point-set-topology

[plain raven]

the objects of this category are like, $V^{\otimes k}\otimes M^{\otimes \ell}$ for natural numbers $k,\ell$. each value of $\ell$ corresponds to one connected component of the category

[gentle osprey]

diligentClerk

[uncut surge]

i think i roughly see your point, yeah, that is a very cool perspective! i'm not quite categorically-minded enough to translate that 1:1 to the de rham setting, but this shouldn't be impossible

[plain raven]

i worked out exactly what this connected component is in the case $\ell=1$ and found that i actually know how to describe its (opposite) category fairly well: the opposite category can be characterized as having as objects the sets $0=\emptyset, 1 = {0}, 2 = {0,1},\dots$ and where maps from $n$ to $k$ are just all injections. An injection $n\to k$ corresponds to a map $V^k\otimes M\to V^n\otimes M$.

[gentle osprey]

diligentClerk

[plain raven]

Call this category $\Delta_{dg}$. so the action of $V$ on $M$ generates a presheaf of $k$ modules over $\Delta_{dg}$. then i found a nice like, canonical way to 'realize' such presheaves as chain complexes, by a kind of cousin of the dold kan functor, and it gives the Koszul complex

[gentle osprey]

diligentClerk

[plain raven]

i'm pretty psyched about this but i'd really like to understand the de rham complex better

[uncut surge]

holy shit lmao that is sophisticated

i feel like everything i could suggest is just bebbi's first mathematics

[plain raven]

yeah i'm into freely generated monoidal categories. and nah lol i have a lot of trouble understanding the shit i'm reading online relevant to this, i found this comment from Mariano alvarez and it makes no sense to me, like i said 'something something operads'

https://mathoverflow.net/questions/12029/algebraic-categorical-motivation-for-the-chevalley-eilenberg-complex

[cedar pebble]

yea so I think the Chevalley-Eilenberg construction is probably the most algebraically natural explanation for the de Rham complex in the smooth setting (without dealing with Hochschild stuff)

it's certainly going to be related to the Hochschild setting in the end because of HKR

[plain raven]

Yeah i just groan a little internally when i think about reading weibel's chapter on hochschild homology, it's like 60 pages

[cedar pebble]

yea that chapter is a pain

I think you already said this above but the de Rham dg-algebra is the CE construction applied to the tangent Lie algebroid TX of X

which is sort of aesthetically satisfying because there's a deeper sense in which most constructions involving dg-algebras are really controlled by the Lie operad (or its higher operad refinement)

[uncut surge]

i guess one obvious, more algebraic perspective is the one of kähler differentials and kähler forms, but the connection to the smooth de rham complex is a little bit loose i think because of garbage like $de^x \neq e^x dx$ in the algebraic setting

[gentle osprey]

Lartomato

[cedar pebble]

yea this is sort of why I refrain from giving this as "the" explanation in the smooth setting. There is a way to make this work properly in the smooth setting, but it's a little funny

like there is a way to make sense of Kahler differentials properly in the setting of smooth algebras and then set up an appropriate HKR theorem that recovers the de Rham complex using Hochschild methods

that's an okay approach too I guess but again it's like

since you're extracting a dg-algebra in the end it should be controlled by some (higher) Lie theory considerations. This is what the tangent Lie algebroid is controlling here.

[plain raven]

Yeah i think that would also be a good setting to consider. I've tried to consider them from different perspectives. in many different settings one can carry out a similar construction but i haven't yet found a situation in where this complex just 'builds itself from scratch', so to speak, i.e. it's the free whaddyacallit generated by such and such etc

[uncut surge]

that is a bit surprising since the de rham differential on higher forms is really just inductively defined from knowing what it does on the lowest level plus leibniz rule, but i get that that might not be easy to make into a "free" thing rigorously

[plain raven]

I'm having a bit of a hard time with the Kahler differentials. just a bit of confusion because there are two ring objects involved. d is of course an R-linear map but not linear with respect to the action of O_X. on the other hand when we form the exterior algebra one can think of it as the exterior algebra of the cotangent sheaf, or sheaf of differentials, over the ring O_X. somehow i don't really get why in the very abstract sense, d should survive that quotienting if it's not a morphism in the category

[cedar pebble]

I mean one thing you can say about freeness is that the de Rham dg-algebra is a semi-free commutative dg-algebra, and then there is a structure theorem that says that any semi-free commutative dg-algebra which is degree-wise finite dimensional has to arise as the CE construction of a Lie \infty-algebroid of finite type

the data in this case is concentrated in degrees 0 and 1 (since you only care to have your differential plus a single bracket), so arises from a Lie algebroid

[uncut surge]

ye the more i think about it, the lie algebroid perspective seems like a really good one to categorically make sense of it

(tho lie algebroids themselves seem like a very high-level starting point already?)

[cedar pebble]

yea definitely

[plain raven]

alright. i think i'll spend some more time learning this stuff, don't know much about lie algebroids at all. was hoping i could find a simple answer that didn't require me to delve too deep into the subject matter. i found a talk by jacob lurie on "Lie Algebras and Homotopy theory" that looks like a pretty good introduction

[cedar pebble]

I mean also one reason why the Hochschild/HKR approach might not be the right one to consider is that the HKR isomorphism does not extend to an isomorphism of dg-algebras here

[plain raven]

he talks about the 'universal' lie algebra, maybe this is a PROP? anyway seems pretty interesting

[cedar pebble]

the definition of Lie algebroids itself isn't so bad it's only when you get to this higher stuff that it gets very homotopically annoying

[plain raven]

ok

let me ask one last question.

https://mathoverflow.net/a/77625

does anybody know what filtration this guy is talking about

I saw somewhere that you define it iteratively by looking at the preimage of iteratively defined sets under the comultiplication but i just don't know anything about hopf algebras other than the bare minimum you learn in like, two weeks of a singular cohomology course

[uncut surge]

you learnt about hopf-algebras for a singular cohomology course

[plain raven]

and i don't know how to describe that filtration more easily so that i can actually work out the associated spectral sequence as he says

[cedar pebble]

yea I was gonna say wow

[plain raven]

eh there was 1 section in the textbook on em, it was 10 pages, i read it. i skipped over the worst of the algebra anyway

[cedar pebble]

I think the filtration here is the one that comes from the canonical filtration on U(g)

[plain raven]

ok cool. so what is that

i know this formua

$C_0$ is defined to be the 'sum of simple coalgebras of $C$'

[gentle osprey]

diligentClerk

[plain raven]

and $C_n = \Delta^{-1}(C\otimes C_0 + C_{n-1} \otimes C)$

[gentle osprey]

diligentClerk

[cedar pebble]

you get a filtration U_0(g)≤U_1(g)≤...≤U(g) where U_0(g) is the base field f and U_n(g) is the k-subalgebra of U(g) generated by products of length n

i.e. by products i(x_1)...i(x_n) where x_1,...,x_n are in g and i:g->U(g) is the canonical embedding

[plain raven]

where I guess $\Delta$ is the free algebra map generated by letting $\Delta(x) = 1\otimes x + x\otimes 1$ for $x\in g$

[gentle osprey]

diligentClerk

[plain raven]

is that right?

ok that's cool, that seems very common-sense. i dig it

[cedar pebble]

I think this is maybe unrelated to what I'm referring to? But yea you get a canonical length filtration on U(g), this passes to a filtration on this simplicial construction

[plain raven]

i thought that might be what he was talking about but i wasn't sure, i didn't even know U(g) was a hopf algebra

mhm

[cedar pebble]

and yea the Hopf algebra structure on U(g) is such that elements of g are Lie like, that is ∆(x)=1⊗x+x⊗1 and this determines it for general elements of U(g)

[plain raven]

ok.

thanks to both of you

[cedar pebble]

it's similar to how the Hopf algebra structure on the group algebra K[G] is defined, where you demand that elements of G are group like, that is ∆(g)=g⊗g

[orchid forge]

@uncut surge do you have a name or reference so I can look up examples of this cohomology

Also, in simple cases does it just recover whatever homology you started with?

Oh nevermind

I think I know where to find it

[uncut surge]

o which cohomology did you mean

the vector field one

that's commonly called gelfand-fuks cohomology, and all sources i know for it are awful, but if you're up for an unpleasant adventure, "Cohomology of Infinite-Dimensional Lie Algebras" by fuks from 1984 is your guide

i also have 50 page notes on that thing which i've been waiting for my supervisor to read for a full year but he doesn't want to

[orchid forge]

I remember that one, but I meant the general idea, derived functors of the sections of some local cohomology sheaf

[plain raven]

yo does anybody know what breaks if you try to do spectral sequences with a sequence of chain complexes $C_0\to C_1\to \dots$ where the maps of chain complexes aren't necessarily injective? This is all in $R$ mod, so everything about the objects is nice

[gentle osprey]

diligentClerk

[plain raven]

like i want to compute the homology of the colimit complex of this sequence

and i know that over R-mod, homology commutes with filtered colimits. so it's not totally implausible that there's a spectral sequence associated to this colimit

[orchid forge]

With injective resolutions that aren't necessarily injective?

[plain raven]

does anything go wrong? what happens if i just pretended like it was a filtration and started trying to compute it

lmao

yes

nah i mean

wait what are you saying exactly

[plain raven]

at that point it might be easier to see if you can get an appointment in their calendar and just lecture at them for an hour about it

[orchid forge]

I thought you were trying to compute something in particular with a resolution, but it seems like you're just trying to see if the computation still makes sense

[uncut surge]

oh those thoughts that i was writing down aren't based on any literature; what comes kinda close to that is what Bredon does in his book "Sheaf Theory", maybe that's worthwhile to look into

he does formulate spectral sequences that come from sheaves of DGAs and when you can say nice things about them

[plain raven]

I am trying to compute something in particular and i'm wondering, first of all, if there's a well-defined spectral sequence associated to a sequence of chain complexes $C_\bullet \to C''\bullet\to C''\bullet\dots$, and second of all, if under good conditions it converges to the homology of the colimit complex $\operatorname{colim}C^{(n)}_\bullet$

[gentle osprey]

diligentClerk

[uncut surge]

hmm does make you wonder what kind of good condition that could be, something like 0 \to C \to 0 of course can't give you anything meaningful

i have no thoughts beyond that tho unfortunately

[orchid forge]

Just in the simplest cases, say X equipped with a good cover, this construction would preserve Cech cohomology wouldn't it?

[uncut surge]

@orchid forge If you mean something like "If S is a sheaf of DGAs, then taking the DGA-cohomology commutes with Cech cohomology", nah, that's generally not true

[tough imp]

Whoops

[uncut surge]

god i'm too stupid to construct examples

it might actually work out for the de rham complex over a smooth manifold? man, it's too late for my brain

[plain raven]

ok thanks. cool

i honestly have no idea what the homotopy colimit of a diagram in Ch(Ab) is.

[uncut surge]

I think if you do this for the de Rham complex over a manifold, then applying de Rham cohomology and then Cech cohomology gives you the Cech cohomology of the constant sheaf (in degree zero of the de Rham cohomology), and if you do it the other way round you get the de Rham cohomology of the manifold (in degree zero of the Cech cohomology)

This doesn't straightforwardly "commute", but if you write this as two spectral sequences, you can argue that they converge to the same limit, and there's your proof that Cech cohomology is identical to de Rham cohomology

[plain raven]

ok.

that makes sense

[plain raven]

Yeah, I understand, I got that by your reference to the 'telescoping trick' in Top

[orchid forge]

I think this is what I meant

[quasi forum]

So I am messing around with this problem trying to see if I can get anywhere.
I think the thing the perplexes me the most is how to deal with the supremum in this problem

That is most likely what causes a restriction on a_i

My initial thought is that since delta= epsilon/|a_n|, we need to force all a_i's to equal one another

However, I am not sure this is a sufficient requirement for continuity, I am not sure it is a necessary one

I could be wrong, but I am pretty sure I proved continuity regardless of the values for a_i

If I am wrong, please correct me. My intuition must be failing somewhere if I am

Oh wait....a supremum for a_i has to exist

[gentle osprey]

Thembossing

Thembossing

[fading vale]

Maybe im missing an easy argument but i dont really see it

[plain raven]

folks i am ????

https://mathoverflow.net/a/77625

i've come back to this a couple times over the course of the afternoon trying to figure out what he's saying but i just for the life of me cannot get the same answer that he does

Perhaps there's some ambiguity in his comments and i'm misinterpreting it.

[fading vale]

Diligent interrupting my incredibly serious and important question on belyis theorem to categorypost... literally quaking

[plain raven]

i'm sorry lol you deleted the original post so i didn't see your profile picture attached to it.

[fading vale]

Huh

Oh texit

[plain raven]

I assumed that dackid posted that message, my eyes just glossed over your name

[fading vale]

Sad!

[plain raven]

yeah i mean you deleted the tex

[fading vale]

(its fine btw i was just joking)

[plain raven]

ok i will ask this question in #groups-rings-fields. surely the sylow group theory people will help me solve this

[fading vale]

Lmfao

actually i will just ping chmonkey

[plain raven]

good idea. @tough imp please help me with my spectral sequences thing and DO NOT READ OTHER MESSAGES IN THE CHANNEL UNTIL THIS GETS RESOLVED

just kidding. help moth. idk i'll go somewhere and try to figure this out

[orchid forge]

not sure what clerk is asking tbh

what's complicated about constructing a filtered simplicial cocommutative coalgebra

jk that's just a lot of syllables

[plain raven]

i mean i think I get what he's trying to say after a lot of help but still

i sat down to do the spectral sequence computation he's talking about and I got that $E_1$ was equal to.... $0$

[gentle osprey]

diligentClerk

[plain raven]

just straight up $0$

[gentle osprey]

diligentClerk

[orchid forge]

what a pleasantly computable homology theory

[plain raven]

yes!

extremely well behaved chain complex

i mean that's a reasonable answer to the question. Like, the spectral sequence is a tool for computing the homology, and the whole 'filtered simplicial cocommutative coalgebra' thing should be a free resolution just by appeal to abstract nonsense. So including the augmentation, the homology should be zero, and ideally if you run the spectral sequence out for far enough you should get 0. What this guy is saying tho is if you compute only the first step of the algorithm, the original complex should simplify down to a simpler complex with the same homology, and this is the standard Chevalley-Eilenberg complex. Somehow i'm already getting that everything vanishes in a puff of smoke in the first step of the algorithm. I can't figure out what I'm doing wrong

[orchid forge]

I do understand the question, I just don't know how to answer it if you want to rubber ducky your way through the argument there's a nonzero chance that one of us will figure it out along the way

[plain raven]

i think i have an educated guess

i just came across a much more sensible interpretation of one of the words he used.

He just said "the filtered complex" and i just made an educated guess as to what filtration to use. But I found a paper which suggests that there's a different filtration which is considered standard (which makes a lot more sense than what I was doing lmao) and so i'm going to try this

[sullen patio]

Hey guys

I know this is basic but can anyone help me find a homeomorphism between (0, 1) and (a, inf)? Every example i come up with breaks when i take a < 0

[plain raven]

Natural logarithm gives a homeomorphism (0,1) <=> (-\infty,0). Then (-\infty, 0) is homeomorphic to (a,\infty) by means of the map x \mapsto -x +a

[sullen patio]

Omg

[plain raven]

I guess a shorter way to say this would be $e^{-x+a} : (a,\infty) \to (0,1)$

[gentle osprey]

diligentClerk

[sullen patio]

I was so closeeee

Thank u 🙏

[plain raven]

np

ok let me spell out my issue.

Let $\mathfrak{g}$ be a Lie algebra. For the sake of argument suppose it's free over a commutative ring $k$. Let $U(\mathfrak{g})$ be the associated universal enveloping algebra, constructed by taking the free tensor algebra $T(\mathfrak{g})$ and modding out by the two-sided ideal given by $x\otimes x, x\otimes y-y\otimes x - [x,y]$.

[gentle osprey]

diligentClerk

[plain raven]

There are two important maps associated to $U(\mathfrak{g})$ here: the 'comultiplication' $U(\mathfrak{g})\to U(\mathfrak{g})\otimes U(\mathfrak{g})$, which sends the monomial $x_1\dots x_n$ to $\prod_{1\leq i\leq n} (1\otimes x_i + x_i\otimes 1)$

[gentle osprey]

diligentClerk

[plain raven]

and the counit $\epsilon : U(\mathfrak{g})\to k$ which sends any non-empty product of elements of $\mathfrak{g}$ to zero; this map is the identity when restricted to the embedded copy of $k$ in the tensor algebra

[gentle osprey]

diligentClerk

[plain raven]

these two structures makes $U(\mathfrak{g})$ into a 'comonoid' in the category of $k$-modules. If you look at the diagram given by freely taking higher tensor powers of $U(\mathfrak{g})$ and all the maps that are induced between these $U(\mathfrak{g})^{\otimes n}$ from the counit and comultiplication, you'll find you've got an augmented simplicial object with $X_n = U(\mathfrak{g})^{n+1}$ and $X_{-1} = k = U(\mathfrak{g})^0$. The face maps are given by applying the unit to one of the elements. The degeneracy maps are given by applying the coproduct.

[gentle osprey]

diligentClerk

[cursive spade]

any intuition behind this definition?

[plain raven]

What's a decomposition / decomposition space?

[cursive spade]

give me a sec

loosely speaking it's a collection of disjoint subsets of a space X

a set is open in D is there union is open in X

[plain raven]

well what book is this and what's the general context

[cursive spade]

stephen willard general topology
he talking about quotient spaces

[plain raven]

There is a filtration on $U(\mathfrak{g})$ given by setting $U(\mathfrak{g})m$ to be the smallest $k$-module containing all products $x_1\dots x{\ell}$ for $\ell\leq m$. We use this to define a filtration on the simplicial object $U(\mathfrak{g})^{\otimes \ast}$ by writing $U(\mathfrak{g})^{\otimes n}_m$ the smallest submodule containing all tensor powers $p_1\otimes\dots p_n$ where the sum of degrees of the $p_i$ is less than $m$. It is not hard to prove that the face and degeneracy maps do not increase the total degree of elements, so this is indeed a filtration of the simplicial object.

There is a spectral sequence associated to this filtration. The first page of this spectral sequence is given by taking quotients $U_{k+1}(\mathfrak{g})^{\otimes\ast}/U_{k}(\mathfrak{g})^{\otimes\ast}$ and computing the homology of that simplicial Abelian group. But for me as far as I can tell the homology of this simplicial Abelian group is always zero

[gentle osprey]

diligentClerk

[hollow harbor]

almost thought this was an answer to bchaotics question lol

[plain raven]

oh bchaotic i see what's going on.

have you ever studied partitions and equivalence relations

[cursive spade]

we already defined the quotient top induced by a map I think this is to somewhat facilitate the transition to quotients defined by ~ equivalence relations

[cursive spade]

yeah

but I don't see the intuition for this semi-continuous definition

[plain raven]

well it's certainly defined in order to make theorem 9.9 true. they want to give a criterion for when the quotient map is closed that can be understood in terms of the equivalence relation/partition on X

have you read the proof of theorem 9.9? maybe it'll give some intuition

[cursive spade]

I skimmed it

[plain raven]

Ok. Write q : X -> X/~ for the quotient map. Then U in X is saturated iff U is of the form q^{-1}(V) for some open V in X/~. They're asking you to focus on the open sets which are in the range of the preimage map q^{-1} : T(X/~) -> T(X).

[cursive spade]

that is focus on saturated open sets

but why

[plain raven]

A decomposition is upper semicontinuous iff, whenever U is open in X and F is an equivalence class contained in X, there's already some V in the image of $q^{-1}$ in between U and V. This has the immediate consequence that, if F is fixed, and U is allowed to vary around all neighborhoods of F, we can always choose some V. So there is a neighborhood basis for F consisting of opens of the form $q^{-1}(V)$, where V is an open nbhd of the point [F] in X/~

[gentle osprey]

diligentClerk

[plain raven]

These distinguished kinds of open sets are cofinal among neighborhoods of F

[cursive spade]

cofinal is new to me

[plain raven]

here it just means that no matter how tightly you draw U around F you can always find some intermediary V of the desired form. Even if you took a descending sequence of opens {U_i} getting closer and closer to F, say the infinite intersection of the U_i was F, you could always find a sequence {V_i} doing the same thing with V_i contained in U_i for each i.

where the V_i are saturated

if you want to see a more "geometric" theorem that uses the same basic arguments as this one / same concepts (but is easier to visualize maybe because the spaces are nicer?) look up the proof that a proper map of locally compact hausdorff spaces is closed

[cursive spade]

it's getting clearer, so would the equivalence relation whose set of equivalence classes are upper semi-continuous are some special type

so these would have some nice properties ?

[plain raven]

yeah closed maps are really nice. i mean, like, in any hausdorff space compact implies closed and the continuous image of a compact set is compact. so this has the property that for any map from a compact Hausdorff space X to a Hausdorff space Y, the map is closed, and it's really useful to know that closed sets get sent to closed sets.

I guess i would stress that if a map f: X -> Y is both closed and open, then since X itself is closed and open, f(X) is closed and open, so f(X) is the union of components of Y

if Y has only one connected component then f(X) must surject onto Y

there are some theorems in complex analysis which say something like "for a holomorphic function f, f is either open or constant"

and if we add in the hypotheses above, like we were doing complex analysis with locally compact hausdorff spaces and proper maps, then the maps would necessarily be closed as well, thus clopen

[cursive spade]

clopen here means both close and open?

[cursive spade]

is there a nice proof of this

[plain raven]

https://math.stackexchange.com/questions/1604210/when-is-the-image-of-a-proper-map-closed

let me also mention some cool immediate corollaries

[cursive spade]

nice this is cool , thanks @plain raven

[plain raven]

np glad i could help

[shadow jolt]

I can get the $i \notin \mathbb{A}$ case, but for the $i \in \mathbb{A}$ case i get $(-1)^{|\mathbb{A}-1|}e_\mathbb{A}e_i$. I mean $(-1)^{x+1} = (-1)^{x-1}$, but I think i'm missing something more drastic. Is it more complicated than counting the swap operations to get $e_i$ on the other side of $e_\mathbb{A}$?

[gentle osprey]

power_transformer

[supple locust]

I was trying to understand how I can get a space of specified fundamental group. So we start with Wedge of circles (depending on the number of generators of the group). And we attach 2 cell along the relations (which give the group by quotienting). Now Van Kampen's theorem is supposed to tell me that the fundamnetal group of resulting space is the specified group. I understand that by Van Kampen the fundamental group of resulting space is free group on some generators quotiented by the fundamental group of the intersection of U and V (U and V) are open sets containing wedge of circles and the two cells ). Why is this group exactly the group generated by the relations ?

[vast estuary]

Hi

How do I prove this?

[vast estuary]

What do they mean by "identify" in the second step?

[empty grove]

define an equivalence relation which makes those things equivalent and quotient by it

ie treat them as the same thing

[vast estuary]

Yeah I know that

but I am not sure how to explicitly write this down

especially since we are just taking "copies" of stuff in R^n

[empty grove]

Do you see the picture?

Are you just asking how to formally write it

[vast estuary]

Yep

[empty grove]

So when we take multiple copies of the same set we just put a label on them which distinguishes them

like if you have multiple copies of S, you just take S x {1}, S x {2} etc

[vast estuary]

Interesting, sure

[empty grove]

Try doing this for the copies and then it'll be good

[vast estuary]

or I can just say S_1, S_2..?

[empty grove]

Yep

[vast estuary]

this is better in terms of defining maps actually

[empty grove]

Well it will be understood that you have done something of that sort when you write S_i

How you distinguish elements does not matter

So we usually don't care about it

[vast estuary]

Right. Let me see how to write this down formally

Although how do we show it's a homeomorphism

because that would need us to know the topology on |K|

and that is not clear from construction

[empty grove]

No, you're taking product with a singleton in what I said

You just put the discrete top on that singleton

S is homeomorphic to S x singleton

[keen urchin]

i count that the following set has 6 connected components. am i right?
$$\mathbb{C} \times \left( \mathbb{C} \setminus {0} \right) \times \left{ z \in \mathbb{C} : 1 \neq |z-1| \neq |z| \neq 1 \right}$$

[gentle osprey]

reking

[empty grove]

Yes

[pearl holly]

Okay I'm misunderstanding something very badly rn

So let's say that $A$ is a retract of $X$. Then I have the long exact sequence
$$\cdots \to H_{n+1}(X, A) \to H_n(A) \to^i H_n(X) \to H_n(X, A) \to H_{n+1}(A) \to^i H_{n+1}$$

[gentle osprey]

Tokidoki ✓

[pearl holly]

Okay and now since i is injective I can just replace H_n+1 with 0 right?

that would give me an isomorphism between H_n(A) and H_n(X)

[empty grove]

no

It just means the the first map is 0

Not that its domain is zero

[pearl holly]

right so the boundary is the zero map, i.e it has image 0 right?

[empty grove]

Yes because the next map has 0 kernel

[pearl holly]

wait let me post a pic of the exact sequence

[empty grove]

Wait your exact sequence is going from n+1 to n to n+1 lol

I hope you're talking about the first n+1

[pearl holly]

Okay so in this pic I would have 0 --> H_n(A) --> H_n(X) --> Hn(X, A) -->etc. right?

[empty grove]

Sure

But that zero is not H_n+1(X,A) necessarily

[pearl holly]

okay and that's because the image of \partial is 0, right?

[empty grove]

Yes

[pearl holly]

wait what?

oh no wait

[empty grove]

H_n+1 → H_n map is the zero map

But that doesn't mean that its domain is 0

[pearl holly]

since the image of \partial is 0 I would have 0 --> H_n(X) --> H_n(X, A) --> etc.

and this is wrong

[empty grove]

Yes you can replace H_n+1 with 0 while keeping exactness after that point

If you have an exact sequence
... → A → B → ...
and you find that the map I've drawn in the middle is the zero map

Then you may change this sequence to
... → A → 0 → B → ...
retaining exactness

But you can't say that A or B is 0

You can take a zero map between any 2 abelian groups

[pearl holly]

ah yeah right so the zero map just factors through the 0

[empty grove]

Yes

[pearl holly]

ahhh yeee okay now I see

ye because I always thought that if the middle map in ... →A →B →... is zero then I could just do this ... →A →0 →... and retain exactness

[empty grove]

F

[pearl holly]

yeah right okay I see now lmao. Thank you for the help!

[empty grove]

[covert inlet]

Hello, if I have two ideals I, J of respective spaces X, Y it is true that if $J \subset I$, then there exists a function $f: X \to Y$ such that for all $A \subset Y$, if $A \in J$ then $f^{-1}(A)\in I$?
My intuition tells me yes and maybe it could be the identity function, but I would like to corroborate it. Thanks.

[gentle osprey]

Samuel Martinez

[vast estuary]

Done

[vast estuary]

Any hints for this though?

[gentle osprey]

Hausdorff

[empty grove]

This isn't what they are saying

And it's not true

It's not a homeomorphism, but am embedding

That's what homeomorphic copy means

[empty grove]

Are these spaces varieties/schemes? Then I think there might be a typo because A ∈ Y and A ∈ J should not be possible at the same time right?

[covert inlet]

No, it was my mistake, I have already edited it, A is subset of Y

[empty grove]

Ah I see

[vast estuary]

Oh yeah makes sense

So I need to show that f is a continuous bijection onto its image

and that it has a continuous inverse

[covert inlet]

I thought about the identity, because if A is in J, then $f^{-1}(A)={a| a\in A}\in J$ then $f^{-1}(A)\in I$. But I don't know if this is correct.

[gentle osprey]

Samuel Martinez

[dusk heron]

I know that you can get a 2-holed torus (genus 2 surface) $\Sigma_2$ by gluing the edges of an octagon appropriately. But is there anybody who has actually given a parametrization of $\Sigma_2$ (as a surface in $\mathbb{R}^3$) with such an octagon as a parameter domain (respecting the identifications needed on the edges)? Is there something else similar to that? The corresponding thing for a torus $\Sigma_1$, with a square $[0,1]^2$ as parameter domain, is pretty easy to come up with by yourself

[gentle osprey]

gustavn64

[dusk heron]

The reason I ask is that I want to make animations illustrating homotopies between some loops in $\Sigma_2$, and such homotopies are easily described in terms of the octagon with glued edges.

[gentle osprey]

gustavn64

[kind cedar]

Can someone help me understand the relationship between "the rank of the Jacobian matrix is not maximal" and "the function fails to be surjective"?

To be more specific, I'm trying to understand critical and regular points on a manifold in an intuitive way, and both definitions are related but I can't see why

[orchid forge]

They are equivalent if f : R^n --> R^m and n >= m

Assuming by "the function" you mean the differential

The Jacobian matrix represents the differential df, which maps tangent vectors to tangent vectors. Its rank is maximal if it equals m, since the rank can't exceed the dimension of the codomain, and it's also maximal if it equals n, since it can't exceed the dimension of the domain

If n >= m and df has maximal rank, then necessarily it has rank m, so the image is all of TR^m

If n <= m and df has maximal rank, then it has rank n so it is injective

You can replace df, R^n, and R^m by any linear map from n dimensions into m here

It's purely linear algebra

[kind cedar]

I guess I got the idea

[kind cedar]

I'm supposed to prove the Torus $\mathbb{S}^{1}\times\mathbb{S}^{1}$ is diffeomorph to $\mathbb{R}^{2}/\mathbb{Z}^{2}$. I have already proven that the the quotient of M by the proper discontinuous action of a group (M/G) is diffeomorph to M, and I understand $\mathbb{Z}^{2}$ is a group with proper discontinuous action on $\mathbb{R}^{2}$ using a simple equivalence p~q iff $(p-q)\in\mathbb{Z}^{2}$.
However I do not know how to link the Torus to the group action... 😦

[gentle osprey]

Necrowizard

[gritty widget]

why not just give an explicit diffeomorphism

[kind cedar]

I'm kinda thinking it's not what my professor wants

based on the previous exercises

[kind cedar]

what explicit diff. could I use? (x,y)=(tan(theta1),tan(theta2)) ?

[orchid forge]

If you restrict to [0,1] x [0,1], you still get a torus

under the projection map, each copy of [0,1] turns into an S^1

[orchid forge]

or if you want an explicit map, for the same reason, $(x,y) \mapsto (e^{2 \pi i x}, e^{2 \pi i y})$

[gentle osprey]

Kogasa

[gritty widget]

what does it mean in (b) to compute?

like for each e_j i've found curves y_j(-\eps,\eps)\to S^3 such that y_j'(0)=e_j

so then dF_1(e_j) = (F\circ y_j)'(0)

but how do i compute this further?

given that (F\circ y_j)'(0) is in T_{F(1)}SU(2) which are derivations

and there's not rly a canonical way to write a derivation right?

[slender plank]

Hello guys I want to show that an open ball B(a,r) is the interior of a closed one B(a,r?

But I font know how to caractérise the fact that an element doesn't belong to the interior

Please help

[unreal stratus]

Well firstly we know that the open ball is contained in the interior of the closed ball

What would happen if we had a point exactly a distance r from a in the interior of a?

Considering that should give you the answer

[slender plank]

For every open ball we build around x we can find a y that doesn't belong to A ?

But I don't get what it's like as a point to not be in the interior

Does it mean that every open in A don't contain such a point

[supple locust]

how is this a normal cover of wedge of two cricles a and b.

like $a ab a^{-1}=a^2ba^{-1}$ does not belong in the group of this covering space

[gentle osprey]

bert

[gritty widget]

i think this should be the same as saying

if p is zero on a dense open subset

then p is zero

but im struggling a bit to see this

any non empty open set is dense in the zariski topology?

and any empty set is the compliment of some closed, hence some algebraic set

[maiden oracle]

im trying to come up with an example of a simply connected space that is not contractible, i know that S^n is not contractible for n > 1 but we haven't proven Brouwers fixed point theorem for N>2 so i can't really use that. However i was wondering whether i could just take [0,1) \cup (1,2] you can contract each component but you can't retract the whole space to a single point, right? i somehow want to show this via contradiction, namely by showing that if this were contractible it would have to be connected but it isn't but i am struggling on proving this

[lunar yoke]

usually a requirement for simple connectedness is that the space is path connected

but [0,1) u (1,2] isnt

[maiden oracle]

we defined simply connected as every fundamental group pi_1(x,X) is trivial and 1-connected as simply connected + path connected

[lunar yoke]

hm then yes that should work

if its contractible it should even be path connected iirc, maybe thats easier to show

[maiden oracle]

yup well that's true, i even saw a proof of this fact a few days ago xd

somehow i didn't think of it

thanks!

[swift fjord]

Why can we take a nbhd of 1 here?

ah I see, yea I think that's it

well something in that vein

close enough

lol

wait right that doesn't make sense actually

I also got confused by the notation

I don't think translates would preserve the distance

like if f(y) is close to f(1)

does that necessarily mean that f(xy) is close to f(x)?

since f is not a homomorphism

oh ok yea I see it now

yea

take the nbhd U of 1 with |f(y)-f(1)|<ep/2, then take some nbhd U_x such that xy falls in U for each y in U_x

from continuity

then U_x is your desired nbhd

Yea I think

that's pretty much the same thing you said but formalised lol

[vast estuary]

Here are we talking about an abstract simplicial complex K, or its topological realization |K|?

[empty grove]

Abstractly it's not a top space so you can't compare it to one

[vast estuary]

So when they say n-simplex, they mean the standard n-simplex

and not a set of n+1 elements, right

[empty grove]

They mean the copy of the standard n-simplex inside |K|

Not the standard n-simplex itself

[stray escarp]

Can we define the chessboard as a metric space?

an N X N tabel

chess board

yes?

I know it will sound stupid but

Can we define a metric over N?

Ohh

Ty man

Okay

[long hornet]

Do you want it to reflect the rules of chess somehow?

I'm not sure you can do that, because they depend on the relative positions of the pieces (most squares are essentially identical anyway).

[stray escarp]

Yap

[long hornet]

It is probably more sensible to define a metric on the "configuration space" of it, by which I mean the space of all possible arrangements of pieces in a chess board (which is huge!)

[stray escarp]

Ohh

Thats bad

[long hornet]

It has 2^192 points (if you don't restrict to "legal" positions)

I think one reasonable way to topologize X (this set) would be as follows. Regard X as a graph, by linking two points iff you can move from one to the other by a legal move. Then X has a natural topology as a graph. Also, the subspace X0 of all legal positions has a simple description: It is just the connected component of p0 (the standard initial position).

[stray escarp]

In theory it should work

[slender plank]

using sequences show that the grip of an open ball is the closed one

I dont know what sequence i should consider

[keen urchin]

what does "in theory it should work" even mean in regards to mathematics

[gritty widget]

can someone pls help with this

[pearl holly]

so if I have a boundary map \partial and I write it like a matrix, is then the dual of that boundary map the transpose of \partial?

[gritty widget]

this is a standard linear algebra fact, nothing special about boundary maps

[pearl holly]

oh shit lmao I didn't know that

I was watching someone on youtube and they called the dual of the boundary map a transpose

why is that tho?

[gritty widget]

the transpose of a matrix represents the dual linear transformation

[pearl holly]

ye right so if I have like vM then this is the same as M^t v?

MTV moment

[gritty widget]

i don't know what this means

[pearl holly]

ye me neither

[pearl holly]

but how do you know this?

[gritty widget]

are you asking how one proves it?

[pearl holly]

man I'm so shitty at this

ye

[gritty widget]

it's just a computation, you can probably find it in any decent linear algebra textbook

eg theorem 2.25 in friedberg

[pearl holly]

okay great thank you so much!

[gritty widget]

ill explain the notation to keep it self contained: $[T]_\beta^\gamma$ means the matrix rep of $T$ with respect to $\beta,\gamma$ (so if $\beta = {v_1,\dots,v_n}$ then its $j$th column is the coordinate rep of $T(v_j)$ with respect to $\gamma$). also $\beta^$ is the dual basis to $\beta$, consisting of $v^j \in V^$ defined by $v^j(v_i) = \delta^j_i$

[gentle osprey]

TTerra

[pearl holly]

okay great!

[gritty widget]

the book used different dual basis notation than me but whatever

upper/lower indices are superior

[wise sigil]

I've just done some calculations and want to confirm if this is true. We're given an embedded submanifold in the $(x,z)$ plane globally parametrized by $(a(t), b(t))$. Is the latitude of the surface of revolution at time $t_0$ a geodesic $\iff a'(t_0)=0?$

[gentle osprey]

Brian485

[gritty widget]

yes i think so

[wise sigil]

Mhm thanks

[abstract pagoda]

dual spaces are collection of functionals

wow

chat update

[gritty widget]

geometrically this corresponds to the tangent to (a(t), b(t)) being parallel to the axis of revolution

i think you can find a proof of this in do carmo's book if you want to check your work

[wise sigil]

right it seems to work out for spheres and what not

ill check it out

[long hornet]

What would a map on S^2 of deg = 2 look like?

[gentle osprey]

Hausdorff

[vast estuary]

n-cell = homeomorphic to open unit disk in R^n

[lunar yoke]

like this?

[vast estuary]

lol why can't i see it

they are homeomorphic to those right

[lunar yoke]

what do you mean, the image wont load or you dont understand?

[vast estuary]

https://www.math.ksu.edu/~hansen/CWcomplexes.pdf

This says open unit disks

[vast estuary]

i don't understand it

this is a cell-complex for me

[lunar yoke]

you start with the 0 cell aka the point in the middle front of the picture

then you attach the two 1 cells that are the two rings

finally you add most of the surface by attaching a 2 cell along the boundary of the two circles

[vast estuary]

interesting

but, this is very vague

i see how that makes sense

[lunar yoke]

well i think the image should give enough information to make it formal

[vast estuary]

vague because don't we have to "prove" that those are indeed n-cells

especially that surface is not a "known" shape

visually true

[vast estuary]

when we say "attaching here" what do we mean btw

is it "attaching along a map"

where you take a disjoint union and quotient?

ok is cell complex = cw complex?

ok cool

is hatcher a good place to read cw complexes (for a beginner)

i just want a reference where these things are explained nicely

so if anyone has any recs, lmk

also here they talk about the attaching map f sending the circle along the path aba^{-1} b^{-1}

what's that? (sending circle)

the circle is one of the two 1-cells we have added right

[pearl holly]

I’m pretty sure the circle is the boundary of the 2 disk here

[vast estuary]

hmm ok maybe i need to think about these things more

[pearl holly]

https://www.youtube.com/watch?v=OodqCBPYsKY&list=PLnLAqsCN_2kcS1uc7FXwQ2Ttvtu55ZdVj&index=2

so I'm watching this video to hopefully gain some intuition but there's a moment in that video that I don't understand

so from 7:40 to 8:20 they try to visualize a cocycle. In 8:20 they say "its coboundary will include these two simplicies" and I don't really understand that and the explanation of it shortly after 8:20. So if we say that $a$ is the edge marked in red then they have a $\sigma \in C^1(X; \mathbb{Z}_2)$ such that $\sigma(a) = 1$. Now $\partial^* (\sigma)$ should be zero because that's what we are considering. But what does it mean for this equation to "include these two simplicies"?

[gentle osprey]

Tokidoki ✓

[fickle marsh]

Can anyone tell me what elements in S1 look like

I can't seem to find it on Google

I don't know if m searching for the wrong thing or what

[pearl holly]

what is S1?

[slender plank]

permutation ?

[fickle marsh]

I believe 2D circles

But I'm not sure how they are represented

[pearl holly]

then it's something homeomorphic to a circle

[fickle marsh]

Like what is -x here

[pearl holly]

the antipodal point of x

[fickle marsh]

Okay but

Of what circle

[long hornet]

It is true that an "abstract" circle doesn't come with an antipodal map

[fickle marsh]

S1 is all circles I thought

[long hornet]

But just pick the unit circle in R^2 and define an antipodal map by transporting it along a homeomorphism

[fading vale]

S^1 is a space

Like bene said this is just the ordinary unit circle

[gritty widget]

$$S^1 = {(x, y) \in \bR^2 : x^2 + y^2 = 1}$$

[gentle osprey]

TTerra

[fickle marsh]

That's very strange

I'm like 99.9% sure in my robotics class it was the set of all possible circles

[gritty widget]

notation isn't universal

[fickle marsh]

Defined as 2 dimensional for polar coordinates

[long hornet]

At least in the exercise you shared, it isn't. I guess you could think of S^1 as the "set" of all possible circles, like the "the number 2 is the class of all pairs" thing, but that's not what is done

[empty grove]

I don't get the definition of the boundary map here

I'm guessing that the left side should be
$d^p((s_{i_0...i_{p}}){i_0...i{p}})$

[gentle osprey]

Moldilocks ✓

[empty grove]

Because we should have a tuple of sections, indexed by the increasing tuples of i's

And they should only go up to i_p and not i_p+1?

But regardless I'm just unable to make sense of the right side 😵‍💫

oh I might just have understood it

Bruh

ty

Would be nice if someone could explain what this complex and its cohomologies would measure tho

Obstruction to gluability and hence to being a sheaf?

[empty grove]

This seems somewhat legit thinking about it, but would be nice if someone could say the extent to which this is true

Also try to guess what the footnote after the well ordering thing says

[orchid forge]

Obstruction to the locally constant sheaf of integers in particular being a sheaf

The definition seems fine. The boundary of a section is the alternating sum of all its restrictions

Oh i see this is for any sheaf, nevermind the first thing

[empty grove]

What is this sheaf?

oh ok

[orchid forge]

Usually cech cohomology is what you get when you let the sheaf be the locally constant sheaf of integers

And you can kind of see what's going on through computations here

[empty grove]

So each F(U) is Z, and restrictions are id?

I'm not familiar with the terminology

hmm I see the definition of a locally constant sheaf

[orchid forge]

Actually, constant sheaf

But the sections are locally constant functions U --> Z

[empty grove]

Would that be constant? Couldn't you have a point with a connected neighborhood and a point without connected neighborhoods?

Or just remove the points from that

Feels like a disconnected open set would have more locally constant functions to Z than a connected one

[plain raven]

i lose my shit when people slap an ordering on I and destroy the functoriality of it

but anyway moldilocks i don't have much intuition for the higher differentials, the most important thing is to understand what the differential looks like in low degree. so the boundary map from C_0 to C_1 sends a family of sections {f_i} to the pairwise differences f_i -f_j on the intersections U_i\cap U_j

[orchid forge]

Sorry i need a few mins

I meant obstruction to gluing sections, like the individual function guys

Not to being a sheaf

[plain raven]

it is obvious then that {f_i} is a cocycle of the boundary map iff all the f_i agree pairwise the intersection.

[orchid forge]

Brb like 5m, and ty clerk

[plain raven]

hey kog

ergo if F is a sheaf then {f_i} glues together to give a unique element of i by the sheaf axiom

so like

you can read the definition of a sheaf of Abelian groups as asking that it's a presheaf P such that for any open cover U = {U_i} the sequence
0 -> F -> C_0(F; U) -> C_1(F;U)

is exact

and it turns out that there's a natural way to extend this sequence all the way out in a way that gives a chain comlpex

that's the short version

there's also a long version.

:)

[empty grove]

Let's go

[plain raven]

sweet

i'm going to make some coffee

[empty grove]

I might fall asleep at any point tho In which case I'd have to read tomorrow

[empty grove]

Sounds like derived functors

[plain raven]

yes and no. it's not directly related to derived functors in the usual sense. imo it's more of a distinct foundation for homological algebra than derived functors which comes with a more flexible notion of what is projective/injective/exact than the usual definition that can be applied in every abelian category. you can think of it like relative homological algebra

[plain raven]

ok. so. let $M$ be a monoid, say in the ordinary sense, just a set theoretic monoid. there are two basic maps associated to $M$, the multiplication $\mu : M \times M \to M$, and the distinguished identity element $e : 1 \to M$. if you look at the whole diagram we get by freely applying the Cartesian product and looking at all the maps induced by $\mu,e$ , identity maps on $M$, etc., you get a diagram of sets called an 'augmented cosimplicial object.'

Speaking precisely there's a category $\Delta$ whose objects are all finite ordinal numbers $0 = \emptyset, 1 = { 0 } , 2 = { 0 < 1 }, 3 = {0 < 1 < 2}$ equipped with the standard linear ordering. The morphisms from $n$ to $m$ are any (weakly) monotonically increasing maps between them. $M$ determines a functor from $\Delta$ to the category of sets which sends $n$ to the $n$-fold cartesian product $M^n$, which I'll call $F$. (Here by the $0$-fold cartesian product I mean the empty product, the terminal object/singleton set.)

The morphisms of $\Delta$ contain certain distinguished maps $\delta_i : n \to n+1$, for $0 \leq i \leq n$, where $\delta_i$ is the unique injective map which skips the element $i$ in $n+1$, and the maps $\sigma_i : n+2 \to n+1$, for $0 \leq i \leq n$, which is the unique surjective map which sends two elements in $n+2$ ($i$ and $i+1$) to $i$ in $n+1$. We know that the maps of the form $\delta_i$ and $\sigma_i$ generate the category $\Delta$ under composition. So to give this functor $F : \Delta \to \mathbf{Sets}$ it suffices to say where $F$ sends $\delta_i$ and $\sigma_i$.

We define $F(\delta_i) : M^{n}\to M^{n+1}$ by $F(\delta_i)(m_0,\dots, m_{n-1}) = (m_0,\dots, m_{i-1}, e, m_i,\dots, m_{n-1})$.

We define $F(\sigma_i): M^{n+2}\to M^{n+1}$ by $F(\delta_i)(m_0,\dots, m_{n+1}) = (m_0,\dots, m_{i-1}, \mu(m_i,m_{i+1}),m_{i+2},\dots, m_{n+1})$.

[gentle osprey]

diligentClerk

[plain raven]

Sweet.

Now it's not hard to see that an analogous construction works, say, in the category of Abelian groups, where we replace $M$ by some ring, $R$. There's a diagram $F : \Delta\to R$ which sends $n$ to the $n$-fold tensor product of $R$ with itself, where the $0$ fold tensor product is understood to be $\mathbb{Z}$. the multiplication of a ring is bilinear so this defines maps out of the diagrma.

[gentle osprey]

diligentClerk

[plain raven]

More generally this works at the level of any monoidal category $C$ endowed with a monoidal product $\otimes$ and a choice of monoid $M$ with respect to that tensor product. $\Delta$ itself is a monoidal category under the tensor product given by ordinal addition (concatenation of the linear orders) with unit the empty set, and has a distinguished monoid $1$ (the multiplication and unit are obvious) and in each case there is a unique functor $F : \Delta\to C$ preserving the monoidal product and the distinguished monoid. This property characterizes $\Delta$ up to equivalence as the free monoidal category generated by a monoid.

Super important property!

[gentle osprey]

diligentClerk

[empty grove]

Ahh I remember you mentioning this before

[plain raven]

yeah it's neat

[empty grove]

You left the last part as an exercise and I did it in a much worse way

[plain raven]

Haha!

Ok.

So.

Let $V$ be a finite dimensional vector space.

[gentle osprey]

diligentClerk

[plain raven]

$V^\ast$ its dual space. Then $V\otimes V^{\ast}$ can be equipped with the structure of an $\mathbb{R}$ algebra in a natural way, with multiplication given by
$\left( v_i \otimes v_i^\ast \right)\left(w_i\otimes w_i^\ast\right) = \sum (w_i^\ast(v_i) \cdot w_i\otimes v_i^\ast$
and identity element given by $\sum e_i\otimes e_i^\ast$, where the $e_i$ is any basis for $V$ and the $e_i^\ast$ denote the standard dual basis.

Another easy way to see this is that $V\otimes V^\ast$ is naturally isomorphic to the vector space $Hom(V,V)$ by the map which sends $v\otimes f$ to the linear functional with one-dimensional image $g_{v,f} : V\to V$ defined by $g_{v,f}(w) = f(w)\cdot v$. With respect to this isomorphism the product structure can be simply described as function composition and the unit element can be characterized as the identity matrix.

[gentle osprey]

diligentClerk

[plain raven]

Make sense?

[empty grove]

I should be able to make sense of it

[plain raven]

if you choose a basis for $V$ this isomorphism basically amounts to the isomorphism between $Hom(V,V)$ and the space of $n\times n$ matrices btw.

[gentle osprey]

diligentClerk

[empty grove]

Is that the dot product?

[plain raven]

it's the action of a scalar on a vector. sorry

maybe concatenation would have been easier to read

[empty grove]

Ah ok yeah makes sense

[plain raven]

So, what's crazy is that the proof that $V\otimes V^{\ast}$ is a monoid goes through at an incredible level of generality, for very general definitions of tensor product, dual, etc. So general that, in fact, if $F : \mathcal{C}\to \mathcal{D}$ is a functor, and it has a right adjoint $G : \mathcal{D}\to \mathcal{C}$, then if you look at functor composition as a kind of 'tensor product', then $G$ is a kind of 'dual' to $F$. And in fact the same proof goes through - $G\circ F$ is a 'monoid' in the monoidal category of functors $[\mathcal{C},\mathcal{C}]$ with 'tensor product' given by composition and unit given by the identity functor.

[gentle osprey]

diligentClerk

[empty grove]

In the first formulation you say standard dual basis That's a thing?

Oh once we fix a basis for V

[plain raven]

for a fixed basis $e_i$ there's a standard dual basis. right.

[gentle osprey]

diligentClerk

[empty grove]

Right

[plain raven]

We say that $G\circ F$ is a 'monad'. Briefly the 'unit' is a natural transformation from the identity functor from $\operatorname{id}\to G\circ F$ which is just the unit of the monoid, the universal arrow $\eta_c : c\to G(F(c))$. The 'multiplication' $\mu_c: GFGFc\to GFc$ is given by applying $G$ to the counit $\varepsilon_{Fc} : FG(Fc)\to Fc$.

[gentle osprey]

diligentClerk

[empty grove]

ah I've seen monads

Just don't know how to use them and monadic functors in a nice way 😵‍💫

[plain raven]

It is a bit easier to bring this back to some concrete reality in the case where, say, $\mathcal{C}$ is the category of Abelian groups, $\mathcal{D}$ is the category of $R$ modules, $F$ is the free $R$ module functor given by tensoring with $R$, $G$ is the forgetful functor. Then the natural transformations $\eta$ and $\mu$ just correspond to the unit maps and multiplication of the ring, $R$.

[gentle osprey]

diligentClerk

[plain raven]

OK. So, if $GF$ is a monoid, then by the grand universal property of $\Delta$, there is a canonical functor from $\Delta$ to the category of endofunctors $[\mathcal{C},\mathcal{C}]$ preserving the tensor product on the nose and sending $1$ to the distinguished monoid $GF$. This gives an augmented cosimplicial object $1 \to GF \to GFGF \dots$. Applying all these functors and natural transformations to some fixed object $c$ in the category gives an augmented cosimplicial object in the category $\mathcal{C}$,
$c \to GFc \to GFGFc \to \dots $

[gentle osprey]

diligentClerk

[abstract pagoda]

fake words wtf

[empty grove]

An augmented cosimplicial object is a functor from Δ to the category?

[abstract pagoda]

augmented like augmentation in reduced homology?

[plain raven]

yes, that's the definition of an augmented cosimplicial object

and yes exactly millionaire

Ok. I have two questions before i go much further. first of all do you know anything about the Dold-Kan correspondence and secondly do you know like, basic homological algebra of ext and tor etc

[abstract pagoda]

i hate my life

i am supposed to know dold kan correspondence

[plain raven]

both of those questions were for moldi

[empty grove]

Nothing about the first, I know that ext and tor are and how to construct them and why the construction lets you extend the exact sequences

[abstract pagoda]

wait

ext functor takes exact sequences to some other category?

[empty grove]

But nothing about there properties beyond what can be easily proved from looking at a resolution

[plain raven]

Ok.

So

[empty grove]

It allows us to extend left exact sequences to longer exact sequences

In a very specific situation

[plain raven]

You want to think of the augmented cosimplicial object $c \to GFc\to GFGFc$ as a cosimplicial 'resolution' of $c$, very much like a cochain complex resolution of $c$ by injectives.

[gentle osprey]

diligentClerk

[plain raven]

ok. so the slogan we want to establish here to summarize all the shit of the past 20 minutes are: Monads give rise to augmented cosimplicial objects. They give us systematic, functorial ways to assign 'injective' cosimplicail resolutions of objects.

and yes in specific cases we can get honest to god injective resolutions out of this.

So let me give a specific example here.

consider the adjunction between $\mathbf{Ab}$ and $\mathbf{Sets}^{\rm op}$. The left adjoint is $\operatorname{Hom}(-, \mathbb{Q}/\mathbb{Z}): \mathbf{Ab}\to \mathbf{Sets}^{\rm op}$. The right adjoint here is given by the functor $X\mapsto \prod_{x\in X}\mathbb{Q}/\mathbb{Z}$.

[gentle osprey]

diligentClerk

[empty grove]

Lol a right adjoint out of Set seems very 😵‍💫

[plain raven]

yes haha but it's just another way of talking about the universal property of the product

If $A$ is any given Abelian group then the unit of this adjunction (i.e. the first map of the functorial resolution of $A$ is of the form
$A\to \prod_{f:A\to \mathbb{Q}/\mathbb{Z}}\mathbb{Q}/\mathbb{Z}$ and sends $a$ to the family of elements $(f(a))_{f: A\to\mathbb{Q}/\mathbb{Z}}$.

this map is indeed injective and the codomain is an injective abelian group. so we've given a functorial embedding of $A$ into an injective abelian group!

[gentle osprey]

diligentClerk

diligentClerk

[plain raven]

thanks odysseus

people say that the proof that Ab has enough injectives depends on the axiom of choice. this is technically true but you don't need it to construct the resolution! you need it to prove that the object $\prod \mathbb{Q}/\mathbb{Z}$ is injective and maybe that the unit map is injective but this construction can still be given and computed with independent of any choice principles.

[gentle osprey]

diligentClerk

[plain raven]

Anyway iterating this construction gives an augmented cosimplicial object whose objects are all injective (other than A). You can turn this into a cochain complex by the standard construction which takes alternating sums of the face maps $\sum (-1)^i d_i$ and this is easily checked to be a cochain complex.

[gentle osprey]

diligentClerk

[plain raven]

so it may sound a little bit weird as to why we're talking about cosimplicial objects when the original question was about cohomology of cochain complex so i just want to say, although this isn't the place to get into this right now, cosimplicial Abelian groups are essentially the same thing as cochain complexes. The Dold-Kan equivalence is the theorem that there is an equivalence of categories between cosimplicial Abelian groups and cochain complexes bounded below by 0.

if you look at contravariant functors from $\Delta$ rather than covariant functors you get an equivalence with the category of chain complexes

[gentle osprey]

diligentClerk

[plain raven]

just to add confusion, the construction i just mentioned (take alternating sums of face maps) is indeed a functor from cosimplicial abelian groups to cochain complexes but it's not the functor which occurs in the dold-kan equivalence theorem lol. the construction i just gave is called the 'unnormalized' complex, the equivalence is given by the 'unnormalized' compelx

ok this has the risk of getting really out of control so let's get it on track. what i've been doing is explaining this very broad framework for constructing cosimplicial resolutions of objects

let's talk about cech cohomology

so what are we trying to do in cech cohomology? we're trying to look at the sheaf as it behaves on the individual open sets of an open cover and consider its behavior. one way to explain what's going on here is that we're like, 'breaking apart' the space into the individual open sets of the cover and look about what happens to the sheaf on this broken-up space.

Specifically if $\mathcal{U}$ is an open cover of $X$, say by ${U_i\right}$ then let's form the space $Y$ which is the disjoint coproduct space $\coprod_{i\in I}U_i$. There's an obvious projection $\pi : Y\to X$ which is the subspace inclusion on each open set $U_i$.

[gentle osprey]

diligentClerk
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[plain raven]

Somehow we're asking: let's simplify the sheaf (presheaf) by pulling it back along the map $\pi$. For $\mathcal{F}$ a presheaf on $X$, and $V$ an open subset of $Y$, let $\pi^\ast\mathcal{F}(V) = \prod_i\mathcal{F}(V\cap U_i)$.

[gentle osprey]

diligentClerk

[plain raven]

I think it it probably the case that if $\mathcal{F}$ is a sheaf than this pullback is also a sheaf.

[gentle osprey]

diligentClerk

[plain raven]

What we're doing here is breaking up $\mathcal{F}$ by looking on it on all these open subsets individidually. The new sheaf on $\mathcal{F}$ is simpler because it doesn't contain or represent information about the global structure of $X$ if the cover $U$ is fine.

[gentle osprey]

diligentClerk

[plain raven]

Does this idea make sense

[empty grove]

Yes, with you so far

[plain raven]

We're destroying a lot of information about $\mathcal{F}$ this way because so much information about $\mathcal{F}$ occurs at the global level but the new sheaf is much more flexible and easy to work with.

[gentle osprey]

diligentClerk

[plain raven]

Now $\pi^\ast$ should have a right adjoint which is given by the direct image functor.

[gentle osprey]

diligentClerk

[plain raven]

Do you believe this?

[empty grove]

Seems natural, but I'm slow with checking adjunctions properly 😵‍💫

[plain raven]

Haha. Well, do it eventually. Just remember that direct image is right adjoint which is the opposite of what you think it is if you draw the map $\pi : Y\to X$ as going from left to right.

[gentle osprey]

diligentClerk

[plain raven]

Mathematics is designed to be confusing! if you keep this in mind you will never fail

ok.

so what the fuck am i talking about?

oh, right.

So look at the composition $\pi_\ast\pi^\ast$ of this pullback and direct image. This gives an endofunctor on the category of sheaves of Abelian groups over $X$.

[gentle osprey]

diligentClerk

[plain raven]

And, in line with what we've been saying, it's a monad. You can think of it as a way to construct canonical, functorial sheaf resolutions, by sheaves that have certain nice properties that make them flexible. Intuitively we might think of them as: flabby, soft, fine

indeed these are all technical terms which have been given to variations on this theme

There are monads which give you flabby sheaves, monads which give you injective sheaves, etc

[empty grove]

[plain raven]

and this monad is like, roughly in the same category as these guys, it's one of their cousins. Sheaves in the image of $\pi_\ast\pi^\ast$ have no interesting information that lives outside any open set $U_i$.

[gentle osprey]

diligentClerk

[plain raven]

They don't have any interesting global information

all interesting information that lives in these sheaves is somehow 'subordinate' to the open cover. All of their real content and structure is truncated down to live within the sets of that open cover.

[empty grove]

Makes sense

[plain raven]

Does that make sense? I'm not sure how well i'm articiculating this. It's kind of like how a projection from a free group onto an arbitrary group doesn't know anything about what the generators are, not how it fits together.

OK. So write $T$ for this monad on sheaves of Abelian groups over X.

[gentle osprey]

diligentClerk

[plain raven]

You can to any sheaf associate in a functorial way this resolution

$\mathcal{F}\to T\mathcal{F}\to T^2\mathcal{F}\to\dots$

[gentle osprey]

diligentClerk

[plain raven]

If you apply global sections to this whole augmented cosimplicial diagram you get an augmented cosimplicial Abelian group

And by taking alternating sums of face maps gives you the Cech complex of the open cover.

[empty grove]

ohh I see it

[plain raven]

In a sense that's the whole explanation. I could say more but only for the sake of making this more concrete by walking you through some of the computations, and it sounds like it's a little late for you

[empty grove]

Well the screenshot I sent was all we have to work with Prof said the remaining stuff about these will be left to assignments and I doubt I'd have gained much from just that

[plain raven]

btw this whole theory of systematic functorial resolutions by monads is often called the 'bar construction' if you want to look up a keyword. It was first used in sheaf cohomology by Godement, and it was used by Eilenberg and Mac Lane to give a resolution for computing group cohomology.

[empty grove]

Staying up was worth it

[plain raven]

cool haha glad to hear it

[empty grove]

This was very cool, thanks a lot. You explain very well, maybe you should start a youtube channel or something 🥸

[plain raven]

🥸

Yeah i did create a real-name discord acct. maybe i should switch to that one. luckily no one would be able to link my real name acct to this acct. certainly they wouldn't find someone ranting about monads for 4 hours and say "Ah, that's definitely Clerk"

[empty grove]

Just make sure to switch accounts before rant

[plain raven]

there is a dual theory for comonads btw - instead of cosimplicial resolutions, simplicial resolutions; instead of cochain complexes, chain complexes; instead of cohomology, homology

The world if comonads gave rise to a cohomology theory and monads gave rise to a homology theory:

attached: utopia.gif

[minor hornet]

I just learnt that in point-set topology, closures behave nicely wrt to subspaces, but interiors don't.

A counterexample would be the x-axis as a subspace of R^2, where for any line segment A on the x-axis, the interior of A in R^2 is empty while the interior of A in the x-axis is an open interval. So when reducing our scope from the space to the subspace, the interior of A "became bigger" as some elements of the boundary of A in R^2 "turned into" elements of the interior of A in the x-axis. This doesn't affect the closure.

My question: I'm wondering if there can be a converse scenario to this example -- where the interior of A "shrinks" while the boundary gains more elements. (All while "preserving" the closure as before)

[true robin]

Let X be a topological space, S a subset of X given the subspace topology and A a subset of S. If U is open in X and a subset of A, it is in particular a subset of S hence is open in S, therefore U is in the interior of A in S, so the interior of A in S is a superset of the interior of A in X.

[long hornet]

One criterion of continuity is f^-1 (int B) being a subset of int f^-1(B), for all subsets B of the codomain. Now let f be the inclusion map, and observe that f^-1 is just intersecting with your subspace.

The second "int" here refers to the interior in the subspace topology

[minor hornet]

Both explanations makes sense. Thanks guys!

[silver umbra]

not rly understanding why a needs to be on the boundary of H^n for this proof to work

any explanation would be greatly appreciated : )

is the inclusion not smooth on all of H^N tho?

not entirely sure what u mean

but ig the interior of H^n is open and open submanifolds have the isomorphic tangent spaces to the main manifold

[quasi forum]

So I have a good idea how to do 4 a). However, I just have a quick question. Are the elements of X* equivalence classes? Otherwise, the inverse of g in the hint does not make much sense.

Nvm, it literally says it in my book -_-

[haughty anvil]

So I'm having some trouble with computing images of lines onto an image plane. For example I have the line (-3,0,-3)+t(0,1,0) from the viewing point (0,-2,1.5) with the image plane being y=0.

I have this 3d calculator thing to help me visualize it but I'm unsure how to map it to the image plane from the perspective of the viewing point.

This is what my friend sent me before he fell asleep but I'm confused by the wording.

[orchid forge]

just to clarify the question, is this what's going on? you have a parametric curve (the line) in space, and you're looking at the projection along the line of sight onto some plane, like a lens?

and you want to find the equation of the curve on that plane

assuming so, here's how you can solve it. Let $P = (0, -2, 1.5)$ be the viewpoint and $\gamma(t) = (-3, t, -3)$ be the line. The line of sight has equation $L(s) = P + (P - \gamma(t))s$. We want to follow this line onto the plane $y=0$, so set the $y$ coordinate of $L(s)$ to 0 and solve for $s$. Then plug that into the $x$- and $y$-coordinates

[gentle osprey]

Kogasa

[haughty anvil]

It's not a ciruve it's just a line.

[haughty anvil]

I'll send you the 3d representation.

[orchid forge]

I know it's a line, but it's given parametrically

I drew a curve because it makes the visualization easier

but everything I said applies, regardless of what the parametric curve or point are

[haughty anvil]

https://www.geogebra.org/3d/cpdzntxk

This is what I've been using.

[orchid forge]

is the problem what i described above?

[haughty anvil]

Okay so in my case I would have L(s) = (0, -2, 1.5) + (0, -2, 1.5) - (-3, t, -3)

Right?

Oh wait. Forgot the s.

[orchid forge]

[haughty anvil]

Is that not right?

[orchid forge]

I'm still trying to make sure that I'm even solving the right problem

[haughty anvil]

Oh lol.

[orchid forge]

so is it

[haughty anvil]

Is what?

[orchid forge]

is the problem I described the kind of problem you are trying to solve

[haughty anvil]

I just need to like map it.

[orchid forge]

[haughty anvil]

Like for example we had to map 2 lines before.

And it looked like this.

[orchid forge]

okay yes

so the problem is what i described

[haughty anvil]

Like we had to give the point of that vanishing point for that problem.

[orchid forge]

you are projecting a given curve along the line of sight onto some plane

[haughty anvil]

Yes.

[orchid forge]

then my solution should be correct

assuming you're only looking at the part of the curve t >= 0 (so it's on the other side of the image plane), then the above gif shows what the solution looks like

(a line segment)

since the line is perpendicular to the image plane, as t --> infinity the resulting curve approaches the ordinary vector projection of P onto the plane

[haughty anvil]

Okay so by the way. What exactly is t and what is s?

[orchid forge]

$t$ is the parameter for the given line $\gamma$, s is the parameter for the line of sight (from P to $\gamma$)

[gentle osprey]

Kogasa

[haughty anvil]

Okay so am I trying to find T(s)?

Sorry I'm still a bit confused.

This is all like super new to me ^^;

[orchid forge]

You draw a line $L(s)$ from the point $P$ to the curve $\gamma$, and find when (for which value of $s$) that line intersects the plane $y=0$

[gentle osprey]

Kogasa

[orchid forge]

The value of $s$ depends on $t$, because the line goes from $P$ to $\gamma(t)$ (a point on the curve). So when you take that value of $s$ and plug it into $L$, you get a curve (depending on $t$) whose $y$ value is $0$.

[gentle osprey]

Kogasa

[haughty anvil]

Wait so is it a line or a curve?

[orchid forge]

both

i'm referring to it as "curve" because there are two lines here, the line you were given and the line of sight from P to the other line

in the above gif, the red lines are the lines of sight $L[s]$ for different values of $t$

[gentle osprey]

Kogasa

[haughty anvil]

Okay so.

Just being sure here.

First things first I do what I was saying but with s so

(0, -2, 1.5) + ((0, -2, 1.5) - (-3, t, -3))s

Or (0, -2, 1.5) +(0, -2s, 1.5s) - (-3s, ts, -3s)

Right?

[orchid forge]

Yeah. For a fixed t, that's the line of sight (red) that goes through both the point P and the given line (blue). So you want to find the s value that makes the y coordinate of that equal to 0.

[haughty anvil]

So it's (3s, -ts - 2s - 2, 4.5s - 1.5)

And I set -ts -2s -2 = 0

Right?

[orchid forge]

Yep. Try to work through the whole solution and see if you can figure out how to check if it's right or not

[haughty anvil]

But wait there's no s where that = 0

[orchid forge]

It will depend on t

[haughty anvil]

So I have to find s and t?

[orchid forge]

as in, for each t, there is a value of s such that -ts -2s - 2 = 0

[haughty anvil]

Oh okay.

[haughty anvil]

Okay so I was able to come up with that s=-2/(t+2)

Is that what I was looking for?

[orchid forge]

Yes, that is the parameter s for which the red line intersects the plane y=0

now if you plug that value of s into L(s), you get a curve (f(t), 0, g(t)) which is what you are looking for

[haughty anvil]

I got this

Forgot the 1.5

[haughty anvil]

Is that right @orchid forge

[long hornet]

Suppose A, B intersect in S^n and let K be their union, and that I know the homology of A, B. Does the Mayer-Vietoris sequence give an explicit description of H_n(K)?

I got an exact sequence of five terms involving these and wasn't able to give any easy description of H_n(K) or H_{n+1}(K)..

I suspected that maybe we need more data

[empty grove]

You'd need to know the homologies of the intersection as well as all the maps between the known homologies in the exact sequence, because for modules you can't just give a dimension argument for the missing terms and need to know the exact kernels etc

[vast estuary]

Hello

I want to formally write down the construction of this CW complex

[gentle osprey]

Hausdorff

[vast estuary]

Is this good? How can I make it better?

I understand the intuitive idea and the process but I suck at constructing CW complexes formally, so please help

[empty grove]

bruh you really don't need to do it this formally

[vast estuary]

lol man i want to, at least once

[empty grove]

You can abstract the details of making the intervals disjoint

[vast estuary]

i want to know i can do it before i stop doing it

[empty grove]

hm ok lol

[vast estuary]

is this ok? how would you make it better

[empty grove]

yes its fine

idk what you mean by make it better

[vast estuary]

like is there a nicer way to do what i did

the presentation looks shit

[empty grove]

not think of the details

I doubt theres a better way to write all this stuff

[vast estuary]

btw is it normal in alg top to not write details?

[empty grove]

Yes because they are trivial 😌

[vast estuary]

because see i also feel bad about just writing "attaching along f" and not actually defining the quotient map and all that

[empty grove]

no but I mean you just worked it out once

and now you never have to think of this again

[vast estuary]

ohh okay cool, this is very different from analysis man

[empty grove]

but you know what that means formally

lol I feel like it is only different because it is hard/infeasible at times to give all the details

[vast estuary]

ok fair, but how do i know when to stop

like someone asks me to show the torus as a CW complex

i can visualize it, but i'm struggling to formalize the details

[empty grove]

right I see

But if you see how the gluing is being done then where are you stuck with that?

[vast estuary]

a 0-cell, two 1-cells, and a 2-cell. my problems are:

1. in 3D space, those 2 1-cells are in orthogonal planes! why do we never formally write that?
2. the 2-cell is homeomorphic to D^2 sure, but it's a very weird shape. so while constructing the torus am i supposed to also mention a homeomorphism from D^2 to that surface?

[empty grove]

1. because we are not necessarily embedding this stuff in 3d space. It is happening abstractly, and it happens to be homeomorphic to the embedding of the torus in 3d. Proof: The embedded torus is homeomorphic to S^1 x S^1, and the CW complex we have is exactly the product cell structure on S^1 x S^1 (show these individually). You can also give the torus a cell structure so that the loops aren't in orthogonal planes fwiw

2. yes and this is where you stop giving details and just say that use any homeomorphism from disk to square because there is no way you are writing such a homeomorphism explicitly

and if you want you can prove generally that instead of pasting an n-simplex at some stage you can attach anything homeomorphic to it

so that you never have to think about that again

[vast estuary]

fair enough. the practice of skipping details is very new to me because in the past two semesters, my analysis and algebra courses have made me be rigorous about everything

thanks tho!

[empty grove]

rather than skipping details its like abstracting them

[empty grove]

it can be just as rigorous as anything else by placing stuff like this inside blackboxes once you prove it

just think of it as a lemma

[vast estuary]

yup, i get the idea now

[uncut surge]

love that this is a valid opener to a math conversation

math is shitposting formalized

[vast estuary]

hello!

so i am paying the price of formally constructing the torus as a CW complex

so i have picture of the unit square in mind, where we identify opposite edges, etc.

[gentle osprey]

Hausdorff

[vast estuary]

What am I missing? How should I complete this?

The problem is that I haven't identified I_1 with I_2, and I_3 and I_4, but my construction ended lol

[uncut surge]

Hmm, I do wonder if the square with identified edges is a good model to make a CW complex out of; 'cus if you wanted to do this, to identify two edges, you'd have to discontinuously jump from one edge to another one while going along the boundary of a single 2-cell... right?

I've not done this rigorously in a while, but as a CW complex I always find it helpful to think of the torus as spanned by the two non-contractible loops you can put into it

But maybe there's a way for the first one, too! I'm just not seeing it immediately

[vast estuary]

If you ask me, the square thingy is more natural. The four corners are 0-cells, the four edges are 1-cells, and the stuff inside is a 2-cell

[vast estuary]

Okay I think I have fixed it

Let me know if this works:

[gentle osprey]

Hausdorff

[empty grove]

Start with 2 copies of I instead of 4

You cannot reduce the number of 1 cells in your complex later

[vast estuary]

Makes sense

[vast estuary]

but the second equivalence relation though?

[empty grove]

what

[gentle osprey]

Hausdorff

[vast estuary]

It basically identifies the two intervals @empty grove

I see your point, should've started with only two intervals

but isn't this fine too

[empty grove]

But weren't you formalizing the cw structure 🤡

In that you can only glue the new n+1 cells in the n skeleton which you have already built

[empty grove]

In this way you could construct structures which aren't cw structures

[finite heath]

So

Closure(A) is closed and contains A

Hence, topological space X (where A is a subset of) --

X - Closure(A) is open

But how is X - Cl(A) contained in X - A

[lunar yoke]

thats just basic set manipulation

if $A \subseteq C$ then $X \setminus C \subseteq X \setminus A$

[gentle osprey]

Phil P

[finite heath]

Oh gotcha, thanks 🙂

[vast estuary]

Fair enough

Should have been careful

Thanks!

Writing it down was totally worth it

[shy moss]

why Zariski topology is not Hausdorff?

[lunar yoke]

so if you just look at the affine line over like C, then the closed sets are exactly the finite sets and the whole line

so all open sets except the empty set are cofinite

so you cant separate points with disjoint open sets

[tough imp]

It’s even easier if you just take an irreducible set

Any two opens intersect

And if you want the scheme version of Zariski topology points aren’t even closed

[maiden oracle]

i have a true/false question that i am confused about: i need to find a topological space such that the fundamental groupoid is not equivalent(in a categorical sense) to a group or show that such a space does not exist. This question doesn't really make sense to me since a group is a groupoid with only one object. So every fundamental groupoid of a space X with more than 1 object(that is points) should not be equal to a group right?

[tough imp]

I don’t know off the top of my head if this is true, but in general an equivalence of categories does not need to respect the number of objects in a category

So you’ll need to further argue if you want to prove that

For example I’m pretty sure a category with 2 objects, the identity being the only self-maps, and then 1 invertible map between the two objects is probably equivalent to the trivial category

No it totally is, that’s its skeleton

[maiden oracle]

hmm ok guess i just won't answer this question in an exam then because i absolutely know nothing substantial about categories xd

ty though

[orchid forge]

The fundamental groupoid is made of a bunch of isomorphic groups within each path component

Different path components need not have the same fundamental group

[obtuse meteor]

can anybody explain the method of acyclic models to me in a way which like

at all gives geometric rhyme or reason to this

[gentle lark]

n >= 5 is not needed here, right?

I've looked around and stack exchange seems to confirm it

At the very least, I feel like I could prove it for n = 4

https://math.stackexchange.com/questions/872028/fundamental-group-of-the-product-of-3-tori-minus-the-diagonal/872200#872200

Like, the proof in this answer very much works, I think?

I'm beginning to think I might have the wrong definition for contractible loop, but what is it if not that its class is trivial in pi_1?

(there's a chance n>=5 is, indeed, not needed, but since Milnor's using Whitney's approximation theorems here to prove it he has his hands tied and has to add an extra hypothesis)

[gentle lark]

It's also pretty trivial for n = 3, right?

[plain raven]

HHHHHHHHHHOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO

i love the method of acyclic models lmao

[obtuse meteor]

like idk

we proved Eilenberg-Zilber in algtop last week

and it feels

fake

[plain raven]

yeah

[obtuse meteor]

like

[plain raven]

it is fake

😦

[obtuse meteor]

there's a bunch of categorical mumbo jumbo

and then like

[plain raven]

eilenberg zilber is fake news

[obtuse meteor]

you use the Delta^k and Delta^k x Delta^ell are contractible

and

[plain raven]

you have hit onto a weird niche thing that i have strong feelings about

[obtuse meteor]

somehow that tells you that you can relate H_n(X) and H_n(Y) to H_n(X x Y)???

like

wtf

[uncut surge]

why is eilenberg zilber fake

oh man you're writing for a very long time

[plain raven]

So i've been trying to figure out the intuition behind eilenberg-zilber for a while.

one thing that's helpful to observe is that it's just an easy fact about the category of pointed spaces that the reduced suspension of the smash product $S(X \land Y)$ is homeomorphic to the reduced join $X\ast Y = X \times I \times Y / ~$, where

$(x, 0, y) ~ (x, 0, y')$ for all $y, y'$

$(x, 0, y) ~ (x', 0, y)$ for all $x, x'$
$(x, t, y_0) ~ (x,1,y_0)$ for $y_0$ the basepoint of $y$ and $t$ is arbitrary
$(x_0, t, y) ~ (x_0 ,0 ,y)$ for arbtirary $t$

[gentle osprey]

diligentClerk

[plain raven]

lol

now if you take, say, two ordinary abstract simplicial complexes $K, L$, they don't have a well-behaved notion of Cartesian product, simplicial complexes don't have products but they do have joins. The join of $K$ and $L$ has as its underlying set of vertices the set $|K| \cup |L|$ and as its $n$-simplices it has all finite subsets sigma of $|K| \cup |L|$ where $\sigma\cap |K|$ is a simplex of $K$ and $\sigma\cap |L|$ is a simplex of $L$. And it's easy to see that the chain complex associated to the join is exactly the tensor product of the two original complexes, up to a dimension shift.

[gentle osprey]

diligentClerk

[plain raven]

Similarly in the category of simplicial abelian groups there is a very natural tensor product, the Day convolution, and via the Dold-Kan correspondence this corresponds exactly to the usual tensor product of chain complexes, up to a dimension shift.

So I think what Eilenberg-Zilber can be thought of as saying is that the "Cartesian product" of chain complexes is homotopy equivalent to the delooping of the join. or conversely the suspension of the cartesian product is homotopy equivalent to the join. and this matches the geometric intuition

however this dimension shift just doesn't appear in the usual statement of the theorem because of the predominant conventions about how the chain complex of a space should be indexed.

as for the method of acyclic models it's incredibly beautiful and i think it's arguably one of the first times you see a need for a kind of homological algebra which is based on the idea that there should be a relative notion of freeness/projectivity; not necessarily projectivity in the ordinary sense of an Abelian category but projectivity with respect to an adjunction (say, if F is left adjoint to U, then any object in the essential image of F is 'free' and any retract of a 'free' object is 'projective')

i wrote a paper on the method of acyclic models so i'm not really sure where to start lol

[obtuse meteor]

:o

[plain raven]

i guess the way i understand the version of the theorem used in eilenberg zilber is that like

[obtuse meteor]

very cool stuff

to me like

it feels weird bc the proof of eilenberg zilber using acyclic models

it doesn't really feel like you're embedding that much topological info if that makes sense?

to say something which seems fairly topological

like the proof that the homology of a suspension is what it is

feels fairly topological

but this doesn't feel that way at all

[plain raven]

let M be a small category, for the sake of argument we can take M to be discrete. Let C be a category, potentially large, and let F : M -> C be a functor, often M is just a small set of objects of C. Now if P : M -> Sets is any functor (uhh actually here let we can extend it along F by left Kan extension, and the left Kan extension construction gives a left adjoint to the 'precompose with F' functor. so this gives an adjunction between [M, Sets] and [C, Sets].
Now use the free-forgetful adjunction Sets <=> Ab to get an adjunction [C, Sets] <=> [C, Ab]. So we hook these two adjunctions together to get an adjunction

[M, Sets] <=> [C, Ab].

In the terminology of the usual statement of acyclic models, a functor X : C-> Ab is said to be 'free' if it's in the essential image of the left adjoint

Similarly one can define a morphism of functors f : X0 -> X1 in [C,Ab] to be 'U-surjective' if U(f) is split epic in [M, Sets]

and using the notion of U-surjective you can define U-acyclic and some other related notions

and this lets you carry out the usual proof that like, projective resolutions of an object are unique up to homotopy in this diagrammatic setting

actually this theorem can be generalized to the level of simplicial objects

there's like, an 'acyclic models' for simplicial objects in an arbitrary category with respect to a certain comonad. and in the case at hand, the comonad is the comonad on [C, Ab] resulting from the adjunction

[modest agate]

let $$X={(a,b,c)\in\mathbb{C}^3\mid a\neq b\neq c}$$ be the set of pairwise distinct triple of points in $\mathbb{C}$. Each triple $(a,b,c)$ determines a triangle $\Delta abc$ in the plane. Also let $$Y=\mathbb{C}\times (\mathbb{C}-{0})\times (\mathbb{C}-{0,1})$$ be the set of triples in $\mathbb{C}^3$ with $y\neq 0$ and $z\notin {0,1}$ give $\mathbb{C}^3$ the usual metric topology and $X,Y$ the subspace topologies. How do I prove that $$f(a,b,c)=(a,b-a,\frac{c-a}{b-a})$$ that $f$ is continuous and admits a continuous inverse $g:Y\to X$ where $f:X\to Y$

[plain raven]

it's pretty cool stuff. idk if that really answers your question.

[obtuse meteor]

hmm

it answers my question that there's like a general method of this that leads to model categories

but that's less

hrm

my question is more philosophical

[gentle osprey]

notsushY

[plain raven]

but i think eilenberg zilber is really a statement about the relation between cartesian product and join, it's just kind of obscured by some notational conventions about indexing

[obtuse meteor]

inceresting

join in what sense?

are we thinking in like simplicial complexes ? I know of the join operator there but haven't worked with it

(like simplicial complexes as our model for hTop)

[plain raven]

yeah i think the join is like, basically just a really fundamental structure when you're thinking about simplicial stuff. The pitch i have for why you should care about the join is the following: Let C denote the subcategory of Top which consists of the objects and maps in the image of the geometric realization functor SSet -> Top. C can be described roughly as follows:

• it is a monoidal subcategory under the join operation which i'll denote * : it contains the empty set (the unit of the join) and the join of two simplicial complexes is again a simplicial complex.
• It contains the singleton space 1, the unique map 0 -> 1, and the unique it contains the unique map 1 * 1 -> 1 (where here 1 * 1 is just the unit interval I )
• it is closed under certain types of colimits.

so i think about simplicial complexes as being built kind of by starting with the singleton set and then taking higher tensor powers with respect to the join

and then gluing those together to form other more complicated structures

the category of simplicial sets is the 'free cocomplete monoidal category on a distinguished monoid" where the distinguished monoid is the terminal object and the monoidal product is the join

the join can be defined for simplicial complexes, simplicial sets and topological spaces, the definition is in chapter 0 of hatcher i think

btw the map that is constructed in eilenberg zilber is like

[obtuse meteor]

:o

very cool

[plain raven]

i think you can already see some of the structure at the level of simplicial sets. i mean like

topologically, just thinking of spaces, there is a canonical map X x Y -> (X * Y)^I which sends the pair (x,y) to the canonical path from x to y in X*Y

and this construction can be carried out at the level of simplicial sets

and you get something which really looks almost exactly like the alexander-whitney map which is used in the eilenberg-zilber theorem, the cup product, cross product

the thing is that abelian groups are like, richer than sets lol so you can represent a family of elements (like a family of simplices that form a homotopy) as a formal sum of elements

and so a lot of this gets crunched down and compressed in a way that is more computationally convenient but

kind of obscures the meaning

@wooden falcon alright so i'm curious

why do you think homology is a cubical concept rather than a simplicial one? to me eilenberg zilber seems really to be about simplicial stuff

[plain raven]

nice

ok

nah, no worries i don't know anything about cubical stuff

so any bare minimum is useful

[dusk heron]

Let $X$ be the union of the unit sphere $S^2\subseteq\mathbb{R}^3$ and the straight line segment going between its north and south poles. Then the fundamental group of $X$ is $\mathbb{Z}$. This can be seen by sliding one of the ends of the line segments until it meets the other one, at which point we have the wedge sum of a sphere and a circle. I agree with it intuitively, but I'm a bit unsure about exactly why (with a semi-rigorous argument) the fundamental group is preserved when the two endpoints of the line segment merge together into one. Why is this the case?

[gentle osprey]

gustavn64

[dusk heron]

In other words, why exactly does this step preserve the fundamental group?

[shy moss]

maybe the first deformation retracts onto the second

[orchid forge]

that's not exactly what they mean

[dusk heron]

The second is not a subspace of the first.

[dusk heron]

What do you mean?

[orchid forge]

X is a sphere with a line through it

you're contracting that line to a point

which is a homotopy equivalence

[dusk heron]

I am not contracting the line to a point.

[orchid forge]

that is what they say to do

[dusk heron]

Really? That is not possible

[orchid forge]

what?

[dusk heron]

If it would be possible to contract the line to a point, the space X would be simply connected, which it is clearly not

[orchid forge]

sorry, draw the line segment along the sphere from the north pole to the south pole

then contract that

[dusk heron]

Contract it how?

[orchid forge]

sliding the north pole down to the south pole

by contracting the (invisible) curve between them