Oct 13, 2025
Please please please send me your best theorems that give criteria for determining that a functor is representable. So far I have
I need more of these to build my intuitions.....
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Please please please send me your best theorems that give criteria for determining that a functor is representable. So far I have
I need more of these to build my intuitions.....
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There are no theorems in category theory.
Emily Riehl, Category Theory In Context
Mathematicians often tell her this; hence the book.
If I had to summarise her views in one sentence, it would be:
Everything is an adjunction.
the stable homotopy category by Cary Malkiewich
There is a particular kind of open-ended idea which I call the blue-sky idea. Namely, someone sitting at his desk with his feet up and looking out the window has come up with some concept, and now hopes that he or someone he knows can come up with a good way of using that idea. At conferences one sometimes encounters mathematicians who specialize in this sort of idea, button-holing everyone who can't manage to avoid them and inflicting their most recent blue-sky idea on them. ... When one looks at the history of mathematics, it may seem that a lot of the most important developments have come out of blue-sky ideas. But in fact, from what I know, this is almost never the case. Good ideas always arise out of some existing line of thought, and it is only after when has put a lot of effort in that one realizes that there is some gradiose general concept that underlies all one's work. For instance, I am pretty sure that Eilenberg and Steenrod didn't sit down together over a beer one day and say, "Wouldn't it be neat to draw up a set of axioms for a thing called a category and then invent a concept called a functor, and then see if these would be useful for anything?" It's pretty clear that in fact they noticed that in algebraic topology the same sorts of situations keep coming up over and over again, and one keeps seeing different theorems with different subject matter, where somehow the proofs always turned out to be more or less the same. So they saw (I believe) the need for a vocabulary and a conceptual framework that would enable mathematicians to talk about all this in a unified way. Ergo: category theory.
Lee Lady, How to do mathematical research
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Functors
I keep confusing myself with these, and references to List and Maybe as Functors is also confusing so I want to clear this up for myself (and provide a reference for myself in the future in case I forget).
Its not untrue that they’re Functors but they are also Not Functors sort of. They’re more like domains that can be lifted to by Functors if that makes more sense? But since there isn’t a name for these Functors, we call it a List Functor or a Maybe Functor I suppose.
So the Functor Law (from the Haskell Wiki - a Quite Reliable Source) says that Functors must follow two Laws. Identity preservations (fairly trivial), and composition which is a little more complicated. I think Id is fairly trivial so I’m going to ignore Id.
First things first, basics. fmap is defined as fmap :: (a -> b) -> f a -> f b. The f is referring to the (a -> b)
So when we talk about List or Maybe being functors what we mean is that we can take some Function f defined as f :: a -> b and then call fmap f (Maybe a) and we will get Maybe b out of it, which means that we can do a -> b without taking it out of the Maybe “wrapper”. In a purely category theory sense, a functor is the transformation the normal types a and b to the wrapped types Maybe a and Maybe b and since its a functor the morphism (function) is also preserved into the Maybe domain, it turns into a Maybe a -> Maybe b.
Now, lets talk about composition (Id is simple, Id should be the same in all domains, if its a -> a then it certainly should be Maybe a -> Maybe b). Composition says composition of morphisms is preserved so
fmap (f . g) Â == Â fmap f . fmap g
I think this is most easy to do by example. Lets even use List as an example so that we can see how functions can effortlessly lift. Composition rule says that if  f :: a -> b and g :: b -> c and you fmap f . g which is the same as fmap g ( f a) you will get the same as if you did the two fmaps separately. That means that  fmap f . fmap g is the same as fmap g (fmap f a). Ok. Confusing. Lets make a list, lst = [1..26] and lets have some f map integers to letters f :: Int -> Char. Obviously if we fmap f lst we will get a list of of letters back (A-Z if we implement f right) but lets say we also implement g :: Char -> String where each character became a word starting with that letter. So there are two different ways of doing it right? fmap f . g will compose f and g together, which means that it goes Int -> Char -> String which becomes Int -> String and then that function is mapped over the list. Alternatively, we can do fmap f . fmap g - first we want to fmap f lst which gives us a list of Chars and then we map over g mapping Strings from the Chars in the list.
6 August 2019, 7pm
I realized how types of structured sets can be thought of as faithful copresheaves.
Let $F$ be a type of structured sets, e.g. the type of groups or the type of topological spaces. For every structured set $A$ in $F$ we have the underlying set $F(A)$ of $A$. And for every structure-preserving map $f : A \to B$ between structured sets in $F$, we have the underlying map $F(f)$ of $f$.
Note that $F(A)$ is not $A$ because $A$ has additional structure. In fact it is possible for two structured sets to be distinct and yet have the same underlying set, since they may differ in structure while having the same elements. In contrast, $F(f)$ does not have any additional structure on it, compared to $f$, besides the additional structure on its domain and codomain; this means that $f$ is uniquely determined by $F(f)$, $A$ and $B$. We may therefore think of $f$ as the ordered triple $(F(f), A, B)$.
Now, two conditions it's reasonable to expect structure-preserving maps to satisfy are as follows:
For any two structure-preserving maps $f : A \to B$, $g : B \to C$, the composition $(F(g) \circ F(f), A, C)$ is also structure-preserving.
For every structured set $A$, the identity map $(1_A, A, A)$ is structure-preserving.
It is easy to prove that if these two conditions are satisfied, then by taking the structured sets in $F$ as objects and the structure-preserving maps between structured sets in $F$ as morphisms, we obtain a category, and then $F$ is a faithful copresheaf on this category.
Of course, we can replace $\mathsf{Set}$ with an arbitrary category $X$ to get a notion of “structured $X$-object”; a type of structured $X$-objects is a faithful functor into $X$.
This also gives us a way to see faithfulness as a natural notion. Faithfulness of a functor $F : X \to Y$ means that the morphisms in $X$ can be thought of as just structure-preserving morphisms between structured $Y$-objects, without any additional structure on the morphisms themselves as opposed to their domains and codomains.