The intersection of $\mathcal{O}_X$-submodules is an intersection as the categorical sense?

Question

I have simple question in studying algebraic geometry.

First, next is the definiiton in the Gortz's Algebraic geometry book (p.174)

Let $\mathcal{F}$ be an $\mathcal{O}_X$-module and let $(\mathcal{F}_i)_{i\in I}$ be a family of $\mathcal{O}_X$-submodules.

The intersection $\cap_{i\in I}\mathcal{F}_i$ of the family $(\mathcal{F}_i)$ is the $\mathcal{O}_X$-submodule of $\mathcal{F}$ defined as the kernel of the canonical homomorphism

$$ \mathcal{F} \to \prod_{i\in I}\mathcal{F}/\mathcal{F}_i .$$

It is the sheaf $U \mapsto \cap_i\mathcal{F}_i(U)$.

Second, next is the definiton in the book, Adamek, Herrlich, 'Abstract and Concrete Categories'( p.201, 11.23 Definition. )

Let $\mathcal{A}$ be a family of subobjects $(A_i, m_i)$ of an object $B$ ( i.e., each $m_i : A_i \to B$ is monomorphism), indexed by a class $I$. A subobject $(A,m)$ of $B$ is called an intersection of $\mathcal{A}$ provided that the following two conditions are satisfied :

(1) $m$ factors through each $m_i$ ; i.e., for each $i$ there exists and $f_i$ with $m= m_i \circ f_i$,

(2) if a morphism $f:C\to B$ factors through each $m_i$, then it factors through $m$

My question is, the intersection $\cap_{i\in I}\mathcal{F}_i$ as the first definition is an intersection as in the second definition?

First, let's consider the simplest case : first consider the case that the index set $I$ has only two element. Then $(\mathcal{F}_i)_{i\in I} = \{\mathcal{F}_1 , \mathcal{F}_2\}$. In the Adameck's book, p.207, exercise 11F, for two subobjects, the intersection is equivalent to pullback square. So, my question is, the next diagram

$$\require{AMScd} \begin{CD} \mathcal{F}_1 \cap \mathcal{F}_2 @>>> \mathcal{F}_1\\ @VVV @VVV \\ \mathcal{F}_2 @>>> \mathcal{F} \end{CD}$$

, is pullback square ?

EDIT : Through consideration, this question seems to be true ; Refer to Edit below.

My first attempt : In the Adamek's book, there is an assoicated problem ( his book, p.208, 11L .) :

11L. Multiple pullbacks A pair $(f, \mathcal{S})$, consisting of a morphism $f:A\to B$ and a source $\mathcal{S} = (f_i : A \to A_i)_{I}$, is called a multiple pullback of a sink $(g_i : A_i \to B)_I$ provided that

(1) $f= g_i \circ f_i$ for each $i\in I$, and

(2) for each pair $(f', \mathcal{S}')$, with $f':A'\to B$ a morphism and $S'=(f_i' : A' \to A_i)_I$ a source for which $f'=g_i \circ f'_i$ for each $i\in I$, there exists a unique morphism $g:A'\to A$ with $f'=f\circ g$ and $f_i'=f_i\circ g$ for each $i\in I$.

(c) Interpret intersections as multiple pullbacks. In particular, show that whenever $\mathcal{R}=((A_i,m_i))_I$ is a family of subobjects of $B$ and $f:A\to B$ is a morphism, then the following are equivalent :

(1) $(A,f)$ is an intersection of $\mathcal{R}$

(2) there exists a (unique) source $\mathcal{S}$ such that $(f,\mathcal{S})$ is a multiple pullback of the sink $ \mathcal{R}=(m_i : A_i \to B)_{i \in I}$.

Now let $f : A:= \operatorname{ker} ( \pi : \mathcal{F} \to \prod_{i\in I} \mathcal{F}/\mathcal{F}_i) \hookrightarrow B:=\mathcal{F}$ be the inclusion, let $f_i : A:=\operatorname{ker} ( \pi : \mathcal{F} \to \prod_{i\in I} \mathcal{F}/\mathcal{F}_i) \hookrightarrow A_i:=\mathcal{F}_i$ be the inclusion ( does this exist ?), and let $m_i : \mathcal{F}_i \hookrightarrow \mathcal{F}$ also be the inclusion.

Then note that $m_i \circ f_i = f$ for each $i\in I$. So, by the above criteria, to show our goal ( i.e. $(A,f)$ is an intersection of the $\mathcal{R}:=( m_i : \mathcal{F}_i \hookrightarrow \mathcal{F})_I$ ), it suffices to show that $(f, \mathcal{S}:=( f_i : A \to A_i:=\mathcal{F}_i)_I)$ is a multiple pullback of the $\mathcal{R}$.

So, let $(f' , \mathcal{S}')$ be a pair such that $f':A'\to B:=\mathcal{F}$ is a morphism and $S'=(f_i' : A' \to A_i:=\mathcal{F}_i)_I$ a source for which $f'=m_i \circ f'_i$ for each $i\in I$.

It remains to show that "there exists a unique morphism $g:A'\to A:=\operatorname{ker} ( \pi : \mathcal{F} \to \prod_{i\in I} \mathcal{F}/\mathcal{F}_i)$ with $f'=f\circ g$ and $f_i'=f_i\circ g$ for each $i\in I$."

• 1. First, note that $\pi \circ f' = \pi \circ ( m_i \circ f'_i) = (\pi \circ m_i) \circ f'_i = 0 \circ f_i' = 0$ (True? : $\pi \circ m_i : \mathcal{F}_i \hookrightarrow \mathcal{F} \to \prod_{i\in I}\mathcal{F}/\mathcal{F}_i$ is really zero morphism? )

So by the universal property of the kernel, there exists a unique morphim $g: A' \to A:=\operatorname{ker}\pi$ such that $f' = f\circ g$.

• 2. Second, note that from $f \circ g = f'$,

$$m_i \circ f'_i = f' = f \circ g = (m_i \circ f_i) \circ g = m_i \circ ( f_i \circ g ) .$$

So, since each $m_i$ is monomorphism, we have $f'_i = f_i \circ g$ and we are done(?).

Correct? EDIT : Bad news. Through communicating with
Daniel Schepler below, $\pi \circ m_i : \mathcal{F}_i \hookrightarrow \mathcal{F} \to \prod_{i\in I}\mathcal{F}/\mathcal{F}_i$ may not be zero morphism.

In fact, this question orginates from the following question :

Q. In the Gortz's Algebraic geometry book, p.373, remark 13.14., he noted that there is an exact covariant functor $M \mapsto \tilde{M}$ from the category of graded $A$-modules to the category of quasi-coherent $\mathcal{O}_{\operatorname{proj}A}$-modules which commutes with direct sums, filtered inductive limits, and tensor products.

My question is, let $N_1 , N_2 \subseteq M$ be graded $A$-submodules. Then $$ \widetilde{N_1 \cap N_2} \cong \tilde{N_1} \cap \tilde{N_2} $$?

If we can view $N_1 \cap N_2$ as the pull back of $N_1 , N_2 \subseteq M$, then since $M \mapsto \tilde{M}$ is exact functor, it preserves pullback (?) and so $\widetilde{N_1 \cap N_2}$ is a pullback of $\tilde{N_1} \to \tilde{M}$ and $\tilde{N_2} \to \tilde{M}$. So, if our question is true, then by the uniqueness of pullback, $ \widetilde{N_1 \cap N_2} \cong \tilde{N_1} \cap \tilde{N_2} $

EDIT : I think that I understand this original question. That is, as I asked in the above question (simplest case), next diagram

$$\require{AMScd} \begin{CD} \mathcal{F}_1 \cap \mathcal{F}_2 @>>> \mathcal{F}_1\\ @VVV @VVV \\ \mathcal{F}_2 @>>> \mathcal{F} \end{CD}$$

, is pullback square. In the Gortz's Algebraic Geometry book, p.206, Example 8.3., he wrote that "In the Category $\hat{\mathcal{C}}$ fiber product always exist" : Let $F$, $G$, $H$ be contravariant functors from $\mathcal{C}$ to $(\operatorname{Sets})$, and let $F \to H$ and $G\to H$ be a morphism of functors. We set for every object $T\in \mathcal{C}$

$$ (F\times_H G)(T):=F(T) \times_{H(T)} G(T),$$ where the right hand side denotes the fiber product in the category of sets (Example 4.12). Then $(F\times_H G) \in \hat{\mathcal{C}}$ and $(F\times_H G)$ forms a fibered product.

Now I show that $\mathcal{F}_1 \cap \mathcal{F}_2 \cong \mathcal{F}_1 \times_{\mathcal{F}} \mathcal{F}_2$. Let $U \subseteq X$ be an open set. Then

$$ \mathcal{F}_1 \times_{\mathcal{F}} \mathcal{F}_2(U):= \mathcal{F}_1 (U) \times_{\mathcal{F}(u)} \mathcal{F}_2(U) \cong \mathcal{F}_1(U) \cap \mathcal{F}_2(U) = ( \mathcal{F}_1 \cap \mathcal{F}_2 )(U). $$

(This is functorial isomorphism? We used that each $\mathcal{F}_1 \to \mathcal{F}$ , $\mathcal{F}_2 \to \mathcal{F}$ are the inclusions. And a presheaf which is isomorphic to a sheaf is also a sheaf) So we are done ( $\mathcal{F}_1 \cap \mathcal{F}_2 \cong \mathcal{F}_1 \times_{\mathcal{F}} \mathcal{F}_2$ )?

And, how about the general case, rather than the simplest case? When the $I$ is finite, this question seems to be true(?). But when $I$ is possibly infinite set? This question remains as curiosty !

Thanks for reading.

Can anyone helps?

Answer 1

I haven't read your (long) question carefully, but here is how you can identify the intersection with multiple pullbacks:

Let me first describe an abstract setup. Start with a category $\mathcal{C}$ which admits limits. Let $I$ be an index set and $(A_i)_{i\in I}$, $(B_i)_{i\in I}$ two families of objects of $\mathcal{C}$ together with morphisms $f_i\colon A_i\to B_i$ for all $i$. Let $C \in \mathcal{C}$ together with morphisms $g_i\colon C\to B_i$ for all $i$.

Lemma: The pullback $(\prod_i A_i)\times_{\prod_i B_i} C$ identifies with the multiple pullback of $(A_i\times_{B_i}C\to C)_{i\in I}$.
Proof: This is just a matter of verifying that both satisfy the same universal property. A map $D\to (\prod_iA_i)\times_{\prod_i B_i}C$ consists of morphisms $a_i\colon D\to A_i$ and $c\colon D\to C$ such that $$ f_i\circ a_i = g_i\circ c, \qquad\text{for all $i$.} $$ But clearly this is the same datum as giving maps $\alpha_i\colon D\to A_i\times_{B_i}C$ such that $$\pi^i_1\circ\alpha_i = a_i,\qquad \text{and}\qquad \pi^i_2\circ\alpha_i = c,\qquad \text{for all $i$,} $$ where $\pi^i_1\colon A_i\times_{B_i}C \to A_i$ and $\pi^i_2\colon A_i\times_{B_i}C \to C$ denote the two projections. Here, the equalities $\pi_2^i\circ \alpha_i = c$ mean that we can view the $\alpha_i$ as morphisms over $C$. Hence, we equivalently obtain a morphism from $D$ into the multiple pullback of $(A_i\times_{B_i}C\to C)_i$. $\quad\square$

Now, we apply the lemma with $A_i = \{0\}$, $B_i = \mathcal{F}/\mathcal{F}_i$ and $C = \mathcal{F}$. Then $\prod_iA_i = \{0\}$, and hence $$ \prod_i A_i \times_{\prod_i B_i} C = \{0\} \times_{\prod_i \mathcal{F}/\mathcal{F}_i} \mathcal{F} = \ker\Bigl(\mathcal{F} \to \prod_i \mathcal{F}/\mathcal{F}_i\Bigr) =: \bigcap_{i\in I} \mathcal{F}_i. $$ On the other hand, we have $$ A_i\times_{B_i}C = \{0\} \times_{\mathcal{F}/\mathcal{F}_i}\mathcal{F} = \ker\bigl(\mathcal{F} \to \mathcal{F}/\mathcal{F}_i\bigr) = \mathcal{F}_i, $$ and hence the lemma shows that $\bigcap_i \mathcal{F}_i$ identifies with the multiple pullback of $(\mathcal{F}_i \to \mathcal{F})_i$.