f^!$ on $D^b(\mathrm {Mod}^0(\Lambda _Y))$.

Proof. Replace $\tilde {\mathcal{V} }$ be an injective complex given in Lemma 4.1. Then $[\tilde {\mathcal{V} }]$ is also an injective complex. By the definition of $f^!$ for $\mathrm {Mod}^0(\Lambda _X)$ and $\mathrm {Mod}^{\mathfrak {I}}(\Lambda _{({\bar {X}}, D_X)})$, we have $[f^!\tilde {\mathcal{V} }]\cong [f^!_K\tilde {\mathcal{V} }]\cong f^!_K[\tilde {\mathcal{V} }]\cong f^![\tilde {\mathcal{V} }]$. This completes the proof.

Proof. First, note that $ {\mathbb {R}} f_! {\mathcal {V}}\simeq f_!( {\mathcal {V}}\otimes K)$ which is deduced from the local consideration. Let ${\mathcal {I}}$ be an injective resolution of ${\mathcal {Y}}$ and $C(\mathrm {Mod}^{\mathfrak {I}}(\Lambda _{({\bar {X}}, D_X)}))$ be the category of bounded complexes of $\mathrm {Mod}^{\mathfrak {I}}(\Lambda _{({\bar {X}}, D_X)})$. Then the left-hand side of the desired equality is

(4.11)\begin{equation} \operatorname{Hom}_{C(\mathrm{Mod}^{\mathfrak{I}}(\Lambda_{({\bar{Y}}, D_Y)})}(f_!({\mathcal{V}}\otimes K), {\mathcal{I}})\cong \operatorname{Hom}_{D^b(\mathrm{Mod}^{\mathfrak{I}}(\Lambda_{({\bar{X}}, D_X)}))}({\mathcal{V}}, f^!{\mathcal{I}}). \end{equation}

We also have a morphism

(4.12)\begin{equation} \operatorname{Hom}_{D^b(\mathrm{Mod}^{\mathfrak{I}}(\Lambda_{({\bar{X}}, D_X)}))}({\mathcal{V}}, f^!{\mathcal{Y}})\rightarrow \operatorname{Hom}_{D^b(\mathrm{Mod}^{\mathfrak{I}}(\Lambda_{({\bar{X}}, D_X)}))}({\mathcal{V}}, f^!{\mathcal{I}}) \end{equation}

coming from the morphism ${\mathcal {Y}}\rightarrow {\mathcal {I}}$. We would like to prove this is an isomorphism. Let us work locally on $Y$. From the construction in Proposition 4.2, the complex ${\mathcal {I}}$ is coming from an injective object $\tilde {\mathcal{I} }$ locally. Hence we have an isomorphism

(4.13)\begin{equation} \mathop{{{\mathcal{H}}}om}\nolimits([\tilde{\mathcal{V}}], f^![\tilde{\mathcal{I}}])\cong [\mathop{{{\mathcal{H}}}om}\nolimits(\tilde{\mathcal{V}}, f^!\tilde{\mathcal{I}})]\cong [{\mathbb{R}\mathcal{H} om}(\tilde{\mathcal{V}}, f^!\tilde{\mathcal{I}})]\cong {\mathbb{R}\mathcal{H} om}({\mathcal{V}}, f^!{\mathcal{Y}}). \end{equation}

Here we used the fact that $[\cdot ]$ is exact and $f^!\tilde {\mathcal{I} }$ is injective. Then Lemma 4.21 completes the proof.

Proof. As usual sheaves, we have a canonical morphism $ {\mathbb {R}} f_*{\mathbb{R} \mathcal{H} om}( {\mathcal {V}}, f^!{\mathcal {Y}})\rightarrow {\mathbb{R} \mathcal{H} om}( {\mathbb {R}} f_! {\mathcal {V}}, {\mathbb {R}} f_!f^!{\mathcal {Y}})$. By the adjunction (Proposition 4.18), we have a morphism ${\mathbb{R} \mathcal{H} om}( {\mathbb {R}} f_! {\mathcal {V}}, {\mathbb {R}} f_!f^!{\mathcal {Y}})\rightarrow {\mathbb{R} \mathcal{H} om}( {\mathbb {R}} f_! {\mathcal {V}}, {\mathcal {Y}})$. We would like to see the composition is an isomorphism. By a local consideration, this can be deduced from the usual case.

Proof. We can apply the same formula for usual sheaves to local representatives. Then we get the desired formula.

Proof. Let ${\mathcal {I}}$ be an injective resolution of ${\mathcal {W}}$ and ${\mathcal {F}}$ be a flat resolution of $ {\mathcal {V}}$. Then we have

(4.15)\begin{align} \operatorname{Hom}_{D^b(\mathrm{Mod}^{\mathfrak{I}}(\Lambda_{({\bar{X}}, D_X)}))}({\mathcal{V}}, {\mathcal{W}})&\cong \operatorname{Hom}_{C(\mathrm{Mod}^{\mathfrak{I}}(\Lambda_{({\bar{X}}, D_X)}))}({\mathcal{V}}, {\mathcal{I}})\nonumber\\ &\cong H^0(\operatorname{Hom}_{\mathrm{Mod}^{\mathfrak{I}}_{\Lambda_{({\bar{X}}, D_X)}}}({\mathcal{V}},{\mathcal{I}})(X))\nonumber\\ &\cong H^0(\mathop{{{\mathcal{H}}}om}\nolimits({\mathcal{F}}, {\mathcal{I}})\otimes \mathbb{k}(X)). \end{align}

Actually $\otimes \mathbb {k}$ is exact, as we will see in the proof of Lemma 6.1. This completes the proof.

Proof. This follows from Lemmas 4.21, 4.10, 4.11, Proposition 4.19.

Proof. The assertions follow easily from Yoneda's Lemma.

Proof. Note that $i$ and $j$ are tame maps. Again, the statement follows from the corresponding statement for usual sheaves and the commutativity results for $[\cdot ]$ proved early in this subsection.

Theorem 5.3 [Reference KedlayaKed11, Theorem 8.2.2]

For a point $x\in Z$, there exists an open neighborhood $U$ of $z$ and a map $f\colon Y\rightarrow U$ which is a proper modification, and at each point of $f^{-1}(z)$, there exists a local covering $\pi$ ramified at $f^{-1}(Z)$ such that $\pi ^*f^*{\mathcal {E}}$ admits a good decomposition at each point of $y\in \pi ^{-1}f^{-1}(Z)$.

Theorem 5.4 [Reference SabbahSab11, Theorem 2.2.1]

The subset $\Phi _x$ is actually a subset of ${\mathcal {O}}(*Z)_z/{\mathcal {O}}_x$. Moreover, there exists a neighborhood $U$ of $x$ such that for any $x'\in U$, $\hat {{\mathcal {M}}}_{x'}$ has a good decomposition and $\Phi _{x'}$ is given by the restriction of representatives of $\Phi _x$.

Theorem 5.5 [Reference SabbahSab13, Theorem 12.5], [Reference MochizukiMoc11, § 3]

There exists an open covering $\{U_i\}$ of a neighborhood of $\varpi ^{-1}(x)$ such that each restriction $(\varpi ^*{\mathcal {M}})|_{U_i}$ is isomorphic to $(\varpi ^*({\bigoplus} _{\alpha \in I}{\mathcal {E}}(\phi _\alpha )\otimes {\mathcal {R}}_\alpha ))|_{U_i}$.

Proof. Since the sheaf is globally presented as a direct sum, the restriction morphism preserves grading. The $\Lambda$-action is given as follows. For $b\in \mathbb{R} _{\geqslant 0}$, we have a canonical morphism

(5.4)\begin{equation} \Gamma_{S\times [{-}a,\infty)}\mathbb{k}_{t\geqslant {\mathop{\mathfrak{Re}}\nolimits}\phi}\rightarrow \Gamma_{S\times [{-}a-b,\infty)}\mathbb{k}_{t\geqslant {\mathop{\mathfrak{Re}}\nolimits}\phi}. \end{equation}

This action gives the action of $T^b$.

Proof. Note that $\operatorname {Gr}^a\Lambda ^\phi _S(U)\cong \Gamma _{U\times [-a, \infty )}(U\times {\mathbb {R}}, \mathbb {k}_{t\geqslant {\mathop {\mathfrak {Re}}\nolimits }\phi })$. This is the kernel of the restriction morphism $\Gamma (U\times {\mathbb {R}}, \mathbb {k}_{t\geqslant {\mathop {\mathfrak {Re}}\nolimits }\phi })\rightarrow \Gamma (U\times (-\infty , -a), \mathbb {k}_{t\geqslant {\mathop {\mathfrak {Re}}\nolimits }\phi })$. Since $U$ is connected, the set defined by $t\geqslant {\mathop {\mathfrak {Re}}\nolimits }\phi$ is also connected. Hence we have $\Gamma (U\times {\mathbb {R}}, \mathbb {k}_{t\geqslant {\mathop {\mathfrak {Re}}\nolimits }\phi })\cong \mathbb {k}$. On the other hand, $\Gamma (U\times (-\infty , -a), \mathbb {k}_{t\geqslant {\mathop {\mathfrak {Re}}\nolimits }\phi })\cong 0$ if and only if $U\times (-\infty , -a)\cap \{t\geqslant {\mathop {\mathfrak {Re}}\nolimits }\phi \}=\varnothing$. This is equivalent to $-a<\inf _U {\mathop {\mathfrak {Re}}\nolimits }\phi$. This completes the proof.

Proof. Let $x$ be a point with $ {\mathop {\mathfrak {Re}}\nolimits }\phi (x)=-d$. The point $x$ is a local minimum if and only if $x$ is in the interior of the closure of $\{-x\,|\,-d< {\mathop {\mathfrak {Re}}\nolimits }\phi (x)\}$. This completes the proof.

Proof. Let $ {\mathcal {V}}\rightarrow {\mathcal {W}}$ be an injective morphism in $\mathrm {Mod}^{\mathfrak {I}}(\Lambda _{(\bar {S}, D_S)})$. We would like to show the induced morphism $ {\mathcal {V}}\otimes \Lambda ^\phi _{(\bar {S}, D_S)}\rightarrow {\mathcal {W}}\otimes \Lambda ^\phi _{(\bar {S}, D_S)}$ is again injective. There exists a covering $\{U_i\}$ of $S$ which is locally finite in $\bar {S}$ such that there exist lifts $ {\mathcal {V}}_i, {\mathcal {W}}_i$ of $ {\mathcal {V}}$ and ${\mathcal {W}}$ over each $U_i$. It is enough to prove the injectivity over each $U_i$.

By Lemma 2.14, one can assume the restriction $ {\mathcal {V}}_i\rightarrow {\mathcal {W}}_i$ is still injective. Since the tensor product commutes with taking stalks, it reduces to show that $ {\mathcal {V}}_x\otimes (\Lambda ^\phi _S)_x\rightarrow {\mathcal {W}}_x\otimes (\Lambda ^\phi _S)_x$ is injective. Since $(\Lambda ^\phi _S)_x\cong \Lambda ^{\phi (x)}$ (Corollary 5.8) is a torsion-free $\Lambda$-module, this completes the proof.

Proof. Since $\max \{0, {\mathop {\mathfrak {Re}}\nolimits }\phi _1- {\mathop {\mathfrak {Re}}\nolimits }\phi _2\}$ is bounded, there exists a large $c\in {\mathbb {R}}$ such that $ {\mathop {\mathfrak {Re}}\nolimits }\phi _2+c\geqslant {\mathop {\mathfrak {Re}}\nolimits }\phi _1$. The nonzero map $\mathbb {k}_{ {\mathop {\mathfrak {Re}}\nolimits }\phi _1\geqslant t}\rightarrow \mathbb {k}_{ {\mathop {\mathfrak {Re}}\nolimits }\phi _{2}+c\geqslant t}$ coming from the inclusion $\{ {\mathop {\mathfrak {Re}}\nolimits }\phi _2+c\geqslant t\}\subset \{ {\mathop {\mathfrak {Re}}\nolimits }\phi _1\geqslant t\}$ induces a morphism $\Lambda ^{\phi _1}_S\rightarrow \Lambda ^{\phi _2}_S\langle c\rangle $ of $ {\mathbb {R}}$-graded $\Lambda _S$-modules. If $\max \{0, {\mathop {\mathfrak {Re}}\nolimits }\phi _2- {\mathop {\mathfrak {Re}}\nolimits }\phi _1\}$ is also bounded, in the same way, we also have a morphism $\Lambda ^{\phi _2}_S\rightarrow \Lambda ^{\phi _1}_S\langle d\rangle $ for some $d\geqslant 0$. The composition $\Lambda ^{\phi _1}_S\rightarrow \Lambda ^{\phi _1}_S\langle c+d\rangle $ is given by $T^{c+d}$. This is the identity of $\Lambda ^\phi _{(\bar {S}, D_S)}$ in $\mathrm {Mod}^{\mathfrak {I}}(\Lambda _X)$. The same for the other direction. This completes the proof of the second part of the statement. We call morphism $\Lambda ^{\phi _1}_S\rightarrow \Lambda ^{\phi _2}_S$ of this kind as well as their scalar multiples standard morphisms. In the below, we will see there are only standard morphisms.

Let $f$ be a nonzero morphism in $\operatorname {Hom}_{\mathrm {Mod}^{\mathfrak {I}}(\Lambda _{(\bar {S}, D_S)})}(\Lambda ^{\phi _1}_{(\bar {S}, D_S)}, \Lambda ^{\phi _2}_{(\bar {S}, D_S)})$. Let us take a lift $\tilde {f}\colon \Lambda ^{\phi _1}_S\rightarrow \Lambda ^{\phi _2+c}_S$ as a morphism of $ {\mathbb {R}}$-graded $\Lambda$-modules locally on $U\subset S$. We can take $c$ such that $c+ {\mathop {\mathfrak {Re}}\nolimits }\phi _2> {\mathop {\mathfrak {Re}}\nolimits }\phi _1$ and replace $\phi _2$ with $\phi _2+c$ We consider $d\in {\mathbb {R}}$ such that the grading $d$-part of $\tilde {f}$ is nonzero. To see this part more explicitly, let us prepare some notation.

Let us set $\mathrm {Int}\overline {\{ x\in U\,|\,-d< {\mathop {\mathfrak {Re}}\nolimits }\phi _i(x)\}}=\sqcup _a S_{d,i}^a$ to be the decomposition into connected components. Since $\operatorname {Gr}^d\Lambda _U^\phi =\mathbb {k}_{\mathrm {Int}\overline {\{ x|-d< {\mathop {\mathfrak {Re}}\nolimits }\phi (x)\}}}$, we have $\operatorname {Gr}^d\Lambda _U^{\phi _i}={\bigoplus} _{a}\mathbb {k}_{S^a_{d,i}}$. We have

(5.7)\begin{equation} \tilde{f}_d\colon \bigoplus_{a}\mathbb{k}_{S^a_{d,1}} \rightarrow \bigoplus_{a}\mathbb{k}_{S^a_{d,2}}. \end{equation}

There exists $d'\in \mathbb{R} _{\geqslant 0}$ such that there exists a connected component $S_i$ of $\mathrm {Int}\overline {\{ x\,|\,-d'< {\mathop {\mathfrak {Re}}\nolimits }\phi _i(x)\}}$ for each $i$ such that $S_i\supset \mathrm {Int}\overline {\{ x\,|\,-d< {\mathop {\mathfrak {Re}}\nolimits }\phi _i(x)\}}$. Then we have a commutative diagram.

(5.8)

Since $S_1$ and $S_2$ are connected, the hom-space between them is one-dimensional. Hence $\tilde {f}_d$ is induced by a standard morphism. This completes the proof.

Proof. For $f\in \operatorname {Hom}_{\mathrm {Mod}^{\mathfrak {I}}(\Lambda _{(\bar {S}, D_S)})}(\Lambda _{(\bar {S}, D_S)}^{\phi _1}, \Lambda _{(\bar {S}, D_S)}^{\phi _2})$, let us take a lift $\tilde {f}\colon \Lambda _{(\bar {S}, D_S)}^{\phi _1}\rightarrow \Lambda _{(\bar {S}, D_S)}^{\phi _2+c}$ as a morphism between $ {\mathbb {R}}$-graded $\Lambda _S$-modules. Since $ {\mathop {\mathfrak {Re}}\nolimits }\phi _2- {\mathop {\mathfrak {Re}}\nolimits }\phi _1$ is negatively divergent, there exists a neighborhood $U$ of $D_S$ such that $ {\mathop {\mathfrak {Re}}\nolimits }\phi _2+c- {\mathop {\mathfrak {Re}}\nolimits }\phi _1$ is negative on $U{\backslash } D_S$. Hence over $U{\backslash } D_S$, the restriction of $\tilde {f}$ is zero there. By Lemma 5.11 and the connectedness of $S$, $f$ is zero everywhere.

Proof. We have $\operatorname {Gr}^a\Lambda _S^{\phi _i}=\mathbb {k}_{\{ x| {\mathop {\mathfrak {Re}}\nolimits }\phi (x)>-a\}}$ for $i=1,2$. Hence we have a map $\operatorname {Gr}^a\Lambda ^{\phi _1+\phi _2}_S\rightarrow \operatorname {Gr}^b\Lambda ^{\phi _1}_S\otimes _k\operatorname {Gr}^c\Lambda ^{\phi _2}_S$ for $a=b+c$. Hence we get a map $m\colon \Lambda ^{\phi _1+\phi _2}_S\rightarrow \Lambda ^{\phi _1}_S\otimes \Lambda ^{\phi _2}_S$. By Corollary 5.8, the stalks of both sides at $x\in X$ are ${\bigoplus} _{-a\leqslant {\mathop {\mathfrak {Re}}\nolimits }\phi _1(x)+ {\mathop {\mathfrak {Re}}\nolimits }\phi _2(x)}\Lambda ^{\phi _1(x)+\phi _2(x)}_S$ or ${\bigoplus} _{-a< {\mathop {\mathfrak {Re}}\nolimits }\phi _1(x)+ {\mathop {\mathfrak {Re}}\nolimits }\phi _2(x)}\Lambda ^{\phi _1(x)+\phi _2(x)}_S$. Hence the kernel and cokernel of $m$ vanishes by multiplying $T^a$ for any $a\in {\mathbb {R}}_{>0}$. Therefore the kernel and cokernel are zero in $\mathrm {Mod}^{\mathfrak {I}}(\Lambda _{(\bar {S}, D_S)})$. This completes the proof.

Proof. One can prove in a similar way as in the proof of Lemma 5.13.

Proof. Let ${\mathcal {I}}$ be an injective resolution of $\Lambda ^{\phi _2}_{(\bar {S}, D_S)}$. We have the following:

(5.9)\begin{align} &\operatorname{Hom}_{D^b(\mathrm{Mod}^{\mathfrak{I}}(\Lambda_{(\bar{S}, D_S)}))}(\mathcal{V}, {\mathbb{R}\mathcal{H} om}(\Lambda^{\phi_1}_{(\bar{S}, D_S)}, \Lambda^{\phi_2}_{(\bar{S}, D_S)}))\nonumber\\ &\quad \cong \operatorname{Hom}_{D^b(\mathrm{Mod}^{\mathfrak{I}}(\Lambda_{(\bar{S}, D_S)}))}(\mathcal{V}\otimes \Lambda^{\phi_1}_{(\bar{S}, D_S)}, \Lambda^{\phi_2}_{(\bar{S}, D_S)})\nonumber\\ &\quad \cong \operatorname{Hom}_{C(\mathrm{Mod}^{\mathfrak{I}}(\Lambda_{(\bar{S}, D_S)}))}({\mathcal{V}}\otimes \Lambda^{\phi_1}_{(\bar{S}, D_S)}, {\mathcal{I}})\nonumber\\ &\quad \cong \operatorname{Hom}_{C(\mathrm{Mod}^{\mathfrak{I}}(\Lambda_{(\bar{S}, D_S)}))}({\mathcal{V}},\mathop{{{\mathcal{H}}}om}\nolimits(\Lambda^{\phi_1}_{(\bar{S}, D_S)}, {\mathcal{I}})). \end{align}

Here we used flatness of $\Lambda ^{\phi _1}_{(\bar {S}, D_S)}$.

First, note that $\mathop {{{\mathcal {H}}}om}\nolimits (\Lambda ^{\phi _1}_{(\bar {S}, D_S)}, {\mathcal {I}})\cong \Lambda ^{-\phi _1}_{(\bar {S}, D_S)}\otimes {\mathcal {I}}$ in $C(\mathrm {Mod}^{\mathfrak {I}}(\Lambda _{(\bar {S}, D_S)}))$. Second, ${\mathcal {I}}$ is locally given by $[\prod _x {\mathcal {F}}_x]$ where ${\mathcal {F}}_x$ is a skyscraper sheaf. Since $\mathop {{{\mathcal {H}}}om}\nolimits (\Lambda ^{\phi _1}_{(\bar {S}, D_S)}, \prod _x{\mathcal {F}}_x)\cong \prod _x\mathop {{{\mathcal {H}}}om}\nolimits (\Lambda ^{\phi _1}_{(\bar {S}, D_S)}, {\mathcal {F}}_x)$, the object $\mathop {{{\mathcal {H}}}om}\nolimits (\Lambda ^{\phi _1}_{(\bar {S}, D_S)}, {\mathcal {I}})$ is also injective. Hence we have

(5.10)\begin{align} &\operatorname{Hom}_{D^b(\mathrm{Mod}^{\mathfrak{I}}(\Lambda_{(\bar{S}, D_S)}))}({\mathcal{V}}, {\mathbb{R}\mathcal{H} om}(\Lambda^{\phi_1}_{(\bar{S}, D_S)}, \Lambda^{\phi_2}_{(\bar{S}, D_S)}))\nonumber\\ &\quad \cong \operatorname{Hom}_{D^b(\mathrm{Mod}^{\mathfrak{I}}(\Lambda_{(\bar{S}, D_S)}))}({\mathcal{V}},\Lambda^{-\phi_1}_{(\bar{S}, D_S)}\otimes{\mathcal{I}})\nonumber\\ &\quad \cong \operatorname{Hom}_{D^b(\mathrm{Mod}^{\mathfrak{I}}(\Lambda_{(\bar{S}, D_S)}))}({\mathcal{V}},\Lambda^{-\phi_1}_{(\bar{S}, D_S)}\otimes^{\mathbb{L}}{\mathcal{I}})\nonumber\\ &\quad \cong \operatorname{Hom}_{D^b(\mathrm{Mod}^{\mathfrak{I}}(\Lambda_{(\bar{S}, D_S)}))}({\mathcal{V}},\Lambda^{-\phi_1}_{(\bar{S}, D_S)}\otimes^{\mathbb{L}}\Lambda^{\phi_2}_{(\bar{S}, D_S)})\nonumber\\ &\quad \cong \operatorname{Hom}_{D^b(\mathrm{Mod}^{\mathfrak{I}}(\Lambda_{(\bar{S}, D_S)}))}({\mathcal{V}},\Lambda^{\phi_2-\phi_1}_{(\bar{S}, D_S)}). \end{align}

This completes the proof.

Proof. This is just from the definition of multi-valued meromorphic functions.