Can someone explain the Yoneda lemma in layman terms? I'm sorry became a brainlet please forgive me I love you

>mapping each g in Hom(A,X) to f composed with g
(this is just a long-winded way of saying that Hom(A,-) preserves the properties of categories, as do all functors)

>yoneda lemma states that the set Nat(Hom(A,-),F) is isomorphic to F(A)
aka, the set of natural transformations between Hom(A,-) and F is in bijection with the set F(A)

injection of morphisms is in bijection with its domain?

>forgetful functor
what is this again?

>What about objects that don't interact with any other objects?
Such an object would be pointless to "study." Category theory is much more about the interactions between objects (morphisms and functors) than about the objects themselves.

I mean objects as in the objects of a category and maps as in the morphisms of the category.

The functor of points of a Scheme is the most significant example I know of.

>automorphism
is an automorphism a ring?

A forgetful functor is one that maps a category to another with "less" structure, simply throwing away the structure.

For example, a group is still a set (but with added structure). A forgetful functor would be the one that maps the category Grp into Set in such a way.

i messed up a lot of that, i'm still learning a lot of this stuff myself
here

>F:A -> Set
should be F:C->Set

>F(A,Y)
should be Hom(A,Y)

by proper set i just meant set, i don't think "proper set" is mathematically any different from "set"
this clarification was needed because some categories are too "large" for their collection of morphisms to be a set, or in the case of enriched category, we explicitly replace the hom-sets with other objects that sometimes aren't sets

why are you trying to learn the yoneda lemma, if you don't mind me asking?
i figured you were an undergrad taking abstract algebra or something, but if you're not, you should definitely brush up on algebra before bothering with something as algebraic as the yoneda lemma
i certainly don't mean to discourage though

Literally this diagram

why is this so complicated and deep into specific field? does it help us with something irl?

I envy you for being so smart user

>pointless to study just because it has no consequences

God damn pleb physicists
Always just thinking about money

>F:A -> Set
>should be F:C->Set
>F(A,Y)
>should be Hom(A,Y)

this mean that given a "thing" of morphisms that have the quality of injection, there is a Set and that Set is isomorphic to F(C) so F(A) is injective to Set. did i get this right?

>why are you trying to learn the yoneda lemma, if you don't mind me asking?
i lost my ability to abstract and create. just trying to get it back.

Knowing an object is the same as knowing what it does

lmao i'm just a sophomore physics student, i've just been killing time in the library between classes by reading math books

i just happen to know a bunch of definitions
when it comes to proofs and the important things, i'm useless

never take a CS professor that expects you to be a mind reader

whats the difference between an endomorphism and an automorphism

kek, I'm far from that

There are plenty of fields that study the objects themselves. But, the whole point of category theory is to look at connections between structures. From a category theory perspective, the objects themselves have little consequence in comparison to morphisms and functors.

an endomorphism is a morphism from an object to itself
an automorphism is an isomorphism from an object to itself
therefore an automorphism is an endomorphism that is also an isomorphism

google it senpai
en.wikipedia.org/wiki/Automorphism

can you draw a picture?

>i lost my ability to abstract and create. just trying to get it back

you started with the yoneda lemma?
i admire your boldness
but mathematics is like mountain climbing
if you don't start at the bottom and acclimate on your way up, you're going to suffocate and freeze to death

if you don't know category theory, you obviously won't understand the yoneda lemma
do not be dissuaded, algebra and category theory is a fun journey
but you must start at the beginning, wherever that is for you

if you're good with algebra, you could probably jump right in to some of the more advanced books like categories for the working mathematician

algebra: chapter 0 is a good book all around, but the exercises can be hard for a true beginner

no no no
i have old notes on this shit. i'm trying to understand my old notes too. but, hours of trying to think just like other people in CS really fucked me up.

Let me make up a hand-wavy analogy - we can get down to business later too.

You just moved into a house. But your gf has a different idea of how to do things with this you space and so you two constantly moving stuff around.

Let h(X,Y) denote the ways in which you can move stuff from X into Y.
You have a kitchen shelf A with a lot space to put stuff (it has some number of free spots), a storage room B (this also has some number of free spots) and a boxes V1, V2, V2... with stuff in it.

So h(V1, A) are the ways in which you can put the things in the box V1 onto the shelf.
(Say V1 contains 5 toys the shelf has 7 spots, well now you can do some combinatorics to see how many possibilities h(V1, A) captures)
Similarly, h(V1, B) are the ways in which you can move stuff from the box V1 into the storage room.
Now say you placed all your stuff on the kitchen shelf A but you guys decide you instead want to move it to the storage room B, in some way. Clearly, the different ways in which you can do this correspond to h(A,B), which are the ways to move stuff from the kitchen to the storage room.
Said again, the ways in which you can change ALL possible strategies h(V1, A) to the strategies h(V1, B) correspond to h(A,B).
And this doesn't depend on the things (the particular box you chose). For all boxes V, the ways in which you can change ALL possible strategies h(V, A) to the strategies h(V, B) correspond to h(A,B). And this is true for kitchen shelves and basement shelves just alike, i.e. holds for all sorts of A,B.

In symbols:
forall V, A, B
h(V, A) -> h(V, B) h(A,B)

A more geometric realization: Consider the Cartesian spaces R^n and R^m. Then e.g. h(R^2,R^3) are all the ways to put a n=2dim sheet into a m=3dim space. Or h(R^1,R^4) are all the ways how to embed a k=1dim line into 4dim space?
How to change a given embedding in h(R^1,R^2) to one in h(R^1,R^4)?

Well, you can just embed your sheet R^2 in your large space R^4, and however the line R^1 was lying on the sheet, after this the line will lie in R^4 somehow. Without having said much about what it means to "change an embedding", let me capture that idea by

forall k, n, m
h(R^k, R^n) -> h(R^k, R^m) h(R^n, R^m)

In category theory, the capital letters objects A,B,... I used here will be the objects or a category C, the individual "ways to put stuff from here to there" are the arrows and the collection of all arrow from A to B is denoted hom_C(A,B) (which is why above I chose the letter h).
A category with no arrows (i.e. where for all A and B, you have hom_C(A,B)={}) is just a set.
A functor acts like a map between categories, the pendant to what a function is for sets. (in fact, you can tweak the definitions to make a functor just be a function)
The most important functor is, for any given object B, the hom-functor hom_C(-,B). As a function maps any object A to the set h(A,B).

The Yoneda lemma (contravariant, and for hom-functors) reads

for all A,B
hom_D( hom_C(-,A) , hom_C(-,B) ) hom(A,B)

It says:
"The ways in which you can "algebraically" translate mappings (of any V) into A to mappings (of any V) into B correspond exactly to the ways how to move from A to B."

Here D is the category in which all arrows are maps between functors, such as hom_C(-,A). D is a "functor category".

Functors as well as the maps between them must be defined as algebraic operation of course, but those are both one-liners.
The hom-functors, like hom_C(-,B), fulfill the requirements of a functor, but they are by far not the only ones.
If F is any functor, then the Yoneda lemma (contravariant) reads

for all A,B
hom_D( hom_C(-,A) , F- ) F(A)

It says:
"Okay, look. I know our definition of a 'functor' was pretty abstract, but to be quite honest senpai, all you could cook up are mostly like hom-functors. Namely, the ways how to "algebraically" translate mappings (of any V) into A whatever the fuck FV turns out to be, this again corresponds exactly to FA, something you know."

>the ways in which you can change ALL possible strategies h(V1, A) to the strategies h(V1, B) correspond to h(A,B)

that's a very good explanation. seems like there are some fascinating implementations of the Yoneda lemma with probability.

Can someone explain derived functors in layman terms? :)

Right derived functor of a left-exact functor is defined by

R^i (F) (M) = H^i (F(J*))

0-->M-->J* is an injective resolution of M

Left derived functor of a right-exact functor is similar, but take projective resolution instead.

sorry still too advanced

Well that is the definition, I will try some motivation.

Given an Abelian category A, i.e. a category that behaves like the category of abelian groups, we can construct another category C(A) of chain complexes.

Exactness is a very important property for sequences of objects in A. Cohomology is a measure of the degree to which a complex is not exact.

Exact functors are functors that preserve exact sequences. Right/Left exact functors preserve exactness up to the leftmost/rightmost object in the sequence.

If we have an exact functor on A, it can be trivially extended to a functor on C(A).

However we can make this functor "nicer".

The derived category D(A) of an abelian category A, is essentially C(A) with complexes taken to be equivalent if their cohomology is equivalent. (i.e. we localize C(A) at quasi-isomorphisms)

(A quasi-isomorphism is a map of complexes s.t. it is an isomorphism on the cohomology)

So we have extended our right/left exact functor to C(A), but now we want to extend it to D(A). i.e. We want the functor to map quasi-isomorphisms to quasi-isomorphisms

To do this we must take the left/right derived functor of the right/left exact functor.
The original functor ends up being a sort 0th order approximation of the derived functor.

ok cool

And, how to apply the Yoneda lemma? There's this thing sometimes called the Yoneda principle:

If [math]\textbf{C}[/math] is a locally small category, [math]y \colon \textbf{C} \to \textbf{Sets} ^{\textbf{C} ^{op}}[/math] is the Yoneda embedding, and [math]A, B[/math] are any objects of the category, then [math]yA \cong yB[/math] implies [math]A \cong B[/math].

Now, to show the firepower of this idea, assume our locally small category is cartesian closed and has coproducts. The claim is that [math](A \times B) + (A \times C) \cong A \times (B+C)[/math]. Using the Yoneda principle, [math]y(A \times (B+C))=\text{Hom}(A \times (B+C), X) \cong \text{Hom}(B+C, X^A) \cong \text{Hom}(B, X^A) \times \text{Hom}(C, X^A) \cong \text{Hom}((A\times B)+(A \times C), X)=y((A\times B)+(A \times C))[/math].

And, by firepower, I mean it makes the proofs very simple sometimes.

>if I can't skim over an explanation of something complicated and understand it, it's too advanced

hey anime guy
Firstly, it's not like the steps in this chain of equalities is obvious. Secondly, this ought to be an application of the embedding but the y's are on the left and right end. It's not even expressed what's being shown here.
To conclude, you just like to write down math you like, some of yoyr personal insights, and nowhere are yoy putting yourself into the learners perspetive. In turn, you make the subject you want to promote eveb more opaque and elitist.
No offence, but think about your actyal goals and how it relates ti what you do

Sorry, I haven't been talking to anyone except sending my homework to a prof, not even posting here, since late December. so my communication skills have degraded quite a lot. I must also admit I assumed the reader to be familiar with these concepts, considering the fact we are discussing the Yoneda embedding here.

1) You know that isomorphism is transitive, that is, [math]X \cong Y[/math] and [math]Y \cong Z[/math] imply [math]X \cong Z[/math]? This now gives that if the stuff between the embeddings are correct, then the embeddings are isomorphic.

2) A category being cartesian closed means [math]\text{Hom}(A \times B, C) \cong \text{Hom}(B, C^A)[/math] for all objects [math]A, B, C[/math]. This is just a definition.

3) The functor [math]\text{Hom}( \cdot , X)[/math] satisfies [math]\text{Hom}(A+B, X) \cong \text{Hom}(A, X) \times \text{Hom}(B, X)[/math]. Here [math]A+B[/math] is the coproduct of [math]A, B[/math], and since the functor is contravariant, it takes coproducts to products, and vice versa.

Now, the first isomorphism (equality) is just what the embedding does. The second isomorphism is the cartesian closedness, the third is contravariance, the four is also contravariance and cartesian closedness. It could be replaced with [math]\text{Hom}(B, X^A) \times \text{Hom}(C, X^A) \cong \text{Hom}(A \times B, X) \times \text{Hom}(A \times C, X) \cong \text{Hom}((A \times B) + (A \times C), X)[/math]. Since the middle stuff is OK, the flanking objects are isomorphic, by transitivity.

is it me or is that more complicated