The Stacks project

76.49 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 76.49.1. 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 67.38.10 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 66.4.8) 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 66.4.9) that $W$ has the desired universal property. $\square$

Lemma 76.49.2. 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 76.49.1 we may assume that $f$ is already unramified; note that this does not destroy assumption (2) or (3). By Morphisms of Spaces, Lemma 67.38.9 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 67.19.2 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 66.4.8) 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 66.4.9. 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 66.4.3 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 76.49.3. 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 76.14.9. First, note that since $p$ is universally closed and $q$ is separated, we see that $f$ is universally closed, see Morphisms of Spaces, Lemma 67.40.6. It follows that any base change of $f$ is universally closed, see Morphisms of Spaces, Lemma 67.9.3. Thus to finish the proof of the lemma it suffices to prove that the assumptions of Lemma 76.49.2 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 67.9.3. Hence $X \times _ Y X \to Z$ is universally closed as a composition of universally closed morphisms (see Morphisms of Spaces, Lemma 67.9.4). This finishes the proof of the lemma. $\square$

Lemma 76.49.4. 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 76.23.6 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 66.4.8) 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 76.49.5. 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 76.49.4 we may assume that $f$ is flat. Note that $f$ is locally of finite presentation by Morphisms of Spaces, Lemma 67.28.9. Hence $f$ is open, see Morphisms of Spaces, Lemma 67.30.6. Let $W \subset Z$ be the open subspace (see Properties of Spaces, Lemma 66.4.8) 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 66.4.9. 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 66.4.3 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 76.49.6. 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 76.49.5 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 76.49.3 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 76.49.7. 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 76.48.11 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 66.4.8) 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 66.4.9) that $W$ has the desired universal property. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 05X7. Beware of the difference between the letter 'O' and the digit '0'.



View Section 76.49 as pdf