Proof. Both of the assertions follow from the fact that $(U,\unicode[STIX]{x1D6E5}|_{U})$ is sub-klt if and only if $(f^{-1}(U),\unicode[STIX]{x1D6E5}_{Y}|_{f^{-1}(U)})$ is sub-klt for any open subset $U\subset X$ (cf. [Reference Kollár and MoriKM98, Lemma 2.30]).◻

Proof. The second equality follows from the fact that $(X,(\unicode[STIX]{x1D6E5}^{{>}0})^{{<}1})$ is klt. Clearly, the inclusion $\operatorname{Nklt}(X,\unicode[STIX]{x1D6E5})\subset \operatorname{Nklt}(X,\unicode[STIX]{x1D6E5}^{{>}0})$ holds. It suffices to prove the opposite one. Let $D$ be a prime divisor contained in $\operatorname{Supp}\unicode[STIX]{x1D6E5}^{{\geqslant}1}$. Since $\operatorname{Nklt}(X,\unicode[STIX]{x1D6E5})$ is a closed subset containing general closed points of $D$, we have that $D\subset \operatorname{Nklt}(X,\unicode[STIX]{x1D6E5})$. In particular, $\operatorname{Supp}\unicode[STIX]{x1D6E5}^{{\geqslant}1}\subset \operatorname{Nklt}(X,\unicode[STIX]{x1D6E5})$.◻

Proof. If $k$ is an infinite field, then the proof of [Reference BirkarBir16, Lemma 9.2] works without any changes. Assume that $k$ is a finite field. Thanks to [Reference PoonenPoo04, Theorem 1.1], we can still make use of Bertini’s theorem. Hence, we can apply the same argument as in [Reference BirkarBir16, Lemma 9.2].◻

Proof. See [Reference Hashizume, Nakamura and TanakaHNT17, Theorem 1.1]. ◻

Proof. See [Reference Hashizume, Nakamura and TanakaHNT17, Proposition 3.5]. ◻

Proof. Let $f:Y\rightarrow X$ and $\unicode[STIX]{x1D6E5}_{Y}$ be as in Proposition 2.10. Since $(X,\unicode[STIX]{x1D6E5})$ is plt, it follows from Proposition 2.10(ii) that $(Y,\unicode[STIX]{x1D6E5}_{Y})$ is also plt. For $S_{Y}:=\llcorner \unicode[STIX]{x1D6E5}_{Y}\lrcorner$, we have that $f^{-1}(S)=f^{-1}(\operatorname{Nklt}(X,\unicode[STIX]{x1D6E5}))=\operatorname{Nklt}(Y,\unicode[STIX]{x1D6E5}_{Y})=S_{Y}$, where the second equality holds by Proposition 2.10(iii). Hence, the induced morphism $S_{Y}\rightarrow S$ has connected fibres. Since $Y$ is $\mathbb{Q}$-factorial and $(Y,\unicode[STIX]{x1D6E5}_{Y})$ is plt, $S_{Y}$ is normal (cf. [Reference Gongyo, Nakamura and TanakaGNT19, Theorem 2.11]). Therefore, $S_{Y}\rightarrow S$ factors through the normalisation $\unicode[STIX]{x1D708}:S^{N}\rightarrow S$:

$$\begin{eqnarray}\displaystyle S_{Y}\rightarrow S^{N}\xrightarrow[{}]{\unicode[STIX]{x1D708}}S. & & \displaystyle \nonumber\end{eqnarray}$$

Then $\unicode[STIX]{x1D708}$ is a finite morphism and has connected fibres and hence $\unicode[STIX]{x1D708}$ is a universal homeomorphism.◻

Proof. We first reduce the problem to the case when $N_{i}\,\cap \,F\neq \emptyset$ for any $i\in I$. Set $I^{\prime }:=\{i\in I\,|\,N_{i}\,\cap \,F\neq \emptyset \}$ and

$$\begin{eqnarray}\displaystyle U_{0}^{\prime }:=U_{0}\bigg\backslash\biggl(\mathop{\bigcup }_{i\in I\setminus I^{\prime }}N_{i}\biggr)=U_{0}\cap \biggl(\mathop{\bigcap }_{i\in I\setminus I^{\prime }}(X\setminus N_{i})\biggr), & & \displaystyle \nonumber\end{eqnarray}$$

where $U_{0}$ is as in (2). Then we have $F\subset U_{0}^{\prime }$. If $U$ is an open subset of $X$ such that $F\subset U\subset U_{0}^{\prime }$, then (2) implies that $N\,\cap \,U$ is connected. Therefore, the problem can be reduced to the case when $N_{i}\,\cap \,F\neq \emptyset$ for any $i\in I$. In what follows, we assume that $N_{i}\,\cap \,F\neq \emptyset$ for any $i\in I$.

Take the decomposition into connected components of $N\,\cap \,F$:

$$\begin{eqnarray}\displaystyle N\cap F=\coprod _{j\in J}\unicode[STIX]{x1D6E4}_{j}. & & \displaystyle \nonumber\end{eqnarray}$$

We also have $N\,\cap \,F=\bigcup _{i\in I}(N_{i}\,\cap \,F)$. We first show that:

1. (iii) for any $i\in I$, there exists an index $j_{i}\in J$ such that $N_{i}\,\cap \,\unicode[STIX]{x1D6E4}_{j_{i}}\neq \emptyset$ and $N_{i}\,\cap \,\unicode[STIX]{x1D6E4}_{j}=\emptyset$ for any $j\in J\setminus \{j_{i}\}$. In particular, it holds that $N_{i}\,\cap \,F=N_{i}\,\cap \,\unicode[STIX]{x1D6E4}_{j_{i}}$ for any $i\in I$.

Fix $i\in I$. We have

$$\begin{eqnarray}\displaystyle N_{i}\cap F=\coprod _{j\in J}(N_{i}\cap \unicode[STIX]{x1D6E4}_{j}). & & \displaystyle \nonumber\end{eqnarray}$$

Since $N_{i}\,\cap \,F$ is non-empty and connected, (iii) holds.

For $j\in J$, set $I_{j}:=\{i\in I\,|\,N_{i}\,\cap \,\unicode[STIX]{x1D6E4}_{j}\neq \emptyset \}$ and $N_{I_{j}}:=\bigcup _{i\in I_{j}}N_{i}$. Then (iii) implies that $I=\coprod _{j\in J}I_{j}$. In particular, we obtain $N=\bigcup _{j\in J}N_{I_{j}}$. Set

(2.12.1)$$\begin{eqnarray}\displaystyle U_{1}:=\mathop{\bigcap }_{j,j^{\prime }\in J,j\neq j^{\prime }}X\setminus (N_{I_{j}}\cap N_{I_{j^{\prime }}}), & & \displaystyle\end{eqnarray}$$

which is an open subset of $X$.

We now show that $F\subset U_{1}$. Pick $j,j^{\prime }\in J$ such that $j\neq j^{\prime }$. For $i\in I_{j}$ and $i^{\prime }\in I_{j^{\prime }}$, we have

$$\begin{eqnarray}\displaystyle N_{i}\cap N_{i^{\prime }}\cap F=(N_{i}\cap F)\cap (N_{i^{\prime }}\cap F)\subset \unicode[STIX]{x1D6E4}_{j_{i}}\cap \unicode[STIX]{x1D6E4}_{j_{i^{\prime }}}=\unicode[STIX]{x1D6E4}_{j}\cap \unicode[STIX]{x1D6E4}_{j^{\prime }}=\emptyset , & & \displaystyle \nonumber\end{eqnarray}$$

where the inclusion follows from (iii). Hence, we have

$$\begin{eqnarray}\displaystyle N_{I_{j}}\cap N_{I_{j^{\prime }}}\cap F=\mathop{\bigcup }_{i\in I_{j},i^{\prime }\in I_{j^{\prime }}}(N_{i}\cap N_{i^{\prime }}\cap F)=\emptyset , & & \displaystyle \nonumber\end{eqnarray}$$

i.e. $F\subset X\setminus (N_{I_{j}}\,\cap \,N_{I_{j^{\prime }}})$. Hence, (2.12.1) implies that $F\subset U_{1}$.

Set $U:=U_{0}\,\cap \,U_{1}$, where $U_{0}$ is as in (ii). Since $F\subset U_{0}\,\cap \,U_{1}=U$, (ii) implies that $N\,\cap \,U$ is connected. We have

$$\begin{eqnarray}\displaystyle N\cap U=\mathop{\bigcup }_{j\in J}(N_{I_{j}}\cap U)=\coprod _{j\in J}(N_{I_{j}}\cap U), & & \displaystyle \nonumber\end{eqnarray}$$

where the last equality follows from (2.12.1). For $j\in J$, we have that

$$\begin{eqnarray}\displaystyle N_{I_{j}}\cap U\supset N_{I_{j}}\cap F\neq \emptyset . & & \displaystyle \nonumber\end{eqnarray}$$

Since $N\,\cap \,U$ is connected, we obtain $|J|=1$, i.e. $N\,\cap \,F$ is connected.◻

Proof. Set $S:=\llcorner D\lrcorner$ and let $S=\sum _{i\in I}S_{i}$ be the irreducible decomposition. We have $\operatorname{Nklt}(X,D)=\bigcup _{i\in I}S_{i}$.

For any sufficiently small open neighbourhood $U$ in $X$ of $f^{-1}(y)$, it follows from [Reference BirkarBir16, Theorem 1.8] that $\operatorname{Nklt}(X,D)\,\cap \,U$ is connected. We apply Lemma 2.12 to $N_{i}:=S_{i}$ and $F:=f^{-1}(y)$. Then it is enough to prove that $S_{i}\,\cap \,F$ is connected. Therefore, after perturbing coefficients of $D$, we may assume that $S=S_{i}$, i.e. $(X,D)$ is plt.

From now on, we treat the case when $\llcorner D\lrcorner =S$ is a prime divisor. In this case, we have $\operatorname{Nklt}(X,D)=S$. If $S\subset \operatorname{Ex}(f)$, then $f$ is a $(K_{X}+D)$-divisorial contraction such that $S=\operatorname{Ex}(f)$. Then the assertion is clear because $f$ has connected fibres. Thus, we may assume that $S\not \subset \operatorname{Ex}(f)$. Since $-(K_{X}+D)$ is $f$-ample, there exists an effective $\mathbb{R}$-divisor $A$ on $X$ such that $A{\sim}_{\mathbb{R},f}-(K_{X}+D)$ and $(X,D+A)$ is plt. Since $K_{X}+D+A{\sim}_{\mathbb{R},f}0$, we have $K_{X}+D+A=f^{\ast }(K_{Y}+D_{Y}+A_{Y})$ for $D_{Y}:=f_{\ast }D$ and $A_{Y}:=f_{\ast }A$. In particular, $(Y,D_{Y}+A_{Y})$ is plt. Then the induced morphism $g:S\rightarrow S_{Y}:=f(S)$ has connected fibres by Proposition 2.11. Since $\operatorname{Nklt}(X,D)\,\cap \,f^{-1}(y)=S\,\cap \,f^{-1}(y)=g^{-1}(y)$ for any $y\in f(S)=S_{Y}$, $\operatorname{Nklt}(X,D)\rightarrow Y$ has connected fibres.◻

Proof. Let $F$ be the sum of the $\unicode[STIX]{x1D711}$-exceptional prime divisors $F^{\prime }$ on $U$ whose log discrepancies are positive: $a_{F^{\prime }}(V,\unicode[STIX]{x1D6E5})>0$. Let $G$ be the sum of the $\unicode[STIX]{x1D711}$-exceptional prime divisors $G^{\prime }$ on $U$ whose log discrepancies are non-positive: $a_{G^{\prime }}(V,\unicode[STIX]{x1D6E5})\leqslant 0$. We set

$$\begin{eqnarray}\displaystyle D_{U}:=\unicode[STIX]{x1D711}_{\ast }^{-1}\unicode[STIX]{x1D6E5}^{\wedge 1}+(1-\unicode[STIX]{x1D716})F+G & & \displaystyle \nonumber\end{eqnarray}$$

for a sufficiently small positive real number $\unicode[STIX]{x1D716}$. Then it holds that:

1. (i) $(U,D_{U})$ is dlt; and

2. (ii) $\operatorname{Supp}\unicode[STIX]{x1D6E5}_{U}^{{\geqslant}1}=\operatorname{Supp}D_{U}^{=1}$.

By Theorem 2.9, there is a $(K_{U}+D_{U})$-MMP over $V$ that terminates:

(2.14.1)$$\begin{eqnarray}\displaystyle U=:X_{0}{\dashrightarrow}\cdots {\dashrightarrow}X_{\ell }. & & \displaystyle\end{eqnarray}$$

For any $i\in \{0,\ldots ,\ell \}$, we define $\unicode[STIX]{x1D6E5}_{X_{i}}$, $F_{X_{i}}$, $G_{X_{i}}$ and $D_{X_{i}}$ as the push-forwards of $\unicode[STIX]{x1D6E5}_{U}$, $F$, $G$ and $D_{U}$ on $X_{i}$, respectively. Then it holds that $K_{X_{i}}+\unicode[STIX]{x1D6E5}_{X_{i}}=\unicode[STIX]{x1D713}_{i}^{\ast }(K_{V}+\unicode[STIX]{x1D6E5})$, where $\unicode[STIX]{x1D713}_{i}:X_{i}\rightarrow V$ denotes the induced morphism. Moreover, for any $i\in \{0,\ldots ,\ell \}$, we get:

$(\text{i})^{\prime }$

$(X_{i},D_{X_{i}})$ is dlt; and

$(\text{ii})^{\prime }$

$\operatorname{Supp}\unicode[STIX]{x1D6E5}_{X_{i}}^{{\geqslant}1}=\operatorname{Supp}D_{X_{i}}^{=1}$.

Step 1. For any $i\in \{0,\ldots ,\ell \}$, it holds that

$$\begin{eqnarray}\displaystyle \operatorname{Nklt}(X_{i},\unicode[STIX]{x1D6E5}_{X_{i}})=\operatorname{Nklt}(X_{i},D_{X_{i}})=\operatorname{Supp}\unicode[STIX]{x1D6E5}_{X_{i}}^{{\geqslant}1}=\operatorname{Supp}D_{X_{i}}^{=1}. & & \displaystyle \nonumber\end{eqnarray}$$

Proof of Step 1.

Fix $i\in \{0,\ldots ,\ell \}$. We obtain

$$\begin{eqnarray}\displaystyle \operatorname{Nklt}(X_{i},D_{X_{i}})=\operatorname{Supp}D_{X_{i}}^{=1}=\operatorname{Supp}\unicode[STIX]{x1D6E5}_{X_{i}}^{{\geqslant}1}\subset \operatorname{Nklt}(X_{i},\unicode[STIX]{x1D6E5}_{X_{i}}), & & \displaystyle \nonumber\end{eqnarray}$$

where the first equality holds by (i)^′ and the second one follows from (ii)^′. Hence, it is sufficient to show that $\operatorname{Nklt}(X_{i},\unicode[STIX]{x1D6E5}_{X_{i}})\subset \operatorname{Supp}D_{X_{i}}^{=1}$. For sufficiently large $b>0$, the $\mathbb{R}$-divisor

$$\begin{eqnarray}\displaystyle A_{X_{i}}:=(\unicode[STIX]{x1D713}_{i}^{-1})_{\ast }\unicode[STIX]{x1D6E5}+(1-\unicode[STIX]{x1D716})F_{X_{i}}+bG_{X_{i}} & & \displaystyle \nonumber\end{eqnarray}$$

satisfies $\unicode[STIX]{x1D6E5}_{X_{i}}\leqslant A_{X_{i}}$. Therefore, we get

$$\begin{eqnarray}\displaystyle \operatorname{Nklt}(X_{i},\unicode[STIX]{x1D6E5}_{X_{i}})\subset \operatorname{Nklt}(X_{i},A_{X_{i}}). & & \displaystyle \nonumber\end{eqnarray}$$

Since $A_{X_{i}}^{\wedge 1}=D_{X_{i}}$, we have $A_{X_{i}}^{{<}1}=D_{X_{i}}^{{<}1}$ and hence $(X,A_{X_{i}}^{{<}1})$ is klt by (i)^′. Thus, it follows from Proposition 2.7 that

$$\begin{eqnarray}\displaystyle \operatorname{Nklt}(X_{i},A_{X_{i}})=\operatorname{Supp}A_{X_{i}}^{{\geqslant}1}=\operatorname{Supp}D_{X_{i}}^{=1}. & & \displaystyle \nonumber\end{eqnarray}$$

Thus, we obtain the desired inclusion $\operatorname{Nklt}(X_{i},\unicode[STIX]{x1D6E5}_{X_{i}})\subset \operatorname{Supp}D_{X_{i}}^{=1}$. This completes the proof of Step 1.◻

Proof of Step 3.

Assume that $\operatorname{Nklt}(X_{i+1},D_{X_{i+1}})\rightarrow V$ has connected fibres. Let $\unicode[STIX]{x1D711}_{i}:X_{i}\rightarrow Y$ be the flipping contraction and let $\unicode[STIX]{x1D711}_{i+1}:X_{i+1}\rightarrow Y$ and $\unicode[STIX]{x1D713}_{Y}:Y\rightarrow V$ be the induced morphisms. Set $N_{Y}:=\unicode[STIX]{x1D711}_{i}(\operatorname{Nklt}(X_{i},D_{X_{i}}))$ and $N_{V}:=\unicode[STIX]{x1D713}_{Y}(N_{Y})$. It follows from Step 1 that $N_{Y}=\unicode[STIX]{x1D711}_{i+1}(\operatorname{Nklt}(X_{i+1},D_{X_{i+1}}))$. By assumption, it holds that the composite morphism

$$\begin{eqnarray}\displaystyle \operatorname{Nklt}(X_{i+1},D_{X_{i+1}})\rightarrow N_{Y}\rightarrow N_{V} & & \displaystyle \nonumber\end{eqnarray}$$

is a surjective morphism with connected fibres. In particular, $N_{Y}\rightarrow N_{V}$ has connected fibres. Since $\operatorname{Nklt}(X_{i},D_{X_{i}})\rightarrow N_{Y}$ has connected fibres by Lemma 2.13, their composition

$$\begin{eqnarray}\displaystyle \operatorname{Nklt}(X_{i},D_{X_{i}})\rightarrow N_{Y}\rightarrow N_{V} & & \displaystyle \nonumber\end{eqnarray}$$

also has connected fibres. This completes the proof of Step 3. ◻

Proof of Step 4.

We have

$$\begin{eqnarray}\displaystyle B_{\ell }:=(\unicode[STIX]{x1D713}_{\ell }^{-1})_{\ast }\unicode[STIX]{x1D6E5}^{\wedge 1}+(1-\unicode[STIX]{x1D716})F_{X_{\ell }}+G_{X_{\ell }}-\unicode[STIX]{x1D6E5}_{X_{\ell }}{\sim}_{V,\mathbb{R}}K_{X_{\ell }}+D_{X_{\ell }}. & & \displaystyle \nonumber\end{eqnarray}$$

Then $B_{\ell }$ is nef over $V$. The push-forward of $-B_{\ell }$ on $V$, which is nothing but the push-forward of $\unicode[STIX]{x1D6E5}_{X_{\ell }}-(\unicode[STIX]{x1D713}_{\ell }^{-1})_{\ast }\unicode[STIX]{x1D6E5}^{\wedge 1}$, is effective. Hence, it turns out by the negativity lemma that $-B_{\ell }$ itself is effective. Since $\unicode[STIX]{x1D716}$ is sufficiently small, it follows that $F_{X_{\ell }}=0$, that is, any $\unicode[STIX]{x1D711}_{\ell }$-exceptional prime divisor $E$ satisfies $a_{E}(V,\unicode[STIX]{x1D6E5})\leqslant 0$. Since $V$ is $\mathbb{Q}$-factorial, it holds that

$$\begin{eqnarray}\displaystyle \operatorname{Ex}(\unicode[STIX]{x1D711}_{\ell })\subset \operatorname{Nklt}(X_{\ell },\unicode[STIX]{x1D6E5}_{X_{\ell }}). & & \displaystyle \nonumber\end{eqnarray}$$

In particular, $\operatorname{Nklt}(X_{\ell },\unicode[STIX]{x1D6E5}_{X_{\ell }})\,\cap \,\unicode[STIX]{x1D711}_{\ell }^{-1}(v)=\unicode[STIX]{x1D711}_{\ell }^{-1}(v)$ holds and this is connected for any closed point $v$ of $V$. This completes the proof of Step 4.◻

Step 2, Step 3 and Step 4 complete the proof of Proposition 2.14. ◻

Proof. Taking the base change to the algebraic closure of $k$, we may assume that $k$ is an algebraically closed field. We now reduce the problem to the case when $K_{X}+\unicode[STIX]{x1D6E5}{\sim}_{\mathbb{R},f}0$. Since $f$ is birational, $-(K_{X}+\unicode[STIX]{x1D6E5})$ is $f$-nef and $f$-big. After replacing $\unicode[STIX]{x1D6E5}$, we may assume that $-(K_{X}+\unicode[STIX]{x1D6E5})$ is $f$-ample. Then there exists an effective $\mathbb{R}$-Cartier $\mathbb{R}$-divisor $A$ such that $A{\sim}_{\mathbb{R},f}-(K_{X}+\unicode[STIX]{x1D6E5})$ and $\operatorname{Nklt}(X,\unicode[STIX]{x1D6E5})=\operatorname{Nklt}(X,\unicode[STIX]{x1D6E5}+A)$. Thus, we may assume that $K_{X}+\unicode[STIX]{x1D6E5}{\sim}_{\mathbb{R},f}0$. In particular, for $\unicode[STIX]{x1D6E5}_{V}:=f_{\ast }\unicode[STIX]{x1D6E5}$, it holds that $(V,\unicode[STIX]{x1D6E5}_{V})$ is a log pair and $K_{X}+\unicode[STIX]{x1D6E5}=f^{\ast }(K_{V}+\unicode[STIX]{x1D6E5}_{V})$.

Let $\unicode[STIX]{x1D711}:V_{1}\rightarrow V$ be a dlt modification of $(V,\unicode[STIX]{x1D6E5}_{V})$ such that $\operatorname{Nklt}(V_{1},\unicode[STIX]{x1D6E5}_{V_{1}})=\unicode[STIX]{x1D711}^{-1}(\operatorname{Nklt}(V,\unicode[STIX]{x1D6E5}_{V}))$ (Proposition 2.10). In particular, $\operatorname{Nklt}(V_{1},\unicode[STIX]{x1D6E5}_{V_{1}})\rightarrow \operatorname{Nklt}(V,\unicode[STIX]{x1D6E5}_{V})$ has connected fibres. Let$f_{1}:X_{1}\rightarrow V_{1}$ be a log resolution of $(V_{1},\unicode[STIX]{x1D6E5}_{V_{1}})$ that factors through $X$. By Proposition 2.14,$\operatorname{Nklt}(X_{1},\unicode[STIX]{x1D6E5}_{X_{1}})\rightarrow \operatorname{Nklt}(V_{1},\unicode[STIX]{x1D6E5}_{V_{1}})$ has connected fibres. Thus, the composite morphism

$$\begin{eqnarray}\displaystyle \operatorname{Nklt}(X_{1},\unicode[STIX]{x1D6E5}_{X_{1}})\rightarrow \operatorname{Nklt}(V_{1},\unicode[STIX]{x1D6E5}_{V_{1}})\rightarrow \operatorname{Nklt}(V,\unicode[STIX]{x1D6E5}_{V}) & & \displaystyle \nonumber\end{eqnarray}$$

has connected fibres and factors through $\operatorname{Nklt}(X,\unicode[STIX]{x1D6E5})$. In particular, also $\operatorname{Nklt}(X,\unicode[STIX]{x1D6E5})\rightarrow \operatorname{Nklt}(V,\unicode[STIX]{x1D6E5}_{V})$ has connected fibres.◻

Proof. It follows from [Reference Gongyo, Nakamura and TanakaGNT19, Lemma 2.22] that (i) implies (ii).

It is enough to show that (ii) implies (i). Assume (ii). Taking the Stein factorisation, the problem is reduced to the case when $f$ is a finite surjective morphism. Since the problem is local on $Y$, we may assume that $X=\operatorname{Spec}\,B$ and $Y=\operatorname{Spec}\,A$. For the induced ring homomorphism $A\rightarrow B$, we have that $W(A)_{\mathbb{Q}}\rightarrow W(B)_{\mathbb{Q}}$ is an isomorphism. Fix a maximal ideal $\mathfrak{m}$ of $A$. We have the following commutative diagram of ring homomorphisms.

By a diagram chase, $\unicode[STIX]{x1D713}:W(A/\mathfrak{m})_{\mathbb{Q}}\rightarrow W(B/\mathfrak{m}B)_{\mathbb{Q}}$ is surjective. On the other hand, $W(A/\mathfrak{m})_{\mathbb{Q}}$ is a field and hence $\unicode[STIX]{x1D713}$ is an isomorphism. In particular, $f^{-1}(\mathfrak{m})$ consists of one point $\mathfrak{n}$. By $W(A/\mathfrak{m})_{\mathbb{Q}}\simeq W(B/\mathfrak{m}B)_{\mathbb{Q}}$ and $W(B/\mathfrak{n})_{\mathbb{Q}}\simeq W(B/\mathfrak{m}B)_{\mathbb{Q}}$, we have $W(A/\mathfrak{m})_{\mathbb{Q}}\simeq W(B/\mathfrak{n})_{\mathbb{Q}}$. Hence, the finite extension $W(A/\mathfrak{m}){\hookrightarrow}W(B/\mathfrak{n})$ of discrete valuation rings is also an isomorphism. Taking modulo $p$ reduction, we have that $A/\mathfrak{m}\rightarrow B/\mathfrak{n}$ is an isomorphism. Thus, (i) holds.◻

Proof. By using Remark 2.2 and the fact that the functor $(-)_{\mathbb{Q}}$ is exact, we obtain the exact sequence (i) by the same argument as in [Reference Berthelot, Bloch and EsnaultBBE07, Proposition 2.2]. The exact sequence (ii) is obtained by (i) and the snake lemma.◻

Proof. Taking the Stein factorisation of the structure morphism $X\rightarrow \operatorname{Spec}\,k$, we may assume that $X$ is of dimension zero. Replacing $X$ by a connected component, we may assume that $X=\operatorname{Spec}\,L$, where $L$ is a finite extension of $k$. Then the assertion is clear.◻

Proof. Clearly, we may assume that $X$ is connected. By the exact sequence

$$\begin{eqnarray}\displaystyle 0\rightarrow W_{n}{\mathcal{O}}_{X}\xrightarrow[{}]{V}W_{n+1}{\mathcal{O}}_{X}\rightarrow {\mathcal{O}}_{X}\rightarrow 0 & & \displaystyle \nonumber\end{eqnarray}$$

and the fact that $X$ is one dimensional, it holds that (i), (ii) and (iii) are equivalent. By [Reference Gongyo, Nakamura and TanakaGNT19, Lemma 2.19], (iii) implies (iv). Moreover, by the exact sequence

$$\begin{eqnarray}\displaystyle 0\rightarrow W{\mathcal{O}}_{X}\xrightarrow[{}]{V}W{\mathcal{O}}_{X}\rightarrow {\mathcal{O}}_{X}\rightarrow 0 & & \displaystyle \nonumber\end{eqnarray}$$

and the fact that $X$ is one dimensional, (iv) implies (i).

The equivalence between (iv) and (v) follows from the fact that $H^{1}(X,W{\mathcal{O}}_{X})$ is a free $W(k)$-module [Reference IllusieIll79, ch. II, Proposition 2.19].◻

Proof. For simplicity, we denote $K=W(k)_{\mathbb{Q}}$, $K^{\prime }=W(k^{\prime })_{\mathbb{Q}}$ and $Y^{\prime }=Y\times _{k}k^{\prime }$ for a $k$-scheme $Y$. We may assume that $X$ is reduced. Let

$$\begin{eqnarray}\displaystyle X^{N}\rightarrow X & & \displaystyle \nonumber\end{eqnarray}$$

be the normalisation of $X$. Thanks to Lemma 2.20, if one of (i) and (ii) holds, then it follows that

$$\begin{eqnarray}\displaystyle H^{1}(X^{N},W{\mathcal{O}}_{X,\mathbb{Q}})=H^{1}(X^{\prime N},W{\mathcal{O}}_{X^{\prime N},\mathbb{Q}})=0. & & \displaystyle \nonumber\end{eqnarray}$$

For the conductor subschemes $C$ and $D$ of $X$ and $X^{N}$, respectively, we have a commutative diagram with exact horizontal sequences:

where $(-)_{K^{\prime }}$ denotes the tensor product $(-)\otimes _{K}K^{\prime }$. As both $\unicode[STIX]{x1D6FC}$ and $\unicode[STIX]{x1D6FD}$ are isomorphisms by Lemma 2.19, so is $\unicode[STIX]{x1D6FE}$ by the five lemma, as desired.◻

Proof. We may assume that $Q$ is a unique non-regular point of $X$. Let $f:Y\rightarrow X$ be a resolution of singularities such that $f(\operatorname{Ex}(f))=Q$. For $E:=\operatorname{Ex}(f)$, we have the exact sequence

$$\begin{eqnarray}\displaystyle 0\rightarrow WI_{E,\mathbb{Q}}\rightarrow W{\mathcal{O}}_{Y,\mathbb{Q}}\rightarrow W{\mathcal{O}}_{E,\mathbb{Q}}\rightarrow 0. & & \displaystyle \nonumber\end{eqnarray}$$

Thanks to the vanishing of $R^{i}f_{\ast }(WI_{E,\mathbb{Q}})=0$ for $i>0$ [Reference Berthelot, Bloch and EsnaultBBE07, Theorem 2.4], it holds that

$$\begin{eqnarray}\displaystyle R^{i}f_{\ast }(W{\mathcal{O}}_{Y,\mathbb{Q}})\simeq H^{i}(E,W{\mathcal{O}}_{E,\mathbb{Q}}). & & \displaystyle \nonumber\end{eqnarray}$$

Therefore, it follows from Lemma 2.21 that (i) and (ii) are equivalent. ◻

Proof. We may assume that $X$ is connected. Moreover, replacing $k$ by $k^{\prime }$ for the Stein factorisation $X\rightarrow \operatorname{Spec}\,k^{\prime }\rightarrow \operatorname{Spec}\,k$, we may assume that $X$ is geometrically connected.

We now show that the assertions (i), (ii), (iii) and (iv) are equivalent. By the exact sequence

$$\begin{eqnarray}\displaystyle 0\rightarrow W_{n}{\mathcal{O}}_{X}\xrightarrow[{}]{V}W_{n+1}{\mathcal{O}}_{X}\rightarrow {\mathcal{O}}_{X}\rightarrow 0 & & \displaystyle \nonumber\end{eqnarray}$$

and the fact that $X$ is one dimensional, it holds that (i), (ii) and (iii) are equivalent. We have that (iii) implies (iv) by [Reference Gongyo, Nakamura and TanakaGNT19, Lemma 2.19]. Moreover, by the exact sequence

$$\begin{eqnarray}\displaystyle 0\rightarrow W{\mathcal{O}}_{X}\xrightarrow[{}]{V}W{\mathcal{O}}_{X}\rightarrow {\mathcal{O}}_{X}\rightarrow 0 & & \displaystyle \nonumber\end{eqnarray}$$

and the fact that $X$ is one dimensional, (iv) implies (i). Thus, (i), (ii), (iii) and (iv) are equivalent.

Thanks to [Reference KollárKol96, ch. II, Lemma 7.5], it holds that (vi) implies (i). Further, (iv) clearly implies (v). Thus, it suffices to show that (v) implies (vi). Lemma 2.21 allows us to replace $X\rightarrow \operatorname{Spec}\,k$ by the base change $X\times _{k}\overline{k}\rightarrow \operatorname{Spec}\,\overline{k}$. Then it follows from [Reference Chatzistamatiou and RüllingCR12, the second last paragraph of § 4.6] that (v) implies (vi), as desired.◻

Proof. Replacing $\unicode[STIX]{x1D6E5}$, we may assume that $-(K_{X}+\unicode[STIX]{x1D6E5})$ is ample. If $X\rightarrow X^{\prime }$ is a birational $k$-morphism of projective normal varieties with $k=H^{0}(X,{\mathcal{O}}_{X})=H^{0}(X^{\prime },{\mathcal{O}}_{X^{\prime }})$, then also $(X\times _{k}\overline{k})_{\operatorname{red}}\rightarrow (X^{\prime }\times _{k}\overline{k})_{\operatorname{red}}$ is birational. Thus, we may replace $(X,\unicode[STIX]{x1D6E5})$ by the end result of a $(K_{X}+\unicode[STIX]{x1D6E5})$-MMP [Reference TanakaTan18b, Theorem 1.1]. Hence, we may assume that one of the following conditions holds:

1. (a) $\unicode[STIX]{x1D70C}(X)=1$;

2. (b) there is a $(K_{X}+\unicode[STIX]{x1D6E5})$-Mori fibre space $\unicode[STIX]{x1D70B}_{1}:X\rightarrow B_{1}$ onto a regular projective curve $B_{1}$ with $(\unicode[STIX]{x1D70B}_{1})_{\ast }{\mathcal{O}}_{X}={\mathcal{O}}_{B_{1}}$.

In what follows, we denote by $Y$ the normalisation of $(X\times _{k}\overline{k})_{\operatorname{red}}$ and denote by $f:Y\rightarrow X$ the composite morphism. By applying [Reference TanakaTan18a, Theorem 1.1] to the regular locus of $X$, we can write

$$\begin{eqnarray}\displaystyle K_{Y}+D=f^{\ast }K_{X} & & \displaystyle \nonumber\end{eqnarray}$$

for some effective $\mathbb{Z}$-divisor $D$.

Suppose (a). Then $Y$ is a projective normal $\mathbb{Q}$-factorial surface such that $\unicode[STIX]{x1D70C}(Y)=1$ [Reference TanakaTan18a, Proposition 2.4(2)]. Since $-K_{Y}$ is ample, $Y$ is a ruled surface. Assume that $Y$ is not rational; let us derive a contradiction. Let $\unicode[STIX]{x1D707}:Z\rightarrow Y$ be the minimal resolution of $Y$. Since $Z$ is an irrational ruled surface, there is a projective morphism $\unicode[STIX]{x1D70B}:Z\rightarrow B$ onto a smooth projective irrational curve whose general fibres are $\mathbb{P}^{1}$. Since $\overline{k}\neq \overline{\mathbb{F}}_{p}$, it follows from [Reference TanakaTan14, Theorem 3.20] that $\unicode[STIX]{x1D70B}$ factors through $\unicode[STIX]{x1D707}$:

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D70B}:Z\xrightarrow[{}]{\unicode[STIX]{x1D707}}Y\rightarrow B. & & \displaystyle \nonumber\end{eqnarray}$$

This is a contradiction to $\unicode[STIX]{x1D70C}(Y)=1$. Thus, we are done for the case (a).

Suppose (b). Since $-(K_{X}+\unicode[STIX]{x1D6E5})$ is ample, there exists an extremal ray $R$ of $\overline{\text{NE}}(X)$ not corresponding to $\unicode[STIX]{x1D70B}_{1}$. By [Reference TanakaTan18b, Theorem 4.4], the extremal ray $R$ induces either a birational morphism or another $(K_{X}+\unicode[STIX]{x1D6E5})$-Mori fibre space $X\rightarrow B_{2}$ onto a curve $B_{2}$. If the former case occurs, then the problem is reduced to the case (a). Therefore, we may assume that there exist two Mori fibre space structures $\unicode[STIX]{x1D70B}_{1}:X\rightarrow B_{1}$ and $\unicode[STIX]{x1D70B}_{2}:X\rightarrow B_{2}$ onto curves $B_{1}$ and $B_{2}$. In particular, any fibre of $\unicode[STIX]{x1D70B}_{i}$ dominates $B_{3-i}$. Let $\unicode[STIX]{x1D70B}_{i}^{\prime }:Y\rightarrow B_{i}^{\prime }$ be the Stein factorisation of the composite morphism:

$$\begin{eqnarray}\displaystyle Y\rightarrow X\times _{k}\overline{k}\xrightarrow[{}]{\unicode[STIX]{x1D70B}_{i}\times _{k}\overline{k}}B_{i}\times _{k}\overline{k}. & & \displaystyle \nonumber\end{eqnarray}$$

Then any fibre of $\unicode[STIX]{x1D70B}_{i}^{\prime }$ dominates $B_{3-i}^{\prime }$. Since $-K_{Y}$ is big, a general fibre of each $\unicode[STIX]{x1D70B}_{i}^{\prime }$ is isomorphic to $\mathbb{P}^{1}$. In particular, $B_{1}^{\prime }\simeq \mathbb{P}^{1}$ and $Y$ is rational.◻

Proof. We may assume that $k$ is separably closed. Since the assertion is well known if $k$ is an algebraically closed field (cf. [Reference TanakaTan15, Fact 3.4 and Theorem 3.5]), the problem is reduced to the case when $k$ is an imperfect field. In particular, $k$ is not algebraic over a finite field. As $X$ is $\mathbb{Q}$-factorial [Reference TanakaTan18b, Corollary 4.11], the assertion follows from Lemma 2.24.◻

Proof. We divide the proof into two steps.

Step 1. The assertion of Lemma 3.4 holds if there exists a projective birational $K(Y)$-morphism

$$\begin{eqnarray}\displaystyle g_{0}:Z_{0}\rightarrow X_{K(Y)} & & \displaystyle \nonumber\end{eqnarray}$$

from a smooth projective $K(Y)$-surface $Z_{0}$.

Proof of Step 1.

Killing the denominators of all the elements of $K(Y)$ defining $g_{0}$, we can find a non-empty open subset $Y^{\prime }$ of $Y$ and morphisms

$$\begin{eqnarray}\displaystyle h^{\prime }:Z^{\prime }\xrightarrow[{}]{g^{\prime }}X^{\prime }:=f^{-1}(Y^{\prime })\xrightarrow[{}]{f|_{f^{-1}(Y^{\prime })}}Y^{\prime } & & \displaystyle \nonumber\end{eqnarray}$$

whose base changes by $(-)\times _{Y^{\prime }}\operatorname{Spec}\,K(Y)$ are the same as

$$\begin{eqnarray}\displaystyle Z_{0}\xrightarrow[{}]{g_{0}}X_{K(Y)}\rightarrow \operatorname{Spec}\,K(Y). & & \displaystyle \nonumber\end{eqnarray}$$

Furthermore, we may assume that:

• ∙ $Z^{\prime }$ is an integral scheme;

• ∙ $g^{\prime }$ is a projective birational morphism; and

• ∙ the composite morphism $h^{\prime }$ is smooth.

In particular, $Z^{\prime }$ is a smooth threefold over $k$. Let $g:Z\rightarrow X$ be a smooth projective compactification of the induced morphism

$$\begin{eqnarray}\displaystyle Z^{\prime }\xrightarrow[{}]{g_{1}}X^{\prime }=f^{-1}(Y^{\prime }){\hookrightarrow}X, & & \displaystyle \nonumber\end{eqnarray}$$

i.e. there are morphisms

$$\begin{eqnarray}\displaystyle g^{\prime }:Z^{\prime }\xrightarrow[{}]{j}Z\xrightarrow[{}]{g}X & & \displaystyle \nonumber\end{eqnarray}$$

such that $j$ is an open immersion, $g$ is projective and $Z$ is an integral scheme smooth over $k$. In particular, $Z$ is a smooth threefold over $k$ which is projective over $X$ and hence over $Y$. We get the composite morphism

$$\begin{eqnarray}\displaystyle h:Z\xrightarrow[{}]{g}X\xrightarrow[{}]{f}Y & & \displaystyle \nonumber\end{eqnarray}$$

whose geometric generic fibre $Z\times _{Y}\overline{K(Y)}$ satisfies the following isomorphisms:

$$\begin{eqnarray}\displaystyle Z\times _{Y}\overline{K(Y)}\simeq Z^{\prime }\times _{Y^{\prime }}\overline{K(Y)}\simeq Z_{0}\times _{K(Y)}\overline{K(Y)}. & & \displaystyle \nonumber\end{eqnarray}$$

In particular, $Z\times _{Y}\overline{K(Y)}$ is a smooth projective rational surface over $\overline{K(Y)}$.

Therefore, we have that

$$\begin{eqnarray}\displaystyle Rf_{\ast }(W{\mathcal{O}}_{X,\mathbb{Q}})\simeq Rf_{\ast }Rg_{\ast }(W{\mathcal{O}}_{Z,\mathbb{Q}})\simeq Rh_{\ast }(W{\mathcal{O}}_{Z,\mathbb{Q}})\simeq W{\mathcal{O}}_{Y,\mathbb{Q}}, & & \displaystyle \nonumber\end{eqnarray}$$

where the first isomorphism follows from the assumption (i) and the third follows from Theorem 3.2. This completes the proof of Step 1. ◻

Proof of Step 2.

Let $K(Y)^{1/p^{\infty }}$ be the purely inseparable closure of $K(Y)$ in the algebraic closure $\overline{K(Y)}$ of $K(Y)$. We fix a projective birational $K(Y)^{1/p^{\infty }}$-morphism

$$\begin{eqnarray}\displaystyle g_{1}:Z_{1}\rightarrow X\times _{Y}K(Y)^{1/p^{\infty }} & & \displaystyle \nonumber\end{eqnarray}$$

from a regular $K(Y)^{1/p^{\infty }}$-surface $Z_{1}$. Note that $Z_{1}$ is smooth over $K(Y)^{1/p^{\infty }}$, since $K(Y)^{1/p^{\infty }}$ is a perfect field. Then there exist a finite purely inseparable extension $L$ of $K(Y)$ and a projective normal $L$-surface $Z_{2}$ with the following projective birational $L_{0}$-morphism:

$$\begin{eqnarray}\displaystyle g_{2}:Z_{2}\rightarrow X\times _{Y}L, & & \displaystyle \nonumber\end{eqnarray}$$

whose base change by $(-)\times _{L}K(Y)^{1/p^{\infty }}$ is isomorphic to $g_{1}$. In particular, $Z_{2}$ is a smooth projective surface over $L$.

Let $Y_{2}$ be the normalisation of $Y$ in $L$ and let $X_{2}$ be the normalisation of $(X\times _{Y}Y_{2})_{\operatorname{red}}$. We get the following commutative diagram of normal $k$-varieties.

Proof of Claim 3.5.

Note that the generic fibre of $X\times _{Y}Y_{2}\rightarrow Y_{2}$ is normal by the geometric normality of $X_{K(Y)}$ (the assumption (iv)). Therefore, (a) holds, since $X_{2}\rightarrow (X\times _{Y}Y_{2})_{\operatorname{red}}$ is the normalisation. It is clear that (b) follows from (a). As $K(Y)\subset L$ is a purely inseparable extension, the assertion (d) holds.

Let us show (c). It follows from the construction that $\unicode[STIX]{x1D6FC}:X_{2}\rightarrow X$ is a finite surjective morphism of normal schemes. In particular, $X_{2}$ coincides with the normalisation of $X$ in $K(X_{2})$. Therefore, it suffices to show that the field extension $K(X)\subset K(X_{2})$ is purely inseparable, which in turn follows from the following equation:

$$\begin{eqnarray}\displaystyle K(X_{2})=K(X_{2}\times _{Y_{2}}K(Y_{2}))=K(X\times _{Y}L), & & \displaystyle \nonumber\end{eqnarray}$$

where the second equality follows from (b). Thus, (c) holds.

Let us show (e). Let $f_{2}:X_{2}\rightarrow Y_{2}^{\prime }\rightarrow Y_{2}$ be the Stein factorisation of $f_{2}$. By (a), there exists a non-empty open subset $Y_{4}$ of $Y_{2}$ such that the induced homomorphism

$$\begin{eqnarray}\displaystyle {\mathcal{O}}_{Y_{2}}|_{Y_{4}}\rightarrow (f_{2})_{\ast }{\mathcal{O}}_{X_{2}}|_{Y_{4}} & & \displaystyle \nonumber\end{eqnarray}$$

is an isomorphism. In particular, $Y_{2}^{\prime }\rightarrow Y_{2}$ is a finite birational morphism of integral $k$-varieties. As $Y_{2}$ is normal, it holds that $Y_{2}^{\prime }\rightarrow Y_{2}$ is an isomorphism and hence we obtain (e). This completes the proof of Claim 3.5.◻

Let us go back to the proof of Step 2. Thanks to (c), (d), (e) and (b) of Claim 3.5, also $f_{2}$ satisfies the same properties (i), (ii), (iii) and (iv) for $f_{2}$, respectively. By Step 1, it holds that $R^{i}(f_{2})_{\ast }(W{\mathcal{O}}_{X_{2},\mathbb{Q}})=0$ for $i>0$. Thanks to (c) and (d) of Claim 3.5, we have that $R^{i}f_{\ast }(W{\mathcal{O}}_{X,\mathbb{Q}})=0$ for $i>0$. This completes the proof of Step 2.◻

Step 2 completes the proof of Lemma 3.4. ◻

Proof. Applying Lemma 3.3 to $f$, we obtain a commutative diagram

that satisfies the properties listed in Lemma 3.3. Since $X$ has $W{\mathcal{O}}$-rational singularities and $\unicode[STIX]{x1D6FC}$ is a finite universal homeomorphism, it holds that $X^{\prime }$ has $W{\mathcal{O}}$-rational singularities. Furthermore, the geometric generic fibre $X_{\overline{K(Y^{\prime })}}$ of $f^{\prime }$ is normal (Lemma 3.3(iii)) and hence it is a normal rational surface by Proposition 2.26. Therefore, it follows from Lemma 3.4 that $Rf_{\ast }^{\prime }(W{\mathcal{O}}_{X^{\prime },\mathbb{Q}})=W{\mathcal{O}}_{Y^{\prime },\mathbb{Q}}$. Thus, we get

$$\begin{eqnarray}\displaystyle Rf_{\ast }(W{\mathcal{O}}_{X,\mathbb{Q}})\simeq Rf_{\ast }R\unicode[STIX]{x1D6FC}_{\ast }(W{\mathcal{O}}_{X^{\prime },\mathbb{Q}})\simeq R\unicode[STIX]{x1D6FD}_{\ast }(W{\mathcal{O}}_{Y^{\prime },\mathbb{Q}})\simeq W{\mathcal{O}}_{Y,\mathbb{Q}}, & & \displaystyle \nonumber\end{eqnarray}$$

where the first and last isomorphisms hold because $\unicode[STIX]{x1D6FC}$ and $\unicode[STIX]{x1D6FD}$ are finite universal homeomorphisms (Lemma 3.3(i)) and the second isomorphism follows from $Rf_{\ast }^{\prime }(W{\mathcal{O}}_{X^{\prime },\mathbb{Q}})=W{\mathcal{O}}_{Y^{\prime },\mathbb{Q}}$.◻

Proof. Fix a resolution of singularities of $X$: $\unicode[STIX]{x1D711}:V\rightarrow X$. We have the following exact sequence induced by the corresponding Grothendieck spectral sequence:

$$\begin{eqnarray}\displaystyle 0\rightarrow R^{1}f_{\ast }(W{\mathcal{O}}_{X,\mathbb{Q}}) & \rightarrow & \displaystyle R^{1}(f\circ \unicode[STIX]{x1D711})_{\ast }(W{\mathcal{O}}_{V,\mathbb{Q}})\rightarrow f_{\ast }R^{1}\unicode[STIX]{x1D711}_{\ast }(W{\mathcal{O}}_{V,\mathbb{Q}})\nonumber\\ \displaystyle & \rightarrow & \displaystyle R^{2}f_{\ast }(W{\mathcal{O}}_{X,\mathbb{Q}}).\nonumber\end{eqnarray}$$

We obtain $R^{1}(f\circ \unicode[STIX]{x1D711})_{\ast }(W{\mathcal{O}}_{V,\mathbb{Q}})=0$, since $Y$ has $W{\mathcal{O}}$-rational singularities. Moreover, we have that $R^{2}f_{\ast }(W{\mathcal{O}}_{X,\mathbb{Q}})=0$, as the fibres of $f$ are at most one dimensional (cf. [Reference Gongyo, Nakamura and TanakaGNT19, Lemma 2.20]). Therefore, it holds that $f_{\ast }R^{1}\unicode[STIX]{x1D711}_{\ast }(W{\mathcal{O}}_{V,\mathbb{Q}})=0$. Thanks to the fact that $\operatorname{Supp}\,R^{1}\unicode[STIX]{x1D711}_{\ast }(W{\mathcal{O}}_{V,\mathbb{Q}})$ is zero dimensional, we get $R^{1}\unicode[STIX]{x1D711}_{\ast }(W{\mathcal{O}}_{V,\mathbb{Q}})=0$.◻

Proof. Replacing $\unicode[STIX]{x1D6E5}$, we may assume that $-(K_{X}+\unicode[STIX]{x1D6E5})$ is $f$-ample.

Proof of Step 1.

In this proof, $X(\mathbb{F}_{p^{e}})$ denotes the number of $\mathbb{F}_{p^{e}}$-rational points on a model $X_{0}$ of $X$ over $\mathbb{F}_{p^{e}}$, i.e. $X_{0}$ is a projective $\mathbb{F}_{p^{e}}$-scheme such that $X_{0}\times _{\mathbb{F}_{p^{e}}}k\simeq X$. We can define $X(\mathbb{F}_{p^{e}})$ if $e$ is a sufficiently divisible positive integer and we fix a model $X_{0}$ (this number possibly depends on the choice of a model $X_{0}$).

We may assume that $Y$ has a unique singular point $y$. Let $g:Y^{\prime }\rightarrow Y$ be a log resolution such that $g(\operatorname{Ex}(g))=\{y\}$. Set $C:=\operatorname{Ex}(g)=g^{-1}(y)$. Let $\unicode[STIX]{x1D711}:W\rightarrow X$ be a log resolution of $(X,\unicode[STIX]{x1D6E5})$ that admits a morphism to $Y^{\prime }$. Then $f^{-1}(y)$ is rationally chain connected [Reference Gongyo, Nakamura and TanakaGNT19, Theorem 4.1]. Hence, by [Reference Gongyo, Nakamura and TanakaGNT19, Theorem 4.8], also $(f\circ \unicode[STIX]{x1D711})^{-1}(y)$ is rationally chain connected. Therefore, its image on $Y^{\prime }$, which is nothing but $C$, is a union of rational curves. In order to prove that $Y$ has $W{\mathcal{O}}$-rational singularities, it is suffices to show that $C$ forms a tree by Proposition 2.23 and [Reference Chatzistamatiou and RüllingCR12, Corollary 4.6.4]. Let $s$ be the number of the vertices and let $t$ be the number of the edges of the dual graph of $C$. Note that since $C$ is connected and simple normal closing, the condition that $C$ forms a tree is equivalent to the condition that $s=t+1$. Then, for a sufficiently divisible $e$,

$$\begin{eqnarray}\displaystyle C(\mathbb{F}_{p^{e}})=s(p^{e}+1)-t & & \displaystyle \nonumber\end{eqnarray}$$

holds because we may assume that each component of $C$ and their intersection are defined over $\mathbb{F}_{p^{e}}$. Hence, the condition $s=t+1$ is equivalent to the condition that

$$\begin{eqnarray}\displaystyle C(\mathbb{F}_{p^{e}})\equiv 1\hspace{0.6em}{\rm mod}\hspace{0.2em}p^{e} & & \displaystyle \nonumber\end{eqnarray}$$

for sufficiently divisible $e$. Therefore, it suffices to show that

$$\begin{eqnarray}\displaystyle Y^{\prime }(\mathbb{F}_{p^{e}})\equiv Y(\mathbb{F}_{p^{e}})\hspace{0.6em}{\rm mod}\hspace{0.2em}p^{e} & & \displaystyle \nonumber\end{eqnarray}$$

for any sufficiently divisible positive integer $e$.

Let $E$ be the sum of all the $\unicode[STIX]{x1D711}$-exceptional prime divisors. We run a $(K_{W}+\unicode[STIX]{x1D711}_{\ast }^{-1}\unicode[STIX]{x1D6E5}+E)$-MMP over $Y^{\prime }$ that terminates. Since $K_{W}+\unicode[STIX]{x1D711}_{\ast }^{-1}\unicode[STIX]{x1D6E5}+E$ is generically anti-ample over $Y^{\prime }$, we end with a Mori fibre space $X_{1}\rightarrow Y_{1}$ over $Y^{\prime }$. Note that the induced morphism $Y_{1}\rightarrow Y^{\prime }$ is birational and hence $Y_{1}$ has $W{\mathcal{O}}$-rational singularities by Lemma 3.7.

Take an arbitrary divisible positive integer $e$. We obtain

$$\begin{eqnarray}\displaystyle Y_{1}(\mathbb{F}_{p^{e}})\equiv Y^{\prime }(\mathbb{F}_{p^{e}})\hspace{0.6em}{\rm mod}\hspace{0.2em}p^{e}, & & \displaystyle \nonumber\end{eqnarray}$$

since $Y_{1}$ and $Y^{\prime }$ are birational and have $W{\mathcal{O}}$-rational singularities [Reference Chatzistamatiou and RüllingCR12, Corollary 4.4.16]. Furthermore, it follows from [Reference Gongyo, Nakamura and TanakaGNT19, Theorem 5.1] that

$$\begin{eqnarray}\displaystyle X(\mathbb{F}_{p^{e}})\equiv W(\mathbb{F}_{p^{e}})\equiv X_{1}(\mathbb{F}_{p^{e}})\hspace{0.6em}{\rm mod}\hspace{0.2em}p^{e}. & & \displaystyle \nonumber\end{eqnarray}$$

On the other hand, $X$ and $X_{1}$ are of Fano type over $Y$ and $Y_{1}$, respectively. Hence, by [Reference Gongyo, Nakamura and TanakaGNT19, Theorem 5.4], we obtain

$$\begin{eqnarray}\displaystyle X(\mathbb{F}_{p^{e}})\equiv Y(\mathbb{F}_{p^{e}}),\quad X_{1}(\mathbb{F}_{p^{e}})\equiv Y_{1}(\mathbb{F}_{p^{e}})\hspace{0.6em}{\rm mod}\hspace{0.2em}p^{e}. & & \displaystyle \nonumber\end{eqnarray}$$

To summarise, we get

$$\begin{eqnarray}\displaystyle Y(\mathbb{F}_{p^{e}})\equiv Y^{\prime }(\mathbb{F}_{p^{e}})\hspace{0.6em}{\rm mod}\hspace{0.2em}p^{e}. & & \displaystyle \nonumber\end{eqnarray}$$

This completes the proof of Step 1. ◻