point-set-topology

[heady skiff]

oh wait but i thought we were assuming that {a} is closed tho

oh wait no

{a} is closed in the standard topology

{a} is open in the subspace topology right or smt

[ebon galleon]

So if a is not a limit point of A, that means that {a} is open in the subspace topology (this if if and only if, btw. And purely topological, this is not special to R)
In particular, if every point of A is not a limit point, then every singleton in A is open

[heady skiff]

o right

[ebon galleon]

This is in the subspace topology yeah. So {a} is definitely not open in R

[heady skiff]

ohh ok got it, then A is open which contradicts A being closed

right?

[ebon galleon]

Still no because this is subspace topology

[heady skiff]

ugh

[ebon galleon]

We gotta use compactness somewhere here

[heady skiff]

oh

OHH

[ebon galleon]

We know A is closed and bounded

[heady skiff]

oh so the unionn of all the singleton sets

is an open cover for A

it must have a finite subcover

but that contradicts A being infinite

?

[ebon galleon]

YES!

[heady skiff]

FUCK YEAH

ok tysm!! that helped a shit ton

[ebon galleon]

No limit points means A would be closed in R (and bounded hence compact) and would have the discrete topology. But the only compact sets with the with the discrete topology are finite

So that contradicts A infinite, like you said

[heady skiff]

ohhh ok sweet

cool thanks!

wait doesn't this question just follow immediately from a bijective map from a compact space to a Hausdorff space being a homeomorphism

oh wait

i have to show it's continuous

[ebon galleon]

I think by map they mean continuous

[heady skiff]

oh

[ebon galleon]

Otherwise being injective from compact to Hausdorff is not sufficient to be continuous in general

[heady skiff]

wait, question; if the complement of A is open does that imply that A is closed?

[queen prism]

what’s the definition of closed

remember that definitions are iff statements, even if they only use “if” or “provided”

because english is a great language

[heady skiff]

ohh right i'm trippin

aight thanks

[narrow cairn]

any tips on showing that the topologists sine curve is not path connected (and also in showing that it is connected)

[gritty widget]

proof by picture

[narrow cairn]

no i totally get the intuition

just dont know how to show that no path could exist

[tawdry widget]

Where did you find this definition… because all I can find is simply sum of signs of crossing, no 1/2 appears…

I guess you define linking number by looking at each knot separately, then take sum, divided by 2
That case each crossing is counted two times exactly, by each knot.

[bitter smelt]

I assume you know where the bad spot is for pconnected

Just assume there is a path, parameterize it, arrive at contradiction

[median sand]

Where to learn the basics of nets quickly, including stuff related to subnets and compactness? In the subject of generalised convergence, is it better to learn nets or filters?

[high hill]

suffer in nlab

[ebon galleon]

[fading vale]

The book topology and geometry by boredom has some stuff

[ebon galleon]

Is it boring because it has nets instead of filters?

[unreal stratus]

by boredom lmao

[red yoke]

There are a few pdfs

[median sand]

Where would one find them?

[ocean narwhal]

i tend to use nets a bit more (how to make a topologist angry 101, essentially), but both are important.

[limber wren]

google "introduction to nets and filters pdf"

look for lecture notes from a university if possible

that's what I do when I need a reference for something, look for university lecture notes like "introduction to X pdf"

[median sand]

I thought he might have something specific in mind.

[eager vigil]

How to see this last relation? (from Hatcher, p 70)

[tribal palm]

five weeks into my top course and i’m still stumbling over the def of bases

not to say i’m not making progress… i’ve certainly cleared up some confusion

[red yoke]

Lifting path to X1 and sending to x2 is same as lifting to X2

[celest plinth]

Hello, I was told this was a good place to discuss n-dimensional mathematics. What am I supposed to think when 3bluebrown talks about hypercube. Like obviously it’s mathematically feasible to have 4 mutually “perpendicular” axis you just can’t draw it. The problem I have is that the motivation came from a 3 dimensional Euclidean geometry analysis, so this means when you start talking about the edges in 4 dimensions, your still talking about an actual shape, but just describing parts of it. Is this understanding correct? Thank you :).

[red yoke]

just describing parts of it

What

If you draw a 3d cross-section, then you indeed only have a part of the shape

But there are also ways to describe 4d shapes without cross-sections

[queen prism]

you can formalize 4D space as the set of all 4-tuples whose coordinates are real numbers
so a point in 4D space can be represented by/is a coordinate (x, y, z, w)
then you can try to figure out how to define cubes in 3 dimensions and see if you can scale that up to 4

[red yoke]

A common representation is using time or color to represent w coordinate

[red yoke]

After that try 4d version of triangle/triangular pyramid and circle/sphere

[celest plinth]

Part of the shape? What shape? Is it a shape immersed in a space with 4 perpendicular axis? Or is it a shape that must be immersed with the 4th dimension as time, temperature etc.. it is then that we are now dealing with a completely different shape, not analogous to a cube in 3 dimensions

[queen prism]

anything that acts like a 4D shape is a 4D shape in a mathematician's eyes

[celest plinth]

But my theory is that we have 4 mutually perpendicular axis, and that is valid in mathematics take 4 basis vectors of 4 demensions. Obviously it’s far more practical if they are modelling things like time, temperature etc. but theoretically we can just say this is a space with 4 perpendicular axis and we can hence have shapes in this space. We just can’t draw it all at once. I think 3blue1brown is talking about a hypercube in this context

[celest plinth]

It’s a different context to be talking about a sphere in terms of euclidean geometry, and then all of a sudden say we are talking about a hypercube with time as 4th dimension. Why didn’t we start with time in the third? Because it’s not a freaking shape and the notion of distance collapses

[queen prism]

you could

that’s a useful visualization

take different slices of a 3D shape and lay them next to each other
you can now visualize a 3D shape as a 2D shape changing in time

[celest plinth]

I don’t think you understand what I’m trying to say :/. Yes you can do what you just described, but you will now have to proceed to re-define your notion of distance. But 3blue1brown uses the same notion of Euclidean distance in all dimensions when it’s not appropriate unless you believe my theory with the 4 perpendicular axis

You can think of my theory visually two, you can visualise all the axis being perpendicular if you take 2 at a time, you just can’t visualise the entire thing

[tidal lynx]

has anyone ever seen the second part

[obtuse meteor]

no but it definitely makes sense that you could formalize this

[ebon galleon]

Yeah I've thought of that before

Should work generally for nets; just freely adjoin a max element

[fair idol]

Why did people even define nets. I get you need them for general statements about continuous maps

[ebon galleon]

sequences are not enough when your space is not first countable

[red yoke]

Nets are just sequences

Intuitively

But general

[steep kite]

Why use directed sets instead of ordinals for nets?

[red yoke]

More general and you can translate to/from filters

[gaunt laurel]

Why are y and f(y) in the same orbit necessarily?

[tawdry widget]

By definition… f deck iff pf=p…

[gaunt laurel]

Sorry, aren't the orbits determined by G here? How does f deck connect with orbits?

Oh

My bad

Sorry

My brain stopped working for a min

Literally is by definition

[tawdry widget]

It happens to everyone

[tribal palm]

eek this took me way too long to realize

isn’t condition (2) here just equivalent to saying “B1 n B2 is open”

for all B1,B2 in the basis

ye ok it definitely is

[alpine oriole]

open wrt the topology generated by B?

[tribal palm]

yes

[alpine oriole]

yah think so

[tribal palm]

it is useful proposition: U is open in T(B) iff for every x in U there is some open A containing x and contained in U

and in particular open A is just the general case of basis element B, then by the def of open sets in T(B) the first is a consequence of the latter

i am looking over my notes from some days ago and it is painfully clear how confused i was… i probably still am but at least now i am aware of it— progress!

[scarlet turtle]

yeah my lecturer just uses this definition which imo makes the "basis" thing more obvious

apologies for the use of two different fonts for the set T and the topology \mathcal{T}. my lecturer has a strange sense of humour

[queen prism]

non-serif characters for variables is unacceptable

[tribal palm]

lol i had to work my way to this def myself, though that is not a def for a basis, but a basis for a specific topology

[scarlet turtle]

that's true tbf

[tribal palm]

i’ve found that for $\mathscr B$ to be a basis for the topology $\mathscr T(\mathscr B)$ which is the set of all unions in $\mathscr B$, we need only two conditions on $\mathscr B$: namely (1) $\bigcup_{B\in \mathscr B} B = X$ and (2) for any finite $\mathscr A\subset \mathscr B$ there is some $\mathscr C\subset\mathscr B$ with $\bigcap_{A\in\mathscr A} A = \bigcup_{C\in\mathscr C} C$

[gentle osprey]

Jens

[tribal palm]

where the second condition is understandably given as the much more handy but equivalent: for any two B1,B2; B1 n B2 is open in T(B) (which in turn is equiv to that it is a union of basis elements, which in turn may be written as that every point in the intersection is contained in a basis element contained in the intersection)

[scarlet turtle]

that's just the same as the definition you have but inducting on the finite A isn't it?

[tribal palm]

yes

but it makes it obvious, especially if you explicitly write out the def of T(B), why it (very forcefully) satisfies the conditions if a topology

and i don’t mention it but you have to keep in mind that any union over an empty family of sets is the empty set, so also ø will be in T(B)

ok but now i am finally through the defs of the first week, time to catch up on the 4 weeks since! byee

[tribal palm]

ok having returned to this, i’ve definitely cleared up much of my confusion and will confidently state T(B) is courser than T(B’) iff B is contained in T(B’)

[limber wren]

that definitely seems true, if B \subset T(B') then T(B) \subset of T(T(B')) = T(B')

[grave sun]

For 2. does the topology matter? I’ve never seen any topologies defined on Z or S^1

This is from Munkres

I'm thinking for each part I can show that the function suggested in 1. is continuous, but I don't know how I'd do that if no topology is specified

[gritty widget]

i would hope the topology matters

Z and S^1 have the subspace topologies from R and R^2, respectively. in particular, Z is discrete

[grave sun]

So do you think those are what the problem expects me to assume the topologies are? Or is the point of the question to define topologies such that each becomes a topological group (in which case I'd think you could use the discrete topology for all of them)?

[gritty widget]

the problem expects you to know what the topologies are

in which case I'd think you could use the discrete topology for all of them
you could, and this should be a hint that it's not the right way to proceed

when have you ever cared about R with the discrete topology?

[limber wren]

You would need to specify the topology in each example, like with S_1 for example, you can use the subspace topology inherited from R^2 with the standard metric topology

or equivalently, the topology on S^1 generated by "open arcs", like open intervals but along the circle

[lime sable]

also that S^1 is the interval [0,1] with the endpoints identified, so it gets a quotient topology from that

when i thought "what's the topology of S^1" that's the first thing i thought of

[tribal palm]

oh wow haha

[pulsar lynx]

I'm reading this book which defines the following for a metric space (M, d):

For any subset S ⊆ M, we define
lim S := {p in M : p is a limit point of S}.
...
For any subset S ⊆ M, we say S is closed set if every sequence in S contains its limit in S.

It's very obvious that lim S is a closed set

But the proof he provides is rather weird and technical

shouldn't it be like 2 lines? I don't get it

[tribal palm]

show

[queen prism]

is that actually how it's worded
seems a bit imprecise

[tribal palm]

a picture plz

[pulsar lynx]

here let me provide the closed set definition, which is even less precise

[queen prism]

munkres is graduate-level?

🇺🇸 moment

[pulsar lynx]

lmao

[queen prism]

what proof did you have in mind

[pulsar lynx]

oh shit I think I'm realizing now

why it's like that lol

just because every point is a limit of some sequence, doesn't mean every sequence contains its limit in the set

[quiet thorn]

i don't think munkres is grad level, but it's not unusual for there to be a first year grad class on pointset

[pulsar lynx]

well, it does, but this must be proven as above

ok ok

[pulsar lynx]

yeah I thought munkres was undergraduate

wait actually

[limber wren]

It's a good rigorous book for pointset topology, you could use it for either grad or undergrad

[pulsar lynx]

what does he mean by "contains all its limits"?

[pulsar lynx]

what is "its"?

[tribal palm]

he calls limits what most people call adherent points of a set

which is basically just the closure

[limber wren]

closed under taking limits, meaning if you have a sequence of points in the set, and that sequence converges, the limit is also in the set

as opposed to something like (0, 1) which doesn't contain the limit of 1/n

[pulsar lynx]

so then doesn't that mean "lim S is closed" has a really trivial proof?

[tribal palm]

though iirc there is some extra care needed if a set has isolated points

[limber wren]

well it's easy but technically you'd have to prove something, like to show lim S is closed, take a point in the compliment, and show that there's an open ball around that point disjoint from limS

but these things kindof just fall out of the definitions

[pulsar lynx]

that's true but closed here is defined as "contains all its limits"

it was proved before that open and closed sets are dual, so you could take the route you mentioned

[limber wren]

oooh true, I was thinking closed in the topological sense

[pulsar lynx]

yeah yeah

[pulsar lynx]

so then I mean

how is the proof not trivial

[queen prism]

limit vs limit point I think

the terminology here is a bit nonstandard

[limber wren]

okay so limS is the set of all limits of sequences in S, and a set is closed if it contains all limits of sequences, lol so yeah

maybe it's pretty trivial

like it's almost tautological

[pulsar lynx]

S is closed if every convergent sequence has its limit contained in S.
lim S = {p in M : p is a limit of S}

hmm

[pulsar lynx]

no yeah lol

[limber wren]

ahh I got the difficulty, if you have a sequence of points in lim(S) and x_n -> x, you need to show x is a limit of points in S

but each x_n is itself a limit of points in S

so there's something to prove

[pulsar lynx]

x is a limit of points in S because it's a limit of {x_n}, for which each x_n is contained in S

I don't see how this isn't trivial

[limber wren]

like you can approximate each x_n by points of S, and you can approximate x by the points x_n, but then show you can approximate x by points in S

lol it's subtle

[pulsar lynx]

isn't each x_n in S tho?

[limber wren]

no

each x_n is a limit of points in S

[pulsar lynx]

wait wth

[limber wren]

since x_n is in lim(S)

[pulsar lynx]

yes that's correct

but S is always a subset of lim S

wait oh oh

shit

[limber wren]

yeaaah

its subtle lol

[pulsar lynx]

ok I'm thinking of like

S = an open ball

lim S = the closed ball

[limber wren]

so you can approximate x by x_n, and each x_n you can approximate by points in S, show you can approximate x by points in S

[pulsar lynx]

and you have x_n lying along the perimeter of the circle

[limber wren]

it's not immediate

[pulsar lynx]

damn wth

no wait hold on hold on

[limber wren]

thats a good example, take the open disc, and a sequence of points on the boundary

[pulsar lynx]

it doesn't matter that x_n aren't necessarily in S

x_n are in lim S

[limber wren]

each point on the boundary is arbitrarily close to the open disk

[pulsar lynx]

which means, by the definition of lim S, that their limit x is in lim S. And thus lim S is closed

?

[limber wren]

but then you need to show the limit of points on the boundary is also arbitrarily close to the open disc

[pulsar lynx]

OK I think I'm starting to understand

[limber wren]

this is kindof the idea, the black points converge to the big black point

each black point has a sequence in the open disk converging to it

[pulsar lynx]

ahh yeah

that makes sense

damn that is subtle

[limber wren]

but then how do you get a sequence of points on inside converging to the big point

yeah

[pulsar lynx]

it almost makes me question my intuition

[limber wren]

lol this should be titled "an exercise in understanding why something isn't obvious"

[pulsar lynx]

like I feel like with this subtlety you could create a pathological set wherein a sequence of limit points doesn't converge to a limit point

[pulsar lynx]

yes lol

[pulsar lynx]

obviously this is not possible, because we just proved it isn't, but

yeah

[eager vigil]

Does anyone know of any place I can find supplementary exercises to Hatcher? I feel like his problems are a tad bit too difficult for me at first; I would like some really simple and easy exercises to first exemplify the theorems and general constructions before I delve into harder problems

[hidden crag]

Rotman has easy exercises

[eager vigil]

Ok, thank you

[feral copper]

Heyo! Is there someone pretty familiar with the Whitney trick in dimension four? I'm willing to understand if there's an obstruction to finding an embedded disk in a special equivariant case

Maybe someone knows about obstructions to what I'm trying to achieve?

So say I have two surfaces F and G in a simply-connected 4-manifold (to make things simpler), and such that they intersect transversely twice in a positive and negative fashion. I have an embedded Whitney disk, but it may intersect F and G in its interior. Say I don't care about intersections with F, so I have G intersects the interior of the disk. Is there a known complete obstruction to finding another disk that doesn't intersect G?

I'm in a special case where F and G are invariant under an involution, so any pair of intersection points itself comes with another such pair. I was wondering if maybe I can use that?

Also, I don't really mind not isotoping F, and going through immersions. That is: I can change F about as much as I want, but G has to remain unchanged

(this is the reason why I don't care about the disk intersecting F in the first place)

[feral copper]

(I know of the framing obstruction, but at the cost of adding new intersection points with F or G, I can always assure that it vanishes so that the normal bundle has a section)

[formal tide]

In a metric space, sequences seem to "be enough", in constrast to the need for nets/filters in other spaces. Is there a precise statement of this?

[unreal stratus]

Metric spaces are first countable.

Well okay I should expand on that like

In a first countable space, continuity of a function is equivalent to sequential continuity (i.e. f( lim x_n) = lim f(x_n) basically)

This isn't true more generally

[queen prism]

i googled first countable
i like this comment https://math.stackexchange.com/questions/1529419/example-of-a-topological-space-which-is-not-first-countable#comment3114772_1529419

[hot locust]

is the product topology of discrete topological spaces, a discrete topology as well?

[umbral panther]

Not to belabor the obvious, but every obstruction to smooth 4-manifolds being like topological 4-manifolds is an obstruction to the existence of Whitney disks. There is certainly no complete obstruction or there would be a lot more positive results that things are diffeomorphic

[feral copper]

Ah, right, makes sense

But again, in my case, I don't really mind changing the surface, as long as its genus remains the same and the self-intersection as well

[unreal stratus]

Yes.

Take an obvious basis for A and B and use that to create a basis for A x B, for example

[umbral panther]

Do you want your disk to be live in the quotient by the group action? If not, how could the group action be relevant?

There is an obstruction that also exists in higher dimensions using the fact that the quotient is not simply connected (assuming the action is free). But you seem to have assumed this obstruction vanishes. If it vanishes, then you can lift to upstairs and have two copies of the same disk. It’s pretty hard for them to interfere with each other

[feral copper]

In my case it's an involution; for a pair +/- of points there is another pair +/-, together with a copy of the disk and its intersection with G. I was wondering: maybe I can use this symmetry and this pari of disks? Also, I can consider an embedding of a disk with 4 points identified on its boundary instead of two, and consider 4 arcs connecting all the pairs of points

[umbral panther]

There is a problem which is that the two disks might intersect each other. So individually they remove intersections, but together they create intersections

The standard strategy is to recurse. You have intersecting points between the Whitney disk and G. So you try to create a new Whitney disk to move them away. Casson / Freedman says that after doing this infinitely many times you succeed (and pass to the topological category), but I think other people created obstructions to doing this after finitely many stages, but I don’t know what they actually did

[coarse night]

how many products?

[gritty widget]

does anyone know of a reference for characteristic classes with a more modern treatment than milnor-stascheff? i cant seem to find anything

[coarse night]

hatcher vbkt

[gritty widget]

thats what ive been looking at but the stuff on char classes uses the k-theory in ch 2 which i dont really want to read all of

[coarse night]

Learn the obstruction theory viewpoint and you're good

[feral copper]

Yeah the issue is I'd like to Stay in the smooth category!
Although I didn't think of possible intersections between the disc and its image

Thanks!

[coarse night]

does it? you can skip chapter 2

you don't necessarily need it

[gritty widget]

iirc it started in ch3 with motivation by talking about classifying maps for vector bundles (which i dont know about)

saying cohomology is easier than the homotpy business of nulhomotopic classifying maps

[coarse night]

chapter 3 is characteristic classes

[gritty widget]

yea

[coarse night]

yes classifying space for complex vb is Gr_k(inf) which is huge

there's also the obstruction theory viewpoint which I really like

[gritty widget]

i figured that was stuff from ch2, is it standard vb material?

[coarse night]

Say you want to construct a nowhere vanishing section of your bundle of rank n, then it's eqv to have a section of the corresponding S^(n-1) bundle. The obstruction for that lies in
H^k(X, π_{k-1}(S^n)), for k=n which is exactly euler class. All the other classes follow same way

[coarse night]

no it's ch 1 stuff

it's tautological if you think hard enough.

[gritty widget]

ok ill go over ch 1 then skip to ch 3

ty!

[coarse night]

yeah ch 1 is necessary

[gritty widget]

ill think ab this some more later lol

yeah i know the stuff in 1.1 just not 1.2

except i guess ik about pullback bundles

basically the stuff thats also in milnor

[coarse night]

it's easy if you triangulate your base space. will explain sometime when you have read enough vb

[coarse night]

all it says that all complex vb arise as a pullback bundle of the canonical bundle Gr_k(inf)

upto homotopy

[coarse night]

this thing

[tiny ridge]

How does this frame fixing work by the way?

Do you kind of spin G around F?

This gives new intersections but seems to undo the twists of the framing curves

Anyway yeah this is hopeless in the smooth category

[tiny ridge]

The example I learnt was if this was always possible, all knots would be topologically slice

Im not 100% sure what it is thats going wrong though

[floral bear]

(de rham) why pre-assume that $d^2=0$, when notation such as $\frac{d^2y}{dx^2}$ is used in 1-var calculus all the time?

[gentle osprey]

cosín™

[floral bear]

what property is this trying to capture, besides the standard homology axioms

why is this a "natural" axiom to make about differential forms

or ig about the d operator

is it the assumption that smooth functions are locally linear?

should i really interpret it as $\frac{d^2y}{dx}$?

[gentle osprey]

cosín™

[gritty widget]

d^2 = 0 is "the boundary of the boundary vanishes"

for the de rham differential it boils down to clairaut's theorem

[floral bear]

..thats $\del$ tho

[gentle osprey]

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

[floral bear]

ah hush

[gritty widget]

it's obviously not the exact same statement

replace "is" with "is analogous to" if it bothers you

[floral bear]

okok sry ty

[gritty widget]

stokes's theorem will make that analogy more precise

[floral bear]

but how so? can you reduce the output to something of the form $\frac{\partial^2 z}{\partial x\partial y}-\frac{\partial^2 z}{\partial y\partial x}$?

[gentle osprey]

cosín™

[gritty widget]

there's a very clear discussion about the de rham differential in arnold's mechanics book. there's probably some nice geometric comment about d^2 = 0 in there

[gentle osprey]

TTeppACAB

[floral bear]

OH. i see, thank you🙏

[tulip bluff]

given the two descriptions of the complex projective line P^1 where the first one is the one given by the Riemann sphere C \cup \infty and the other one given by the set of lines through the origin modulo the equivalence relation that two points are the same if they're on the same line how can i show that these two are equivalent?

[feral copper]

A line in C² is uniquely determined by its slope, and a vertical one has slope oo. That gives an obvious map from lines to C u {oo}

Here, 'vertical' is supposed to be between quotes ofc, but you get the idea

[tidal lynx]

∞

[tulip bluff]

im sorta confused with the notion of the points in the latter space are they of the form [l] for a line in C^2 through the origin or [z_0 : z_1] with (z_0,z_1) in C^2 \ {0}? im not sure how to get the slope from [z_0 : z_1]

[feral copper]

They are 1-dimensional subspaces of C²

The slope is simply z0/z1

[feral copper]

This is if you refer to a point [z0:z1] in homogeneous coordinates

[tulip bluff]

uhmmm im not sure what are homogeneous coordinates i take that [z_0 : z_1] isn't just notation?

[feral copper]

It's a notation for the equivalence class of the point (z0,z1) under the relation 'allow re-scaling non-zero points'

[quiet thorn]

[z_0 : z_1] is the equivalence class of (z_0, z_1) ~ (k z_0, k z_1) for k \ne 0

and of course you can extend this to [z_0 : ... : z_n]

[tulip bluff]

so we can map [z_0 : z_1] to z_0/z_1 but this is still an element in C^2 and not in C which we would want for C u {∞}?

[umbral panther]

If what were always possible?

If topological Whitney disks could always be smoothed, then topological side would imply smooth slice. And it doesn’t, but that just kicks the can down the road to prove some other distinction between the categories

But if you’re saying something about the failure to construct topological Whitney disks, that sounds a lot crazier. It is conjectured, or at least open, that there are enough topological Whitney disks to prove the surgery statements. I’ve never heard anything about the impossibility of topological Whitney disks

[feral copper]

[1:0] is mapped to oo

Also, this z0/z1 is a ratio of complex numbers, that's a complex number

[tulip bluff]

oh okay. do we require it to have modulus 1 to be on the sphere?

[feral copper]

It certainly won't have modulus one!

What sphere would this be on? It happens that thus manifold is diffeomoephic to a 2-sphere, but that's different

[tulip bluff]

isn't the riemann sphere actually a sphere or am i just missing something in the definition?

[feral copper]

It is a sphere, yes!

But the ratio z0/z1 doesn't always have modulus one; that'd correspond to the equator (it's a circle)

[tiny ridge]

A knot in S^3 certainly bounds an immersed disk in D^4.

Take the immersed disk. If the topological Whitney trick was possible you would obtain a topological embedded disk in D^4 bounding your knot

But there are knots which are not TOP slice, so this is a contradiction

[umbral panther]

Yeah, ok, but there are unsliced knots in higher dimensions, too, so you can express some obstructions in generic terms

[tiny ridge]

Probably this is a framing issue?

I don’t quite see it

[umbral panther]

I think it must be the fundamental group obstructing framing

[tiny ridge]

The framing given by such a disk would be the Seifert framing. But instead we have the blackboard framing or something?

This is probably why you always see the Whitney trick being done when your surface is intersecting something else. The other surface can be leveraged to correct the framing

“By spinning around it”

[hot locust]

finitely many so it worked out ig

but hey, say it's an arbitrary product. Would it still work out the same way?

[limber wren]

No, I think there are counterexamples like the product of {0, 1} over N, but I don't remember all the details

[hot locust]

oh cool, I'd check that out. Thanks.

[coral pivot]

What are some common topologies put on the fundamental groupoid of a space? Are any of these etale?

[umbral panther]

If you have a reasonable (slsc) space, then there are only two choices: discrete, or give the objects the topology of the original space. This is probably etale in some sense

[fickle elm]

What is the meaning of etale here? The topology ? or some morphism?

[umbral panther]

Maybe that the map from morphisms to objects give by the source is etale, ie, a local homeomorphism

[distant lichen]

Exactly

[coral pivot]

Yeah that's what I'm thinking. The c star algebra of such groupoids have nice theory

And I had an idea to study the c star alg of this

[umbral panther]

Say X has universal cover Y. Applying the general procedure for producing a groupoid from a map Y->X should produce the fundamental groupoid. But the C* algebra for this should be Morita equivalent to X

[hot locust]

Since quotient maps imply continuous fns then can I just argue based on the continuity of f and gof? Is that fine?

no wait, argh nvm I need sleep.

how do I show that this is a quotient map based on g being continuous?

okay so is it enough to show surjection?

[narrow cairn]

getting mildly stuck on this problem: "a space is locally compact and hausdorff iff it is homeomorphic to an open subspace of a compact hausdorff space." I get that for one direction you can use the one point compactification, but i dont get the other directoon

[ebon galleon]

Well, for the other direction you want to show that open subspaces of a compact Hausdorff space are locally compact Hausdorff (LCH).
In particular, compact ==> locally compact. So you can show that an open subspace of a LCH space is again LCH (just to highlight the key assumptions for the proof).

[opaque scroll]

Should it maybe be a locally compact Hausdorff space?

[ebon galleon]

I don't remember the exact details, but for a point in your subspace, it has a compact neighborhood in the larger space.
There seems to be a canonical choice for your compact neighborhood in the subspace then, no?

[opaque scroll]

Guess you would have to show that open neighborhoods contain closed neighborhoods. Maybe using that the boundary is compact.

[ebon galleon]

That also works, and is probably a better proof in retrospect since it's an important property of LCH spaces

[narrow cairn]

sorry yes

typo

[ebon galleon]

This is what jagr is talking about

[opaque scroll]

This is what I would take as the definition of locally compact, but I guess it's a theorem here

[ebon galleon]

If you take this as the definition of locally compact, isn't the question trivial then?

[opaque scroll]

No cause you have to show that compact Hausdorff spaces satisfies this

[ebon galleon]

I guess showing compact Hausdorff ==> LCH (with this definition) is the exercise then yeah

[ebon galleon]

What definition are you taking for locally compact? As jagr points out, there are a few

(all equivalent for Hausdorff spaces though, but that requires a proof in itself, of )

[hot locust]

okay maybe a bit off topic but is a T3 space necessarily compact?

[ebon galleon]

No. Consider the real line

[hot locust]

ahh, okay

[opaque scroll]

Any metric space is T3

[ebon galleon]

T6 even

[hot locust]

what would be a nice counterexample to 'Continuous image of a Hausdorff space is a Hausdorff space'

[unreal stratus]

Uh if you want a minimal example like

Take the Sierpinski space S (so {0,1} with the open sets empty, {0},{0,1}), which is the smallest non hausdorff space

[gritty widget]

identity map from discrete topology to indiscrete topology

[unreal stratus]

You can view it as a quotient of [0,1] by identifying all points > 0 with 1 I think

Okay nice tterra oops

Another is uhhh

Take a hausdorff space X and crush a non-closed subspace A to a point

Then if you pick a limit pt of A which is not in A, you can't separate it from A by opens

My other example is a special case of this lol

[narrow cairn]

sorry as i said i mistyped and meant to say LCH not just LC

forgot to edit but its edited now

[narrow cairn]

doesnt work if the spaces have less than 2 points

[gritty widget]

who cares

[ebon galleon]

Right, I know. But still, it's an exercise in and of itself to show the definitions are equivalent. Which is why it's relevant.

[hot locust]

ahh, gotcha. Thanks.

[narrow cairn]

sure, then my book gives "a space is locally compact if for every point there is a compact set containing an open set containing the point"

[ebon galleon]

Right, that's the one I would take to be the definition; it's the weaker definition.
Clearly, every compact space is locally compact with this definition; take the entire space

[ebon galleon]

So you'll want to show this stronger characterization of LCH for this exercise

[opaque scroll]

It really scratches my gears to call something "locally X" if it's not a local property

[ebon galleon]

That's a local property still

[opaque scroll]

To me a local property should be true for open neighborhoods

Otherwise you're relying on global structure

[ebon galleon]

Fair

But all spaces are hausdorff

[unreal stratus]

Algebraic topologists: 😄
Alg geometers:

[ebon galleon]

Jkjk, but I see what you mean. It should be inherited by open subspaces

[opaque scroll]

Sticking with the "all spaces are Hausdorff" slogan, still seems like an argument to pick the strong definition.

Like if the definitions are equivalent anyway, might as well pick the definition that makes sense.

[ebon galleon]

I don't write the textbooks

[quiet thorn]

i'd read ryx's textbook

[heady skiff]

what would the open sets of the identification space be if the elements are the subsets of X? i know they would be union of the subsets but which subsets

[ebon galleon]

It needs to be a partition of X. You can't take all subsets of X.

[heady skiff]

oh yea my b i meant the disjoint subsets of X whose union is X

by points do they mean open sets ig is my question

[gentle osprey]

Ryx (Home for flowers)

[ebon galleon]

this is just explicitly writing out what pi^{-1}( O ) is

[heady skiff]

got it

thanks bro bro

[queen prism]

quotients

[ebon galleon]

"bro bro"

[heady skiff]

wait

is identification space synonymous w/ quotient space

[queen prism]

i hope it is

you know how the quotient Z/3Z identifies 0, 3, 6, 9, ... as just one element?

[ebon galleon]

Yes. "Identification" seems to be more common for older texts.

[heady skiff]

ah i see

ayo

the intuition goes crazyyy

[queen prism]

nani

oh that's a different book

[heady skiff]

yea i just googled "identification spaces" for intuition

how does this show that the topology is the largest for which f is continuous? i can't think of anything at all lol... is it just because f^{-1}(U) open implies U is open and that has to do with the quotient map or smt?

i'm also trying to see how the quotient topology is the largest topology for which pi: X --> Y is continuous... for if this topology was coarser than some topology, our open sets in Y wouldn't necessarily have an open preimage..hm

oh

S not contained in the quotient topology implies that it's preimage is not open

here, what is I?

[ebon galleon]

[0,1] with standard (subspace) topology

[heady skiff]

ah ok thanks

how do we add tv + (1 - t)x if (1 - t)x is in R^2 and tv is in R^3? do we just embed (1 - t)x into R^3 and then add 'em?

[queen prism]

give x last coordinate 0

[fervent root]

if A = {1,2,3,4} in the finite compliment topology on Z, would these be right?

Int(A) is A
Cl(A) =Z
bd(A) = {1,2,3,4},
A’ = A

[eager vigil]

Could someone please help me with finding all connected covering spaces of RP^2 wedge RP^2?

I have the following, but I'm so confused about the the quotienting to get the orbit space

I can't see how to get type three and four

[tawdry widget]

It’s just direct calculation… elements of G are of three kinds: A: of the form a(ba)^n, B of the form b(ab)^k, C: of the form (ab)^m or (ba)^m
For a subgroup H of G, H trivial you get H={1} or type 1)
H contains only elements of one kind among A,B,C you get type 2) or 3)
H contains more than one kinds of elements then it will contain all three kinds of elements you will get type 4)

[eager vigil]

Yeah, but how would one go about finding the covering spaces for those subgroups?

I.e. how would one go about figuring out how the orbit spaces are like?

[tawdry widget]

Any subgroup H there exists Y->X corresponding H

Any subgroup H of π1(X, x0), let Y be the quotient set of paths f: I -> X, such that f(0)=x0, over equivalence relation : f equivalent to g if f(1)=g(1) and class of fg^-1 is in H

Topology of Y see gtm 119 rotman chapter 10

[alpine bolt]

Hi, I am trying to under classifying bundles. In particular I just read that the bundle Gr_{n}R^{\infty} \times R^{\infty} ---> Gr_{n}R^{\infty} is the universal bundle and classifies real n-plane bundles.

So my question this bundle is universal in the sense that every real $n$-dimensional vector bundle $ E_{1} ---> X$ that admits a bundle map to a $n$-dimensional vector bundle E_{2} ----> Y. This map will uniquely factor through a bundle map $ E_{1} ---> X$ into Gr_{n}R^{\infty} \times R^{\infty} ---> Gr_{n}R^{\infty} ?

Like is the universal bundle a colimit or limit or something in the homotopy category of rank n vector bundles?...
How do you understand this?

[unreal stratus]

You can think of the construction sending X to the set of iso classes of rank n vector bundles over X as a functor C^op -> Set where C is, say, category of paracompact spaces with maps given by homotopy classes of continuous maps

Then the universal bundle E -> B has the property that B represents the above functor

With the natural isomorphism between the two functors given by pulling back as you desctibe

[fickle elm]

Let X be a sufficiently nice space (e.g. a manifold) and $E\rightarrow X$ a real vector bundle of rank n. Take an open cover $U_i$ of X, the transition functions $U_i\cap U_j\rightarrow GL_n(\mathbb{R})$ completely determines the vector bundle. By some argument of cocyle conditions on the transition map (gluing consistently), we know this gives us a class in the first Cech cohomology group $\check{\mathrm{H}}^1(X,GL_n(\mathbb{R}))$. Since the base space is nice, Cech cohomology can be identified with singular cohomology on X valued in the group $GL_n(\mathbb{R})$. By the general theory of cohomology, we know that the cohomology functor is represented by the Eilenberg-Maclane space $K(GL_n,1)=BGL_n$. This is why you can get a space to represent the functor sending X to the isomorphism classes of real vector bundles over X.

To understand why it is universal, best to think in the setting of $GL_n(\mathbb{R})$-principal bundle associated to the real vector bundle. The universal cover $EGL_n\rightarrow BGL_n$ admits a free action of $GL_n$ by deck transformation. Every $GL_n$-principal bundle over X can be obtained by a free action of $GL_n$ on the pullback bundle $X\times_{BGL_n} EGL_n$, which is determined by the homotopy type of the map $X\rightarrow BGL_n$.

To see why $BGL_n$ is the Grassmannian $Gr_{n,\infty}$. Use the definition that Grassmannian is obtained from the Stiefel manifold $V_n$ (it is all then-frames while Grassmannian should be n-frames modulo the $GL_n$ actions). It gives a fiber sequence $GL_n\rightarrow V_n\rightarrow Gr_n$. Compare with the fiber sequence $GL_n\rightarrow EGL_n\rightarrow BGL_n$. Note that both $V_n$ and $EGL_n$ are contractible, so it gives the same space in the homotopy category.

[gentle osprey]

Dong_Valentino

[fickle elm]

After writing it down, it feels really complicated. But I really cannot think of an easier explanation. I mean the vector bundle funtor is represented by the grassmannian seems really very mysterious.

Maybe think of the vector bundle as projective modules over the C-infty(X). There may be a purely algebraic argument to classify the projective module but I guess you still need to use some cohomology or homotopy theory. Perhaps.

[umbral panther]

Here is a really slick construction. Fix your base space X. Consider the infinite dimensional vector space V of continuous sections of this vector bundle. At every point you can evaluate a section to the fiber at that point. This is surjective if X is a reasonable space. Define Gr(n,V) as the space of n-dimensional quotient of V. This gives a map from X to Gr(n,V).

Show that the original vector bundle is isomorphic to the pullback of the canonical vector bundle on this grassmannian. This shows that every vector bundle is the pullback of the canonical vector bundle on some infinite dimensional grassmannian. Then you have to prove that all infinite dimensional grassmannians are close enough to the same that all vector bundles are the pull back from a single infinite dimensional grassmannian. (You probably have to restrict the definition of reasonable space more to do that). Finally you have to prove that if two vector bundles are isomorphic their classifying maps are homotopic. By construction, their classifying maps are the same, but you have to worry that you messed this up when transferring from the boutique grassmannian to the single canonical grassmannian

[unreal stratus]

Hot

[quiet thorn]

cold

[soft zephyr]

Goldilocks

[narrow cairn]

how might i show that the gluing lemma holds if the functions are defined on elements of a locally finite closed cover?

[unreal stratus]

Because continuity is local, the locally finite case should reduce immediately to the finite case

:)

Okay I am new to stuff with groups acting on spaces but was wondering - let's say X is a CW cpx and a group G acts on X. How nice an action does it have to be for X/G to have a natural CW complex structure? I would assume if G is finite and acts on a finite CW complex via cellular maps then it'd be fine; idk what happens more generally

[umbral panther]

If it acts by cellular maps, that’s practically assuming you already have a CW complex structure on the quotient.

If the action is smooth or piecewise linear, it’s ok. I don’t think there’s anything more you can say

[narrow cairn]

lol im dumb thats so obvious

thanks

okay im stuck again this time on showing that in a LCH space or a complete metric space, every closed countable set has an isolated point

so far all i have is that such a set would have to be infinite and have dense complement

[unreal stratus]

Dw

What's funny is I don't remember ever proving that myself cause never needed to, but it looks simpler in hindsight lol

[unreal stratus]

Sure cool thanks

[heady skiff]

would this work? since compact subsets of a Hausdorff space is closed, we know that A and B contain their limit points. so for each a in A we can find a neighborhood of a that doesn't intersect B. so we can form the union of all these neighborhoods which would be a neighborhood for A which doesn't intersect B, and we can do the same for B

[unreal stratus]

Those needn't be disjoint

You thickened up A in a way that doesn't hit B and thickened up B in a way that didn't hit A

But those thickenings could touch

Also, your argument doesn't use compactness - it only uses the fact the subsets are closed. But this theorem is actually false for closed instead of compact subspaces I'm p sure

Hm I can't think of a counterexample though

[heady skiff]

oh yea that's a good point

[heady skiff]

true - i thought the point of this exercise was to use the fact that compact subsets of a Hausdorff space are closed

[unreal stratus]

Ah ok

Now I am trying to find a counterexample lol

Gr

[heady skiff]

for this?

[unreal stratus]

Yeahh

[heady skiff]

hm

hmm maybe we could tweak the method used for showing that a compact subset of a Hausdorff space is closed

fix some b in B, then for each a in A find open neighborhoods of b and A that are disjoint, then keep on doing this process for each b?

idk

wait nvm i think this is equivalent to my previous argument oops

welp

contradiction it is

[narrow cairn]

?

[umbral panther]

Right. This separation property for closed subsets is the T4 axiom. There are Hausdorff (T2) spaces that are not T4. There are even T3 spaces that are not T4. T3 is the axiom exactly halfway between T2 and T3, that you can separate a point from a closed set using open sets

[unreal stratus]

Yeah sure

[heady skiff]

could I have a hint for this question please?

i'm currently fooling around with contradiction and assuming that every neighborhood of A and B are nonempty

this has led me to consider the intersection of all neighborhoods O_i(A) \cap O_i(B) and take their union, which form an open cover for A U B which we know is compact

and i have no clue if this is going anywhere lol

[heady skiff]

i'm confused, so we use S^n to denote the n-dimensional sphere whose points are that of R^{n + 1} - so if we were to think of S^2, it would be a 3 dimensional sphere, visually, but we would define it to be a 2d sphere?

also how do you visually verify that S^1 x S^1 is a torus?

like i'm having trouble

[queen prism]

S^2 is the surface of a 3D ball

[heady skiff]

seeing how the cartesian product turns out be the torus

oh

[queen prism]

you can draw a line in 1, 2, 3, 4, 5 dimensions and it will still essentially be 1-dimensional

[heady skiff]

wdym by that

[queen prism]

you can even bend it to form a circle but it will still be 1-dimensional

[heady skiff]

well okay so we have S^1 x S^1 - wouldn't that be a subset of R^2, how would we visualize a torus that way?

[heady skiff]

sorry i'm not really seeing your point

[hidden crag]

S^1 x S^1 is not in R^2

[heady skiff]

why not

oh

wait so S^1 is the 1-dimensional sphere whose points are that of R^2 which have distance 1 from the origin right

[queen prism]

let’s say you take a sheet of paper in space
it’s basically a 2D object, yeah?

[heady skiff]

yea

[queen prism]

now fold it to form a cube
is it 2D or 3D?

[heady skiff]

3D i guess

[queen prism]

but the original shape was 2D

[heady skiff]

i'm just a bit confused by what they mean by "1-dimensional" sphere - do they mean that it's of the form (x, 0) for x in R but that it's embedded in R^2?

[heady skiff]

oh okay

i guess i kind of see your point?

[queen prism]

S^1 is just a circle

it’s called a sphere because circles, spheres, hyperspheres are all surfaces of balls

but it’s a 1 because locally the circle looks like a line
it’s like how you can approximate C^1 functions by lines if you zoom in enough

[heady skiff]

so S^1 looks like this

[queen prism]

yes

[queen prism]

ok this second point is kinda misleading because all we need in this context is continuity, not differentiability

[heady skiff]

okay, so the "1-dimensional" is in reference to zooming in really close and it looks like a line

[queen prism]

but imagine just looking at the upper half of the circle

if you were to flatten it out, you’d get a line

[heady skiff]

ohh o k i see

i don't know how you visualize S^1 x S^1 at all/how it comes about if both points are from R^2

so technically wouldn't it be four dimensional?

cuz like

wait no

wait

say we're considering say

((.8, .6), (-.8, -.6)) on the unit circle

but that doesn't make sense because the components are 2-dimensional

sorry i'mm so lost

if we define the torus to be the cartesian product S^1 x S^1 i can't see how that works

[gritty widget]

a point on the torus is specified by two angles. an element of S^1 x S^1 is two angles

[queen prism]

a

[heady skiff]

damn

alr

was i supposed to know that or some shit

because as far as I'm aware the cartesian product o f two sets A x B is the set of all (a, b) such that a is in A and b is in B

so I'm just using the definition of S^1 as a set

damn

[gritty widget]

an element of S^1 is an angle

[heady skiff]

i

okay

alr i accept that

so

just to be clear

S^1 is not the (x, y) such that x^2 + y^2 = 1

it's just angles

[gritty widget]

oh no that's definitely what S^1 literally is

don't interpret my messages too literally

[heady skiff]

bruh

[gritty widget]

(cos θ, sin θ)

[heady skiff]

ok

thanks

[unreal stratus]

S^1 x S^1 is a donut

[bitter smelt]

(A hollow donut)

[heady skiff]

if f is an open map is that just saying that its inverse is continuous?

[queen prism]

not necessarily

f may not have an inverse

but if f is invertible then yes

[heady skiff]

oh right

how is the inverse of the projection map well-defined? do we just always assume that the second factor is the second topological space or smt?

[lime sable]

this is "preimage" not inverse

[heady skiff]

yea ig also how is the preimage well defined right

wait

oh

nvm

[lime sable]

if you have a function f: X -> Y between sets, you can write f(U) and f^{-1}(V) for image/preimage of subsets U of X and V of Y

[heady skiff]

makes sense i'm being dumb

thanks

[chrome ridge]

Explicit contraction of the dunce cap ?

[narrow cairn]

please

[feral copper]

Wouldn't a subset without isolated points look like a dense subset? Which, ofc, cannot be closed if proper

In any case, no isolated points should contradict countability

[narrow cairn]

intuitively i agree but having a hard time proving

[opaque scroll]

This is essentially the Baire category theorem

So assume your LCH space is countable, closed and without isolated points.

Let r1, r2, ... be an enumeration of the points and let Ui be the open set where you have removed the first i points.

Let x1 be a point in U1, and let B1 be a compact neighborhood of x1 contained in U1. Since x1 is not an isolated point B1 has infinite interior. Since U2 is cofinite B1\cap U2 has infinite interior, so pick a point x2 and a compact neighborhood B2 contained in the interior of B1\cap U2.

Now the Bi are a countable chain of closed compact sets, so has nonempty intersection, but it is contained in the intersection of the Ui, which is empty. Contradiction.

[narrow cairn]

wait, how do we guarantee a compact nbhd in U1 exists? i forget

[opaque scroll]

It's locally compact

Didn't you prove that an open subspace of a locally compact Hausdorff space is locally compact earlier?

[trail charm]

in the proof of excision theorem (Hatcher 122) this is kinda confusing me. Y is a convex set, LC_n(Y) denotes subgroup of C_n(Y) generated by linear chains, λ is a generator of LC_n(Y)

b_λ is the image of the barycenter b of Δ^n under λ, and b_λ is the cone operator

I kind of just don't really understand what T does in general, even after reading the geometric motivation

i.e., I get that it defines a chain homotopy, but geometrically what does it do

the last line it says that it takes the image of this subdivision under the projection. first off, what even is the point of this whole process if we just take the projection? why not just take the barycentric subdivision on Δ^n, since collapsing Δ^n x I to Δ^n would just give you the same thing (i think)

does that mean T is the projection mapping? if so, I don't get how that is a mapping LC_n to LC_n+1

(this is me trying to show that projection Δ^1 x I to Δ^1 is the same as barycentric subdivision on Δ^1)

[opaque scroll]

I think the point is you divide Delta^n x I into several copies of Delta^n+1 . Then for each of them you get a n+1-chain by first projecting down to Delta^n x {0}, then applying the map given by lamda. Then T is just the sum of these chains.

So yeah T is basically just composing with the projection Delta^n x I -> Delta^n, just modulo the fact that Delta^n x I is not Delta^n+1

[trail charm]

OHHHHHHHHHH

im ngl i dont like hatchers proof

maybe im just bad at visualizing this stuff

[distant lichen]

Hatcher tends to be particularly bad about relying on visualisation

[chrome ridge]

Is there a way to write an explicit contraction of the Dunce cap ?

[distant lichen]

The dunce cap?

[chrome ridge]

Dunce hat? Triangle whose sides are identified non-cyclically

[ebon galleon]

https://core.ac.uk/download/pdf/81112833.pdf This I presume. Take a simplex, give an ordering to each side as indicated and then quotient.

[gentle osprey]

grothendieckfan1 (Ryx)

[narrow cairn]

okay im stuck:
show that a connected, locally euclidian, paracompact hausdorff space is second countable

[lime sable]

found this on wikipedia for paracompact, it might be helpful

A topological space is metrizable if and only if it is a paracompact and locally metrizable Hausdorff space.

[narrow cairn]

hmm, not sure what extra conditions are needed for metrizability to imply second countability

[umbral panther]

Metrizable plus countable dense set implies second countable
But I feel like using that theorem is circular

[narrow cairn]

im thinking of just skipping the last few exercises here

since theyre so hard

[opaque scroll]

How about this?

||Take a cover by Euclidean open sets||

||Refine to a locally finite Euclidean cover||

||Let x1 be a point in your space and U1 an open neighborhood that only intersects finitely many sets in your cover.||

||let E1, E2, ... be the sets in the open cover that intersect U1, and for each rational point in each of them pick a neighborhood that only intersects finitely many sets in the cover, and take a union of all of them. Since the rationals are dense this union contains the Eis, and since it's a countable union of things that only intersects finitely many things each it only intersects countably many Euclidean sets.||

||Call this union U2, and define U3, U4, ... inductively||

||Letting U be the union of all the Uns, note that U is open. If there is a point not in U, then it would be contained in some Euclidean from the cover. But if E intersects U, then it must intersect Un for some n, but then E would be contained in Un+1. So U is closed.||

||Since our space is connected U is everything. But U contains only countably many of the Euclidean sets, so our space is the countable union of Euclidean neighbors||

And since Euclidean space is second countable that should do the trick

I never used Hausdorff, so maybe there's a mistake somewhere...

[lime sable]

||and since it's a countable union of things that only intersects finitely many things each it only intersects finitely many Euclidean sets.||
i don't follow, ||why does it intersect finitely many Euclidean sets?||

[opaque scroll]

I meant to write ||countably many||

[narrow cairn]

how the fuck was i expected to think of that

[lime sable]

i guess one of the main ideas is getting countable things from the rational points

and then the other idea is this trick to show it's connected

[opaque scroll]

Yeah, the first idea you think of should really be "how can I write my space as a countable union of Euclidean neighborhoods"

[narrow cairn]

you know what i did like 27 out of 34 problems im just moving on to the next chapter

thanks for the help yall

[heady skiff]

can i have a hint for this question please?

[narrow cairn]

thats a fun one but its weird

cant write out a hint rn, on phone sorry

[robust drum]

Here’s a hint (or a sub problem)

Show that for any compact set C and any point x not in C, there are disjoint neighborhoods of C and x respectively

Use that sub problem to show the full problem

[heady skiff]

all good

[heady skiff]

oh okay so lemma then LOL

i'll try that

[robust drum]

YEAH that’s the word

[heady skiff]

LMAO

okay cool thanks

[narrow cairn]

(a hint for this lemma: try to construct an open cover for C where every element of the cover is associated with some disjoint neighborhood of x)

any intuition behind the definition of a cell decomposition there? seems super random

@ me if you answer

[heady skiff]

how do you pronounce munkres

monk-ers

or monk-rees

[tidal lynx]

mon in monday plus kers in bonkers

well damn

bro has the primary source too

[heady skiff]

yea munkrehs seems accurate

in my head i've always sadi monk-rees

[heady skiff]

i'm a little bit confused, how could we make sense of a "one point set" if we're working in R^2?

oh

wait i'm dumb

do they mean (x, y) for {x x y}

[silver spruce]

yes: {(x,y)}

for any point in interior of [0, 1] x [0, 1], it should be left "untouched" (that is, not identified with other points) by the identifications that are being made here

so its a one point set

[heady skiff]

i see

thanks

[tribal palm]

man it’s so nice to have a lecturer who is both properly knowledgeable and practical

[trail charm]

:( do u have any book recs to supplement hatcher

and or lecture notes, videos, etc

[tribal palm]

god i’m flying through these problems

https://tenor.com/view/chiquichico-gif-26004262

[queen prism]

god i'm flying

[narrow cairn]

?

this doesnt make any sense, why does the map have to restrict to a homeomorphism from the boundary to all lower-dimensional cells?

ryx writing the bible over here

[ebon galleon]

The idea behind cell decomposition is that you have a space which can be built up by attaching cells (disks) repeatedly, in order of increasing dimension. So you start out with a bunch of points with the discrete topology (the 0-cells), and to this you attach lines (in the form of [0,1]) along their boundary ({0} and {1}), so what you end up getting looks like a bunch of vertices with edges between them; that is, a graph. Then to this, you attach 2-disks along their boundary (the circle) to the graph, and then....
So the requirement that the characteristic map takes boundary(D^n) into the cells with dimension strictly less than n formalizes this notion of attaching in order of increasing dimension: The boundary of the n-disk gets identified with the lower dimension, and it cannot take values in dimensions greater than n (for example, you can't attach a line to a 2-disk). And the requirement that the characteristic map restricts to a homeomorphism on the interior does 2 things:

1. It forces you to get a nice topology in the end (it's always T_4) because the topology ends up depending largely on the disks' topologies, and
2. It basically tells you that you're attaching only the boundary of the disk to lower dimensions (for example, when you attach a 2-disk, it ensures that you can't attach the upper half of the disk in addition to the boundary, since that would give some pathological shit)

/rant but it's basically just building a nice space in layers

[narrow cairn]

wait, but say you attach one line to one point and then a circle to that line, and pick a different point and attach a line to that point, why does the boundary of the disk have to be 'attached' to both lines if one of the lines sprouts off of a different point

[ebon galleon]

The boundary doesn't have to be attached to all cells of lower dimension, if that's what you mean. Just into the union of them.

[narrow cairn]

okay i still dont get it

i get it up to the graph part, you're just drawing lines between points

but where exactly is the circle attached?

[tribal palm]

i so regret not taking that Veblen book my library was giving away

[ebon galleon]

It's really something you gotta think about pictorially to get an intuition lol. So like you might have something like in the first drawing, where you attach part of the circle to one line and another part to a second line; you end up getting something that looks like a 2-disk. Or it might be like something in the second picture where you "collapse" the circle to a single point; you end up getting something that looks like a 2-sphere. [or you can get some wacky shit too, these are just intuitive examples]

(shit drawing skills but whatever)

[narrow cairn]

ah i see, so its like "shading in" parts of the graph

?

[ebon galleon]

The first example looks kinda like that, but I wouldn't describe the sphere example that way (what would be shaded in there?)

[narrow cairn]

im so confused

[ebon galleon]

Uhh okay sorry I maybe didn't do the best at explaining it lol

[narrow cairn]

no i think im just not getting it

probably not your fault

i totally get the first image

but like, how is there any relationship between the 2 cell in green and the purple 1 cell?

[ebon galleon]

There isn't. So the characteristic map for the green 2-cell would map the boundary to the red and blue 1-cells (which are then contained in the union of all 1-cells and 0-cells)

[bitter smelt]

(A 2 cell doesn't have to be attached to every 1 cell in the 1-skeleton)

[narrow cairn]

oh, shit, i just misread the definition

thought that the restriction to δD had to be a homeomorphism to the union of all lower dimensional cells

sorry about that, i think i understand to some extent now

[ebon galleon]

it happens
yeah the homeomorphism part is just the open cells (interiors)

[heady skiff]

bro got honorable

[ebon galleon]

ya idk who thought that one was a good idea

[gritty widget]

if you speak semi-authoritatively on any sort of math in the advanced channels you're bound to get honourable

it also helps to have friends in the honourable/mod team

[ebon galleon]

damn I don't even have to be correct in what I'm saying

[gritty widget]

it used to be lgbt = honourable but new mods are homophobes

[ebon galleon]

isn't everyone on this server lgbt pretty much

[hidden crag]

I failed, ryx got yellow anyway...

[gritty widget]

most are

[ebon galleon]

:o

[narrow cairn]

i wonder what the statistics are

idk if i'd say most, probably most within people that are actually active but if you include the thousands of people here who just use the help channels then idk

[unreal stratus]

lol

[narrow cairn]

there seems to be a considerably higher concentration of lgbt people on discord (or at least, in the majority of the servers im active in) than in the general population or on some other platforms

[tribal palm]

{lgbt} is dense in {people | discord}

[narrow cairn]

well, i think theres a certain type of confirmation bias that can sometimes factor into these things- in online spaces like this, many people are openly lgbt an put it in their profiles, whereas people rarely put "im straight and cisgender" in their profiles lol

[tribal palm]

(someone plz correct me in how this does not work out topologically)

[narrow cairn]

well you'd have to find an open set containing only straight people

[gritty widget]

everyone's a little gay, so that's not possible

[unreal stratus]

Propaganda

[tribal palm]

this is obvious, i agree and admit my mistake

[unreal stratus]

Lol I was about to type something out but didn't want to disturb the enlightening topological conversation

jk dw i no longer need to ask (i think)

[narrow cairn]

are servers the open sets?

[tribal palm]

no the space is spanned by the people and people are just connected regions of behaviour

[narrow cairn]

well, youve lost me now

[tribal palm]

smh read more Butler (no i am just not making sense and will go back to typing up this topology assignement now)

[narrow cairn]

the gay function is strictly positive?

whats the integral of gay dpeople

we (by that i mean I) should stop shitposting in the topo channel

[tidal lynx]

I like men btw

[gritty widget]

what is your gender

[tidal lynx]

M

[gritty widget]

gamma any%

wr

[narrow cairn]

gamma?

[gritty widget]

[tidal lynx]

who’s gamma

[ebon galleon]

Gamma is just propaganda. Pay no attention.

[tidal lynx]

I gotta meet this gamma guy

[tribal palm]

ok back on topic this seems suspiciously easy so i will just double check in am not off course: if i want to show for any two basis elements B1, B2, that if x lies in the intersection B1 n B2, then there is some basis element B with x in B subset B1 n B2... if i can show B1 is a subset of B2, then i can simply take B = B1 and we are done, right?

[tidal lynx]

how do you define basis?

[tribal palm]

basis for a topology

[tidal lynx]

Yeah there’s different definitions. One of them actually requires the condition you are trying to prove

[tribal palm]

what

oops

what's another definition than my condition

the ones i've seen have all been equivalent

[tidal lynx]

Yeah they’re equivalent (your problem is proving one direction of the equivalence probably)

I’m just asking which definition you’re working with

[tribal palm]

i'm not interested in proving that equivalence right now... i only want someone to verify if the last part of the above is sufficient to satisfy the condition i started with in the message (before and after the "...")

[ebon galleon]

If B1 is a subset of B2 then sure but just recognize that's generally not true

[tribal palm]

ok thanks

[tidal lynx]

Bro did not want to tell me his definition 😭

[tribal palm]

i prefer the def of a basis: fancy B subset P(X) with (1) B covers X, and (2) if B1,B2 are in fancy B, there is a family A in fancy B with the intersection B1 n B2 = union of the elements in A

then we define the top generated by fancy B to be the set of unions in fancy B

[tidal lynx]

According to that definition your original claim is true by definition no?

Or… maybe you were trying to show a certain collection IS a basis

[tribal palm]

...yes

i could've been clearer in that

[tidal lynx]

Ah that makes sense. Sorry

[heady skiff]

if X = A U B and X is open, must it be that A and B are also open?

i want to know if it's true or not before starting to prove it just in case it's false so i don't waste time ykwim

[narrow cairn]

what? absolutely not, take (a, b] and [b, c) in R

[heady skiff]

hm

[gritty widget]

a lot of math is wasting time, get used to it

[narrow cairn]

all of math is a waste of time

[gritty widget]

so true

[tribal palm]

wasting time is a good way to learn something

[narrow cairn]

especially topology

[tribal palm]

[heady skiff]

so if X = A U B with X clopen is either A or B clopen?

[heady skiff]

not when you have a pset due the same day

but other than that

yes

[narrow cairn]

are you in an actual class or just a reading group

[tribal palm]

this is unrelated to learning something

[heady skiff]

actual class

[tribal palm]

[ebon galleon]

Still no, take (-infinity, 0], (0, infinity). R is clopen in R.

[heady skiff]

hmm, i'm a bit confused then. i'm assuming that the only subsets on a space X which are both open and closed are X and the empty set and trying to show that X cannot be expressed as the union of two disjoint nonempty sets

i've been following my prof's proof

suppose X = U \cup V with U and V open, intersection empty

[narrow cairn]

disjoint nonempty open sets?

[heady skiff]

two disjoint

nonempty sets

open

my b

[narrow cairn]

well take U, V disjoint nonempty open with their union as X

then V = X\U right?

[heady skiff]

O

[narrow cairn]

so what can you say about V

[heady skiff]

that it's closed

wait

[narrow cairn]

im assuming youre doing connectedness

[heady skiff]

so then it's either X or the empty set by assumption

yeah

[narrow cairn]

when did your class start?

[heady skiff]

uhh like a little over a month ago?

we're going over topological groups and quotient spaces rn

but this week's hw is on connectedness lmfao

[narrow cairn]

what class is it, just point-set?

because if so thats a fast class, what are you gonna cover in like december lol

[heady skiff]

algebraic topology

[narrow cairn]

ah

[heady skiff]

i have no clue lol

[narrow cairn]

so you're taking algtop but you havent taken pointset? (disclaimer i have never been to a university class i have no idea the structuring)

[ebon galleon]

wait alg top with no prior topology?

wtf

[narrow cairn]

does a pointset class generally include CW complexes?

[gritty widget]

post syllabus

[heady skiff]

the syllabus doesn't have the course schedule on it lol

[ebon galleon]

No list of topics?

[narrow cairn]

what is pointset anyways, like chapters 2-4 or 2-5 from Lee ITM?

[gritty widget]

cw complexes are not covered in the average general topology class

[narrow cairn]

what class are they generally covered in?

[gritty widget]

algebraic topology

[narrow cairn]

huh

[ebon galleon]

Thanks, I'll now be showing up to your classes

[heady skiff]

damn i'll start going to lec again then

[narrow cairn]

?

[gritty widget]

first 5 or 6 chapters of munkres

[ebon galleon]

So continuity, compactness & connectedness, quotient spaces, fundamental group, triangulation, simplicial homology, and degree/lefschetz number

That's a good amount

[narrow cairn]

1 semester and all that? damn

[ebon galleon]

my dumb class barely covers half of that

[narrow cairn]

im slow asf

[heady skiff]

nah imma die fr

[narrow cairn]

can you send your last pset?

im wondering how hard problems usually are for a university class

[heady skiff]

oh it's just problems from armstrong, the one that's due two days from now is just 2 problems 💀

pg. 60, #31 and #34 from armstrong

[narrow cairn]

can you send the problems

[heady skiff]

sure

give me one sec

[narrow cairn]

thanks

[heady skiff]

also if ur gonna discuss the answers or smt if you could spoiler that would be great 😭

[narrow cairn]

fun problems, someone check my work on the first one
||(a) the finite complement topology should be connected and (b) for the half open intervals the components should be singletons||

not confident for (b)

id have to double-check

problems seem easy tho im glad that i wont be asked to do like lee 4-31 or anything similar

[opaque scroll]

Is the half open interval topology the one with basis [a, b)?

[unreal stratus]

ye ||true for any infinitee set||

[unreal stratus]

Yeah

[ebon galleon]

In general, ||co-κ topology (sets with cardinality < κ are closed) is connected on a set with cardinality ≥ κ, and discrete on a set with cardinality < κ||

[narrow cairn]

(just to be clear this is only an actual topology iff κ infinite)

[ebon galleon]

ofc

[gritty widget]

cock topology

[unreal stratus]

Lol

[heady skiff]

why is it that each point of a discrete topological space is a component of the space? is it because if I have {x}, then {x, y} is not connected since the closure of {x} doesn't intersect {y} and similarly the closure of {y} doesn't intersect {x}?

[unreal stratus]

Hm not how I would show that space isn't connected

That seems less direct

Just note {x}, {y} are open and cover it

[heady skiff]

wait how does that show they're each components

[unreal stratus]

It shows {x,y} isn't connected

But also like this whole idea isn't enough to show that {x} is a component

You need to show that any subset set of cardinality >= 2 is disconnected

Proof works the same tho

[heady skiff]

wait sorry how does covering it shot it's not connected?

potato over my gun, i move in silence cause fellas be clocking my funds

[unreal stratus]

What is your definition of connected

I am going straight from the usual definition

Actually this may be a weird munkres thing lol

[heady skiff]

A space $X$ is connected if whenever it is decomposed as the union $A \cup B$ of two nonempty subsets then $\bar{A} \cap B \neq \varnothing$ or $A \cap \bar{B} \neq \varnothing$

[gentle osprey]

okeyokay

[unreal stratus]

Oh lol

[queen prism]

I remember learning that def from real analysis

[unreal stratus]

I am going from the one that X cannot be written as a union of two disjoint open subsets

Which is equivalent

[heady skiff]

ye

oh

i see what you mean, so {x} U {y} = {x, y} where {x} and {y} are open which means that {x, y} is not connected

crazy proposition: the open sets in the finite complement topology are infinite

the converse is not true however

[queen prism]

finite sets:

[narrow cairn]

funny story, there are no sets in which cofiniteness is equivalent to infiniteness

[heady skiff]

am I trippin, or are all subsets of the real numbers under the finite complement topology components? bc consider an arbitrary subset $X \subseteq \mathbb{R}$. if $X$ were to be expressed as $X = U \cap V$ with $U$ and $V$ open, then we must have either $U$ or $V$ equal to $\mathbb{R} - A$ where $A$ is some finite subset of $\mathbb{R}$. Say $V = \mathbb{R} - A$. but then we would have to have $U = A$ which is finite and hence closed under the finite complement topology

[narrow cairn]

you mean all sets are connected right

[heady skiff]

oh yea

[narrow cairn]

you cant have all sets being components

[heady skiff]

roight

is that true tho

all sets are connected

under finite complement topology

[narrow cairn]

well all you care about is that the set itself is connected lol

and no, not all sets would be connected

[gentle osprey]

okeyokay

[narrow cairn]

take for instance {0, 1}. (R \ {0}) cap {0, 1} = {1} open and (R \ {1}) cap {0, 1} = {0} open so {0, 1} disconnected

by similar logic all finite sets are disconnected and all infinite sets are connected

[heady skiff]

(R \ {0}) cap {0, 1} = {1} open

wait how did u get this

because doesn't that imply that {0, 1} is open in the finite complement topology

oh

i see ur point

[narrow cairn]

no, this is the definition of the subspace topology

[heady skiff]

you unioned them

wait

nvm

oh right

[heady skiff]

so this is saying that {0, 1} is disconnected in the subspace topology right

[narrow cairn]

yes

[heady skiff]

but not in the original finite complement topology

[narrow cairn]

in order to determine if a set is connected in a space we use the subspace topology

otherwise [0, 1] wouldnt be connected lol

since it cant be the union of two open sets

clearly

[ebon galleon]

Yes, connectedness is a property of the space as a whole; whether a not a subset is connected is determined by regarding it as a topological space in its own right with the subspace topology

[narrow cairn]

same with compactness/paracompactness and a lot of similar qualities

[heady skiff]

i see

alr i'll take that as my hint for finding the components of R under the finite complement topology then

thanks

[narrow cairn]

well, you can define compactness and paracompactness equivalently a different way that doesnt use the subspace topo but

nobody does

[heady skiff]

if the infinite sets are the connected sets under the finite complement topology, how do I check for maximality then? is it a cardinality argument?

[narrow cairn]

wym maximality?

R is infinite

[ebon galleon]

If you have an infinite set, there's clearly a maximum infinite subset

[queen prism]

wait where did maximality come from

[heady skiff]

cuz a component is a maximal connected subset

[queen prism]

o that

[heady skiff]

uh so are the components just the infinite sets of R?

[narrow cairn]

bruh

the components cant be the infinite sets because they dont form a partition

R is connected with this topology

what do you know about the component of a connected space

[distant lichen]

I like Bredon

[trail charm]

sick 😎 i have that one i’ll use it more

[opaque scroll]

What's the biggest subset you can think of? is that connected?

[queen prism]

in case no one has made it clear to you yet (because no one made it clear to me), when we say that a set is "larger" than another, we mean it contains the other set
so a set with some property is maximal if it is not contained in any other set with that property

[unreal stratus]

Dumb q, if X,Y are nice spaces with nice actions of G and we have a G-equivariant homotopy equivalence h: X -> Y (with the homotopies showing it is invertible being G equivariant) then it induces a homotopy equivalence X/G -> Y/G right? Like the homotopies showing h is an equivalence should just factor through ig

[umbral panther]

No

G=Z
X=R
Y=pt

[unreal stratus]

I don't see how there is a G-equivariant homotopy equivalence

Like you can't even have an equivariant map Y -> X right since no point of X is fixed by G

[umbral panther]

Oh, sorry, I missed your definition. Yeah, then the fixed points are a functor

[unreal stratus]

Nice, thanks

Was just worried about some pointset technicalities in smth i was writing but it is easier than i thought after writing it out :)

[fair idol]

Where can I find a simplicial structure of something like a torus but with three 1-dimensional holes and one 2-dimensuonal holes. I wanna compute the simplicial homology and verify I get the stuff I expect.

I know the regular torus with one 1- dimensional hole and one 2-dimensuonal hole comes from identifying pairs of sides of a square and twisting.

How can I generalized this to get the shape I want

[languid patrol]

a torus doesn't have a 2-dimensional hole, depending on what you mean by "hole"

what is the mathematicially rigorous version of what you are asking?

[fair idol]

What's the last nonzero homology group

[languid patrol]

just \pi_1

[fair idol]

The hollow interior part

[languid patrol]

the torus has a universal cover which is contractible, so it can't have higher homotopy groups by the homotopy lifting theorem

ah sorry you said homology

Is the rigorous version that you want a space with H_1 = Z^3, H_2 = Z?

[fair idol]

Yeah exactly

[languid patrol]

do you care about how nice the space is? do you want it to be compact and without boundary?

[fair idol]

The shape is still hallow but now h1=3

I don't know why such a space would have to be so nice. I don't consider the regular torus nice

[languid patrol]

what is not nice about the torus?

[fair idol]

I thought you meant something about the quality like smoothness.

[languid patrol]

Well there are lots of smooth structures on the torus, although only one equivalence class

[fair idol]

Okay. Well such an object that looks like a torus but with 3 coffee mug handles on it.

[languid patrol]

that's a different question

[fair idol]

How do you get a simplicial complex structure on that. I thought the point of a simplicial complex is that I can ignore homeomorphism. That's the same question. A torus is a coffee mug

Wait no sorry. A coffee mug isn't hallow

[languid patrol]

Oh so you don't want it to be hollow? You're thinking of the torus as a 3-manifold with boundary?

[unreal stratus]

Wdym by "ignore homeomorphism"

Not sure how that relates to simplicial complexes

[tidal lynx]

Is a topology on a set X determined by the class of continuous functions X -> X it admits?

I ask this because I read somewhere that "if the same sequences converge then the topologies are the same"

ok nvm lol

[ebon galleon]

Well, I mean it kinda (but not really) depends on what topologies are put where but yes

[tidal lynx]

Wait really?

It holds for general topological spaces?

[tidal lynx]

I see why it holds for "sequential spaces", which require this to hold

[ebon galleon]

Like you need to specify what topologies go with what space (like are you considering a fixed topology T on the codomain, and varying over all topologies on the domain? Or do you mean just for that topology, (X,T) --> (X,T)?)

If it's the first, then just consider identities lol. If it's the latter then no; all functions (X,T) --> (X,T) are continuou for both the indiscrete topology and the discrete topology, but these are ofc distinct when |X| > 1

[tidal lynx]

Ok I see

I was considering the latter btw

[ebon galleon]

Yeah I realized that after what i first said lol

[tribal palm]

i'm loving this assignment

especially since i am actually managing to work through these with relative ease (i suppose these are all rather elementary problems)

[tribal palm]

i have never really had a look at subspaces before now though, so i've reached problem 13, but am struggling to make sense of the open sets in the subspace A

does, say, the fact that (-1/2,1/2) is open in R make the subset U = {0} u {1/n | n > 1} open in A ?

[unreal stratus]

Yes

In fact there is a nice basis of opens for A

Which you are close to having

[tribal palm]

that feels very weird

in a lack for better words

oh wow i just turned blue, i dunno if i should be happy about this

[queen prism]

asphyxiation

[unreal stratus]

Hm well like

[tribal palm]

can i be pink rather