Well, in general I believe the second stronger definition of constant is better. However, it looks like in this case I may have been led astray by that general fact, because what we’re doing is rephrasing stuff about equivalence relations in terms of their quotients, and the quotient that corresponds to the full relation is the support of an object, which is subterminal and not necessarily terminal. So in this case, I think you’re probably right to use the weaker notion.

]]>What do you have against terminal objects?

It’s personal ;-) I’d rather not be dependent on them, because I have the intuition that ’being constant’ has nothing to do with ’being terminal’. I didn’t look it up earlier, because I simply accepted your definition of factoring through the terminal object, but your reference to constant morphism helped a lot. I like the first definition much better than the second, and it also fits better when combining it with joint-monicity.

Why do you feel that it would be better for it to be terminal? I understand this gives you more/better properties, but I only want those properties if they also follow intuitively from what I think a fraction should be. In other words, why ask for a strong definition if the weak definition is sufficient to capture what you want? To me that goes against Occam’s razor, so to speak.

]]>In general, $X\times X$ being the kernel of $e$ is the same as saying that for any two morphisms $u,v$ with codomain $X$ we have $e u = e v$, i.e. that $e$ is a constant morphism in the weak sense. If $e$ has an image, this is the same as saying that that image is subterminal, and I think it would be better for it to be terminal. But note that at constant morphism there is given a way to phrase the strong sort of “constancy” without assuming the existence of a terminal object, so if you really happen to be in a category without a terminal object (but with pullbacks and products? I can’t think of too many categories like that) then you could use that.

]]>What do you have against terminal objects?

]]>Some more thought…

So given maps $f : X \rightarrow Y$ and $g : X \rightarrow Z$ the intuition is that whenever $e = hf = kg$, the map $e$ equates whatever is equated by $f$ or $g$. If we want to state that $e$ equates ’everything’, we could state that as $e$ is constant, but this requires the existence of a terminal object. Alternatively, we could perhaps also state that $X \times X$ is the kernel of $e$?

**Definition 3**: a map $g : X \rightarrow Z$ is a fraction of $f : X \rightarrow Y$ if 1) $(f,g)$ is jointly monic, 2) $g$ is epic, and 3) any map $e$ that factors through $f$ and $g$ has $X \times X$ as its kernel.

Pity I still have to assume existence of the product $X \times X$, but imho better than using a terminal object (is there a difference whenever both exist?). Also a pity that $g$ being epic does not simply follow from a “smallest solution” condition, and that I do not immediately see under which conditions we get uniqueness of fractions. I agree with you, however, that this may in general be unreasonable to assume, but should work out in reasonable conditions.

]]>Now, I’ll start studying the factoring through a terminal object as a way to obtain isomorphism of fractions under reasonable conditions :-)

Firstly, if I’m not mistaken, using the existence of pushouts is not necessary, because the terminal object is supposed to *be* the pushout… so

**Definition 2**: a map $g: X \rightarrow Z$ is a fraction of $f : X \rightarrow Y$ if 1) $(f,g)$ is jointly monic, 2) $g$ is epic, and 3) for any $h: Y \rightarrow A$ and $k : Z \rightarrow A$ with $hf = kg$, we find $hf$ is constant (i.e. factors through the terminal object).

Following your idea (I think) the extension of $f$ and $g$ into a commuting square represents the “union, closed under symmetry and transitivity” of the equivalences represented by the original maps $f$ and $g$. That sounds okay to me. If this union is constant, i.e. factors through the terminal object, that implies that “all information is thrown away” by the new equivalence, i.e. everything is equated.

That last part is close, but not the same I think, as my intuition that: whenever $f$ makes a distinction, then $g$ should try to avoid making this distinction.

The tricky part is that in some categories, $g$ may not be able to avoid making distinctions. In my category of choice, a terminal object exists, but nonetheless I’m not sure that $g$ is free to avoid making distinctions (okay perhaps that just means the distinctions will show up in the terminal object then… true… but this is subtle). I’d have to think about quite examples for a while to make this more concrete though. Furthermore, I don’t like to have to use the terminal object as part of my definition, because I think that having a constant map *implies* that all information is thrown away, rather than that it is *the same as* saying all information is thrown away. In a sense, using the terminal object makes the definition stronger than I intend it to be.

I’ll give it a shot as soon as I have the time :-)

]]>using the sentence “factors through” without an indication of left or right is ambiguous

Technically, yes. Although that would be true regardless of which of $f$ or $h$ is said to “factor through” the other. But if $f$ and $h$ share a domain but not a codomain, or a codomain but not a domain, as is usually the case, then the ambiguity is resolved as only one choice is possible.

Does something like that deserve a wiki-page or comment according to you?

Yes, I would probably generalize it to a statement about one *family* of morphisms factoring through another jointly-monic family, and add it in a “Properties” section to jointly monic family. The special case when both families are single morphisms is already recorded at monomorphism#properties: if $g f$ is a monomorphism, so is $f$.

Back to your definition:

Definition 2: Given a map $f: X \rightarrow Y$, a fraction of f is an epimorphism $g : X \rightarrow F$ such that (1) if $e : X \rightarrow Z$ is an epimorphism such that $f$ and $g$ both factor through it, then $e$ is an isomorphism; and (2) the pushout of $f$ and $g$ is constant (factors through the terminal object).

In a category where epi+mono implies iso, this is equivalent to saying

- Definition 2b: Given a map $f: X \rightarrow Y$, a fraction of f is an epimorphism $g : X \rightarrow F$ such that (1) if $e : X \rightarrow Z$ is an epimorphism such that $f$ and $g$ both factor through it, then $e$ is a monomorphism; and (2) the pushout of $f$ and $g$ is constant (factors through the terminal object).

Which in a category where products exist comes down to

- Definition 2c: Given a map $f: X \rightarrow Y$, a fraction of f is an epimorphism $g : X \rightarrow F$ such that (1) $(f,g)$ is a joint morphism; and (2) the pushout of $f$ and $g$ is constant (factors through the terminal object).

That definition is closer to my goal of treating compositions as relations between objects, and under mild conditions coincides with your intuition according to my previous post. Nice.

Now, I’ll start studying the factoring through a terminal object as a way to obtain isomorphism of fractions under reasonable conditions :-)

]]>Thanks… But then, when $f = hk$, do you also say $f$ factors through $k$? In that case, using the sentence “factors through” without an indication of left or right is ambiguous…

Now I’ll reread the posts above to see if they make sense again/still to me ;-)

The little theorem is what it is, of course. But I guess it is also pretty standard for people working with joint morphisms often. I guess so standard that there is no reference for it… Does something like that deserve a wiki-page or comment according to you? There is value in such ’trivia’, but there is also a big risk of a wiki become just a collection of such trivia then…

]]>Ah, I see the confusion. When $f = h k$ we say that $f$ factors through $h$ (since writing $f = h k$ is “factoring” $f$, analogously to $6 = 2\cdot 3$), not that $h$ factors through $f$.

]]>Ah, or maybe I misunderstood what you mean by “factors through”. I couldn’t find a formal definition… I’ll elaborate on what I found and make it self-contained, including the standard definitions. (Is there a way to put diagrams in posts?)

- Definition: a map $f : X \rightarrow Y$ is monic if for any two $i,j : A \rightarrow X$ with $fi = fj$ we find $i = j$.
- Definition: a pair $(f,g)$ of maps $f : X \rightarrow Y$ and $g : X \rightarrow Z$ is jointly monic if for any two $i,j : A \rightarrow X$ with $fi=fj$ and $gi=gj$ we find $i = j$.
Definition: a product of $Y$ and $Z$ is an object $Y \times Z$ with projections $k : Y \times Z \rightarrow Y$ and $l : Y \times Z \rightarrow Z$ such that for any pair $f : X \rightarrow Y$, $g : X \rightarrow Z$ there is a unique map $h : X \rightarrow Y \times Z$ such that $kh = f$ and $lh = g$.

Theorem 1: given a pair $(f,g)$ of jointly monic maps $f : X \rightarrow Y$ and $g : X \rightarrow Z$, let $h : X \rightarrow B$ be such that there exist $k : B \rightarrow X$ and $l : B \rightarrow Y$ with $f = kh$ and $g = lh$, then $h$ is monic.

Proof: assume two $i,j : A \rightarrow X$ such that $hi = hj$, then $fi = khi = khj = fj$ and $gi = lhi = lhj = gj$, so by joint monicity $i = j$. Therefore, $h$ is monic.

Theorem 2: assuming products exist, let $(f,g)$ be a pair of maps $f : X \rightarrow Y$ and $g : X \rightarrow Z$. Furthermore, assume that for any $h : X \rightarrow B$ with $k : B \rightarrow X$ and $l : B \rightarrow Y$ such that $f = kh$ and $g = lh$ we find $h$ is monic. Then $(f,g)$ is jointly monic.

Proof: Observe that in particular $h : X \rightarrow Y \times Z$ is assumed to be monic, with $k : X \times Y \rightarrow X$ and $l : X \times Y \rightarrow Y$ the natural projections from the product. Now, let $i,j : A \rightarrow X$ be a pair of maps such that $fi = fj$ and $gi = gj$. Then by definition of product there is a unique arrow $u : A \rightarrow Y \times Z$ such that $ku = fi$ and $lu = gi$, as well as a unique arrow $u' : A \rightarrow Y \times Z$ such that $ku' = fj$ and $lu' = gj$. Because $fi = fj$ and $gi = gj$ we find $u = u'$, and because $khi = fi$ and $lhi = gi$ we find $u = hi$ and similarly $u' = hj$. Therefore, $hi = hj$. We assumed $h$ to be monic, therefore $i = j$, which concludes the proof that $(f,g)$ is jointly monic.

in a category in which all products exist $(f,g)$ is jointly monic if and only if any map $m$ that factors through $f$ and $g$ is monic.

That doesn’t sound right. The map to the terminal object always factors through $f$ and $g$, but isn’t generally monic. Maybe I misunderstood what you mean?

]]>Trying to relate your definition to mine, I discovered the following (which is probably already obvious to you).

Theorem: in a category in which all products exist $(f,g)$ is jointly monic if and only if any map $m$ that factors through $f$ and $g$ is monic.

That any epi that factors through $(f,g)$ is iso would then be true in any category where mono+epi implies iso.

]]>I mean an isomorphism $Y\cong Y'$ *not* making the obvious triangle commute.

Mike, I haven’t been following this thread, and so I may be misunderstanding or maybe I’m being dense, but: what do you mean? I consider two quotients $q: X \to Y$ and $q': X \to Y'$ to be isomorphic if there is an isomorphism $Y \to Y'$ making the obvious triangle commute. In that case, $q$ and $q'$ correspond to the same equivalence relation. So I don’t know what you’re asking.

]]>@Todd, thanks! Do you know if anything is known about the relation of “having isomorphic quotients” on that lattice?

]]>My thinking was that if $f$ is also epi, then the pushout should *be* the terminal object, since that would correspond to the union of the kernel pairs of $f$ and $g$ in the lattice of equivalence relations being the full relation. If $f$ is not epi then whatever extra stuff is in its codomain will still be there in the pushout, but the map from $X$ to the pushout factors through the image of the image of $f$, hence through the terminal object.

Definition 2: Given a map f:X→Y, a fraction of f is an epimorphism g:X→F such that (1) if e:X→Z is an epimorphism such that f and g both factor through it, then e is an isomorphism; and (2) the pushout of f and g is constant (factors through the terminal object).

Thanks Mike, I already felt actually creating the kernel pairs was cumbersome. I’m still chewing on the ’constant’ part, which is a new trick for me.

]]>Classically, the lattice of equivalence relations on a set forms a geometric lattice or a matroid lattice. I don’t think I’ve ever considered the situation for more general regular (or exact) categories.

]]>

Definition 2: Given a map $f : X \rightarrow Y$. Let $k,k' : R \rightarrow X$ be the kernel pair of $f$ (the pullback of $f$ over itself). An object $F$ is afractionof $f$ if there exists an epimorphism $g : X \rightarrow F$ for which the kernel pair $h,h' : S \rightarrow X$ (the pullback of $g$ over itself) has the property that every $m : X \rightarrow Z$ with $h m = h'm$ and $k m = k'm$ is a monomorphism.

Close! But I think the inclusions go the other direction: $h m = h' m$ means that the kernel pair of $m$ *contains* $(h,h')$, not conversely. Also there should be some kind of maximality condition, e.g. that their union in the poset of equivalence relations is the full relation, i.e. they are complements in that poset. Finally, an object in category theory should be equipped with a morphism rather than just be such that a morphism exists (e.g. a cartesian product is equipped with projections $A\leftarrow A\times B \to B$). So if we turn *all* the equivalence relations into epimorphisms, I would state the definition something like this:

**Definition 2**: Given a map $f : X \rightarrow Y$, a **fraction** of $f$ is an epimorphism $g:X\to F$ such that (1) if $e:X\to Z$ is an epimorphism such that $f$ and $g$ both factor through it, then $e$ is an isomorphism; and (2) the pushout of $f$ and $g$ is constant (factors through the terminal object).

However, with this definition as stated, your first conjecture

Fractions are unique up to isomorphism.

is false in complete generality. Consider any lattice $P$ as a thin category and let $x\in P$ be an object that has more than one distinct complement, i.e. $y,y'$ with $x\wedge y = x\wedge y' = 0$ and $x\vee y = x\vee y' = 1$. Then since every morphism in a thin category is epi, $0\to y$ and $0\to y'$ are nonisomorphic fractions of $0\to x$. This is also a counterexample to your second conjecture, since $0 = x\times y = x\times y'$.

However, a more reasonable formulation would be to add conditions on the category and/or restrict to some well-behaved subclass of epimorphisms like strong, extremal, or regular ones. Of course it’s still false if you ask for the isomorphism to respect the morphisms from $X$, but absent that condition I don’t yet know the answer (though there could still be an easy one, I haven’t thought about it very hard).

Moreover, your third conjecture:

Given a natural projection $f : X \times Y \rightarrow X$, we find $Y$ is a fraction of $f$.

is almost true, in good cases. It’s not quite true even in $Set$ because $X\times Y \to Y$ may not be an epimorphism (consider $X=\emptyset$), but it is true in $Set$ if both $X$ and $Y$ are nonempty. More generally, I believe it is true in any regular category if instead of “epimorphism” we say “regular epimorphism” (or assume that all epis are regular, as in a pretopos) and assume that $X$ and $Y$ are well-supported, i.e. $X\to 1$ and $Y\to 1$ are regular epi. (The proof I have in mind uses the internal language, though, and I haven’t tried to compile it out in terms of arrows.)

I feel like the structure of the lattice of equivalence relations (or quotients) of an object in a regular category must have been studied before, but I don’t know where to find it. Perhaps stackexchange might help.

]]>Yes, that might work too, I suppose. (According to kernel you actually mean kernel pair, right? The pullback of $f$ along itself.)

After a nights sleep, I figured out how to adapt my own definition a bit as well, so now I have two definitions which both seem reasonable (for the time being).

**Definition 1**: Given a map $f : X \rightarrow Y$, an object $F$ is a*fraction*of $f$ if for every $g : X \rightarrow Z$ with $(f,g)$ jointly monic, there exists an $h: Z \rightarrow F$ such that $(f,hg)$ is jointly monic and $hg$ is an epimorphism.**Theorem 1**: Fractions are unique up-to isomorphism.Proof: let $F$ and $F'$ both be fractions of $f$. Observe that $(f,1)$ is jointly monic. Applying the definition of fraction gives us maps $g : X \rightarrow F$ and $g' : X \rightarrow F'$ such that $(f,g1)$ and $(f,g'1)$ are jointly monic, and $g = g1$ and $g' = g'1$ are epimorphisms. Applying the definition of fraction again gives us maps $h$ and $h'$ such that $hg = g'$ and $h'g' = g$. From this we obtain $1g = g = h'g' = h' h g$, and because $g$ is epic, $1 = h' h$. Similarly, $1 = h h'$. So $h$ and $h'$ are isomorphisms between $F$ and $F'$.

**Conjecture**: Given a natural projection $f : X \times Y \rightarrow X$,*if*$f$ has a fraction,*then*it is isomorphic to $Y$.**Conjecture**: Given a natural projection $f : X \times Y \rightarrow X$, we find $Y$ is a fraction of $f$.

And my first quick attempt to formalize Mike’s suggestion, while keeping it ’comparable’ to the first definition. (I’m not sure if I got the translation into arrows entirely right, please have a look).

**Definition 2**: Given a map $f : X \rightarrow Y$. Let $k,k' : R \rightarrow X$ be the kernel pair of $f$ (the pullback of $f$ over itself). An object $F$ is a*fraction*of $f$ if there exists an epimorphism $g : X \rightarrow F$ for which the kernel pair $h,h' : S \rightarrow X$ (the pullback of $g$ over itself) has the property that every $m : X \rightarrow Z$ with $hm = h'm$ and $km = k'm$ is a monomorphism.**Conjecture**: Fractions are unique up-to isomorphism.**Conjecture**: Given a natural projection $f : X \times Y \rightarrow X$,*if*$f$ has a fraction,*then*it is isomorphic to $Y$.**Conjecture**: Given a natural projection $f : X \times Y \rightarrow X$, we find $Y$ is a fraction of $f$.

I’ll get back later to seeing whether I can prove any of the conjectures, or find conditions under which they can be proven.

]]>Here’s another thought: surjective functions with domain $X$ are the same as equivalence relations on $X$, and although your $f$ isn’t assumed surjective, it only matters up to its image, hence up to its kernel pair which is the equivalence relation it determines. On this side, I think that what you are looking for is, given an equivalence relation $R$ on $X$ (the kernel of $f$), you want another equivalence relation $S$ on $X$ such that $R\cap S = \Delta_X$, the diagonal relation, and which is maximal with this property.

]]>Thanks, Mike. That was indeed a typo, should have had $Z$ be $Y$ instead, so I’ve edited.

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