## 100.9 Immersions of algebraic stacks

Immersions of algebraic stacks are defined as follows.

Definition 100.9.1. Immersions.

A morphism of algebraic stacks is called an *open immersion* if it is representable, and an open immersion in the sense of Section 100.3.

A morphism of algebraic stacks is called a *closed immersion* if it is representable, and a closed immersion in the sense of Section 100.3.

A morphism of algebraic stacks is called an *immersion* if it is representable, and an immersion in the sense of Section 100.3.

This is not the most convenient way to think about immersions for us. For us it is a little bit more convenient to think of an immersion as a morphism of algebraic stacks which is representable by algebraic spaces and is an immersion in the sense of Section 100.3. Similarly for closed and open immersions. Since this is clearly equivalent to the notion just defined we shall use this characterization without further mention. We prove a few simple lemmas about this notion.

Lemma 100.9.2. Let $\mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $\mathcal{Z} \to \mathcal{Y}$ be a (closed, resp. open) immersion. Then $\mathcal{Z} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X}$ is a (closed, resp. open) immersion.

**Proof.**
This follows from the general discussion in Section 100.3.
$\square$

Lemma 100.9.3. Compositions of immersions of algebraic stacks are immersions. Similarly for closed immersions and open immersions.

**Proof.**
This follows from the general discussion in Section 100.3 and Spaces, Lemma 65.12.2.
$\square$

Lemma 100.9.4. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $W$ be an algebraic space and let $W \to \mathcal{Y}$ be a surjective, flat morphism which is locally of finite presentation. The following are equivalent:

$f$ is an (open, resp. closed) immersion, and

$V = W \times _\mathcal {Y} \mathcal{X}$ is an algebraic space, and $V \to W$ is an (open, resp. closed) immersion.

**Proof.**
This follows from the general discussion in Section 100.3 and in particular Lemmas 100.3.1 and 100.3.3.
$\square$

Lemma 100.9.5. An immersion is a monomorphism.

**Proof.**
See Morphisms of Spaces, Lemma 67.10.7.
$\square$

Lemma 100.9.6. If $f : \mathcal{X} \to \mathcal{Y}$ is an immersion, then $|f| : |\mathcal{X}| \to |\mathcal{Y}|$ is a homeomorphism onto a locally closed subset. If $f$ is a closed, resp. open immersion, then $|f|$ is closed, resp. open.

**Proof.**
Omitted.
$\square$

The following two lemmas explain how to think about immersions in terms of presentations.

Lemma 100.9.7. Let $(U, R, s, t, c)$ be a smooth groupoid in algebraic spaces. Let $i : \mathcal{Z} \to [U/R]$ be an immersion. Then there exists an $R$-invariant locally closed subspace $Z \subset U$ and a presentation $[Z/R_ Z] \to \mathcal{Z}$ where $R_ Z$ is the restriction of $R$ to $Z$ such that

\[ \xymatrix{ [Z/R_ Z] \ar[dr] \ar[rr] & & \mathcal{Z} \ar[ld]^ i \\ & [U/R] } \]

is $2$-commutative. If $i$ is a closed (resp. open) immersion then $Z$ is a closed (resp. open) subspace of $U$.

**Proof.**
By Lemma 100.3.6 we get a commutative diagram

\[ \xymatrix{ [U'/R'] \ar[dr] \ar[rr] & & \mathcal{Z} \ar[ld] \\ & [U/R] } \]

where $U' = \mathcal{Z} \times _{[U/R]} U$ and $R' = \mathcal{Z} \times _{[U/R]} R$. Since $\mathcal{Z} \to [U/R]$ is an immersion we see that $U' \to U$ is an immersion of algebraic spaces. Let $Z \subset U$ be the locally closed subspace such that $U' \to U$ factors through $Z$ and induces an isomorphism $U' \to Z$. It is clear from the construction of $R'$ that $R' = U' \times _{U, t} R = R \times _{s, U} U'$. This implies that $Z \cong U'$ is $R$-invariant and that the image of $R' \to R$ identifies $R'$ with the restriction $R_ Z = s^{-1}(Z) = t^{-1}(Z)$ of $R$ to $Z$. Hence the lemma holds.
$\square$

Lemma 100.9.8. Let $(U, R, s, t, c)$ be a smooth groupoid in algebraic spaces. Let $\mathcal{X} = [U/R]$ be the associated algebraic stack, see Algebraic Stacks, Theorem 94.17.3. Let $Z \subset U$ be an $R$-invariant locally closed subspace. Then

\[ [Z/R_ Z] \longrightarrow [U/R] \]

is an immersion of algebraic stacks, where $R_ Z$ is the restriction of $R$ to $Z$. If $Z \subset U$ is open (resp. closed) then the morphism is an open (resp. closed) immersion of algebraic stacks.

**Proof.**
Recall that by Groupoids in Spaces, Definition 78.18.1 (see also discussion following the definition) we have $R_ Z = s^{-1}(Z) = t^{-1}(Z)$ as locally closed subspaces of $R$. Hence the two morphisms $R_ Z \to Z$ are smooth as base changes of $s$ and $t$. Hence $(Z, R_ Z, s|_{R_ Z}, t|_{R_ Z}, c|_{R_ Z \times _{s, Z, t} R_ Z})$ is a smooth groupoid in algebraic spaces, and we see that $[Z/R_ Z]$ is an algebraic stack, see Algebraic Stacks, Theorem 94.17.3. The assumptions of Groupoids in Spaces, Lemma 78.25.3 are all satisfied and it follows that we have a $2$-fibre square

\[ \xymatrix{ Z \ar[d] \ar[r] & [Z/R_ Z] \ar[d] \\ U \ar[r] & [U/R] } \]

It follows from this and Lemma 100.3.1 that $[Z/R_ Z] \to [U/R]$ is representable by algebraic spaces, whereupon it follows from Lemma 100.3.3 that the right vertical arrow is an immersion (resp. closed immersion, resp. open immersion) if and only if the left vertical arrow is.
$\square$

We can define open, closed, and locally closed substacks as follows.

Definition 100.9.9. Let $\mathcal{X}$ be an algebraic stack.

An *open substack* of $\mathcal{X}$ is a strictly full subcategory $\mathcal{X}' \subset \mathcal{X}$ such that $\mathcal{X}'$ is an algebraic stack and $\mathcal{X}' \to \mathcal{X}$ is an open immersion.

A *closed substack* of $\mathcal{X}$ is a strictly full subcategory $\mathcal{X}' \subset \mathcal{X}$ such that $\mathcal{X}'$ is an algebraic stack and $\mathcal{X}' \to \mathcal{X}$ is a closed immersion.

A *locally closed substack* of $\mathcal{X}$ is a strictly full subcategory $\mathcal{X}' \subset \mathcal{X}$ such that $\mathcal{X}'$ is an algebraic stack and $\mathcal{X}' \to \mathcal{X}$ is an immersion.

This definition should be used with caution. Namely, if $f : \mathcal{X} \to \mathcal{Y}$ is an equivalence of algebraic stacks and $\mathcal{X}' \subset \mathcal{X}$ is an open substack, then it is not necessarily the case that the subcategory $f(\mathcal{X}')$ is an open substack of $\mathcal{Y}$. The problem is that it may not be a *strictly* full subcategory; but this is also the only problem. Here is a formal statement.

Lemma 100.9.10. For any immersion $i : \mathcal{Z} \to \mathcal{X}$ there exists a unique locally closed substack $\mathcal{X}' \subset \mathcal{X}$ such that $i$ factors as the composition of an equivalence $i' : \mathcal{Z} \to \mathcal{X}'$ followed by the inclusion morphism $\mathcal{X}' \to \mathcal{X}$. If $i$ is a closed (resp. open) immersion, then $\mathcal{X}'$ is a closed (resp. open) substack of $\mathcal{X}$.

**Proof.**
Omitted.
$\square$

Lemma 100.9.11. Let $[U/R] \to \mathcal{X}$ be a presentation of an algebraic stack. There is a canonical bijection

\[ \text{locally closed substacks }\mathcal{Z}\text{ of }\mathcal{X} \longrightarrow R\text{-invariant locally closed subspaces }Z\text{ of }U \]

which sends $\mathcal{Z}$ to $U \times _\mathcal {X} \mathcal{Z}$. Moreover, a morphism of algebraic stacks $f : \mathcal{Y} \to \mathcal{X}$ factors through $\mathcal{Z}$ if and only if $\mathcal{Y} \times _\mathcal {X} U \to U$ factors through $Z$. Similarly for closed substacks and open substacks.

**Proof.**
By Lemmas 100.9.7 and 100.9.8 we find that the map is a bijection. If $\mathcal{Y} \to \mathcal{X}$ factors through $\mathcal{Z}$ then of course the base change $\mathcal{Y} \times _\mathcal {X} U \to U$ factors through $Z$. Converse, suppose that $\mathcal{Y} \to \mathcal{X}$ is a morphism such that $\mathcal{Y} \times _\mathcal {X} U \to U$ factors through $Z$. We will show that for every scheme $T$ and morphism $T \to \mathcal{Y}$, given by an object $y$ of the fibre category of $\mathcal{Y}$ over $T$, the object $y$ is in fact in the fibre category of $\mathcal{Z}$ over $T$. Namely, the fibre product $T \times _\mathcal {X} U$ is an algebraic space and $T \times _\mathcal {X} U \to T$ is a surjective smooth morphism. Hence there is an fppf covering $\{ T_ i \to T\} $ such that $T_ i \to T$ factors through $T \times _\mathcal {X} U \to T$ for all $i$. Then $T_ i \to \mathcal{X}$ factors through $\mathcal{Y} \times _\mathcal {X} U$ and hence through $Z \subset U$. Thus $y|_{T_ i}$ is an object of $\mathcal{Z}$ (as $Z$ is the fibre product of $U$ with $\mathcal{Z}$ over $\mathcal{X}$). Since $\mathcal{Z}$ is a strictly full substack, we conclude that $y$ is an object of $\mathcal{Z}$ as desired.
$\square$

Lemma 100.9.12. Let $\mathcal{X}$ be an algebraic stack. The rule $\mathcal{U} \mapsto |\mathcal{U}|$ defines an inclusion preserving bijection between open substacks of $\mathcal{X}$ and open subsets of $|\mathcal{X}|$.

**Proof.**
Choose a presentation $[U/R] \to \mathcal{X}$, see Algebraic Stacks, Lemma 94.16.2. By Lemma 100.9.11 we see that open substacks correspond to $R$-invariant open subschemes of $U$. On the other hand Lemmas 100.4.5 and 100.4.7 guarantee these correspond bijectively to open subsets of $|\mathcal{X}|$.
$\square$

Lemma 100.9.13. Let $\mathcal X$ be an algebraic stack. Let $U$ be an algebraic space and $U \to \mathcal X$ a surjective smooth morphism. For an open immersion $V \hookrightarrow U$, there exists an algebraic stack $\mathcal Y$, an open immersion $\mathcal Y \to \mathcal X$, and a surjective smooth morphism $V \to \mathcal Y$.

**Proof.**
We define a category fibred in groupoids $\mathcal Y$ by letting the fiber category $\mathcal{Y}_ T$ over an object $T$ of $(\mathit{Sch}/S)_{fppf}$ be the full subcategory of $\mathcal{X}_ T$ consisting of all $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ T)$ such that the projection morphism $V \times _{\mathcal X, y} T \to T$ surjective. Now for any morphism $x : T \to \mathcal X$, the $2$-fibred product $T \times _{x, \mathcal X} \mathcal Y$ has fiber category over $T'$ consisting of triples $(f : T' \to T, y \in \mathcal{X}_{T'}, f^*x \simeq y)$ such that $V \times _{\mathcal X, y} T' \to T'$ is surjective. Note that $T \times _{x, \mathcal X} \mathcal Y$ is fibered in setoids since $\mathcal Y \to \mathcal X$ is faithful (see Stacks, Lemma 8.6.7). Now the isomorphism $f^*x \simeq y$ gives the diagram

\[ \xymatrix{ V \times _{\mathcal X, y} T' \ar[d] \ar[r] & V \times _{\mathcal X, x} T \ar[r] \ar[d] & V \ar[d] \\ T' \ar[r]^ f & T \ar[r]^ x & \mathcal X } \]

where both squares are cartesian. The morphism $V \times _{\mathcal X, x} T \to T$ is smooth by base change, and hence open. Let $T_0 \subset T$ be its image. From the cartesian squares we deduce that $V \times _{\mathcal X, y} T' \to T'$ is surjective if and only if $f$ lands in $T_0$. Therefore $T \times _{x, \mathcal X} \mathcal Y$ is representable by $T_0$, so the inclusion $\mathcal Y \to \mathcal X$ is an open immersion. By Algebraic Stacks, Lemma 94.15.5 we conclude that $\mathcal{Y}$ is an algebraic stack. Lastly if we denote the morphism $V \to \mathcal X$ by $g$, we have $V \times _{\mathcal X} V \to V$ is surjective (the diagonal gives a section). Hence $g$ is in the image of $\mathcal{Y}_ V \to \mathcal{X}_ V$, i.e., we obtain a morphism $g' : V \to \mathcal{Y}$ fitting into the commutative diagram

\[ \xymatrix{ V \ar[r] \ar[d]^{g'} & U \ar[d] \\ \mathcal{Y} \ar[r] & \mathcal{X} } \]

Since $V \times _{g, \mathcal X} \mathcal Y \to V$ is a monomorphism, it is in fact an isomorphism since $(1, g')$ defines a section. Therefore $g' : V \to \mathcal Y$ is a smooth morphism, as it is the base change of the smooth morphism $g : V \to \mathcal{X}$. It is surjective by our construction of $\mathcal{Y}$ which finishes the proof of the lemma.
$\square$

Lemma 100.9.14. Let $\mathcal X$ be an algebraic stack and $\mathcal{X}_ i \subset \mathcal X$ a collection of open substacks indexed by $i \in I$. Then there exists an open substack, which we denote $\bigcup _{i\in I} \mathcal{X}_ i \subset \mathcal X$, such that the $\mathcal{X}_ i$ are open substacks covering it.

**Proof.**
We define a fibred subcategory $\mathcal{X}' = \bigcup _{i \in I} \mathcal{X}_ i$ by letting the fiber category over an object $T$ of $(\mathit{Sch}/S)_{fppf}$ be the full subcategory of $\mathcal{X}_ T$ consisting of all $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ T)$ such that the morphism $\coprod _{i \in I} (\mathcal{X}_ i \times _{\mathcal X} T) \to T$ is surjective. Let $x_ i \in \mathop{\mathrm{Ob}}\nolimits ((\mathcal{X}_ i)_ T)$. Then $(x_ i, 1)$ gives a section of $\mathcal{X}_ i \times _{\mathcal X} T \to T$, so we have an isomorphism. Thus $\mathcal{X}_ i \subset \mathcal{X}'$ is a full subcategory. Now let $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ T)$. Then $\mathcal{X}_ i \times _{\mathcal X} T$ is representable by an open subscheme $T_ i \subset T$. The $2$-fibred product $\mathcal{X}' \times _{\mathcal X} T$ has fiber over $T'$ consisting of $(y \in \mathcal{X}_{T'}, f : T' \to T, f^*x \simeq y)$ such that $\coprod (\mathcal{X}_ i \times _{\mathcal X, y} T') \to T'$ is surjective. The isomorphism $f^*x \simeq y$ induces an isomorphism $\mathcal{X}_ i \times _{\mathcal X, y} T' \simeq T_ i \times _ T T'$. Then the $T_ i \times _ T T'$ cover $T'$ if and only if $f$ lands in $\bigcup T_ i$. Therefore we have a diagram

\[ \xymatrix{ T_ i \ar[r] \ar[d] & \bigcup T_ i \ar[r] \ar[d] & T \ar[d] \\ \mathcal{X}_ i \ar[r] & \mathcal{X}' \ar[r] & \mathcal{X} } \]

with both squares cartesian. By Algebraic Stacks, Lemma 94.15.5 we conclude that $\mathcal{X'} \subset \mathcal{X}$ is algebraic and an open substack. It is also clear from the cartesian squares above that the morphism $\coprod _{i \in I} \mathcal{X}_ i \to \mathcal{X}'$ which finishes the proof of the lemma.
$\square$

Lemma 100.9.15. Let $\mathcal X$ be an algebraic stack and $\mathcal X' \subset \mathcal X$ a quasi-compact open substack. Suppose that we have a collection of open substacks $\mathcal{X}_ i \subset \mathcal X$ indexed by $i \in I$ such that $\mathcal{X}' \subset \bigcup _{i \in I} \mathcal{X}_ i$, where we define the union as in Lemma 100.9.14. Then there exists a finite subset $I' \subset I$ such that $\mathcal{X}' \subset \bigcup _{i \in I'} \mathcal{X}_ i$.

**Proof.**
Since $\mathcal X$ is algebraic, there exists a scheme $U$ with a surjective smooth morphism $U \to \mathcal X$. Let $U_ i \subset U$ be the open subscheme representing $\mathcal{X}_ i \times _{\mathcal X} U$ and $U' \subset U$ the open subscheme representing $\mathcal{X}' \times _{\mathcal X} U$. By hypothesis, $U'\subset \bigcup _{i\in I} U_ i$. From the proof of Lemma 100.6.2, there is a quasi-compact open $V \subset U'$ such that $V \to \mathcal{X}'$ is a surjective smooth morphism. Therefore there exists a finite subset $I' \subset I$ such that $V \subset \bigcup _{i \in I'} U_ i$. We claim that $\mathcal{X}' \subset \bigcup _{i \in I'} \mathcal{X}_ i$. Take $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}'_ T)$ for $T \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$. Since $\mathcal{X}' \to \mathcal{X}$ is a monomorphism, we have cartesian squares

\[ \xymatrix{ V \times _\mathcal {X} T \ar[r] \ar[d] & T \ar[d]^ x \ar@{=}[r] & T \ar[d]^ x \\ V \ar[r] & \mathcal{X}' \ar[r] & \mathcal X } \]

By base change, $V \times _{\mathcal X} T \to T$ is surjective. Therefore $\bigcup _{i \in I'} U_ i \times _{\mathcal X} T \to T$ is also surjective. Let $T_ i \subset T$ be the open subscheme representing $\mathcal{X}_ i \times _{\mathcal X} T$. By a formal argument, we have a Cartesian square

\[ \xymatrix{ U_ i \times _{\mathcal{X}_ i} T_ i \ar[r] \ar[d] & U \times _{\mathcal X} T \ar[d] \\ T_ i \ar[r] & T } \]

where the vertical arrows are surjective by base change. Since $U_ i \times _{\mathcal{X}_ i} T_ i \simeq U_ i \times _{\mathcal X} T$, we find that $\bigcup _{i \in I'} T_ i = T$. Hence $x$ is an object of $(\bigcup _{i\in I'} \mathcal{X}_ i)_ T$ by definition of the union. Observe that the inclusion $\mathcal{X}' \subset \bigcup _{i \in I'} \mathcal{X}_ i$ is automatically an open substack.
$\square$

Lemma 100.9.16. Let $\mathcal X$ be an algebraic stack. Let $\mathcal{X}_ i$, $i \in I$ be a set of open substacks of $\mathcal{X}$. Assume

$\mathcal{X} = \bigcup _{i \in I} \mathcal{X}_ i$, and

each $\mathcal{X}_ i$ is an algebraic space.

Then $\mathcal{X}$ is an algebraic space.

**Proof.**
Apply Stacks, Lemma 8.6.10 to the morphism $\coprod _{i \in I} \mathcal{X}_ i \to \mathcal{X}$ and the morphism $\text{id} : \mathcal{X} \to \mathcal{X}$ to see that $\mathcal{X}$ is a stack in setoids. Hence $\mathcal{X}$ is an algebraic space, see Algebraic Stacks, Proposition 94.13.3.
$\square$

Lemma 100.9.17. Let $\mathcal X$ be an algebraic stack. Let $\mathcal{X}_ i$, $i \in I$ be a set of open substacks of $\mathcal{X}$. Assume

$\mathcal{X} = \bigcup _{i \in I} \mathcal{X}_ i$, and

each $\mathcal{X}_ i$ is a scheme

Then $\mathcal{X}$ is a scheme.

**Proof.**
By Lemma 100.9.16 we see that $\mathcal{X}$ is an algebraic space. Since any algebraic space has a largest open subspace which is a scheme, see Properties of Spaces, Lemma 66.13.1 we see that $\mathcal{X}$ is a scheme.
$\square$

The following lemma is the analogue of More on Groupoids, Lemma 40.6.1.

Lemma 100.9.18. Let $\mathcal{P}, \mathcal{Q}, \mathcal{R}$ be properties of morphisms of algebraic spaces. Assume

$\mathcal{P}, \mathcal{Q}, \mathcal{R}$ are fppf local on the target and stable under arbitrary base change,

$\text{smooth} \Rightarrow \mathcal{R}$,

for any morphism $f : X \to Y$ which has $\mathcal{Q}$ there exists a largest open subspace $W(\mathcal{P}, f) \subset X$ such that $f|_{W(\mathcal{P}, f)}$ has $\mathcal{P}$, and

for any morphism $f : X \to Y$ which has $\mathcal{Q}$, and any morphism $Y' \to Y$ which has $\mathcal{R}$ we have $Y' \times _ Y W(\mathcal{P}, f) = W(\mathcal{P}, f')$, where $f' : X_{Y'} \to Y'$ is the base change of $f$.

Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks representable by algebraic spaces. Assume $f$ has $\mathcal{Q}$. Then

there exists a largest open substack $\mathcal{X}' \subset \mathcal{X}$ such that $f|_{\mathcal{X}'}$ has $\mathcal{P}$, and

if $\mathcal{Z} \to \mathcal{Y}$ is a morphism of algebraic stacks representable by algebraic spaces which has $\mathcal{R}$ then $\mathcal{Z} \times _\mathcal {Y} \mathcal{X}'$ is the largest open substack of $\mathcal{Z} \times _\mathcal {Y} \mathcal{X}$ over which the base change $\text{id}_\mathcal {Z} \times f$ has property $\mathcal{P}$.

**Proof.**
Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Set $U = V \times _\mathcal {Y} \mathcal{X}$ and let $f' : U \to V$ be the base change of $f$. The morphism of algebraic spaces $f' : U \to V$ has property $\mathcal{Q}$. Thus we obtain the open $W(\mathcal{P}, f') \subset U$ by assumption (3). Note that $U \times _\mathcal {X} U = (V \times _\mathcal {Y} V) \times _\mathcal {Y} \mathcal{X}$ hence the morphism $f'' : U \times _\mathcal {X} U \to V \times _\mathcal {Y} V$ is the base change of $f$ via either projection $V \times _\mathcal {Y} V \to V$. By our choice of $V$ these projections are smooth, hence have property $\mathcal{R}$ by (2). Thus by (4) we see that the inverse images of $W(\mathcal{P}, f')$ under the two projections $\text{pr}_ i : U \times _\mathcal {X} U \to U$ agree. In other words, $W(\mathcal{P}, f')$ is an $R$-invariant subspace of $U$ (where $R = U \times _\mathcal {X} U$). Let $\mathcal{X}'$ be the open substack of $\mathcal{X}$ corresponding to $W(\mathcal{P}, f)$ via Lemma 100.9.7. By construction $W(\mathcal{P}, f') = \mathcal{X}' \times _\mathcal {Y} V$ hence $f|_{\mathcal{X}'}$ has property $\mathcal{P}$ by Lemma 100.3.3. Also, $\mathcal{X}'$ is the largest open substack such that $f|_{\mathcal{X}'}$ has $\mathcal{P}$ as the same maximality holds for $W(\mathcal{P}, f)$. This proves (A).

Finally, if $\mathcal{Z} \to \mathcal{Y}$ is a morphism of algebraic stacks representable by algebraic spaces which has $\mathcal{R}$ then we set $T = V \times _\mathcal {Y} \mathcal{Z}$ and we see that $T \to V$ is a morphism of algebraic spaces having property $\mathcal{R}$. Set $f'_ T : T \times _ V U \to T$ the base change of $f'$. By (4) again we see that $W(\mathcal{P}, f'_ T)$ is the inverse image of $W(\mathcal{P}, f)$ in $T \times _ V U$. This implies (B); some details omitted.
$\square$

## Comments (2)

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