The Stacks project

75.49 When is a morphism an isomorphism?

More generally we can ask: “When does a morphism have property $\mathcal{P}$?” A more precise question is the following. Suppose given a commutative diagram

\[ \xymatrix{ X \ar[rr]_ f \ar[rd]_ p & & Y \ar[ld]^ q \\ & Z } \]

of algebraic spaces. Does there exist a monomorphism of algebraic spaces $W \to Z$ with the following two properties:

  1. the base change $f_ W : X_ W \to Y_ W$ has property $\mathcal{P}$, and

  2. any morphism $Z' \to Z$ of algebraic spaces factors through $W$ if and only if the base change $f_{Z'} : X_{Z'} \to Y_{Z'}$ has property $\mathcal{P}$.

In many cases, if $W \to Z$ exists, then it is an immersion, open immersion, or closed immersion.

The answer to this question may depend on auxiliary properties of the morphisms $f$, $p$, and $q$. An example is $\mathcal{P}(f) =$“$f$ is flat” which we have discussed for morphisms of schemes in the case $Y = S$ in great detail in the chapter “More on Flatness”, starting with More on Flatness, Section 38.20.

Lemma 75.49.1. Consider a commutative diagram

\[ \xymatrix{ X \ar[rr]_ f \ar[rd]_ p & & Y \ar[ld]^ q \\ & Z } \]

of algebraic spaces. Assume that $p$ is locally of finite type and closed. Then there exists an open subspace $W \subset Z$ such that a morphism $Z' \to Z$ factors through $W$ if and only if the base change $f_{Z'} : X_{Z'} \to Y_{Z'}$ is unramified.

Proof. By Morphisms of Spaces, Lemma 66.38.10 there exists an open subspace $U(f) \subset X$ which is the set of points where $f$ is unramified. Moreover, formation of $U(f)$ commutes with arbitrary base change. Let $W \subset Z$ be the open subspace (see Properties of Spaces, Lemma 65.4.8) with underlying set of points

\[ |W| = |Z| \setminus |p|\left(|X| \setminus |U(f)|\right) \]

i.e., $z \in |Z|$ is a point of $W$ if and only if $f$ is unramified at every point of $X$ above $z$. Note that this is open because we assumed that $p$ is closed. Since the formation of $U(f)$ commutes with arbitrary base change we immediately see (using Properties of Spaces, Lemma 65.4.9) that $W$ has the desired universal property. $\square$

Lemma 75.49.2. Consider a commutative diagram

\[ \xymatrix{ X \ar[rr]_ f \ar[rd]_ p & & Y \ar[ld]^ q \\ & Z } \]

of algebraic spaces. Assume that

  1. $p$ is locally of finite type,

  2. $p$ is closed, and

  3. $p_2 : X \times _ Y X \to Z$ is closed.

Then there exists an open subspace $W \subset Z$ such that a morphism $Z' \to Z$ factors through $W$ if and only if the base change $f_{Z'} : X_{Z'} \to Y_{Z'}$ is unramified and universally injective.

Proof. After replacing $Z$ by the open subspace found in Lemma 75.49.1 we may assume that $f$ is already unramified; note that this does not destroy assumption (2) or (3). By Morphisms of Spaces, Lemma 66.38.9 we see that $\Delta _{X/Y} : X \to X \times _ Y X$ is an open immersion. This remains true after any base change. Hence by Morphisms of Spaces, Lemma 66.19.2 we see that $f_{Z'}$ is universally injective if and only if the base change of the diagonal $X_{Z'} \to (X \times _ Y X)_{Z'}$ is an isomorphism. Let $W \subset Z$ be the open subspace (see Properties of Spaces, Lemma 65.4.8) with underlying set of points

\[ |W| = |Z| \setminus |p_2|\left(|X \times _ Y X| \setminus \mathop{\mathrm{Im}}(|\Delta _{X/Y}|)\right) \]

i.e., $z \in |Z|$ is a point of $W$ if and only if the fibre of $|X \times _ Y X| \to |Z|$ over $z$ is in the image of $|X| \to |X \times _ Y X|$. Then it is clear from the discussion above that the restriction $p^{-1}(W) \to q^{-1}(W)$ of $f$ is unramified and universally injective.

Conversely, suppose that $f_{Z'}$ is unramified and universally injective. In order to show that $Z' \to Z$ factors through $W$ it suffices to show that $|Z'| \to |Z|$ has image contained in $|W|$, see Properties of Spaces, Lemma 65.4.9. Hence it suffices to prove the result when $Z'$ is the spectrum of a field. Denote $z \in |Z|$ the image of $|Z'| \to |Z|$. The discussion above shows that

\[ |X_{Z'}| \longrightarrow |(X \times _ Y X)_{Z'}| \]

is surjective. By Properties of Spaces, Lemma 65.4.3 in the commutative diagram

\[ \xymatrix{ |X_{Z'}| \ar[d] \ar[r] & |(X \times _ Y X)_{Z'}| \ar[d] \\ |p|^{-1}(\{ z\} ) \ar[r] & |p_2|^{-1}(\{ z\} ) } \]

the vertical arrows are surjective. It follows that $z \in |W|$ as desired. $\square$

Lemma 75.49.3. Consider a commutative diagram

\[ \xymatrix{ X \ar[rr]_ f \ar[rd]_ p & & Y \ar[ld]^ q \\ & Z } \]

of algebraic spaces. Assume that

  1. $p$ is locally of finite type,

  2. $p$ is universally closed, and

  3. $q : Y \to Z$ is separated.

Then there exists an open subspace $W \subset Z$ such that a morphism $Z' \to Z$ factors through $W$ if and only if the base change $f_{Z'} : X_{Z'} \to Y_{Z'}$ is a closed immersion.

Proof. We will use the characterization of closed immersions as universally closed, unramified, and universally injective morphisms, see Lemma 75.14.9. First, note that since $p$ is universally closed and $q$ is separated, we see that $f$ is universally closed, see Morphisms of Spaces, Lemma 66.40.6. It follows that any base change of $f$ is universally closed, see Morphisms of Spaces, Lemma 66.9.3. Thus to finish the proof of the lemma it suffices to prove that the assumptions of Lemma 75.49.2 are satisfied. The projection $\text{pr}_0 : X \times _ Y X \to X$ is universally closed as a base change of $f$, see Morphisms of Spaces, Lemma 66.9.3. Hence $X \times _ Y X \to Z$ is universally closed as a composition of universally closed morphisms (see Morphisms of Spaces, Lemma 66.9.4). This finishes the proof of the lemma. $\square$

Lemma 75.49.4. Consider a commutative diagram

\[ \xymatrix{ X \ar[rr]_ f \ar[rd]_ p & & Y \ar[ld]^ q \\ & Z } \]

of algebraic spaces. Assume that

  1. $p$ is locally of finite presentation,

  2. $p$ is flat,

  3. $p$ is closed, and

  4. $q$ is locally of finite type.

Then there exists an open subspace $W \subset Z$ such that a morphism $Z' \to Z$ factors through $W$ if and only if the base change $f_{Z'} : X_{Z'} \to Y_{Z'}$ is flat.

Proof. By Lemma 75.23.6 the set

\[ A = \{ x \in |X| : X\text{ flat at }x \text{ over }Y\} . \]

is open in $|X|$ and its formation commutes with arbitrary base change. Let $W \subset Z$ be the open subspace (see Properties of Spaces, Lemma 65.4.8) with underlying set of points

\[ |W| = |Z| \setminus |p|\left(|X| \setminus A\right) \]

i.e., $z \in |Z|$ is a point of $W$ if and only if the whole fibre of $|X| \to |Z|$ over $z$ is contained in $A$. This is open because $p$ is closed. Since the formation of $A$ commutes with arbitrary base change it follows that $W$ works. $\square$

Lemma 75.49.5. Consider a commutative diagram

\[ \xymatrix{ X \ar[rr]_ f \ar[rd]_ p & & Y \ar[ld]^ q \\ & Z } \]

of algebraic spaces. Assume that

  1. $p$ is locally of finite presentation,

  2. $p$ is flat,

  3. $p$ is closed,

  4. $q$ is locally of finite type, and

  5. $q$ is closed.

Then there exists an open subspace $W \subset Z$ such that a morphism $Z' \to Z$ factors through $W$ if and only if the base change $f_{Z'} : X_{Z'} \to Y_{Z'}$ is surjective and flat.

Proof. By Lemma 75.49.4 we may assume that $f$ is flat. Note that $f$ is locally of finite presentation by Morphisms of Spaces, Lemma 66.28.9. Hence $f$ is open, see Morphisms of Spaces, Lemma 66.30.6. Let $W \subset Z$ be the open subspace (see Properties of Spaces, Lemma 65.4.8) with underlying set of points

\[ |W| = |Z| \setminus |q|\left(|Y| \setminus |f|(|X|)\right). \]

in other words for $z \in |Z|$ we have $z \in |W|$ if and only if the whole fibre of $|Y| \to |Z|$ over $z$ is in the image of $|X| \to |Y|$. Since $q$ is closed this set is open in $|Z|$. The morphism $X_ W \to Y_ W$ is surjective by construction. Finally, suppose that $X_{Z'} \to Y_{Z'}$ is surjective. In order to show that $Z' \to Z$ factors through $W$ it suffices to show that $|Z'| \to |Z|$ has image contained in $|W|$, see Properties of Spaces, Lemma 65.4.9. Hence it suffices to prove the result when $Z'$ is the spectrum of a field. Denote $z \in |Z|$ the image of $|Z'| \to |Z|$. By Properties of Spaces, Lemma 65.4.3 in the commutative diagram

\[ \xymatrix{ |X_{Z'}| \ar[d] \ar[r] & |Y_{Z'}| \ar[d] \\ |p|^{-1}(\{ z\} ) \ar[r] & |q|^{-1}(\{ z\} ) } \]

the vertical arrows are surjective. It follows that $z \in |W|$ as desired. $\square$

Lemma 75.49.6. Consider a commutative diagram

\[ \xymatrix{ X \ar[rr]_ f \ar[rd]_ p & & Y \ar[ld]^ q \\ & Z } \]

of algebraic spaces. Assume that

  1. $p$ is locally of finite presentation,

  2. $p$ is flat,

  3. $p$ is universally closed,

  4. $q$ is locally of finite type,

  5. $q$ is closed, and

  6. $q$ is separated.

Then there exists an open subspace $W \subset Z$ such that a morphism $Z' \to Z$ factors through $W$ if and only if the base change $f_{Z'} : X_{Z'} \to Y_{Z'}$ is an isomorphism.

Proof. By Lemma 75.49.5 there exists an open subspace $W_1 \subset Z$ such that $f_{Z'}$ is surjective and flat if and only if $Z' \to Z$ factors through $W_1$. By Lemma 75.49.3 there exists an open subspace $W_2 \subset Z$ such that $f_{Z'}$ is a closed immersion if and only if $Z' \to Z$ factors through $W_2$. We claim that $W = W_1 \cap W_2$ works. Certainly, if $f_{Z'}$ is an isomorphism, then $Z' \to Z$ factors through $W$. Hence it suffices to show that $f_ W$ is an isomorphism. By construction $f_ W$ is a surjective flat closed immersion. In particular $f_ W$ is representable. Since a surjective flat closed immersion of schemes is an isomorphism (see Morphisms, Lemma 29.26.1) we win. (Note that actually $f_ W$ is locally of finite presentation, whence open, so you can avoid the use of this lemma if you like.) $\square$

Lemma 75.49.7. Consider a commutative diagram

\[ \xymatrix{ X \ar[rr]_ f \ar[rd]_ p & & Y \ar[ld]^ q \\ & Z } \]

of algebraic spaces. Assume that

  1. $p$ is flat and locally of finite presentation,

  2. $p$ is closed, and

  3. $q$ is flat and locally of finite presentation,

Then there exists an open subspace $W \subset Z$ such that a morphism $Z' \to Z$ factors through $W$ if and only if the base change $f_{Z'} : X_{Z'} \to Y_{Z'}$ is a local complete intersection morphism.

Proof. By Lemma 75.48.11 there exists an open subspace $U(f) \subset X$ which is the set of points where $f$ is Koszul. Moreover, formation of $U(f)$ commutes with arbitrary base change. Let $W \subset Z$ be the open subspace (see Properties of Spaces, Lemma 65.4.8) with underlying set of points

\[ |W| = |Z| \setminus |p|\left(|X| \setminus |U(f)|\right) \]

i.e., $z \in |Z|$ is a point of $W$ if and only if $f$ is Koszul at every point of $X$ above $z$. Note that this is open because we assumed that $p$ is closed. Since the formation of $U(f)$ commutes with arbitrary base change we immediately see (using Properties of Spaces, Lemma 65.4.9) that $W$ has the desired universal property. $\square$


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