point-set-topology

[ebon galleon]

by definition of function

uwu

[unreal stratus]

note that f(f^-1(A)) ins't necessarily A

but is contained in A

if that is what is causing some confusion

[heady skiff]

ohhh i see what you mean

[ebon galleon]

ff^{-1}B = B iff B is a subset of fX

[heady skiff]

bc the function is defined for every point of x

bruh

[high hill]

just expand everything to their set theoretical defns

[ebon galleon]

f^{-1}fA = A iff f injective (at least in general)

[high hill]

and things follo

[ebon galleon]

when in doubt, do this

[heady skiff]

yea my set theory is so shit

[ebon galleon]

yeah you'll need it for topology so uhhhhh

[heady skiff]

all good!

i can name 3 sets

the rational numbers

integers

natural numbers

so i'm chillin

[broken nacelle]

brother...

you could at least add in complex numbers

[ebon galleon]

[broken nacelle]

wait

where are the reals?

[heady skiff]

i was aboutta say positive integers

[broken nacelle]

anywho

[high hill]

rnt these fake

[broken nacelle]

is this sort of argument common?

where you embed your space into one whose fundamental group is easier to calculate

[void tapir]

yeah (because negative numbers are fake)

[broken nacelle]

and then use some algebra to get the fundamental group you wanted?

[heady skiff]

wait

[broken nacelle]

they're about as real as the real numbers

[heady skiff]

but what if f is not injective

then how could we take the preimage

cuz like

[high hill]

bruh

[heady skiff]

wait

wait

let me write out the definition

of preimage

lol

[high hill]

f : A -> B, Y sub B
f^-1(Y) := {x in A : f(x) in Y}

[heady skiff]

OHHHHHHHHHHHHHHHHHH

[ebon galleon]

[heady skiff]

LMFAOOOOOOOOOOOO

[ebon galleon]

my guy

[broken nacelle]

lmao

[ebon galleon]

just write it out

[high hill]

do some blumming set theory when

[ebon galleon]

it aint hard

[heady skiff]

nah writing it out takes too long

[broken nacelle]

does anyone know?

[coral pivot]

You dont really gain a lot of info from embedding interms of pi1

like, things with really bad pi1 can embedd into e.g. euclidean space

[heady skiff]

how would I check that f is a homeomorphism? It's pretty clear to me that f and g are bijections, and i'm trying to check the open set property. If V is an open set in R, then this amounts to showing that f^-1(V) is an open set in R, or that g(V) is an open set in R. but I don't know how to do that if we're talking about the topology definitions, I'm pretty sure I have an idea how to do it if we're talking about open set has a ball about each of its points definition

or is it because the open sets of R are the open intervals and g(V) forms an open interval

or am i trippin

[coral pivot]

I mean both definitions are equivalent, you should convince yourself of the equivalence first

[heady skiff]

yeah i probably should

[narrow cairn]

okay so im stuck on this problem: Let $G$ be a topological group and let $\Gamma$ be a subgroup of $G$. For some $g \in G,$ let $L_g$ represent the left action of $g$ on $G$. Let $f : G \rightarrow G/\Gamma$ be the map that sends each point to its left coset in $G/\Gamma$. Show that there exists a homeomorphism $\theta_g : G/\Gamma \rightarrow G/\Gamma$ such that $\theta_g \circ f = f \circ L_g$

my idea was to just have $\theta_g(h\Gamma) = gh\Gamma,$ which is well defined, but I cant seem to show its a homeomorphism. Having a hard time wrapping my head around topological groups.

[gentle osprey]

most likely to honorable

[narrow cairn]

obviously if i can show its continuous it has the obvious inverse theta_{g^-1} which would have to be continuous as well

so i just need to show that its continuous

[ebon galleon]

If you have a quotient map f: X --> X/~, then any continuous map g: X --> Z that's constant on fibers of f induces a continuous map f':X/~ --> Z

[narrow cairn]

oh right passing to the quotient

seen this one

[ebon galleon]

ah, okay you've seen it then

[narrow cairn]

its just an application of the characteristic property of the quotient topo

[ebon galleon]

yurr

[narrow cairn]

okay but how do i apply that here

[ebon galleon]

anything you do with quotients, think of it lol

[ebon galleon]

You know f o L_g is continuous

and we know that theta_g o f is equal to f o L_g

should just be an application of the quotient property?

I think?

[narrow cairn]

ahh i see

[heady skiff]

to show that the half space of R3 with positive third coordinate is open, you have to use the archimedian property right

[unreal stratus]

uhh not really no

[heady skiff]

oh really

[unreal stratus]

idk why you would?

[heady skiff]

i let (x1, x2, x3) x3 > 0 be arbitrary

and then i choose a ball with radius 1/n < x3

idk

maybe i'm tripping

[unreal stratus]

i mean you can just pick radius x_3/2

lol

[heady skiff]

oh

[queen prism]

draw a picture 🧊

[heady skiff]

oh yea you're so right

[limber wren]

you're right that you need radius 0 < r < x_3 but yeah, you don't need it to be 1/n or anything

[heady skiff]

yea, and in order to show that the ball is contained in the half open space I need to show that for $(y_1, y_2, y_3) \in B((x_1, x_2, x_3), \epsilon))$, $\sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2 + (x_3 - y_3)^2} < \frac{x_3}{2} \implies y_3 > 0$ right

[gentle osprey]

okeyokay

[unreal stratus]

how are you defining the topology actually, via a metric?

yes though that is (hopefully) straightforward and i wouldn't bother to write some tedious proof dw

[heady skiff]

im not sure, i was just trying to verify this

[tidal lynx]

bro hates the idea of sticking to one textbook

[heady skiff]

nah this textbook is assss

nah jk who am i to judge

[tidal lynx]

on to the next

[heady skiff]

i just thought that the next chapter on homeomorphisms was kinda hand-wavy

i looked up some reviews and some ppl agree so i supplemented with munkres

[heady skiff]

ok i don't know how to do this i forgot my algebra

[tidal lynx]

minimum y3 can be here without violating the inequality is x3/2 (then y1 = y2 = 0)

[tidal lynx]

so you can say “assume y3 < x3/2” and get a contradiction

[heady skiff]

ah

that seems easier

[heady skiff]

how did u get that

[unreal stratus]

i mean hopefully clear geometrically right lol

[heady skiff]

i guess but like

yea

like intuitively obvious i suppose

but like

algebra hard

[unreal stratus]

the way i'd do it is like

if $y_3 \le x_3/2$ then $\sqrt{\sum_i(x_i - y_i)^2} \ge |x_3 - y_3| \ge x_3/2$

[gentle osprey]

potet

[heady skiff]

damn that's crazy

you're so right

[unreal stratus]

Well kinda think about the geometric intuition like

the distance is smollest when the only thing that's changed is the x_3 ig

[heady skiff]

yeah ig that makes sense

also like

is my complaint that this part of the textbook is kinda hand-wavy valid

like ok visually speaking yes h is clearly one-to-one and onto

but like if you haven't defined the function (i mean ig it would be a good exercise to do so) then you haven't proved it rigorously or at least in the way that you would normally see how functions are one-to-one or onto if that makes sense

also wtf is the construction of U in E^{n + 1}, like idk what half lines are, it would be helpful if he at least defined a set or something

or gave a precise description/definition

or at least came witih a picture

ok rant over idk

oh okay so let me see if I get this right; so say we have a topological space with a collection of open sets {O_i}. then if a set N is open in the topology (open in the sense that its a neighborhood of each of its points) then we can take the union of the which are given by this topology about each of the points in N to get N. conversely, N is an open set by definition? like i'm confused; there's the open sets that define the topology right? and then if some set happens to be open in the ball sense or whatever then its the union of some balls which make up that defining topology

like is the only time that the two notions of open sets (the one that defines the topology and the one that's the ball definition) coincide is when the topological space happens to be a metric space? what if there's no notion of distance?

Oh wait I think I see now

[ebon galleon]

The notion of a "ball" only makes sense when you have a metric

[ebon galleon]

iirc, my original comment was just to explain that a set is open iff it's a neighborhood of each of it's points. So the first bit you mentioned shows that, if a set is a neighborhood of each of its points, then it is open. And the converse is pretty much just by definition of what it means for a set to be a neighborhood of a point

"A set U is a neighborhood of x in U if there is an open set O with x in O subseteq U" well okay if U is open, then clearly we can take O = U in this definition, since for any point x in U, we have x in U subseteq U

[heady skiff]

And the converse is pretty much just by definition of what it means for a set to be a neighborhood of a point

that it's an open set?

[heady skiff]

i see

[silver spruce]

Half lines here are rays originating from p

since you also exclude p from U, it follows that U is open (try with E^2 or E^3 if its not clear why U is open)

[heady skiff]

yeah I guess I was just pretty confused by this. "O is one of the original open sets, then by definition a neighborhood of each of its points in X" "U an open set, so by definition its a neighborhood of each of its points of X"

[heady skiff]

ok, i'll look up what the definition of a ray is lol

[silver spruce]

its the same one from highschool geometry

[heady skiff]

yea i don't remember that shit

[silver spruce]

line segment but you let one of the ends go to infinity for a lack of better vocab coming to my mind atm

e.g. y = x for x >= 0

[heady skiff]

ohhh ic so the open sets here have the property that each one of them contains an open set

where the notion of an open set is the ones defined by the topology

i think

[heady skiff]

ah ok

[ebon galleon]

what

[heady skiff]

bro idk i'm so lost

anyone here a cartographer

[queen prism]

every metric space X has a distance function
this generates an object called an open ball, and now you can define a subset U of X to be “open” if every point p in U has a ball centered at p and contained in U
it turns out when you define open sets this way, the collection of open sets happens to satisfy the rules of a topology—namely, the empty set and X are open, unions of open sets are open, and finite intersections of open sets are open
and as a result you can rightfully call your open sets, open sets
(this is kind of putting the cart before the wheel but the point stands)

not every collection of sets satisfying the rules of a topology comes from a metric, however
for example, if X has more than one point, then the indiscrete topology on X cannot arise from a metric, even though the indiscrete topology certainly plays by the rules of a topology

[silver spruce]

also btw be careful jumping around w/ books if youre new to pointset cause sometimes ppl mean open neighborhoods when they say neighborhoods (munkres does this iirc)

[ebon galleon]

new bladewood pfp

[heady skiff]

yea that makes a lot of sense, thanks for typing tha tout

when do the open sets given by a topology coincide with the definition of open sets being a neighborhoods of each of its points. like you distinguish some subsets of a topological space X, and call them open sets in the sense that they form your topology. but you can only define them in terms of this way right, or are we defining the open sets which form the topology to have this property that they're a neighborhood of each of their points

sorry i should probably go to office hours for this, don't want to hog this channel

[queen prism]

nonono keep talking I don’t want to do my work

[heady skiff]

llol that's basically all i have to say/ask

plus anal is calling

[ebon galleon]

always, that's basically what this is outlining

i know at least 4 or 5 equivalent ways to define what a topology on a set is

[heady skiff]

alright, i'll just stop thinking about it now then lol

thanks

[ebon galleon]

probably not worth stressing yourself over anyways

Take the open set definition as foundational. It's the most useful

[tidal lynx]

can you list them

[ebon galleon]

The others, including the notion of neighborhood, altho they work, just aren't as nice

[ebon galleon]

Sure

[heady skiff]

wait sorry open set definition as the collection of subsets of X that form the topology or being a neighborhood of each of its points 💀

[silver spruce]

former

[heady skiff]

ok ok

and i'll just assume that they have this property that they're a neighborhood of each of their points without thinking about why because this induces a lack of intelligence/response

[queen prism]

an open set is an element of a topology
a topology is a collection of open sets

[heady skiff]

ye

[ebon galleon]

1. The usual definition with open sets
2. Their complements in terms of closed sets
3. Assigning to each point x in X, a collection of subsets N(x), the neighborhoods of x, subect to a few conditions (idk the precise formulation off the top of my head)
4. A closure operation cl : P(X) --> P(X) satisfying the Kuratowski axioms (S subset clS for all S, cl(emptyset) = emptyset, cl(A U B) = clA U cl B, and clclA = clA)
5. a subframe of the powerset P(X) (i think this one works? perhaps with one or two restrictions)
   and there's probably a couple more

Oh, 6) an interior operation int: P(X) --> P(X) satisfying the dual of the Kuratowski operations

oh then we could also view them as special cases of pseudotopological spaces/convergence spaces

So 7) a map FX --> P(X) (or \beta X --> P(X)), where FX is the set of filters on X (\beta X the set of ultrafilters on X) subject to a variety of conditions

which I will not elaborate on because convergences suck

but again, you should mostly just stick to 1 and 2

[tidal lynx]

I see convergence space, are any of these uniform or cauchy spaces?

[ebon galleon]

uhh idk cauchy spaces, but uniform spaces don't generalize topological spaces

[tidal lynx]

Oh

[ebon galleon]

Realistically, there should be some corresponding definition for each category of "spaces" that Top (category of topological spaces) embeds into

So in the above, 4 is for cech closure spaces (6 is for "interior" spaces, which is not a real notion but idk what else to call them), 5 is just some order theory nonsense, 7 is the fact that Top embeds into PseudoTop and Convergence

I know that in the case of Uniform, the category of completely regular spaces fully faithfully embeds, but not all of Top

[tidal lynx]

is every guy above an equivalent formulation or are some full on generalizations?

[tidal lynx]

Because you say “embeds” here which makes me feel like some are the latter

[ebon galleon]

Cech closures (equivalent to pretopological spaces), convergence spaces, pseudotopological spaces all are. The one I gave for closures is the formulation for topological spaces tho

(cech closure operators don't need clclS = clS, which is idempotency. In order words, the closure of a set needs not be closed for arbitrary ones, and the topological spaces are precisely the ones for which all closures of sets are in fact closed)

[tidal lynx]

Ah

[ebon galleon]

It looks online like cauchy spaces are a generalization of Hausdorffness in some sense (even as general as Hausdorff convergences, not just topologies)

anyways, realistically you will not frequently need all of these generalizations (very rarely in fact lmao)

I just happen to know because I looked into them for (part of) my summer research

[tidal lynx]

It is certainly cool to know of them, but yea I don’t know much actually about them lol

Thanks for curating the list

[ebon galleon]

yeah idk too much more than basic properties and some definitions but it is kinda interesting

[unreal stratus]

Hm

If X is, say, a closed manifold minus finitely many poitns, do we have that $\tilde{H_i}(X^+) \cong H_i^{\text{Borel-Moore}}(X)$? Idk how nice the space has to be for this

[gentle osprey]

potet

[narrow cairn]

is the product of quotient maps a quotient map

i.e. if you have four spaces A, B, C, D and quotient maps p: A -> C and q: B -> D, is p x q: A x B -> C x D a quotient map?

[ebon galleon]

nope

[unreal stratus]

Not necessairly, no, though with some local compactness hypotheses it is true

[ebon galleon]

try to come up with an example. As potet mentions you need them to not be locally compact Hausdorff

[narrow cairn]

okay, that makes this problem harder then

was gonna use that to show that a quotient group of a topological group is a topological group

the inverse i got i think but multiplication is hard

continuity i mean

[ebon galleon]

Spefically, iirc you need either (A and D) or (B and C) to be exponentiable (read this as locally compact Hausdorff, but more generally this holds iff they are "core compact")

In particular, any finite space is core compact I believe, so you'll need them to be infinite I think

(exponentiability here meaning some abstract, categorical nonsense, so don't worry about it)

[narrow cairn]

any small hints?

[ebon galleon]

ah, for checking multiplication is continuous?

oh yeah

you already said that

i can't read

[narrow cairn]

lol

[ebon galleon]

lemme think if i know

yeah nothings coming to me off the top of my head, don't know top groups super well

[red yoke]

No but products of open maps are open

||And G → G/H is open||

[ebon galleon]

||Is this true for topological groups?|| Big if true

[narrow cairn]

hmm why

[red yoke]

||Yes if H closed||

[ebon galleon]

Cool

[narrow cairn]

i want a hint to be clear

not the answer

[red yoke]

This be hint

[narrow cairn]

right

ill give it a shot

okay so let U be open in G, then we want to show q(U) open in G/H. q(U) = {gH : g ∈ U} we want to show that q^-1(q(U)) is open, which is {a : aH = gH for some g ∈ U}
i honestly have no clue where to go from here

i think i just need a small nudge but maybe im way off

[gritty widget]

try a familiar example and see if you can come up with any nice expression for q^{-1}(q(U))

[narrow cairn]

its like all of the equivalence classes that have a nonempty intersection with U

[gritty widget]

i would suggest writing out the condition "aH = gH for some g in U" in more detail

"ah = g for some g in U and some h in H" to "ah is in U for some h in H..."

that kind of thing

[narrow cairn]

wait can we consider this somehow as G acting on H

is that the answer or do i need to pursue a different line

or do i need to do orbit shenanigans

[gritty widget]

i think that you're thinking too hard

q^{-1}(q(U)) is the set of a such that a is in hU for some h in H

what is this set and why is it open

[narrow cairn]

ah, could you say that this is the union of all hU, each of which is open since the left action of a group on itself is homeomorphic? so we have a union of open sets which is open

does that work?

[gritty widget]

it is exactly what i had in mind

[narrow cairn]

epic

thanks for the hints

i would never have gotten that

now its time to see if i can solve the rest

oh i see you just conclude that q^2 is open, continuous and surjective and thus a quotient map, and then let m be the multiplication map G^2 -> G, and pass q o m to the quotient q^2

clever

[narrow cairn]

passing to the quotient is so useful lol

[tidal lynx]

In the second part of 4(b), any hints on what the construction would look like? I can’t just union over all the topologies that are contained in all the \Tau_\alpha… because I saw in 4(a) that the union of topologies is not necessarily a topology

well hmm? if Tau is the unique topology, then it certainly contains the union of all such topologies, and also must be equal to it (since it itself is included in the union)

[queen prism]

since it itself is included in the union
why is that

[tidal lynx]

I mean it’s supposed to satisfy the condition right

[queen prism]

oh yes

so what happens in 4a

[tidal lynx]

I just thought of something: if \Tau_1 \subset \Tau_2 \subset \Tau_3 \subset …, then the answer here is \Tau_1, which is the intersection of each \Tau_\alpha

[queen prism]

what fails when you try to show that the union is a topology

[tidal lynx]

You can have a set contained in the union that isn’t included in every topology you’re unioning over

Ok the answer is definitely the intersection of all \Tau_alpha

[tidal lynx]

I’m still trying to grapple where (if any) the error in my reasoning is here

[queen prism]

oh my brain derped and associated something in that with something else

ok so T contains the union of all the topologies in {T_a}

are you claiming that T is contained in {T_a} as well?

[tidal lynx]

T here is the unique maximal topology, and T_a are the topologies satisfying the condition?

[queen prism]

{T_a} is the family of topologies you are given and T is your maximal topology

[tidal lynx]

So if {T_b} is the family of topologies satisfying the condition, my claim was that since T contains each T_b, and T is a member of {T_b}, the union over all {T_b} is equal to T

[queen prism]

what condition are we talking about
that each T_b contains all of the T_as?

[tidal lynx]

each T_b is contained in all of the T_a’s

is what I mean as the condition

[queen prism]

oh

I'm very, very smol brain and I thought you meant the first part

[tidal lynx]

nah second part is what’s tripping me up

[tidal lynx]

I’m fairly certain the answer to the second part is “the intersection over all {T_a}”, but OTOH, the answer should also be “union over all such {T_b}” by the quoted message, so I’m trying to reconcile if this second approach leads to the same answer or if there is some error

[queen prism]

I like the idea of using the intersection

so let T be the intersection of all the T_as
what can you say about the relation between T and the T_bs?

oh

you already did that

well in that case I'd say you're done

comparing your T to the T_bs is really the "unique" part, I think

showing that it's maximal, out of all the T_bs that are contained in every T_a

[tidal lynx]

Do you know why taking the union over all {T_b} would have also worked (as said before, unions of topologies are not necessarily topologies, but here, they magically are)

I guess that after one has shown that the intersection over all {T_a} is the unique maximal topology, then it is a corollary that this topology is equal to the union over all {T_b}… but that’s not very satisfying.

[queen prism]

why do you know that cup{T_b} is a topology again

[tidal lynx]

By my last comment

But that’s the unsatisfying reason

[queen prism]

yeah
can you do that without using the intersection over the T_a

if you don't invoke cap{T_a} then I'm not sure if you can show that cup{T_b} is a topology

[tidal lynx]

Yeah I just wanted to see if anyone knew how to

[queen prism]

if nothing else it gives another characterization of {T_a}, which is cool

[silver umbra]

https://math.stackexchange.com/questions/2741557/hatcher-exercise-3-2-3-cup-product-on-h1-mathbbrpn-mathbbz-2

i dont know if this is incredibly simple

but im struggling to understand why h* necessarily sends the generator alpha to the generator beta

[limber wren]

random question I thought of: take a permutation, and draw it over the circle like below, identifying each pair of connected points. What is the fundamental group, and can it be described in terms of the symmetric group?

lol I don't know the answer in general, just thought of it

[obtuse meteor]

There’s only two choices (H^1 is Z/2Z). It sends alpha to 0 or to beta. If it sends to 0 this is a trivial map

[feral copper]

This is the same as a wedge of circles. The fundamental group will be a free group, and the number of generators will depend on the number of enclosed regions. That also happens to depend on where you place the points and how you connect them. That boils down to a combinatorics problem now!

I asked myself the same question about connecting roots of unity on the circle, and it happens to be not-so-interesting, because essentially the fundamental group of 1D CW-complexes is kinda boring

[limber wren]

ahh yeah that makes sense, lol I was hoping for some characterization in terms of the permutation, but we can't always get what we want can we

[feral copper]

I mean, if you look at n-th roots of 1 and connect a root to its image under a fixed permutation, the fundamental group of this helps describing the permutation. But it's just a re-phrasing of something not very deep 🙂

(Here I take this example of roots of 1 because you somehow don't choose how to connect the dots)

[floral bear]

why is continuity of a map (say from A to B) determined by the pre-image of open(B) sets, as opposed to the image of open(A) sets?

i just feel like in most other categories, relations among subsets of the source object are preserved by a homomorphism, so it seems odd that it's "backwards" in a sense for Top

(also, is there any meaning to maps that preserve A's open sets?)

[limber wren]

it matches the normal definition of continuity from analysis using epsilons and deltas

[ebon galleon]

there are a variety of reasons why, but one very simple on is that open maps (ones where the image of open sets are again open) generally aren't as nicely behaved, at least in the way we would want (and don't correspond with the analysis notion of eps-delta continuity as kryojyn points out). In part, this is because the inverse image actually behaves much nicer than forward image does

[floral bear]

right, that.

[limber wren]

like that's the most important motivating exercise, convince yourself that the e-d definition is equivalent to the topological one

[floral bear]

in what ways does the inverse image "behave nicely" (btw dont feel obliged to answer if this is dumb lol)

[ebon galleon]

the inverse image preserves unions, intersections, and complements

forward image only preserves unions

[limber wren]

no its a good question

[floral bear]

ohhh okay

the inverse image sort of behaves like a surjection then, i guess

[ebon galleon]

idk whether that's really a reason why continuity is defined in terms of inverse image, but it's a nice connection nonetheless

[floral bear]

that actually makes sense ty

[ebon galleon]

But this is probably a better intuitive reason why

[floral bear]

thank yall🙏

[queen prism]

examples https://math.stackexchange.com/questions/75589/open-maps-which-are-not-continuous

[limber wren]

one way of looking at it is that continuity allows you to control how close points in the image of a map are by controlling how close points in the preimage are, so by saying the preimage is open when the image is open means you can find neighborhoods in the preimage to control neighborhoods in the image

lol thats hand wavy but it's kindof the intuitive idea

[unreal stratus]

A paper i'm reading talks about a geneerator of $\tilde{H}_{2r}(K(\mathbb Z;2);\mathbb Z)$ which is dual to the "rth power of the fundamental class in cohomology". What does this duality actually mean here?

[gentle osprey]

potet

[unreal stratus]

(I do understand what the rth power of fundamental class in cohomology should be, just don't understand how they're moving to homology)

I guess you can use universal coefficients (which'll just give you an iso) and then consider the preimage under the canonical inclusion of a vector space into its dual

[tidal lynx]

wtf, on R, the upper limit topology is strictly finer than the K-topology (okay fine), but the same isn’t true for the lower limit topology (they are incomparable)

weird

[gritty widget]

upper limit topology and K topology and lower limit topology are memes

[tidal lynx]

I thought K topology was used in K theory

[red yoke]

Well the K-topology isn't symmetric

[tidal lynx]

Yea if K were like 1/n for any integer n then both results being the same would check out

but it’s only the positive guys so I guess it makes sense

[gritty widget]

[tidal lynx]

I just made that shit up ngl

But they both start with K so it must be

[gritty widget]

it would be funny if it were true

[narrow cairn]

this basically, there are open and closed maps (which are not equivalent) but there's no analogous concept for continuous maps since replacing the word 'open' with 'closed' in the definition is equivalent

[red yoke]

If you want to think of morphisms as preserving relations, think of limits of nets

Not open subsets

[dawn sage]

I don't know how to prove this. I proved that if $Z$ is an irreducible closed subset of $X$ (that meets $U$), then $Z\cap U$ is an irreducible closed subset of $U$. But the assignment, $Z\mapsto Z\cap U$ does not seem to be a bijection.

[gentle osprey]

Potitov06

[dawn sage]

For example, if $X=\mathbb{R}$ and $U=(0,1)$, then $[-2,2]$ and $[-3,3]$ are mapped to the same thing. So the map is not injective.

[gentle osprey]

Potitov06

[fickle elm]

In your example, if you assume the topology on $\mathbb{R}$ is Hausdoff, then the only possible irreducible closed subsets are singletons. It is easy to see that the statement is true under the Hausdoff condition.

[gentle osprey]

Dong_Valentino

[fickle elm]

For the moment I cannot come up with a complete argument why this is true in a general topological space. I will think about it.

[fickle elm]

For the other way around, let $Z_1$ and $Z_2$ be two irreducible closed subsets of X such that $Z=Z_1\cap U=Z_2\cap U$ is irreducible and closed in $U$. Now take the closure of $Z$ in X and $\overline{Z}$ is irreducible closed in X. Moreover, $\overline{Z}$ is contained in both $Z_1$ and $Z_2$, since they are irreducible, we can conclude that $\overline{Z}=Z_1=Z_2$.

[gentle osprey]

Dong_Valentino

[floral bear]

categorical or topological limits?

[red yoke]

Nets are topological

[somber phoenix]

In munkres, ch2 section 15.2, he's defining a subbasis for the product topology... But, the topology created by the subbasis is not at all comparable with the topology from the product basis..

All the elements in our subbasis are weird cross like sets. And when we take finite intersections of these we can end up with disjoint subsets and extremely weird looking topology

While with the topology from the normal basis looks like just squares

Yet he compares them...

Only a few elements of S belongs in the topology...

[gritty widget]

for those of us who don't have a copy of munkres immediately available could you post screenshots of what you're referring to

[gritty widget]

what happens when you intersect two subbasis elements

[somber phoenix]

This would happen

[gritty widget]

if it helps, pi_1^{-1}(U) = U x Y and pi_2^{-1}(V) = X x V, but you should be able to see this from the picture

[somber phoenix]

Actually no

That'd be empty

I see now.. M

[ebon galleon]

That's clearly open in the product topology

[gritty widget]

this is not a picture of two subbasis elements

[somber phoenix]

It is, but the intersection is wrong

No actually the intersection is correct

[gritty widget]

a picture of two subbasis elements is here. this is not just one

[ebon galleon]

No that's just one

[somber phoenix]

This is two subbasis elements, the dotted area and the one with full lines

[ebon galleon]

Cause it's union of π^-1 of an open set on X and an open set in Y

[gritty widget]

the union of two subbasis elements is not necessarily a subbasis element

[somber phoenix]

Yes, so the union of two of those gives us a cross

[ebon galleon]

No, you're right it's not

Look at the description of the subbasis

It is written as a union

Oh

No I see now

[somber phoenix]

Yes. A pillar on x axis, union, I pillar on the y axis

[ebon galleon]

I think Mallot and I had the same confusion

[gritty widget]

probably

[ebon galleon]

S, the subbasis, is the union of those two collections of sets

[somber phoenix]

Oh

[ebon galleon]

The inverse image of opens in X and in inverse image of opens in Y

Now, what you had in mind, with the crosses, is also a subbasis btw

[somber phoenix]

So we get a bunch of rectangles as elements in S

[ebon galleon]

Which is why I didn't question it lmao

[gritty widget]

and what is the intersection of two of these rectangles

[somber phoenix]

No wait, I'm so confused

[gritty widget]

the subbasis consists of two kinds of sets, U x Y for U open in X, and X x V for V open in Y

the intersection of U x Y and of X x V is what

[somber phoenix]

Oh OK, so, each element in just a pillar then?

But the union kinda breaks it?

[ebon galleon]

They can be vertical or horizonal

[somber phoenix]

Hm...

[ebon galleon]

But yeah

[unreal stratus]

Ye unions needn't be of the same form

[gritty widget]

[somber phoenix]

No nvm

The individual elements in the S elements don't mix

Just putting y and x pillars in same box

[ebon galleon]

Yeah

[somber phoenix]

Mm ok, then the vast majority of the finite intersections are empty? I think I get it now

[ebon galleon]

I mean we can't really quantify it like that, but if it helps you to think of it that way, sure

(since both collections of empty and nonempty intersections have same cardinality)

[somber phoenix]

Yeah, I mean it's "hard" for subsets to have intersections

[ebon galleon]

Exercise for when you're confident that subbasis works, convince yourself that the subbasis of crosses also works (solution: ||clearly the crosses are open in the topology generated by the columns. But every column is a "cross" by taking one of the π^-1 to be of the empty set||)

[limber wren]

yeah horizontal strips and vertical strips are both open for that reason, and intersecting them gives the normal basis elements for the product topology, so by having intersections of subbasis elements you can still get the product topology

[somber phoenix]

I had convinced myself that the weird type of topology I had drawn up was indeed a topology, but I hadn't started to write proof... Because I realized it couldn't be compared with ordinary

[ebon galleon]

Oh a subbasis always generate a topology

But in this case the topology generated by the crosses is in fact the usual topology on R^2

Altho I agree it's hard to see at first

[somber phoenix]

That something like this, is a collection in the topology, which union of singular elements in that collection indeed also is an element in the larger collection... But I haven't thought much about intersect criteria

[dawn sage]

Thank u

[ebon galleon]

These are not irreducible

[ebon galleon]

This statement should also be true when we restrict to irreducible components I believe? In which case it's clearer? (Since they partition the space, you can't have two distinct closed subsets of X that have the same intersection with a subspace)

[unreal stratus]

Hm does anyone have a reference for this?

(screenshotted from hatcher but he omits the proof lol)

[silver umbra]

im trying to figure out why the inclusion i: CP^1 -> CP^n induces an isomorphism

i_*: H_2(CP^1) -> H_2(CP^n)

[umbral panther]

Do you know that both groups are Z and you’re trying to figure out what the map is? Or are you trying to compute it in the first place?

If you’ve computed the groups, run through the computation again checking that they’re compatible

Or you could just use Poincaré duality

[feral copper]

@silver umbra The 2-skeleton of CP(n) is CP(1). More generally, the 2k-skeleton of CP(n) is CP(k).
You only add cells of dimension at least three to construct CP(n) from CP(1), that tells you the inclusion is an isomorphism

(sorry bw, no need for fancy Poincaré technology nor knowing the groups :P)

[silver umbra]

wait why does the fact that you can only add cells of at least three tell us that the inclusion is an isomorphism?

[feral copper]

Actually it's because it's cells of dimension at least 4 in this case, sorry

2-cells give generators (of H_2), 3-cells relations

I'm using cellular homology here (which is the same as singular one)

[silver umbra]

im still not quite fully understanding

[feral copper]

You can re-do the definition of cellular homology. The C_2 is free abelian whose generators are 2-cells.
Here, all odd-degree chain groups are trivial (do you know the cell decomposition of CP(n)?). So C_2 is identified with H_2. But now, the 3-skeleton of CP(n) has the same H_2 than the whole!

So yeah, you do need that the (2k+1)-skeleton of CP(n) is CP(k)

This Wiki page explains it

[gentle osprey]

Matplotlib

[narrow cairn]

i know theres probably a simple trick to this but i cant seem to find it:
show that if X is a compact space and {F_n | n ∈ N} is a nested set of countably many closed sets in X. show that the intersection across all F_n is nonempty

all i have so far is that all F_n are compact

[ebon galleon]

There is something missing here. At least, you also need the assumption that each F_n is nonempty ofc

[narrow cairn]

oh right ofc

obviously

[ebon galleon]

Have you seen compactness described in terms of the finite intersection property

[narrow cairn]

i dont think so?

[ebon galleon]

perhaps this could be a useful tool for this then

[shadow charm]

Take the complement of your intersection using de Morgan

[narrow cairn]

oh huh

hang on let me try that

[gentle osprey]

Ryx (Home for flowers)

[ebon galleon]

hope that's right now

If you're able to show this, then your specific question will follow as a special case since they are nested 🙃

[feral copper]

I guess you have sequential compactness? (Extracting convergent subsequences)
If so, try constructing a sequence (x_n) with x_n being in all F_1, ..., F_n

[feral copper]

Otherwise, this is the way

[ebon galleon]

hmm

[feral copper]

(the => way is your exercise)

[ebon galleon]

unless metric space or some other assumptions, compact does not imply sequential compact

[feral copper]

Ikr

[ebon galleon]

[feral copper]

I was thinking about this

[ebon galleon]

oh i see

[feral copper]

No okay now I'm being dumb

I confused myself, I need to stop x')

Sorry x')

[ebon galleon]

[feral copper]

[ebon galleon]

anyways, I want to say this should hold for any chain of nonempty closed subsets in a compact space

not just F_0 >= F_1 >= ...

[feral copper]

Btw, I brute-forced using sage and knot tables, and I found what I was looking for the other day! So I'm posting this as a little 'puzzle' (I'm not asking, I have the solution)
Try showing that the knots 9_34 and 9_47 are distinct, without doing any sort of computation (no polynomial invariants, no seifert matrices, nothing very involved)

[somber phoenix]

Isn't this a contradiction?

R(-inf, a) U (b, inf), suggests that subset is closed, because [a, b] is open as a topological space. By the definition of open sets for a topology

[umbral panther]

What if you have a space that is compact but not sequentially compact?

[nocturne basalt]

not sure what you mean by [a, b] is open as a topological space, it certainly isnt open using the stanrad topology on R

[somber phoenix]

For each x in X, there exists some B in basis s.t. B subset of U.

And since it's topology, there is no order

[nocturne basalt]

if your basis is open balls, there are no basis elements which contain a or b and are contained in [a, b]

[somber phoenix]

But the topology doesn't use open balls, since there is no order

[tidal lynx]

Example 1 is using the standard topology on R, which does use balls as a basis

[somber phoenix]

But that's on ordered sets, it does not mention this...

[tidal lynx]

Which topology do you think it’s talking about

[somber phoenix]

The topology in the beginning, which was very loosely defined

[tidal lynx]

Where in the beginning?

[somber phoenix]

In this book, ch 2, section 13

[limber wren]

to be specific, [a,b] is a closed subset of R. It's obviously open in itself ( if you ignore the fact that it's a subset of R and just treat it as a topological space instead of a subspace of R)

[tidal lynx]

In the bottom of pg 81 in Munkres, he writes that if a topology isn’t specified on R, then it is implicit he is talking about the “standard topology”, the topology generated by the basis of all open intervals

I should add that everyone more or less follows this convention. It’s not unique to Munkres

[somber phoenix]

Hm, I see.

[tidal lynx]

Later in Section 14 Munkres states that the order topology on R (using the usual ordering) is the same as the standard topology

Well he doesn’t really state it, it’s one of the exercises

[tiny ridge]

Is this a colorability exercise

Eg one is tricolorable but the other isn't?

[feral copper]

Almost! There's a catch

It's this but more subtle

[languid patrol]

||different number of three colorings!||

[narrow cairn]

(fyi i ended up using this it was better than whatever the FIP nonsense was)

[ebon galleon]

(fyi this is how you prove the FIP nonsense anyways)

though i agree the FIP definition is not exactly enlightening

[tiny ridge]

@feral copper Gotcha

[ebon galleon]

but it is something standard you'll probably see in a topology course?

[heady skiff]

Why is it that $\bar{\bar{A}} \subseteq \bar{A}$? If $x \in \bar{\bar{A}}$ and $x \in \bar{A}$ then obviously $x \in \bar{A}$. But if $x$ is a limit point of $\bar{A}$ then every neighborhood of $x$ intersects $\bar{A}$. But this doesn't imply that every neighborhood of $x$ intersects $A$, right? For instance, one neighborhood could just have intersection of a limit point of $A$ and not an actual point of $A$

[gentle osprey]

okeyokay

[queen prism]

the closure of a set is closed, right?
what’s the closure of a closed set?

[heady skiff]

ohhhh

it would just be itself

wait but to my point like

if i wanted to be like rigorous or whatever what was the problem with my argument

[ebon galleon]

Also if every neighborhood of x intersections A bar, then necessarily it intersections A

[heady skiff]

oh shit maybe i'm forgetting demorgan or something

[ebon galleon]

Since then it's a neighborhood of that point in A bar. And by definition then, that point is either in A, or it's a limit point of A, so then the neighborhood intersections A.

[heady skiff]

ohhh

[ebon galleon]

or something like that idk I worded it poorly

[heady skiff]

okay that makes sense then

yea

got it thanks

[ebon galleon]

now

What if we take away the requirement that the closure of a set is closed

[unreal stratus]

What if we take away the requirement that a topological space is a topological space

[ebon galleon]

jk let's not

[ebon galleon]

pseudo-quasitopological spaces

[queen prism]

i remember doing this with the limit-point definition in abbott and it was a nightmare
it’s so much easier with the topological definition

intersection of all closed sets

[ebon galleon]

yeah then it's obviously closed lol

[heady skiff]

yea i asked my prof why he decided to go with bartle and sherbert for this semester of analysis and he said the same thing i'm pretty sure

how he didn't like how abbott defined closed sets

[queen prism]

i think it's intuitive, it's just harder to prove stuff with

[heady skiff]

oh yeah it's def intuitive

but i mean for a closed set all you have to show is it's complement is open iirc

[ebon galleon]

depends, there are a few instances where the limit point definition is easier. But good to keep both in mind

[heady skiff]

which seems like a lot less of a pain in the ass

[tawdry widget]

x in cl(cl(A)). For any open neighborhood U of x, U intersection cl(A) is not empty: so there exists y in U intersection cl(A).
You notice just a simple fact: now U contains y so U is also an open neighborhood of y.

[ebon galleon]

yurr

[tawdry widget]

(Simply because an open neighborhood of this x is also an open neighborhood of a point in cl(A))

[heady skiff]

okay I'm a little bit confused probably about definitoins and such. trying to show $\bar{A \cup B} \subseteq \bar{A} \cup \bar{B}$, assuming $x$ is a limit point of $A \cup B$. now every neighborhood of $x$ intersects either $A$ or $B$, but doesn't that imply that $x$ need not be a limit point of $A$ or $B$? because of the or, you could have some neighborhood which intersects $B$ but not $A$

[gentle osprey]

okeyokay

[ebon galleon]

perhaps this is one where (intersection of all closed subsets containing A) is a good definition of clA

[unreal stratus]

It is the morally correct definition

[ebon galleon]

Or, suppose you had a point in cl(A U B) such that there were neighborhood U and V of x, with U n A = empty and V n B empty

[heady skiff]

ah

[ebon galleon]

Could you use that to get a contradiction

[heady skiff]

contradiction

yeah i mean should be pretty clear

let me try

[median sand]

What's are some acclaimed AT texts besides Hatcher?

[tawdry widget]

I read gtm119 rotman I think it’s very detailed

[heady skiff]

in general when you want to show that a function f is a homemorphism, is it okay to just show that f and it's inverse are continuous on their given domains, or should you use that if O is an open set then f(O) is an open set and same with f^{-1}

[tawdry widget]

I think both are fine, as long as you have already shown it’s bijective

Same thing, since f(O) is (f^-1)^-1(O)

[heady skiff]

Ah right so maybe the latter is easier then

Only have to show one thing

[queen prism]

well they're equivalent

[ebon galleon]

bijective continuous map is homeomorphism iff it is open iff it is closed

[heady skiff]

Ye ik

Oh wait so to show homeomorphism assuming f is bijecgive

You only have to show O open implies f(O) open?

[ebon galleon]

no, it also has to be continuous

open is not sufficient

[heady skiff]

oh

[ebon galleon]

consider the identity map on a space, with the domain having indiscrete topology (only empty set and entire space are open), and codomain has discrete topology (everything is open)

this map is open but not continuous

[heady skiff]

hmm okay i'll think about that

[narrow cairn]

whats the other definition

[ebon galleon]

cl A = A union {limit points of A} = X - int(X - A)

[narrow cairn]

ah that sucks

[ebon galleon]

last one is pretty much never useful

[narrow cairn]

alternatively just use the sequence definition and pretend non first countable spaces dont exist

[heady skiff]

yo

there's a theorem for two bases generated the same topology right

what was it again

[queen prism]

think rectangles and disks

[heady skiff]

ye i know that

like my book gave that example visually

but like what's the formal theorem

ah

nvm

[tidal lynx]

If B_1 and B_2 are your two bases, they generate the same topology iff every guy in B_1 is a union of guys from B_2 and vice versa

i.e. they can “simulate” each other

[ebon galleon]

or in other words

if the basis elements are open in the other topology

[tidal lynx]

True

[ebon galleon]

just a more intuitive way to think of it imo

[tidal lynx]

I think it’s also good to know: the topology generated by B_1 is finer than that of B_2 iff every guy in B_2 is a union of guys from B_1 (is open in the topology generated by B_1)

This implies the fact above

[unique cove]

Hello

Is this the proper channel for dimension theory?

[quiet thorn]

#algebraic-geometry ?

[broken nacelle]

I don't get how this solves the issue of the cell decomposition being far from the space's topology

like, for example, consider S^2 as the one point compactification of the plane

and then, just like the example just above the highlighted text, take the vertical lines as your 1 cells

and the point at infinity as our only 0-cell

is this not a valid cell decomposition of S^2?

like, how does requiring the existence of a characteristic map remedying this exactly?

[feral copper]

You'd need two 2-cells to finish the S² here

The char map ensures you're gluing cells by their boundary

If you'd allow gluing in the interior, you'd have weird things you don't like. Take a twisted example of gluing a 1-cell to a 0-cell at each rational number

I don't even know if that'd be continuous or not, but it's twisted enough that you can believe there's weird shenanigans that could happen

[broken nacelle]

huh

why

and even if that was true

the middle part is still just meridanal lines

I can give an even simpler example

consider $I \times I / {0, 1} \times I$ and let the 1 cells be the image of $I \times {x}$ under the quotient map

[gentle osprey]

DarQ

[broken nacelle]

and the 0-cells are the image of {0} \times I and {1} \times I

note that this space is literally a 2 cell

[broken nacelle]

Oh wait, the quotient is collapsing {0} x I and {1} x I to a single point instead of distinct ones

Eh, w\e, it still works it's just not a 2-cell

[feral copper]

I don't understand what you meant xD so the y-axis is a 1-cell that you glue to the 0-cell being the point at infinity. That's literally a circle. You need to cap it off with two disks (the two hemispheres) to obtain a sphere!

[high hill]

itd be helpful if u describe it geometrically. or better draw dq

like describe the entire thing start to end

[thick terrace]

i'm having a hard time unpacking this lemma, why isn't it the case that V, not its closure, is a subset of U?

[novel acorn]

You can write it as V subset Vbar subset U

[thick terrace]

bruh

[novel acorn]

So V is a subset of U
But the important thing is that the closure of V is also contained in U

[thick terrace]

what would this visually look like though, i can't at all visualise it

[thick terrace]

makes sense

so like, it's saying that for a point x there is an open set U, which is a subset of another open set?

i don't see how this translates to regularity

[umbral panther]

Yeah, something’s wrong with this definition. It would allow a cell decomposition in which every point is a 0 cell.

I guess you could fix it by requiring that for every n the topology of the subspace consisting of the interiors of all the n cells is the disjoint union of the interiors

[broken nacelle]

yes but like

not just the y-axis

all vertical lines on the plane

[broken nacelle]

I think I figured it out

I think cell decompositions aren't supposed to be well-behaved

unlike CW decompositions which requires the set of cell to be coherent with the topology of the space

[unreal stratus]

No a cell complex should just be synonymous with a CW complex

[broken nacelle]

not in lee I suppose

he makes the distinction very explicit

[unreal stratus]

Oh really lol

[unreal stratus]

Okay sure

I wonder why you would care about ones that aren't cw complexes lol

[ebon galleon]

depends on the context

Sometimes cell complex is used for a space built by gluing cells without regard to the order of dimension

Such spaces are homeomorphic to CW complexes I believe (this should just be colimits commuting with colimits?)

But the general notion of a cell complex is useful because the cofibrant objects in the standard Quillen model structure on Top are precisely the retracts of cell complexes

And so for these purposes, it is much more convenient if we can just disregard the order of the dimensions added

(i think)

but idk if this is the distinction that darq is talking about in Lee

[ebon galleon]

(in particular, it works nicer for the purposes of cofibrant generation of the model structure - I.e., for the small object argument - if we consider the more general notion)

(here comes bw to tell me i'm wrong )

[umbral panther]

I don’t think such a cell complex is homeomorphic to a cw complex, only to a retract of one. (But that’s good enough for most purposes) For example, if you glue a 2 disk to a disk along space filling curve, I don’t think that’s a cw complex

[ebon galleon]

that is what my supervisor told me

[umbral panther]

I panicked and crossed out my example, but I think it was right

[ebon galleon]

lemme think on this

this should be true up to at least homotopy

[ebon galleon]

I wanna say that this is what Proposition A.2 is in Hatcher? I am not super familiar with the point-set definitions of cell complexes and CW complexes since I have mostly worked with the categorical/colimit definitions. But that is also what https://math.stackexchange.com/a/2946892 seems to indicate

[umbral panther]

That’s homotopy equivalence, not homeomorphism

[umbral panther]

Oh, I missed this question

As you said, retracts of cell complexes are cofibrant, and thus homotopy equivalent to other cofibrant objects they are weakly equivalent to

[unreal stratus]

fair enough lol

what are the advantages of cell complexes

i am coming at this with homotopy-theoretic leanings which is probably the problem lol

[umbral panther]

Cell complexes arise naturally from model categories. They are cofibrant. Maybe a push out of cell complexes is along a closed inclusion is automatically cell?

[ebon galleon]

I'm still not convinced they aren't all homeomorphic to CW complexes, but as you said, up to (weak) homotopy equivalence it doesn't really matter

[ebon galleon]

Some variation of this should be true, since the pushout of a cofibration is a cofibration (in particular, this is also true if we restrict to relative cell complexes). So if A --> B is a relative cell complex, C is a cell complex, then B \cup_A C should be a cell complex

And in particular, A --> B is a closed (T_1) inclusion, as you mentioned

[feral copper]

Ah, but iirc you need at most countably many cells

Also, you'd end up with only a wedge of circles, not a sphere

Because you need to glue the 1-cells together

What you described is something we call a foliation

[broken nacelle]

that's what I was saying!

[feral copper]

Yeah but you only allow cells to be glued on their boundary to cells of lower dimension

[broken nacelle]

stated as is a cell decomposition would allow you to do that

[feral copper]

No it won't

[feral copper]

This

You can't glue a 1-cell to a 1-cell on something that's not its boundary; here you're trying to glue the interior of the 1-cell too

That's just not a cellular space. I'm not saying it doesn't give the 2-sphere, I'm saying it's not a CW-structure on it

As I said, it's a foliation

[broken nacelle]

I don't understand

[broken nacelle]

well yea

my question this morning was regarding this definition of a cell decomposition

not a CW decomposition

[feral copper]

Yeah I'm calling them either interchangeably, but I mean the same thing

You could do the same with parallel planes to describe the 3-sphere, and that's again a foliation, not a cell structure

[broken nacelle]

I'm sorry I just don't think we're on the same wavelength lel

I understand what you're saying

[feral copper]

Ask again your question, sorry if I'm unclear

[broken nacelle]

yea so for context

here's the paragraph from lee I was asking about

and basically lee says that, yes, decomposing the plane into vertical lines doesn't tell use much about the topology of the plane

and to remedy this he requires the existence of the characteristic map in the definition of a cell decomposition

[feral copper]

Okay, so that's not correct then, it does tell you something interesting. In this case, it's even a geodesic foliation that admists a section

It's just that the definition of cellular stuctures we like is as stated because this allows to do stuff, like homology for instance

[broken nacelle]

I see so lee is straight up just wrong

[feral copper]

Not wrong

It's just that it's not of interest for this purpose

But I can assure you that foliations are very useful xD

[broken nacelle]

I meant wrong for this

[feral copper]

The phrase

without some restriction on how the cells fit together
is more correct. The characteristic map is one way of asking for nice behaviour. There may be others. This one is useful, so that's the one we call a cell structure.

[broken nacelle]

I see xD

ok thankyu

[feral copper]

Don't over-think it. If you're convinced that gluing cells on their boundary to lower-dimensional cells is something natural enough to do, then you're good!
Btw, did you understand the CP(1) / CP(n) thingy?

[broken nacelle]

I don't remember

I think you mean this?

yea I still don't understand this paragraph as much as I would like

[feral copper]

The fact that if $X^{(k)}$ is the $k$-skeleton of $X$, then $H_k(X)=H_k(X^{(k+1)})$

[gentle osprey]

Matplotlib

[broken nacelle]

oh uhhh

did I ask about this?

[feral copper]

Not specifically, but the question you asked about how to prove that $H_2(CP^2)=H_2(CP^n)$ was a special case of this

[gentle osprey]

Matplotlib

[broken nacelle]

I don't remember asking about this

[feral copper]

No biggie 😉

Was checking up on you

Because for once that's the kind of things I can answer about, not those bundley-fibrationey-bubbly-thingies

[broken nacelle]

I will ping you when I get here

[feral copper]

Ask me about non-orientable surfaces in 4-manifolds all day, but if there's a fibration, I'm out

[heady skiff]

i see why X takes the value 1 on points of A and -1 on the points of B, but i'm struggling to see why X takes values strictly between +-1 on points of X - (A \cup B)

can anybody explain why?

nvm

i think i got it

[tawdry widget]

Because A and B are closed therefore d(x,A)=0 iff x in A. So f(x)=1 iff x in A, similar for the another one

[narrow cairn]

having an issue understanding regular coordinate balls

any intuition regarding that?

[gritty widget]

what's a regular coordinate ball

[heady skiff]

can somebody help me understand how a) implies that f and g agree on C?

[gritty widget]

what happens when n gets big in (a)

[heady skiff]

well it converges to g(x), so |f(x) - g(x)| <= 2^nM/3^n, but how does that imply that they have the same value on C?

if |f(x) - g(x)| < \epsilon for every \epsilon > 0 they would agree on C right

or am i smokin

[queen prism]

lim 2^n M / 3^n = ?

[heady skiff]

oh wait LOL

yea

u right

this is kind of a sick argument tho

i love constructive proofs

seems like there's a lot more of that in analysis, altho cauchy's theorem in group theory is so fucking cool

[narrow cairn]

exactly

what lee gives

,rccw

[gritty widget]

did you draw a picture

[narrow cairn]

i hate drawing

but fine

like i understand the words but why do we care

[gritty widget]

you are literally reading a book on one of the most visual subjects there is

you must draw

[narrow cairn]

i understand the definition but finding it hard to apply

ill read the theorem a couple more times

[narrow cairn]

fair enough lol

[ebon galleon]

perchance drawing helps (it always does)

[gaunt laurel]

Would this generally be regarded as sufficient for a solution or how does one typically make these homotopical arguments rigorous?
I have the same trouble in (d) and (e). It seems like I can see how the solution would work, but I don't know how to make it rigorous.

Similarly for this solution

[broken nacelle]

contractible is the wrong term

you must mean nullhomotopic

contractible is for spaces, nullhomotopic is for maps

[gaunt laurel]

Ah, yes, sorry

[broken nacelle]

yea, ok, you do have the idea but that's not how you'd prove it

you can't just say unknot the circle lmao

[gaunt laurel]

I know. That's why I'm asking. Cause it feel like I haven't been equipped with any machinery for that sort of stuff in the book (Hatcher)

[broken nacelle]

you have

you just need to think of it the right way

here's a hint

what do you need to do to view $\gamma$ as an element of $\pi_1(D^2 \times S^1)$?

[gentle osprey]

DarQ

[gaunt laurel]

I would need a map from I into $D^2 \times S^1$ giving the circle, right?

[gentle osprey]

Uncle Fyodor

[broken nacelle]

remember, \gamma is a map from I to the circle A

that's what a loop means

[gaunt laurel]

And thus an element of $\pi_1 (D^2 \times S^1)$, right?

[gentle osprey]

Uncle Fyodor

[gaunt laurel]

since A is a subspace of it

Sorry, not sure I understand.

[broken nacelle]

not formally

[gaunt laurel]

Oh

[broken nacelle]

elements of $\pi_1 (D^2 \times S^1)$ are homotopy classes of loops $f\colon I \to D^2 \times S^1$

[gentle osprey]

DarQ

[gaunt laurel]

Ahh, so we compose with the inclusion?

[broken nacelle]

whereas $\gamma$ is a loop $\gamma\colon I \to A$

[gentle osprey]

DarQ

[broken nacelle]

yep

well

[broken nacelle]

I forgot to mention that these loops start at a basepoint

but it doesn't matter here coz our space is path connected

just keep that in mind in the future

[gaunt laurel]

Makes sense. But I'm most of all confused about how one is supposed to argue the actual retraction. Should I find a parametrization for the circle and then a map that gives the retraction?

[gaunt laurel]

I will. Thank u

[broken nacelle]

again

wrong choice of words

retraction has a very specific meaning that doesn't apply here

[gaunt laurel]

Oh ye

I meant argue the nullhomotopy

[broken nacelle]

ok, so now that we see \gamma as an element of the fundamental group

and we know what elements of the fundamental groups look like

you do know what $\pi_1(D^2 \times S^1)$ looks like, right?

[gentle osprey]

DarQ

[gaunt laurel]

I believe so. All loops mapping into $D^2 \times S^1$, so $I \to D^2 \times S^1$ with equal start and end point

[gentle osprey]

Uncle Fyodor

[broken nacelle]

yea, but like

[gaunt laurel]

Just out of curiosity...

[broken nacelle]

did you calculate the group?

[gaunt laurel]

Calculate the group? What do you mean?

[broken nacelle]

like when hatcher calculated the fundamental group of the circle

[gaunt laurel]

Oooh, is that what one should do for each of these types of problems? I hadn't even considered that the calculations he did there are generalized

Intuitively it feels like the fundamental group should be isomorphic to that of S1 here and thus Z

[broken nacelle]

yes, that's the first thing you should be looking for

[broken nacelle]

yep

but how do you prove that rigorously?

it's not very hard

if you know the right definition to use

[gaunt laurel]

Oohh, I see

One could use the product property

[broken nacelle]

oh chrew

[gaunt laurel]

and then use $\pi_1 (D^2) = 0$ and $\pi_1 (S^1) = \mathbb{Z}$

[gentle osprey]

Uncle Fyodor

[gaunt laurel]

But I guess that wouldn't teach me anything

[broken nacelle]

I was thinking that the filled in torus deformation retracts to its central circle

but this works too

[gaunt laurel]

Ahhh, I see. That makes sense

[broken nacelle]

ok, so now we know that the fundamental group is generated by the loop that goes around the torus once, right?

[gaunt laurel]

Yes

[broken nacelle]

and \gamma is an element of said fundamental group

[gaunt laurel]

Yes

[broken nacelle]

so we can write gamma in terms of 1-loop

[gaunt laurel]

It seems like it would be $a a^{-1}$ which should be equivalent to $1$, the constant loop at the basepoint

[gentle osprey]

Uncle Fyodor

[gaunt laurel]

But the intersection with itself seems weird to me

Ahhh, but that doesn't matter for the actual map

[broken nacelle]

you can do that coz gamma is a map from I ..

yep

you got it lmao

[gaunt laurel]

Makes so much more sense this way

[broken nacelle]

like, if you regarded the actual space A

you wouldn't be able to do this

intersecting A with itself would give an entirely different space

not even a homootpy equivalent space since even its fundamental group changes

[gaunt laurel]

I see. Yes

Oh wow, so for example in problem (d) here, the approach would be to try to show that the fundamental group is Z x Z?

[broken nacelle]

it's not Z x Z lel

it's more complicated since it's not abelian

[gaunt laurel]

Oh christ

[broken nacelle]

but you don't have to calculate the entire fundamental group

first notice one thing

what's the fundamental group of D^2 v D^2?

[gaunt laurel]

The trivial group?

0

Since it is star shapes with respect to the connecting point, or...?

So any loop would be homotopic to the constant loop by the linear map, I think

[hidden crag]

alternatively just say it deformation retracts onto the point where the two disks are glued

(the star shaped argument works as well)

[broken nacelle]

yea exactly

so you need only prove that its boundary's fundamental group cannot be trivial

(sry for disappearing, my internet is fucking with me)

[gaunt laurel]

Totally fine. Thank you a bunch. I wrote out what I now feel like perhaps could be considered a formal proof. Mind checking if u have some time :))?

[surreal burrow]

i think this is 5 am brain but i happen to be working on the same pset as fyodor can someone please explain why the first sentence doesn't mean i_* is bijective?

@broken nacelle pretty please 🥺

[unreal stratus]

Why should it

[hidden crag]

recall the definition of bijective

[unreal stratus]

Having a left inverse doesn't imply there is a right inverse

[surreal burrow]

oh sheesh

u right

[hidden crag]

you should go to sleep my friend

[unreal stratus]

Basically if this were true then like

Every retract would be homeomorphkc to the original space

[surreal burrow]

every retract would be a deformation retract no?

but yeah i get what ur saying

[hidden crag]

even stronger, as potato said

[surreal burrow]

damn i'm getting flamed x4

[surreal burrow]

ah i see because in the deformation retract they just need identity up to homotopy

ok ok ok

[hidden crag]

don't waste brainpower on thinking what would happen if left inverse implied having a right inverse

or maybe not right now

[surreal burrow]

yeah i'm trying to figure something out tho

[ebon galleon]

All injections are surjections

[surreal burrow]

basically i have an injective i_* but this doesn't have a corresponding well defined r_*

but i'm not able to wrap my head around it

i_* in particular is the map Z->Z taking x to 2x

[surreal burrow]

I GET IT OK

IM DUMB 😭😭

[ebon galleon]

Oh I wasn't making fun of you

[high hill]

hi, tell me

[ebon galleon]

I just thought that was a funny thing timo said

[surreal burrow]

what the fuck

why is everyone out of the woodwork just to roast me today

[hidden crag]

you do have a retract

by assumption

[surreal burrow]

yeah

i'm trying to understand why that's a contradiction

[unreal stratus]

Why what is a contradiction

[hidden crag]

what

[surreal burrow]

i'm saying that like

fuck

wait let me collect my thoughts

basically the problem is || show mobius strip doesn't retract to boundary circle ||

[unreal stratus]

Sure

[surreal burrow]

so the induced map you get from the inclusion is x->2x

how does that lead me to contradiction

[unreal stratus]

Well I think you've answered it yourself

[surreal burrow]

wait what

[unreal stratus]

Do you see why this doesn't have a left inverse

[surreal burrow]

u asked why it's a contradiction 😭😭😭

[surreal burrow]

yeah

[unreal stratus]

So that's it right

[surreal burrow]

yeah ur right but my question is like

if i have a map between two spaces that have homotopy group Z

[unreal stratus]

Yes

Fundamental group but yeah

[surreal burrow]

does it have to have been induced by a homotopy between the spaces?

[ebon galleon]

No

[unreal stratus]

Uh you seem a lil confused

What do you mean by homotopy between spaces

Homotopy equivalence?

[surreal burrow]

yeah sorry

[unreal stratus]

Then certainly not

[surreal burrow]

i just need an example lol

[ebon galleon]

A continuous map f: X --> Y always induces a homomorphism piX --> piY. This can be an isomorphism even if f is not a homotopy equivalence

[unreal stratus]

Wait so like

Okay this question is vague lol

[tiny ridge]

Step 1: State your question.

[surreal burrow]

ok

[unreal stratus]

You are talking about a map between spaces being induced by a Homotopy equivalence

That doesn't make sense

[tiny ridge]

Let them state their question potato

We'll see what they have in mind

[surreal burrow]

if i have a homomorphism between fundamental groups which both happen to be Z, then do the corresponding spaces have to be homotopy equivalent.

[ebon galleon]

Nope

[surreal burrow]

wait a sec i missed my own question i know that to be obviously not the case

ok i'm trying to get too general when specific is probably better

[tiny ridge]

There is a variable in your question which is irrelevant to the rest of it. Between any two groups there is always a homomorphism.

[surreal burrow]

let me try again

[tiny ridge]

Namely, the trivial homomorphism

So that seems redundant doesn't it?

[surreal burrow]

yeah exactly

[unreal stratus]

The question that is most interesting is probably like
Let f : X -> Y be a (pointed) map of spaces which induces an isomorphism on fundamental groups. Is it a homotopy equivalence

[surreal burrow]

i already understood that

[tiny ridge]

OK, so reformulate

[surreal burrow]

so in this particular case, i have a space which has fundamental group Z and a subspace which has fundamental group Z. there is no retract between the spaces (to me it seems) because the induced map between these two copies of Z is not an isomorphism. i'm asking basically that if there is a retract, does it have to be a deformation retract. that is, is what "seems to me" to be true actually true

[tiny ridge]

A retract may exist yet there may be no deformation retract, in general, if that is your question.

[unreal stratus]

The issue is not the map not being an isomorphism actually

The issue is it admitting no retraction

Which is simpler actually

[tiny ridge]

Here is an example which may satisfy you. Consider S^2 v S^1. This retracts to S^1, the second wedge factor. There is no deformation retract.

[unreal stratus]

Here is an example of a retraction that is not a deformation retract: include a point into any non-empty space. There is always a retraction (the constant map!) But it usually cannot be a deformation retract

[surreal burrow]

yeah i know of this case, like i know deformation retract != retract in the general case, i'm asking specifically in the case where both the space and the subspace have fundamental group Z

[surreal burrow]

what is the map between the fundamental groups in this scenario?

actually let me think of it for a bit

[tiny ridge]

Sure

[unreal stratus]

Knowing about fundamental groups is virtually never enough to conclude anything about homotopy equivalence

[tiny ridge]

Potato's comment and my example are related in fact

[unreal stratus]

This is an important point

Well, in the positive sense I mean aha

Like the fundamental group can provide obstructions

But knowing a map induces an isomorphism on pi1 isn't enough to conclude it is a homotopy equivalence

[surreal burrow]

i mean will higher homotopy groups help with that, if my understanding is correct here?

[tiny ridge]

Yes. But that is a deep theorem.

See "Whitehead's theorem"

[unreal stratus]

And this is only true for nice enough spaces I should add but yeah

[hidden crag]

all spaces are nice enough

[ebon galleon]

If you believe hard enough this is true

[broken nacelle]

ah yes

all spaces are homeo a point

[unreal stratus]

Except the empty set

[broken nacelle]

lmao

[hidden crag]

always that one kid in class

[unreal stratus]

Me?

[broken nacelle]

[unreal stratus]

Rippies.

I don't think I am too pedantic

[surreal burrow]

ok i fail to see how the map between the fundamental group of S^1 V S^2 and the fundamental group of S^1 is not an isomorphism 😭

[unreal stratus]

It is

[surreal burrow]

wait but if it is, then S^1 V S^2 deformation retracts to S^1 and that doesn't seem to be true

[unreal stratus]

But the latter doesn't deformation retract

[broken nacelle]

no

that's not true

[unreal stratus]

Why should it being an iso mean it deformation retracts

Remember a map inducing an iso on pi1 does not imply it is a homotopy equivalence

[surreal burrow]

sorry i meant the induced homomorphism by the inclusion map

[unreal stratus]

That is what I am referring to too

[surreal burrow]

oh wait

i'm a troller

[unreal stratus]

Lol

[surreal burrow]

I LITERALLY FLIPPED THE CONDITION

[tiny ridge]

The induced map by the inclusion is an isomorphism on pi_1, yet the big space doesn't deformation retract to the smaller one.

[unreal stratus]

[surreal burrow]

everything works out when if is the same as iff

[unreal stratus]

For the sake of your original question this is overcomplicating stuff

But these are good questions

[surreal burrow]

me when i'm scared to ask questions in class so i ask my beloved math discord server instead

[unreal stratus]

Noice

No I mean I feel if you ask a q in class it is likely others will have the same q anyway but yes fair enough too