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Stream: learning: questions

Topic: When is every point-surjective morphism an epimorphism?

M Nestor (Aug 25 2024 at 17:56):

A morphism f: X -> Y in a category with terminal object 1 is called point-surjective if for every y: 1 -> Y there exists x: 1 -> X such that fx = y. Is every point-surjective morphism an epimorphism?

Rémy Tuyéras (Aug 25 2024 at 18:04):

This kind of properties would not hold for objects with multiple sorts/levels. For example, point-surjective functors are not epimorphisms because the terminal object only really captures the objects of your categories. In $Cat$, point-surjective functors are functors that are surjective on objects

Madeleine Birchfield (Aug 25 2024 at 18:25):

Morphisms out of the terminal object do not always define the points. For example, in the category of abelian groups, the points are morphisms out of the integers $\mathbb{Z}$, which is the tensor unit in Ab, rather than the trivial group $0$, which is the terminal object in Ab. See [[pointed object in a monoidal category]].

M Nestor (Aug 25 2024 at 18:38):

yea, it seems to be false. In the category of pointed sets let X = {a, b, *} and Y = {c, *}, where * is the distinguished point. Define f: X → Y by f(a) = f(b) = f(*) = *. The f is point-surjective not an epimorphism: define g, h: Y → {d, e, } where: g(c) = d, g() = * h(c) = e, h(*) = *. Then gf = hf but g ≠ h. (excuse formatting)

Madeleine Birchfield (Aug 25 2024 at 18:51):

M Nestor said:

yea, it seems to be false. In the category of pointed sets let X = {a, b, *} and Y = {c, *}, where * is the distinguished point. Define f: X → Y by f(a) = f(b) = f(*) = *. The f is point-surjective not an epimorphism: define g, h: Y → {d, e, } where: g(c) = d, g() = * h(c) = e, h(*) = *. Then gf = hf but g ≠ h. (excuse formatting)

For the category of pointed sets, points are point-preserving functions out of the set with two elements (i.e. $\mathbb{2} = \{0, 1\}$, where the point is given by $0$). So a point-surjective morphism would be a morphism $f:X \to Y$ such that for all $y:\mathbb{2} \to Y$ there is $x:\mathbb{2} \to X$ such that $f \circ x = y$. But $\mathbb{2}$ is not the terminal object in the category of pointed sets - the terminal object is a zero object in the category of sets and is the set with one element, i.e. $\{0\}$.

Jean-Baptiste Vienney (Aug 25 2024 at 18:52):

@M Nestor. I think your question is still relevant. You can still think about what are the categories that satisfy this property. This property is satisfied by $\mathbf{Set}$. Maybe it is also satisfied by every (elementary) topos?

Jean-Baptiste Vienney (Aug 25 2024 at 18:58):

What we already know (thanks to the category $\mathbf{Ab}$) is that being cartesian or even cartesian closed is not sufficient.

Madeleine Birchfield (Aug 25 2024 at 19:02):

What about being a pretopos? Or if that isn't sufficient, a Heyting pretopos?

Jean-Baptiste Vienney (Aug 25 2024 at 19:05):

Maybe it is simpler than that. What about the categories in which every object is a coproduct of (copies of) the terminal object?

Madeleine Birchfield (Aug 25 2024 at 19:09):

Jean-Baptiste Vienney said:

Maybe it is simpler than that. What about the categories in which every object is a coproduct of the terminal object?

My point in my original two comments on this thread is that in categories like $\mathrm{Ab}$ and $\mathrm{Set}_*$ the points are not morphisms out of the terminal object, but rather morphisms out of the tensor unit, so @M Nestor's definition of a point-surjective morphism is wrong. So I don't think it really has anything to do with the terminal object.

So maybe, it is in categories in which every object is a coproduct of (copies of) the tensor unit.

Jean-Baptiste Vienney (Aug 25 2024 at 19:18):

But if your monoidal category has finite products, then it is also a cartesian monoidal category. And the unit of this cartesian monoidal category is the terminal object. So being the unit of a monoidal category doesn't ensure that morphisms out of the monoidal unit should correspond to the intuitive notion of point.

I think the that the reason why the morphisms out of the monoidal unit in some monoidal categories correspond to the intuitive notion of point is that in these monoidal categories the monoidal unit is somehow built from the one-point set, the terminal object in $\mathbf{Set}$. I say this because in $\mathbf{Mod}_R$, a point of $M$ corresponds to a linear map $R \rightarrow M$ where $R$ is the free module generated by the one-point set.

Jean-Baptiste Vienney (Aug 25 2024 at 19:23):

Maybe we could look at the monoidal categories which have a strong monoidal functor to them out of $\mathbf{Set}$.

Jean-Baptiste Vienney (Aug 25 2024 at 19:24):

That's the case of $\mathbf{Mod}_R$ (and so of $\mathbf{Ab}$ for $R=\mathbb{Z}$), the functor is the free-module one. And the free module on $X \times Y$ is the tensor product of the free module on $X$ with the free module on $Y$.

Jean-Baptiste Vienney (Aug 25 2024 at 19:25):

And as I said above, the free module on $*$ is $R$.

Jean-Baptiste Vienney (Aug 25 2024 at 19:26):

But thinking about how we can generalize what's going on in $\mathbf{Set}$ (by generalizing to toposes or something else) is also interesting.

Jean-Baptiste Vienney (Aug 25 2024 at 19:28):

Jean-Baptiste Vienney said:

That's the case of $\mathbf{Mod}_R$ (and so of $\mathbf{Ab}$ for $R=\mathbb{Z}$), the functor is the free-module one. And the free module on $X \times Y$ is the tensor product of the free module on $X$ with the free module on $Y$.

Maybe for this phenomenon we need something stronger like a left adjoint in the $2$-category of monoidal categories, strict monoidal functors and monoidal natural transformations of a strict monoidal functor to $\mathbf{Set}$.

Jean-Baptiste Vienney (Aug 25 2024 at 19:30):

To sum up, to me, they are at least two interesting separate questions to examine here.

Madeleine Birchfield (Aug 25 2024 at 19:32):

Jean-Baptiste Vienney said:

But if your monoidal category has finite products, then it is also a cartesian monoidal category. And the unit of this cartesian monoidal category is the terminal object. So being the unit of a monoidal category doesn't ensure that morphisms out of the monoidal unit should correspond to the intuitive notion of point.

Another example of this situation: if your monoidal category has finite coproducts, then it is a cocartesian monoidal category. And the unit of this cocartesian monoidal category is the initial object. But morphisms out of the initial object are unique and don't represent the points. The issue with additive categories like Ab is that their terminal objects are zero objects and coincide with the initial objects.

Ralph Sarkis (Aug 25 2024 at 19:39):

In these examples, the special object you consider is special because it is the representing object for a faithful forgetful functor $U: \mathbf{C} \to \mathbf{Set}$. Let us call this object $F\mathbf{1} \in \mathbf{C}$, it satisfies $\mathrm{Hom}(F\mathbf{1}, -) \cong U$.

Now, your point-surjective property for a morphism $f: X \to Y$ says precisely that $\mathrm{Hom}(F\mathbf{1},f) = f \circ - : \mathrm{Hom}(F\mathbf{1},X) \to \mathrm{Hom}(F\mathbf{1},Y)$ is surjective, hence epic in $\mathbf{Set}$. Since faithful functors reflect epimorphisms, and $\mathrm{Hom}(F\mathbf{1},-) \cong U$ is faithful by hypothesis, $f$ is epic whenever it is point-surjective.

Madeleine Birchfield (Aug 25 2024 at 19:39):

Jean-Baptiste Vienney said:

And as I said above, the free module on $*$ is $R$.

I wonder what's underlying all of this is that, given a category $C$ with a free-forgetful adjunction with the category of sets $\mathrm{Set}$, points in $C$ are morphisms out of the free object in $C$ on the terminal object $1$ in Set.

Madeleine Birchfield (Aug 25 2024 at 19:40):

Ralph Sarkis beat me to the point.

Madeleine Birchfield (Aug 25 2024 at 19:41):

Hopefully @M Nestor is still around to see this.

Jean-Baptiste Vienney (Aug 25 2024 at 19:56):

Consider a category $\mathcal{C}$ with a terminal object $1$, then, we have a functor $Hom(1,-):\mathcal{C} \rightarrow \mathbf{Set}$. I'm wondering what is the relation between this functor being faithful and every point-surjective morphism being an epimorphism.

Jean-Baptiste Vienney (Aug 25 2024 at 19:57):

I'm trying to go into the opposite direction of what has been said by @Ralph Sarkis.

Jean-Baptiste Vienney (Aug 25 2024 at 19:58):

Outch, we should replace $1$ by something else probably.

Jean-Baptiste Vienney (Aug 25 2024 at 20:00):

Anyway, I'm wondering if we can get somehow a functor like Ralph's one from every category which satisfies the property.

M Nestor (Aug 25 2024 at 20:00):

Madeleine Birchfield said:

Hopefully M Nestor is still around to see this.

yea I'm curious if there is a necessary and sufficient condition but my trivial counterexample has already convinced me to stop trying to prove it in general haha.

M Nestor (Aug 25 2024 at 20:01):

the title question has already been answered so maybe I should rename it to "When is every point-surjective morphism an epimorphism?"

Jean-Baptiste Vienney (Aug 25 2024 at 20:05):

We should look at the categories with an object $\square$ such that being an epimorphism is equivalent to begin $\square$-point surjective where we say that a morphism $f:X \rightarrow Y$ is $\square$-point surjective if for every $y:\square \rightarrow Y$, there exists $x:\square \rightarrow X$ such that $f \circ x=y$.

Jean-Baptiste Vienney (Aug 25 2024 at 20:05):

Since we realized that $\square$ is not necessarily a terminal object.

Jean-Baptiste Vienney (Aug 25 2024 at 20:07):

The question was about whether $\square$-surjective implies epimorphism when $\square$ is a terminal object. What about the other direction. Does being an epimorphism implies being $\square$-surjective when $\square$ is a terminal object?

Madeleine Birchfield (Aug 25 2024 at 20:16):

Ralph Sarkis said:

In these examples, the special object you consider is special because it is the representing object for a faithful forgetful functor $U: \mathbf{C} \to \mathbf{Set}$. Let us call this object $F\mathbf{1} \in \mathbf{C}$, it satisfies $\mathrm{Hom}(F\mathbf{1}, -) \cong U$.

Now, your point-surjective property for a morphism $f: X \to Y$ says precisely that $\mathrm{Hom}(F\mathbf{1},f) = f \circ - : \mathrm{Hom}(F\mathbf{1},X) \to \mathrm{Hom}(F\mathbf{1},Y)$ is surjective, hence epic in $\mathbf{Set}$. Since faithful functors reflect epimorphisms, and $\mathrm{Hom}(F\mathbf{1},-) \cong U$ is faithful by hypothesis, $f$ is epic whenever it is point-surjective.

I wonder if this is still true in the infinity-category case, where $C$ is an $(\infty, 1)$-category and $U$ is a forgetful $(\infty, 1)$-functor into the $(\infty, 1)$-category of $\infty$-groupoids.

Madeleine Birchfield (Aug 25 2024 at 20:26):

Jean-Baptiste Vienney said:

We should look at the categories with an object $\square$ such that being an epimorphism is equivalent to begin $\square$-point surjective where we say that a morphism $f:X \rightarrow Y$ is $\square$-point surjective if for every $y:\square \rightarrow Y$, there exists $x:\square \rightarrow X$ such that $f \circ x=y$.

Now I'm beginning to understand why category theorists talk about "[[generalised elements]]" - because in many categories of mathematical structures, set-theoretic elements of the mathematical structure $X$ do not correspond to the [[global elements]] of $X$. The global elements are the morphisms from the terminal mathematical structure into $X$.

M Nestor's used "point" as a synonym of "global element" while I used "point" as a synonym of "set-theoretic element".

Kevin Carlson (Aug 26 2024 at 02:14):

Ah, yes, glad somebody caught that ambiguity! As y’all have already seen, it’s exceptionally rare that maps which are surjective with respect to maps from the terminal object must be epis, but much more common that there be some object satisfying that property. In particular as Ralph says if there’s an object $x$ whose representable is faithful, which is sometimes called a generating object, then maps which are surjective for $x$-indexed generalized elements are always epi.

Kevin Carlson (Aug 26 2024 at 02:16):

There are at least two different interesting directions to take this from here: (1) what if there is no (obvious) such $x$? In lots of categories, for instance most toposes, it’s easier to find a family of $$$x$’s such that the product of all their representables together is faithful. Such a family is sometimes just called a generator (or separator) or less risking, a generating or separating set. It might be fun to think of some natural generating set for some of your favorite toposes that aren’t too much like Set. When you have a generating set, you know that any morphism surjective on generalized elements for all the generators must be epi.

Kevin Carlson (Aug 26 2024 at 02:18):

(2) What about the converse? When is it true (sometimes/always/never) that in a category with a generator, the epis are precisely the point-surjective maps? To be clear, I know that this question has an interesting and reachable answer, at least a partial one, just don’t want to spoil it too soon.