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I'm trying to train myself in the category theory way of thinking, but I'm running into a wall translating a rather easy result into the abstract category setting.

Part 1: the setup

The result (and all my confusion) takes place in a category $\mathcal{C}$ with two properties:

Property 1: direct products exist

Property 2: Intersections of subobjects of an object $X$ in $\mathcal{C}$ exist as an object in $\mathcal{C}$ and in particular as a subobject of $X$.

These are not outrageous demands: the categories of sets, groups, vectorspaces, topological spaces etc all satisfy this demand. (Assuming the empty set counts a a topological space.)

In these examples it is also very clear what we mean by a subobject and an intersection: a subobject is a subset that is also an object and an intersection is just an intersection.

In an abstract category where the objects are not necessarily sets however we need to take some care to define these terms. I will do that first, and then get to my actual question. I am pretty sure that I got this part right and that what I write here is completely standard, but I write it out anyway so that we are all on the same page. (And of course: if my translation of the notions of 'subobject' and 'intersection' into category language is actually wrong, then that might be the source of the confusion and I would be happy to hear it!)

Definition 1: By a subobject of an object $X \in \mathcal{C}$ we mean an equivalence class of monomorphism into $X$, where $m_1, m_2 \colon Z \to X$ are equivalent when there exists an invertible automorphsim $f$ of $Z$ such that $m_1 = m_2 \circ f$, $m_2 = m_1 \circ f^{-1}$.

Definition 2: The set of all monomorphsims into $X$ is partially ordered by the ordering '$\subseteq$' which we define by $$i \subseteq j \textrm{ iff there exists a monomoprhism } k \textrm{ such that } i = j \circ k.$$

(This feels strange, talking about morphisms rather than objects, but hey, that's category theory.)

This ordering descents nicely to equivalence classes and the reason to work with equivalence classes rather than actual monomorphisms is that on the equivalence class level we have that $i = j$ if and only if $i \subseteq j$ and $j \subseteq i$. (That said I will write as if I am just talking about ordinary monomorphsims rather than classes.)

Now we can reformulate the properties we assume that $\mathcal{C}$ has:

Property 1: Direct products exist

Property 2: for every non-empty collection of subobjects of a given object $X \in \mathcal{C}$, there exists a unique greatest common lower bound for the subobjects in the collection w.r.t. to the $\subseteq$ ordering.

And with property 2 under our belt we can clean up the language a bit by writing:

Definition 3: This greatest common lower bound of a collection of subobjects will be called the intersection of these subobjects

I believe this perfectly catches in the abstract setting what Property 1 and 2 were saying above. (But please correct me if I'm wrong.)

Part 2: the actual question

We need one more definition, this time not standard but of my own making. (If this concept already has a name, please let me know!)

Definition 4: Let $X, Y, Z$ be objects in $\mathcal{C}$ and let $i \colon Z \to X \times Y$ be a monomorphism, so that (formally) $i$ or (informally) $Z$ is a subobject of $X \times Y$. A $Z$-box is a monomorphism $f$ into $X \times Y$ satisfying

  • $f$ can be written as $f_1 \times f_2 \colon X' \times Y' \to X \times Y$ for monomorphisms $f_1 \colon X' \to X$ and $f_2 \colon Y' \to Y$
  • There exist a monomorphism $j \colon Z \to X' \times Y'$ such that $i = f \circ j$.

So in other words: a $Z$-box is a subobject of $X \times Y$ that contains the subobject $Z$ but (possibly unlike $Z$) can be written itself as a direct product of subobjects of $X, Y$ respectively.

The following result is not hard to show. (In fact I typed two proofs and then removed them because the question had become too long, they can be read by looking at the revision history of this post.)

Theorem 1: let $\mathcal{C}$ be a concrete category possessing properties 1 and 2 above and let $X, Y, Z$, $i \colon Z \to X \times Y$ be as in Definition 4. Then the intersection of all $Z$-boxes is a $Z$-box itself.

Here by 'concrete' I mean that all objects in $\mathcal{C}$ are also sets, the underlying set of the direct product is the cartestian product and subobjects are just subsets that are also objects. I believe 'concrete' is the standard term for this.

Now my question is this:

Question 1: is the following variant of Theorem 1 also true?

Claim 2: let $\mathcal{C}$ be any category possessing properties 1 and 2 above and let $X, Y, Z$, $i \colon Z \to X \times Y$ be as in Definition 4. Then the intersection of all $Z$-boxes is a $Z$-box itself.

From attempting to translate my two proofs of Theorem 1 into the abstract category setting and failing (the details can be found in the revision history), I finally concluded that the answer to Question 1 is most likely 'no'. (After expecting it to be 'obviously' yes for a long time.)

If the answer to Question 1 is indeed 'no', I have two followup questions:

Question 2: Can you give a counterexample to Claim 2, so a category with properties 1 and 2 in which the intersection of $Z$-boxes is not a $Z$-box for some $X, Y, Z$?

Question 3: What is the weakest, but still natural sounding, 'Property 3' that we could demand of our category so that Claim 2 becomes true again? 'Being concrete' works (as shown by Theorem 1), but maybe something weaker is possible?

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    $\begingroup$ Hello, Vincent. This post is quite long, and it is unlikely to get answers (it is maybe as long as a short blog entry, which some people may not want to read.) Can you please try to fix this? For example, you can try to start with Q1 only, and leave the other two for a later post, while making the post shorter nonetheless. $\endgroup$ – Pedro Tamaroff May 13 at 10:35
  • $\begingroup$ Yes this makes sense. I will think about a good way of doing this and edit it later $\endgroup$ – Vincent May 13 at 16:21
  • $\begingroup$ @PedroTamaroff What do you think of this version? It is still long, but much shorter and easier to grasp, I think $\endgroup$ – Vincent May 13 at 22:10
  • $\begingroup$ The concrete result implies the general result, since you can use the Yoneda embedding to reduce from the general case to the concrete case. $\endgroup$ – Zhen Lin May 13 at 22:31
  • $\begingroup$ @ZhenLin can you make this into an answer and perhaps add some details? I am not quite sure how this would play out $\endgroup$ – Vincent May 14 at 8:58
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First a comment: You have to be a little bit careful with the translation from the concrete case, since the categorical definition of subobject (in terms of monic arrows) may not coincide with "subset which is also an object", even in a concrete category. Also, even if subobjects have their "usual" meaning in a concrete category, it seems Theorem 1 may not hold. See the counterexample at the end of the answer, which takes place in a concrete category.

Here's a simpler version of your question:

Let $X$ be an object, $m\colon X'\hookrightarrow X$ a subobject, and $f\colon Z\to X$ an arbitrary arrow. We say $f$ factors through $X'$ if there is an arrow $f'\colon Z\to X'$ such that $m\circ f' = f$. Note that since $m$ is monic, $f'$ is unique if it exists.

Question 1': Let $C$ be a category with Properties 1 and 2 (products and intersections of subobjects). Let $X$ be an object, $(X_i)_{i\in I}$ a family of subobjects of $X$, and $X_\cap$ their intersection. Let $f\colon Z\to X$ be an arbitrary arrow which factors through each subobject $X_i$. Does $f$ factor through $X_\cap$?

I claim that Question 1' is equivalent to your Question 1.

Suppose Question 1' has a positive answer. Then for any arrow $f\colon Z\to X$, we can define $\text{im}_f(Z)$ to be the intersection of the family of all subobjects $X'\hookrightarrow X$ such that $f$ factors through $X'$. Then $f$ factors through $\text{im}_f(Z)$, so we have $Z\to \text{im}_f(Z)\hookrightarrow X$.

Now given $m\colon Z\hookrightarrow X\times Y$ as in Question 1, define $f = \pi_X\circ m\colon Z\to X$ and $g = \pi_Y\circ m\colon Z\to Y$. Then it's straightforward to show that $\text{im}_f(Z)\times \text{im}_g(Z)\hookrightarrow X\times Y$ is a $Z$-box, and moreover it is below every other $Z$-box, so it is the intersection of all $Z$-boxes.

Conversely, suppose Question 1 has a positive answer. Let $X$ be an object, $(X_i)_{i\in I}$ a family of subobjects of $X$, $X_\cap$ their intersection, and $f\colon Z\to X$ an arrow which factors through each $X_i$. Note that $f\times \text{id}_Z\colon Z\to X\times Z$ is a monic arrow, and for any subobject $m\colon X'\hookrightarrow X$ such that $f$ factors through $X'$, we have that $m\times \text{id}_Z\colon X'\times Z\hookrightarrow X\times Z$ is a $Z$-box. By Question 1, the intersection of all $Z$-boxes is a $Z$-box, which has the form $g\times \text{id}_Z\colon X'\times Z\hookrightarrow X\times Z$. Then $f$ factors through $X'$, and $g\colon X'\hookrightarrow X$ is below $X_i$ for all $i\in I$, hence $X'$ is below $X_\cap$, and thus $f$ factors through $X_\cap$.

Now in "normal" situations, the intersection $X_\cap$ of a family of subobjects will actually be the limit of the diagram consisting of all the monic arrows $m_i\colon X_i\hookrightarrow X$ (this is a fiber product / pullback of a family of monic arrows). In this case, Question 1' has a positive answer, by the universal property of the limit. Conversely, if this diagram has a limit, then the limit object will be a subobject of $X$, and hence will be the intersection of the family.

So a natural answer to Question 3 is: Claim 2 is true in any category which has fiber products of families of monic arrows.

On the other hand, to answer Question 2: We can refute Question 1' by exhibiting a family of subobjects which has an intersection but fails to have a fiber product. Consider the category of all groups which do not have size $2$. This category has products (since no product of groups of size $\neq 2$ has size $2$) and intersections of subobjects: If a family of subgroups of a group has a set-theoretic intersection of size $2$, then their category-theoretic intersection is the trivial group.

Now let $X = C_2\times \mathbb{Z}\times \mathbb{Z}$, let $X_1 = C_2\times \mathbb{Z}\times \{0\}$, and let $X_2 = C_2\times \{0\}\times \mathbb{Z}$. Let $Z = \mathbb{Z}$, and let $f\colon Z\to X$ be the map $f(n) = (\overline{n},0,0)$, where $\overline{n}\in C_2$ is the reduction of $n$ mod $2$. Then $f$ factors through $X_1$ and $X_2$, but it fails to factor through the category-theoretic intersection of $X_1$ and $X_2$, which is the trivial group.

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  • $\begingroup$ Aaah this is great! I will think a bit more about it and perhaps ask a followup question $\endgroup$ – Vincent May 14 at 17:01
  • $\begingroup$ Reading and rereading I find that the core of the matter lies in your sentence 'Now in "normal" situations, the intersection $X_\cap$ of a family of subobjects will actually be the limit of the diagram consisting of all the monic arrows $m_i\colon X_i\hookrightarrow X$ (this is a fiber product / pullback of a family of monic arrows).' So the answer to question 3, the existence of such fiber products, is not so much a 'Property 3' as it is 'What property 2 should have been all along'. Somehow the definition of intersection as a greatest lower bound is not strong enough, and we should have... $\endgroup$ – Vincent May 14 at 19:54
  • $\begingroup$ ...defined it as a limit in the category sense in order to truly catch our intuition about intersections. Is that correct? I am still struggling to grasp how these two concepts (greatest common lower bounds and fiber products) differ precisely and maybe come back to it. Many many thanks for this great answer, either way! $\endgroup$ – Vincent May 14 at 19:57
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    $\begingroup$ @Vincent I agree with this analysis - that the fiber product is, for most purposes, a "better" definition of intersection for subobjects. I decided at the end to leave such a statement out of my answer, because the condition that the poset of subobjects of $X$ has meets seems interesting on its own, even if these meets don't necessarily play well with the non-monic arrows into $X$. $\endgroup$ – Alex Kruckman May 14 at 20:02

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