point-set-topology

[shadow charm]

It’s continuous

[empty grove]

I see

[hollow harbor]

Umm

Hmm

Maybe?

Im thinking of it being unbounded variation on any neighborhood

[empty grove]

Long line is also connected right

[hollow harbor]

But maybe thats not enough

True

[shadow charm]

Where is the long line used

[hollow harbor]

Long line has many nontrivial path connected components though

[gritty widget]

The fact that topologists sine curve is connected means it has any of its neighbourhood contains a connected neighbourhood, so if a retraction existed it would be path-connected

So it's not ANR

[shadow charm]

Yeah I feel like any continuous function will have a path connected graph

[hollow harbor]

Yes

You're right

images of path connected sets are path connected

[shadow charm]

Right

[hollow harbor]

My intuition for "connected but not path connected" being "things are infinitely far from each other" has been shattered

[marsh forge]

I think you need much nastier constructions to make the two disagree

but its 6am

[shadow charm]

Well topologist’s done curve is easy to généraliser

[marsh forge]

so take everything i say with a few pounds of salt

[hollow harbor]

[shadow charm]

Just take a function with a nasty wiggly singularity

[hollow harbor]

Yeah i thought my functions had wiggly enough singularities everywhere

[shadow charm]

And artificially add a point at that singularity

[hollow harbor]

Whatever just sum topo sine curves shifted to every rational

And scaled like 1/2^n

[marsh forge]

topologists sine curve is such a bad name

we dont claim that shit

[shadow charm]

Ig the use for connectedness is that it’s a weaker property that’s easier to conclude and so can easily give us spaces not continually mapped to or smth

[marsh forge]

Oh I mean

[shadow charm]

But idk still sounds a little pathological to me

[marsh forge]

connectedness is super useful

in and of itself

for example like, if you know P holds locally on a connected space and you show P holds anywhere, then P holds everywhere

Er maybe you need not P to be local too

[shadow charm]

Hmm fair enough

[marsh forge]

Also you know that continuous functions send connected components to connected components even without paths

which shows up fairly regularly

[shadow charm]

Yeah I can see that

[marsh forge]

like i said though, most topologists don't study things where the two notions differ

and some even define connectedness via homotopy groups

(which only see the path version)

[arctic relic]

What’s the general idea behind why CW complexes are Hausdorff? The proof in Hatcher is not intuitive to me.

Never mind I’m stupid

[gritty widget]

Connectedness is also useful in that a set between connected set and its closure is connected

Which can fail for path-connectedness

[shadow charm]

Not sure what you mean here since every path connected space is connected

[gritty widget]

?

[shadow charm]

If X retracts on a subspace A, is the morphism induced by the inclusion of A in X on the abelianizations of pi_1(A) to pi_1(B) injective?

I need this to be true but I don’t see any reason why it should be haha

[shadow charm]

Nvm I see what you mean

[empty grove]

Retract means that there is an equation

Apply functor to equation and it remains true

[shadow charm]

What do you mean by equation?

A commuting diagram for a monomorphism?

[empty grove]

ri = 1

[marsh forge]

the defn of a retract can be used with functoriality of pi_1 to get what you want

technically the functoriality of pi_1 and abelianization

[shadow charm]

See I was thinking of doing this

But then I confused myself

[marsh forge]

unconfuse yourself 🙂

[shadow charm]

If you can do this with functors

Then can’t you say the same to show that abelianization preserves injections?

Which it doesn’t

Ohhh wait

The left inverse of an injective homomorphism isn’t necessarily a homomorphism

Alright I see

So this means if the left inverse of an injective homomorphism is a homomorphism then abelianization preserves that injection? (And surjection for that matter)

Functoriality too strong ngl

[empty grove]

Yes. It's the left invertibility that is strong, not functoriality here lol

[shadow charm]

Yeah fair enough

It just kinda feels like diagram black magic

[empty grove]

😌

[shadow charm]

Hatcher exercises are so much fun

[empty grove]

There's also a weaker notion of left cancellability

[shadow charm]

These fundamental group computations are really cute

[empty grove]

and that isn't preserved by functors

[empty grove]

[shadow charm]

What is it

[empty grove]

Left cancellability lol

fg = fh → g = h

Retraction says that i is left invertible with left inverse r

So it is like domain vs field

[shadow charm]

Are the maps here continuous 🤔

[empty grove]

Maps in whatever category

All maps in category of topological spaces are continuous

[shadow charm]

So f here doesn’t have a left inverse but left cancels?

[empty grove]

Left cancellable morphisms are called monic

And left invertible split monic

[empty grove]

Ye

You might have heard monomorphism

[shadow charm]

And i suppose it left cancels because it has an inverse map that’s not in the category you’re considering

Like in Set or smth

[shadow charm]

Yeah

[empty grove]

It left cancels because we can show it lol

You could go the Set route and use the faithful forgetful functor

But that would require you to do strictly more

[shadow charm]

I see

So I just finished an exercise where we show that the orientable surface of genus g retracts onto a circle that doesn’t separate it but not on a circle that does

Does this generalize to non-orientable surfaces?

And to manifolds of higher dimension?

[fair idol]

Suppose there was a continuous map from S^n-> T^2. I want to show that this map is null homotopic if n>=2.

I feel like this has something to do with lifting properties but I'm not sure. Does anybody have a lead on how I might try to think about this problem?

I know that the fundamental group of S^n is trivial and the fundamental group of T^2 is ZxZ

[marsh forge]

How much covering theory do you know

[fair idol]

I know of a Galois correspondence

[marsh forge]

That is enough

[shadow charm]

pick x_0 in the image of S^n and a loop at x_0 in that image. There is a loop in S^n that maps to that loop injectively. Take a nullhomotopy of that loop in S^n and compose it with your map. That will be a nullhomotopy in T^2 so that the image is simply connected and hence the map is null-homotopic? Im not sure if what im saying makes sense

[marsh forge]

You are right that this follows from lifting properties

[marsh forge]

this is incorrect

[shadow charm]

yeah imagined that much

[marsh forge]

in particular inducing the 0 map on pi_1 does not imply the map is nullhomotopic

[shadow charm]

ah okay

[marsh forge]

@fair idol You should look through the lemmas you have about lifting, depending on your text

[fair idol]

Okay I'll review the lemmas from hatcher and see if I come up with anything then thank you

[shadow charm]

may i ask why this is the case ? It's always been my intuition for homotopies of maps

[fair idol]

Is there a common covering space for the torus?

[empty grove]

You can get it from how the torus is glued from a polygon

[shadow charm]

I see the issue here: confusing simply connected and contractible

[marsh forge]

The identity map S2->S2 is not nullhomotopic but induces 0 on pi1

Pi1 barely sees any homotopical data

For the same reason that pi1X=0 does not imply X is contractible

[fair idol]

Is there a way to calculate the fundamental group of R^2 minus two points?

[shadow charm]

Yeah exactly it was just me making a dumb confusion

[shadow charm]

It deformation retracts onto a wedge of two circles

[fair idol]

TYSM!

[rancid umbra]

does this generalize to R^2 minus k points deformation retracting onto a wedge of k circles?

[shadow charm]

Yeah

[rancid umbra]

neat. what about R^n minus k points? just change circles to n-1 spheres?

[shadow charm]

R^n - k points is simply connected

In fact R^n - a discrete subspace is simply connected

But I think yeah you change it to n-1 spheres

Sounds reasonable to me

[rancid umbra]

for n >2 right?

[shadow charm]

Yeah

Here’s a neat exercise for you that I did the other day if you find this neat

Compute the fundamental group of R^3 - X where X is the union of n lines through the origin

[rancid umbra]

this feels like the analogue of R^2 minus n points

just in R^3

[shadow charm]

It’s harder

[rancid umbra]

hmm. i don’t have any technical tools at my disposal rn, i just don’t know any theorems or a lot of terminology so i’m kind of intuiting/visualizing everything rn

i can’t really say what happens for more than two lines

i’ll remember this exercise for later tho

[gilded crane]

how can a topological space be considered a space if it has no notion of distance?

[cedar pebble]

Because it still possesses a notion of which points are “near” other points

Even if there isn’t defined by some numerical value of distance

[gilded crane]

when i was googling this question i did stumble upon "nearness" a lot

but... doesn't "nearness" still require distance?

[cedar pebble]

not necessarily

[gilded crane]

how can you describe something as near to something else through just set theory? zero geometry

[cedar pebble]

right so you lose a bit of geometry as you say

but the picture to keep in mind is like

[gaunt linden]

It's just conventional to use the word "space" about this particular kind of structure. Don't think that word choice encodes any deep truth.

[cedar pebble]

if you have any set X there are kinda to extremes of a topological space you can define from this

one is the discrete topology, where every subset of X is open

you think of this as just like, a discrete cloud of points, no two points are "near" each other

the other is the codiscrete topology or the trivial topology, where only the empty subset and X are open

so in the other extreme you think of this as like

all the points in the set X totally cohered together, every point is "near" every other point

in general any topology on X is between these two extremes

[gilded crane]

ah, i see how they are extremes, all topologies on X would be somewhere in between the two

[cedar pebble]

it's sort of like a general topology is more "spread out" than the trivial topology

if you "spread out" the points as much as possible you just get a discrete cloud

it is true that some spaces in between these extremes are in fact metrizable and you can attach an actual metric function to refine this notion of nearness

and there are definitely things that are more "geometric" that require this extra structure

[gilded crane]

ok so... take three points a,b,c. are you telling me they are near each other if {a,b,c} is an open set, but not near each other if {a},{b},{c} are open sets instead?

[cedar pebble]

something like this

although note that in general points don't have to be closed 🙂

[gilded crane]

yes! they are clopen

i convinced myself of this earlier today

(in a discrete topology)

[cedar pebble]

I mean more generally like

there are interesting topological spaces where some points {x} are not closed subsets

a good example is the Zariski topology

here there's definitely no metric but you can still kinda like

draw things

and they are fairly geometric still

[gilded crane]

oh jeez, i only just started my topology course, this has stuff in it that is still beyond me

[cedar pebble]

there are some examples that aren't so bad
if you have a commutative ring R then you define a topological space Spec(R) whose points are the prime ideals of R with the Zariski topology

the idea of the definition being like

intersection and union of opens has something to do with addition and multiplication of ideals

[gilded crane]

👀

[cedar pebble]

for example if you just take the ring of integers Z the prime ideals are (p) for p prime, and (0)

the ideals (p) are maximal, so they correspond to closed points

whereas (0) is not maximal, and it corresponds to a point that is not closed

it's kinda like

smeared out across the whole space

in particular the closure of that point is the whole space

nothing like this exists for like metric spaces

but you can still kinda visualize it as some kinda like

"ghostly" point that is smeared out across the whole picture

[rancid umbra]

can u do an interpretation in R with nearness defined in terms of open sets? just for concreteness?

[gilded crane]

what is the closure of a point tho?

[cedar pebble]

so the closure of any subset is like

the smallest closed subset of your space containing that subset

[gilded crane]

yeah, that makes sense to me, but if we are just talking about a point, wouldn't its closure be the empty set?

[cedar pebble]

well for closed points the closure is just the point itself

[gilded crane]

oh ok

[cedar pebble]

so like in metric spaces this is what you always see, points are always closed and you don't have these more general points

[fading vale]

if you take a point x then you have open balls B_x(r) = {points y with d(x, y) < r} about x. so we replace our usual heuristic for "closeness" (d(x, y) is small) with "x and y share many open balls"

[cedar pebble]

the point is you can have weird stuff like this happen and it makes it impossible to have any kind of notion of distance in your space

but things can still be pretty geometric

like in the case of the Zariski topology you can still draw things

so long as you add in these weird non-closed points into your idea of what geometry is

[rancid umbra]

but that’s true for any two points haha

[gilded crane]

only on the usual topology on R

[fading vale]

"share many" not "share a"

This is not precise from a cardinality POV but its the heuristic

[gilded crane]

thanks by the way, im slowly chewing through the wiki page for the zarinski topology and its helping me a lot, and i kinda understand why they call them discrete and indiscrete topologies now 🙂

[rancid umbra]

give me two points x and y that don’t share uncountably many open balls

[fading vale]

"this is not precise from a cardinality POV"

[rancid umbra]

then what is many lol

[gilded crane]

wait actually how is it not precise? isn't the cardinality of (0,1) the same as the cardinality of all of R?

[rancid umbra]

“many” is ambiguous. and that’s what i’m trying to understand by annoying moth (sorry moth)

[fading vale]

Its ambiguous because this is an intuitive motivation for why open sets distinguish a notion of distance and why topologies generalize metric spaces

[gritty widget]

I have a difficult question.

[fair idol]

Can someone tell me if my reasoning about this is correct

Say there is a continuous map f:S^n-> T^2 with n>=2. \pi(S^n)=0. Further the universal cover of T^2 is R^2 and as a universal cover \pi(R^2)=0. Therefore the induced maps f_(pi(S^n))\subset p_(pi(R^2)) is trivial true. Therefore the lifting criterion is satisfied so that there is a lift of f to f':S^n->R^2

[gritty widget]

Define ~ on R by x ~ x iff x - y is rational. Then ~ is an equivalence relation and ****(R/~, u~) is not Hausdorff.****

I need help with that second part.

The eq. rel. part is easy.

[marsh forge]

yep

[fair idol]

Surely this has something to do with the density of the equivalence classes of rationals

[marsh forge]

you might want to point out that pi_1 S^n is 0

[fair idol]

Okay somehow I think I can use this to show that f is nullhomotopic. But f=pf' so maybe showing the rhs is nullhomotopic is easier

[gritty widget]

Oh... But aren't there only two equivalence classes?

x ~ y iff x - y is rational.

[fair idol]

For some reason this reminds me of the proof of a non measurable set

[gritty widget]

Hmm...

[gritty widget]

But wouldn't this relation produce only two classes?

[fair idol]

I don't think there will be only two, but I could be wrong idk. Why do you think there are only two classes?

[gritty widget]

No, you're right.

The classes are one for each rational?

I mean, is that how the relation works?

I'm just confused as to how equivalence classes work, lol.

sorry

But I'll look that up now as well.

Q as in Q?

So QU{r}?

[gaunt linden]

$\mathbb Q+r$ is a shorthand notation for ${x+r\mid x\in\mathbb Q}$.

[gentle osprey]

Troposphere

[gritty widget]

Ah.

[gaunt linden]

This is a certain set of real numbers. Sometimes different choices can give the same set -- for example r=sqrt(2) and r=sqrt(2)+1 give us the same set out at the end.

[gritty widget]

Ah, okay.

[gaunt linden]

r=pi gives us a different set of reals, though.

R/~ is then the set of all these equivalence classes.

You equip it with the quotient topology. Can you figure out now what the quotient topology is here?

[gritty widget]

Uh.

Well, it's all of the classes...

Is it just equivalent to all real numbers?

[gaunt linden]

No, each equivalence class is one point of the quotient space.

[gritty widget]

Oh, so all irrationals?

irrational + rational

[gaunt linden]

Recall that sqrt(2) and sqrt(2)+1 are in the same equivalence class.

[gritty widget]

Since it's of that form.

[gritty widget]

So is it a particular real + any ration number?

[gaunt linden]

Yes sort of -- but for each equivalence class there are many different reals that each create that class, so you can't point to any very "particular" real it comes from.

One of the equivalence classes is Q itself.

[fair idol]

This might be too broad of a statement

[gaunt linden]

Consider the identity map from S^2 to itself.

Right.

Since the sphere isn't contractible.

[sinful cloak]

No there are uncountably many since $\mathbb{R} = \bigcup_{[x] \in \mathbb{R}/\mathbb{Q}} [x]$. If there was only countably many then the reals would be a countable union of countable sets.

[gentle osprey]

M8732

[gritty widget]

You helped a lot and I think I found my solution.

Thanks!

I used the denseness of Q.

I think it works.

[sinful cloak]

You meaning someone else? I just wanted to clarify this bit.

[hollow harbor]

yes

yes

the homotopy is f_t(x, y) = (tx, ty)

[gaunt linden]

A map with a contractible codomain is always homotopic to a constant.

[marsh forge]

You should be able to reason this out yourself from definitions

[marsh forge]

Well my point is more, there is only so much of your work I can do for you without hurting you

sometimes you just gotta work it out

[marsh forge]

I am trained in both math education and mathematics. My job is to teach math.

[glossy pine]

fajitas got fucked

[marsh forge]

Perhaps I should at least explain: this is not a difficult exercise, and if you do not see how to prove it, you really need to review the more basic notions from hatcher ch0. If I explain how to prove it, I rob from you the chance to actually internalize these definitions on your own, and I risk allowing you to continue through the book without fixing this gap. Inevitably that will lead to more questions in this channel later on that could have been prevented now.

[marsh forge]

The overall purpose is to help you learn topology, not to answer any and every question.

[gritty widget]

did you figure it out yet

it comes straight from definitom when you apply it

you can draw out diagrams to help

[arctic relic]

Trying proving this theorem if you’re still a little shaky on definitions: A map is nullhomotopic iff it factors through a contractible space

[gritty widget]

It's cool that just from connectedness and local connectedness we can get arc-wise connectedness and local arc-wise connectedness

It seems kind of amazing

[arctic relic]

Arcwise?

What’s the difference between arcwise and path connected in Hausdorff spaces?

[coarse night]

what's arcwise connectedness?

like some arc shaped path connectedness ?

[gritty widget]

probably injective paths

[empty grove]

I was under the impression that it is exactly path connectedness

[coarse night]

okey

[coarse night]

ngl I'm still under same impression

[empty grove]

5 people guessing and no one willing to google it

[coarse night]

why google when I can bother some honorables

[gritty widget]

google is more reliable

[hollow harbor]

,w arcwise connectedness

[empty grove]

Arcwise means that the witnessing paths connecting any 2 points be embeddings

[hollow harbor]

oh well~ guess it doesn't exist

when in the world is that not just path connectedness

[hollow harbor]

[empty grove]

Closest guess 🍪

[hollow harbor]

ah i see

[coarse night]

why call it arcwise then, it's a misnomer

[hollow harbor]

it's silly

[coarse night]

exactly

[empty grove]

There's a copy of the arc connecting the 2 points

Arc being defined as an embedding of the closed interval

[arctic relic]

Is it F_2n+1

[empty grove]

Yes

[coarse night]

ig I was picking examples from locally compact, connected, locally connected, metrizable space that's why couldn't find any difference

[arctic relic]

I looked at the n=1,2 cases, drew loops in different homotopy classes, and then approximated graph as a CW complex, drew the relations and elements, and that’s how I got F_2n+1. Is there a more rigorous way to do it?

[empty grove]

x ↦ x/|x|

[arctic relic]

That’s directEd towards me right?

[empty grove]

Yes

[arctic relic]

Thanks

[shadow charm]

2n-1 *

[arctic relic]

Damn

[shadow charm]

It deformation retracts on a sphere with 2n holes

[arctic relic]

Oh I counted wrong that’s my bad

[shadow charm]

Give it a CW complex structure by putting these holes in a bouquet

And adding a loop around

[arctic relic]

Since n=1 is F_1, n=2 is F_3 I just got the recurrence relation wrong

[empty grove]

Sphere with 2n holes is plane with 2n-1 holes

By stereographic projection

[shadow charm]

Omg

How didn’t I think of that lol

Well whatever my method works

[arctic relic]

Is that the x-> x/|x| thing?

[empty grove]

That's the deformation retraction onto the sphere with 2n holes

[gritty widget]

My prof calls it path-connectedness, but usually path-connectedness doesn't require the path to be an embedding

I think

[empty grove]

Ye it don't

[gritty widget]

So sources like wikipedia calls this arcwise connectedness

The remarkable part imo is that you can get intervals just using connectedness and local connectedness

There is a characterization of an interval as a continuum with only two noncut points

You can use it to obtain arcs in your space

[arctic relic]

That stereo graphic projection thing is genius

[empty grove]

[gritty widget]

Yep

You can thank Riemann for that

[arctic relic]

Moldilocks is my Riemann

[empty grove]

[gritty widget]

Speaking of cool constructions

I've recently read about a thorem in my book which is basically Bernstein set

And I read about those not a long ago

So it was nice

[crude vector]

Fellas, I was hoping someone could look over this proof for me, since the one I found online is different, and I have doubts about the last part. Could someone let me know if this proof works? $\$

$\emph{Theorem.}$ Every connected metric space with at least two points is uncountable. $\$

$\emph{Proof.}$ Let X be a connected metric space with at least 2 points. Pick $p_0, p_1 \in X$ such that $p_0 \neq p_1$. Let r = $d(p_0, p_1)$. Pick $\delta \in \mathbb{R}$ such that $0 < \delta < r$. $\$

Consider the two sets $A_{\delta} = {p \in X \mid d(p_0, p) < \delta}$ and $B_{\delta} = {p \in X \mid d(p_0, p) > \delta}$. Clearly, $A_{\delta} \subset X$ and $B_{\delta} \subset X$.

By (c), we know that $A_{\delta}$ and $B_{\delta}$ are separated. Therefore, since X is connected, $X \neq A_{\delta} \cup B_{\delta}$. Since both sets are subsets of X, this implies that there exists a q $\in$ X such that q $\notin A_{\delta}$ and q $\notin B_{\delta}.$ i.e. $d(p_0, q)$ is not $< \delta$ nor is it $> \delta$. By trichotomy of $\mathbb{R}$, we have that $d(p_0, q) = \delta$. $\$

This means that for each $\delta \in$ (0, r), there exists a q $\in$ X such that $d(p_0, q) = \delta$. Since there are uncountably many $\delta \in$ (0, r), it follows that there are uncountably many such corresponding q $\in X$. Hence, X is uncountable. $\qed$

[gentle osprey]

strad

[empty grove]

Nicely written 😌

[crude vector]

Many many thanks, dear Moldi

Can I take that to mean we're good?

[empty grove]

Yeah, if you feel that there is more to say then you could say that a point can't have distance delta and delta' at the same time lol

[crude vector]

Yeah that's exactly what I felt like I skipped, maybe I should add that for completeness' sake

Hey now this sounds familiar

[viral yoke]

Compute $\pi_1(\mathbb{R} \mathrm{\textbf{P}}^2 \vee \mathbb{R} \mathrm{\textbf{P}}^2)$

Unless something tricky is happening here, isn't this just the free product on Z/2Z with itself?

[gentle osprey]

eM

[cedar pebble]

Yup fundamental group of wedge sum is free product of fundamental groups

[viral yoke]

Got it. I thought some sorcery was going to pop it like the fundamental group of RP^n is Z/2Z for all n > 1.

Still freaks me out lol because I still can't see why

[orchid forge]

it may be helpful to think about what nGroupoid said as a categorical statement

wedge product is the coproduct in the category of pointed topological spaces, and the fundamental group(oid) is a left adjoint (to what?), so it preserves colimits

or, if that sounds more confusing instead of less, just apply SvK

[marsh forge]

Pi1 is not a left adjoint and does not preserve colimits

[marsh forge]

Why what exactly?

[peak crystal]

any suggestion for the topology lectures?

[marsh forge]

I think there are some on YouTube that follow hatcher

I remember them being okay

[orchid forge]

It is a left adjoint when viewed as a functor from the homotopy category of pointed CW complexes

Only a small concession but I did leave it open ended by suggesting the question "to what?"

What I'm describing follows essentially from thinking about how to categorify (in a non-technical sense) SVK

[marsh forge]

Which in particular is not what you said lol

(You said pointed topological spaces)

[orchid forge]

Uhh

Oh I did

Woops

But anyway, I wanted to add that it's not necessarily straightforward how to encode the conditions of svk into the choice of category, but that is how it be

And if you studied category theory before topology it might be a nice way to see the theorem

[peak crystal]

I'm not able to find

[marsh forge]

I searched hatcher topology lecture and found a few

One by Pierre Albin

[peak crystal]

I need introduction level

algebraic is different from basi topology?

[marsh forge]

Oh sorry did you mean point set?

I’d search for lectures on the book munkres

[peak crystal]

yes for this one

I need first 8 chapter's of Munkres

[viral atlas]

There's a topology playlist by ICTP that uses Munkres

[peak crystal]

Thank you

[gentle osprey]

Hausdorff

Hausdorff

Hausdorff

[unreal stratus]

Isn't Cantor set a counterexample or smth

I mean obviously embedded in C though

[gentle osprey]

Hausdorff

[vast estuary]

Does this help?

[gentle osprey]

Hausdorff

[gritty widget]

@vast estuary if you remove the copy of Cantor set from \mathbb{C}, it's still (in fact path) connected

[vast estuary]

That makes sense, thank you

[unreal stratus]

(phew, nearly deleted my post xd)

Is there a nice way to show the complement is path connected or is that obvious

[gritty widget]

@unreal stratus if you're in [0, 1] x {0} you can just move upwards or downwards

But $\mathbb{C}\setminus [0, 1]\times {0}$ is path-connected

[gentle osprey]

Blitz

[gritty widget]

So in fact, if you remove any subset of [0, 1] x {0} then it's path-connected

[unreal stratus]

Oh sure, is it just cause we're in R^2 lol fair

[gritty widget]

This set is union of four open sets C_1, ..., C_4 each homeomorphic to C, and C_i intersects with C_(i+1), so it's connected (for example)

So path-connected

Alternatively, cases

[native raptor]

Wait there’s a different channel for diff geo and topology now

Where does knot theory/3-/4-manifolds go?

[marsh forge]

Either really

[light rivet]

So Im having trouble understanding chain maps. Let $X, Y$ be a top space and $f: X \to Y$ a continuous map. We have $(C(X) = \oplus C_n(X), \partial_C)$, $(C(Y) = \oplus C_n(Y), \partial_D)$ be chain complexes where $C_n(X)$ is the free abelian group generated by continuous maps from $\Delta^n \to X$, where $\Delta^n$ is the $n$-dim simplex, and similarily defined for $C_n(Y)$. Consider the map $g: C(X) \to C(Y)$ which sends a generator $\sigma: \Delta^n \to X$ of $C(X)$ to $g(\sigma) = f\circ \sigma$. I certainly see it maps $\sigma$ to an generator of $C(Y)$, but I don't see why it is obvious this map $g$ is a homomorphism

[cedar pebble]

uhh

you already said free Abelian group

so the operation is addition

[gentle osprey]

MasakaBakana

[empty grove]

To specify a map on a free abelian group, it is enough to specify that map on a basis

ie every set map from a basis to another abelian group extends uniquely to a map from the free abelian group to that other abelian group

[arctic relic]

If X is a graph with (x_i) a sequence of points in X with distinct edges, why can’t (x_i)_n converge? My reasoning was along the lines of supposing that it did converge -> every edge leads to a single vertex -> use the def of weak topology -> ???

By graph I mean one dimensional CW complex

-> the set U around the limit of x_n is closed since it contains its limit points but U intersect X^0 is open

Is my guess

Since tHe 0 skeleton has discrete topology U intersect X^0 is open

[sterile owl]

Anybody want to help me with proving homotopic equivalence?

[marsh forge]

Go ahead and post

[vast estuary]

if a set A (subset of C^n) is not discrete

[gentle osprey]

Hausdorff

[empty grove]

Yes

If an a has a neighborhood with finitely many points, it has one with no other points

[sterile owl]

"prove that a discrete space conditions of 'm' points is homotopy equivalent to a discrete space consisting of 'n' points if and only if 'm'='n'"

Kinda new to topology, not sure how to formulate a proof tbh

[marsh forge]

You could try to show that the number of connected components is preserved by homotopy

[uncut stratus]

Does anyone know if the group presented by:

<a,b|aba(b^-1)>

Is a known group?

I was calculating the fundamental group of the Klein bottle and I got this bad boy, but I don't know if there is a more explicit way to write it

[empty grove]

Infinite non abelian group I don't think it's any of the basic ones like D_∞ etc

[uncut stratus]

Oh
Cool
Guess I'll leave it written like that then

Thank you kind sir

[empty grove]

[gritty widget]

Hmm... well, could as well post it in #groups-rings-fields since the problem is group theory at this point

[uncut stratus]

I googled it up as well and it seems people agree with Moldi
I don't wanna bother another channel😅

[grave egret]

Well I think it's well-known... as the fundamental group of the Klein bottle

Another presentation for it is <x, y | x^2y^2>

Also another way to describe it would as an HNN extension for the inversion automorphism on Z

[empty grove]

No one will be bothered by a question lol

[swift fjord]

I will be

[empty grove]

No one worth not bothering will be bothered by it lol*

Was that the right number of negatives

[swift fjord]

Moldi ur always so mean to me

[empty grove]

😼

[uncut stratus]

I wouldn't be so sure about it
Anyway, now I want to ask

[uncut stratus]

I like your funny words magic man
Damn I should have know that self - proclaiming as "worthy of accessing the advanced math channels" was a risky move

Or I could just look up on Google what HNN is

Maybe I know it

[empty grove]

My dog is worthy of accessing the advanced math channels

[uncut stratus]

Big oof

[grave egret]

https://en.m.wikipedia.org/wiki/HNN_extension

[gritty widget]

Or you could just ignore the things you don't know

[empty grove]

They should make math 2 with less things that I don't know

[uncut stratus]

Jk, never heard of that
But that page mentions the SVK theorem, which I'm currently studying, so maybe you are right

[uncut stratus]

Well looks like someone discovered the foundation of my academic career

[gritty widget]

it's okay to black box things

[grave egret]

Indeed, though you do need to at least understand the words in the statement of a theorem in order to black-box it

[coarse night]

i literally black box everything and prove them a year later

[uncut stratus]

Hey dudes it's algTopologyLover89 here with another sweet question for y'all

I'm currently studying coverings (are they even called like that in English? basically the classic image of S^1 with the big infinite spiral above it)

And they sucks
But that's not all

[coral pawn]

What are the generators of the fundamental group oooo (where o is S^1)

[uncut stratus]

I don't understand why the preimage of a point (you call it the fiber in English?) always has the discrete topology

[uncut stratus]

You mean 4 circumferences with one point in common?
There are 4 generators and the foundamental group is ZZZ*Z (free product)

[marsh forge]

both are fine, and cover is also fine

[coral pawn]

That's because there is a neighborhood around each point whose preimage is a disjoint union is sets that each map homeomorphically to the original neighborhood

Thus, the fibre, when endowed with subspace topology can be written as the union (one sheet intersect the fibre)

Where the union is taken over all sheets

One sheet intersected with the fibre is always a single point

Thus, each singleton is open

So you get the discrete topology on the fibre

[coral pawn]

No that's not what I meant

Take 4 circles

Join first 2 at a point

[uncut stratus]

That's why
Thanks so much, I got it

[coral pawn]

Join the second and third at a different point

[uncut stratus]

It's the same thing

You mean like the Audi logo

They are homotopically the same

[coral pawn]

Got it

What will the generators be?

[marsh forge]

intuitively each circle. specifically it depends on your choice of basepoint

[coral pawn]

Say at the point between the first and second circle

Here's what I think

[marsh forge]

basically the generators are each circle combined with a path to your choice of basepoint

[uncut stratus]

It's kinda hard to write, I'll make a quick drawing

[coral pawn]

1. Go around the first circle
2. Go around the second circle
3. Go around the second and third circle
4. Go around the second, third, and fourth circle

[marsh forge]

this ends up working too i think

[coral pawn]

Alright. Don't need a proof of it. I trust you guys

I hope my professor does too

[marsh forge]

Well you should probably prove this

I think my basis is easier to prove

and yours follows from mine

by inverting the second circle's generator and recovering the third, and then the fourth

[uncut stratus]

Oh, yes, it was easier written than I thought

[marsh forge]

interesting that you both came up w that basis

that surprises me lol

[coral pawn]

We're the same person

[marsh forge]

it seems far less obvious to me

[coral pawn]

What's your basis?

[marsh forge]

each element is one circle

[coral pawn]

Well technically

[marsh forge]

then just draw a path to your basepoint

[coral pawn]

Our basis are different

Because he includes the first circle in all of them

[uncut stratus]

This works too
It makes clearer that there are 4 circumferences and it's because of this "4 holes" the group is what it is

I like it

[marsh forge]

It is also easy to prove

I can give a hint if you'd like

[coral pawn]

Wait Max

What's your basis?

[marsh forge]

i have written it three times now

please just read my old messages lol

[coral pawn]

Whoops

I was asking this question because we had to compute the image of this thing as a cover of the wedge product of circles

On my exam

And I didn't have the time to think too hard about what the generators were so I just went with my intuition

[high hill]

contradiction

[carmine star]

Heyys guys

Do i need to learn abstract algebra first before starting topology or i could doing it right away ?

( or these two don't connect at all ? )

[marsh forge]

you need the union for alg top

but can start with either

[quartz edge]

I'm reading Hatcher and he gives an example of SVK on torus knots. For the most part, the explanation is intelligible. But he does this thing where he writes as follows:

[gentle osprey]

grist bundle

[marsh forge]

oh boy

this one

[quartz edge]

I'm not really sure

I get this

[marsh forge]

this is in my opinion

the worst part of hatcher lol

[quartz edge]

Ok I will skip it

[marsh forge]

i would honestly skip this example

[quartz edge]

It seems bad

[marsh forge]

The idea for that paragraph I think

is you take a mapping cylinder connecting the times-m map to the times-n map

like

take the mapping cylinder for both

glue along the unmodified side

thats the basic premise

[carmine star]

Do you guys usually use Munkres ?

[quartz edge]

Yes, for point-set

[marsh forge]

For point set its the book I usually recommend

[carmine star]

I see, I'll go for that one then

[marsh forge]

I would not use munkres as a cookbook, however

[quartz edge]

This is what i was assuming

[carmine star]

I'm self-taught, I'd prefer sth easy to understand, do you have one in mind ?

[quartz edge]

but the way he builds that does not seem uhh

well-defined LOL

[quartz edge]

munkres is the easiest

there are faster books

(actually, i have a better recommendation.)

try to read munkres but when you inevitably get impatient, use this

https://pi.math.cornell.edu/~hatcher/Top/TopNotes.pdf

hatcher wrote this

it's meant to get you ready to read his book on algtop

[marsh forge]

oh i forgot about the hatcher notes

yeah

use those

[marsh forge]

it is

the bad part isnt that

its the like

next part iirc

where you have to visualize real hard

[quartz edge]

well, i'm okay with that

[marsh forge]

i spent like 30min on it trying to figure out what was going on

[quartz edge]

the trouble is i do not understand what the parametrization of the cylinder is

i guess the second param is the length

but the first param... i assume it's S^1

but like

what's the period

how far does z have to go to make a single loop?

is z a real parameter? or is it in the complex circle |z|=1

[marsh forge]

2i

2pi

[quartz edge]

so then why does he slap an exp() on it

[marsh forge]

this is the parameterization of S^1 as the unit circle of C

e^itheta parameterizes the circle

as theta goes from 0 to 2pi

this is the like, super canonical parameterization of the circle

you can think of e^2pi/nz as rotating z by 2pi/n

[quartz edge]

uh

oh, ok

z is actually on the unit complex circle

in retrospect this is obvious

[carmine star]

Thank you !

[peak crystal]

Is it possible that arbitrary intersection of open set is neither open nor closed?

[empty grove]

Yes

(-1/n, 1) for each n

[peak crystal]

Got it thank you

Then how to prove that $\mathbb{Q}$ is can’t be written in countable intersection of open set in $\mathbb{R}$

[gentle osprey]

Giyu

[peak crystal]

Ok

[gritty widget]

I need reference for continuous image of relatively compact set is relatively compact

Okay I see

https://math.stackexchange.com/a/1857410/476484

I've used the same equality for subsets of compact spaces before

[tardy meadow]

must be losing the plot - I realised that I have been thinking of limit points and accumulation points as the same thing since i started learning topology

then again i can't remember the last time I used a topological space that isn't a metric space, or at least closures in them

(for those interested https://math.stackexchange.com/questions/409317/when-is-an-accumulation-point-not-the-limit-of-some-sequence-in-a-topological-sp)

[arctic relic]

What’s the fundamental group of (RP^2 x RP^3) v RP^4?

Presentation*

[marsh forge]

Do you know how to compute fundamental groups of products and sums?

[arctic relic]

Yeah I’m having trouble finding the relations

[marsh forge]

What relations?

[arctic relic]

I know it’s Z/2Z x Z/2Z x Z/2Z

Like for the loops

[marsh forge]

Not quite

The wedge sum does not give a product it gives a free product

[arctic relic]

<a,b,c| >?

<a,b,c|a^2,b^2,c^2 something something>, right?

[marsh forge]

huh?

Do you know what a free product is?

[arctic relic]

Yes

[marsh forge]

Okay, do you know how to write a presentation for RP^2 x RP^3

[arctic relic]

Just to make sure we’re both on the same page: the fundamental group of the Klein Bottle is <a,b| ab=ab^-1>

[marsh forge]

yes

[arctic relic]

No

I have an idea but it’s clearly not right

Give me a sec to look through Hatcher

[marsh forge]

Lets start there

[arctic relic]

Yeah having issue with that

[marsh forge]

Do you know what it is as a group?

[arctic relic]

Is it not Z/2Z?

[marsh forge]

the fundamental group of this?

[arctic relic]

I’m thinking of S^2/x~-x product with S^3/x~ -x, which results in Z/2Z x Z/2Z I believe

[marsh forge]

Great

Do you know what the relations there are?

[quartz edge]

When you build cell complexes, do the n-cells have to be attached in a continuous fashion to the rest of the space?

[marsh forge]

yes

[arctic relic]

No clue

[quartz edge]

gotcha. hatcher does not really make this clear

[marsh forge]

every function you see in hatcher is continuous

[quartz edge]

yes, but his definition leaves this detail out

maybe it's in the appendix but not ch0

oh

wait

[arctic relic]

Oh is the relation just 1=a^2

[quartz edge]

i see what you mean. he just uses 'map' to mean 'continuous map'

right, thanks

[marsh forge]

this is the only relation in Z/2

but Z/2xZ/2 has 2 generators

lets call them a and b

we know we have a^2 and b^2

Is this enough?

[arctic relic]

ab

[marsh forge]

ab=?

[arctic relic]

Identity

[marsh forge]

no

(a,1)*(1,b)=(a,b)=/=e

[arctic relic]

Ok that makes sense

[marsh forge]

ab does have a relation, though

[arctic relic]

abab = e

[marsh forge]

hm

that does work

but its not the conceptual answer I was looking for

namely ab=ba

this is ofc equivalent to aba^-1b^-1=e

and bc we know a^2=b^2=e

its equivalent to what you wrote

[arctic relic]

In terms of concepts, why is ab=ba more profound than abab = e?

[marsh forge]

abab=e only works due to the accident that a=a^-1

the condition you want to impose is that the two generators a and b commute

to make the group abelian

[arctic relic]

Ok I didn’t even realize that

[marsh forge]

Oh I see where your relation came from now

hahah

thats funny

anyway

that is all the relations for the first two spaces

Can you finish it off?

(call the third generator c or something)

[marsh forge]

if you are not sure why this is true let me know

[arctic relic]

Give me like 30 seconds

Just add c and c^2

The latter as the relation

[marsh forge]

yep

[arctic relic]

What’s a good heuristic for why that’s all the relations in the first 2 spaces?

Like in more complicated spaces it’s going to be a lot harder to find the relations, no?

[marsh forge]

Well, the a^2 relations come from however you computed the pi_1 of RP^n

The commutativity relation is just an artifact that if I take GxH the elements purely from G commute with those purely from H

https://math.stackexchange.com/questions/2376694/group-presentation-of-the-direct-product

this post seems to discuss the idea

[arctic relic]

Thank you!

[obsidian dune]

what exactly does a function from topological spaces do

like does it map sets to sets, elements of the topologies, or does it map points to points

[quartz edge]

depends on the wielder.

so you are right to be confused

because those functions are usually given by their point-wise mappings

they map points to points

but sometimes you'll see that when speaking abstractly about functions or spaces, the authors will write as if they just map sets to sets

the only really interesting functions are continuous

those, given an open set in the target space, have a pre-image of an open set in the source space

when you're working with particular functions you often need to use the point mapping notion

but when dealing with functions in general you usually can just work without point mappings nicely enough

and not even recall that they do such things

[obsidian dune]

right, i'm trying to show the statement "every map from an indiscrete space is continous" is false and i know it has something to do with taking the inverse image of a set in in the topology of Y and show that it's not in the topology of X i. e. the indiscrete topology

[quartz edge]

you don't need the point definition

for that

to show that's false, you just provide 1 counter-example

does your name say oktofabus

oktafapus

[obsidian dune]

i don't know what they map, like do i map subsets to subsets?

it's octofungus. close tho

[quartz edge]

yes

ah i see

[quartz edge]

yes

yeah so

[obsidian dune]

okay great! so i can just choose a mapping that sends some closed set (i. e. not in the indiscrete) to a topology containing that subset

[quartz edge]

yeah

not quite

construct a mapping for which the inverse maps open sets to not open sets

(it's easier to just build an injective function and not have to worry about weirdness with the inverse)

(if injective, then you're right)

(but "closed" is not right, you need to be sure it's not open. sets can be open and closed at the same time)

(in fact all closed sets in the indiscrete topology are open)

(and vice versa)

(so that does not work at all)

[obsidian dune]

so mapping the indiscrete to the discrete

[quartz edge]

indeed

you want a not-open set to map to an open set

and you might like it to be a one-to-one mapping

[obsidian dune]

gotcha, well thank you for all your help

[next spade]

For this 1st part of the proof

Is there an explanation as to why we can find a polygonal path from $q_i$ to D

[gentle osprey]

bla bla

[marsh forge]

Okay maybe this works: start by drawing an arc from q to the curve C, and then follow C until you hit D. Because C is closed you are guaranteed this will happen eventually

@next spade gonna go to sleep soon if you have any questions

[next spade]

Ohh and we can do this because C is simple closed right got it..

[heady grove]

Anyone know how to do the if direction of part (b)

[coarse night]

if X is finite, can you say anything about the singleton set {x}?

[heady grove]

Its in tau

If X is finite is tau just the set of singleton elements

[coarse night]

(there unions too not just the singletons)

so can you separate 2 distinct points?

[heady grove]

no

i mean yes

[coarse night]

[heady grove]

i see

thanks

[crude vector]

does this proof work, ladies and gents?

wait

Scratch that, I'll look it over again first

[crude vector]

$\emph{Theorem.}$ Every separable metric space has a countable basis. $\ \
\emph{Proof.}$ Let X be a separable metric space and let E be a countable dense subset of X. For each p $\in$ E, define $V_p \coloneqq {B_r(p) \mid r \in Q)} \$
Since there are countably many r $\in \mathbb{Q}$, we have that $V_p$ is countable, and since there are also countably many p $\in$ E, we have that T = $\cup_{p \in E} V_p$ is countable. $\ \$
We claim that T is a basis for X. To show this, let x $\in$ X and let G $\subset$ X be open such that x $\in$ G. $\$
Since G is open, there exists an open ball $B_s(x)$ centered at x, for some $s > 0$, such that $B_s(x) \subset G$. Now consider the open set $B_{\frac{s}{4}}(x)$. By density of E in X, there exists a p $\in B_{\frac{s}{4}}(x)$. By density of $\mathbb{Q}$ in $\mathbb{R}$, there exists a rational number q such that $\frac{s}{4} < q < \frac{s}{2}$. We have $B_q(p) \in T$, $B_q(p) \subset B_s(x) \subset G$ and $x \in B_q(p)$. Therefore, T is a countable basis for X.

[gentle osprey]

strad

[crude vector]

Does this look fine now?

[little hemlock]

yea. the s/4 lower bound on q is unnecessary, but everything checks out. If you're writing this up for homework, you might want to justify the inclusion B_q(p) \subset B_s(x)

[crude vector]

Oh, it is? But how can I be sure that B_q(p) contains x then?

Also yeah that inclusion's been bugging me, I'll work on justifying that, thank you

oh wait is it just triangle ineq.

[little hemlock]

ah yea true, you would need that for B_q(p) to contain x

@crude vector

[crude vector]

Ah sweet, got it

Thanks a bunch for your help 😄

[little hemlock]

npnp

[blissful yoke]

here, we've only defined interior points to be any point which contains an open ball contained in the set; i'm not entirely sure how the fact that A is comprised of interior points implies that any sequence in A complement cannot converge to something in A

nvm i got it

[idle night]

What does "conjugate" mean here

[gritty widget]

https://mathworld.wolfram.com/Conjugation.html

[idle night]

What's the group?

Nvm

[obsidian dune]

im stuck on even starting this problem, "prove that the boundary of the inverse image of B is a subset of the inverse image of the boundary of B for all subsets B of Y if the map f between topological spaces X and Y is continuous"

[gritty widget]

pick a point in the first set. show that it's in the second

that's how you start

[obsidian dune]

should of said that i already have that part

i guess

its what to do after that, as in showing it's actually in the second, that's confusing. i know image of the closure of A is a subset of the closure of the image of A since f is continuous but i'm not sure how to use that

[rancid umbra]

just checking open intervals is fine because they form a basis for the standard topology on R (to check that the map is open). but if you just check closed intervals, how does that tell you that, for example, f(C) is closed, where C is the cantor set?

wikipedia it. it is closed but has no intervals

thats not the only special thing tho

[viral yoke]

Let $p: \tilde{X} \to X$ and $q: \tilde{Y} \to Y$ be covering spaces. Show that $\tilde{X} \times \tilde{Y} \to X \times Y$ is also a covering space. Further, use this result to construct 2022 covering spaces for the 2-torus $T^2$.

[gentle osprey]

eM

[viral yoke]

The first part was fairly straightforward. However, I'm a little confused on the wording on the second part.

My guess is write down a covering map such that the preimage of an open neighborhood of a point produces 2022 open sets

[coarse night]

"2022 covering space"

[viral yoke]

Hmm. Is it constructing a 2022-fold covering map?

[marsh forge]

What does the problem say, exactly

[viral yoke]

Exactly that

[marsh forge]

then it wants 2022 covering spaces that are different

[viral yoke]

Ew

[marsh forge]

I think its not hard

in particular the choice of 2022 is just cute

not important

[viral yoke]

Okay

so

Regarding $S^1$ as in the complex plane, $p: S^1 \to S^1$ given by $z \mapsto z^{2022}$ and $q: S^1 \to S^1$ given by $w \mapsto w^{2 \times 2022}$ are both covering spaces for $S^1$, respectively, we have $(p \times q): S^1 \times S^1 \to S^1 \times S^1$ given by $(p \times q)(z,w) = (z^{2022}, w^{2 \times 2022}$ is a covering space for $S^1 \times S^1$, and therefore $T^2$.

[gentle osprey]

eM

[marsh forge]

that is 1/2022

[gritty widget]

Im watxhing this lecture series by Ravi Vakhil and he is saying something weird

If you take the set of all genus n riemann surfaces

they make a manifold

[viral yoke]

Then taking powers would just do it then

And make sure I count to 2022 correctly

Hopefully

[marsh forge]

technically finding 2023 is still a valid proof

so just overshoot

[viral yoke]

"-10 for overshooting"

[marsh forge]

any grader who does that deserves a firing squad

[gritty widget]

nvm he said it doesnt exist

[marsh forge]

thats like

fully not a joke lol

if you remove a single point for overcounting you should lose your job

[gritty widget]

Oh he was talking about moduli spaces

[gritty widget]

Suppose I have a paracompact topological space X

U is open in X×[0,infty) and U contains X×{0}

how do I show that there exists a map f : X to (0,infty) such that y ≤ f(x) implies (x,y) in U

[gritty widget]

One idea would be to define $F:X\Rightarrow [0, \infty)$ by $F(x) = {r\geq 0 : {x}\times [0, r]\subseteq U}$ and try to use Michael selection theorem, the problem is though that this doesn't give us a function into $(0, \infty)$ (indeed, clearly if we just wanted a function to $[0, \infty)$ we'd just take $f \equiv 0$).

[gentle osprey]

Blitz

[gritty widget]

I have another idea though. What if you take $$g(x) = \sup {1\geq r > 0 : {x}\times [0, r]\subseteq U}$$ and try to prove that it's lower-semicontinuous. Then I think you can prove there is a continuous $f$ with $0< f< g$.

I think this still holds in the general setting of paracompactness

[gentle osprey]

Blitz

[gritty widget]

@gritty widget I could try to attempt the proof if you're interested

[gritty widget]

Yeah i have tried this

[gritty widget]

hmmm

[gritty widget]

Yeah sure

Oh wait

I think you can just partition of unity this mf

actually I'm not sure, is your space assumed Hausdorff?

Yup

that's important

my definition needs paracompact to be hausdorff

Yeah

gg

so clearly

pi_X(U) gives a open covering

What you want to do is take partition of unity subordinated to $U_q = {x\in X : 0 < q < g(x)}$

[gentle osprey]

Blitz

[gritty widget]

for rational number q

yeah

and then damp that

with 1/2^k(n) factor

so that y ≤ f(x) implies (x,y) in U

With appropriate k(n) for each n

you then sum them all (multiplying by the q-th function by q)

and this is your function, more or less

yeah that works too

you still would have to prove g is lower semi-continuous

hm the construction i have in mind does not require to do any of that

wew

Let $X$ be paracompact Hausdorff and $f, g:X\to\mathbb{R}$ be such that $g < f$, $f$ is lower-semicontinuous, $g$ is upper-semicontinuous.
For $q\in\mathbb{Q}$ set $$U_q = {x\in X : g(x)<q }\cap {x\in X: f(x)>q}$$ which is open. Let $\mathcal{U} = {U_q : q\in\mathbb{Q}}$. There is $E\subseteq \mathbb{Q}$ such that $q\mapsto U_q$ is a bijection from $E$ to $\mathcal{U}$. Let $\varphi_q$, $q\in E$ be partition of unity correspodning to the open cover $\mathcal{U}$ and set $\varphi = \sum_{q\in E} q\varphi_q$. Then $\varphi$ is continuous with $g<\varphi<f$.

[gentle osprey]

Blitz

[gritty widget]

👍

Can't we omit semi continuity

lemme see

semi-continuity is so that $U_q$ is open

[gentle osprey]

Blitz

[gritty widget]

I'm sorry, but you'll have to put additional work in this somehow if you want to prove it

there's no running away from it

[lunar yoke]

i think the following might work as well:
since $[0,\infty)$ is paracompact and locally compact, $X \times [0,\infty)$ is paracompact and in particular normal.
We can use Urysohn to get a function $g : X \times [0,\infty) \to [0,1]$ so that $g^{-1}(0) = X \times 0$ and $g^{-1}(1) = (X \times [0,\infty)) \setminus U$ since these are disjoint closed subsets. Note that if $g(x,y) < 1$, then $(x,y) \in U$.

[gentle osprey]

Phil

[lunar yoke]

i thought of then defining $f : X \to (0,\infty)$ as something like $f(x) = min(1, \inf {y \in [0,\infty) | g(x,y) > 1/2})$,

[gentle osprey]

Phil

[lunar yoke]

since U is open and contains X x 0 i feel like this infimum should always be positive, and that f is continuous, but my brain isnt really working rn so could be wrong

anyway then y <= f(x) would mean in particular that g(x,y) <= 1/2 < 1 hence (x,y) in U

[gritty widget]

ugh how do we show its continuous

Nice construction btw

[lunar yoke]

it would certainly involve continuity of g

but im not so sure either xd

imagine manually checking that things are continuous

its been too long

[gritty widget]

[gritty widget]

Also I'm pretty sure U_q not being disjoint will mess this up

9k

Ok

Gg boys @lunar yoke @gritty widget

Heres my proof

[gritty widget]

This was my idea

Which I'll continue

Gimme a sec

So given $r\in X$ , there exists $h(r)>0$ and a neighborhood $U(r)$ of r such that
$$U(r)\times [0,h(r)]\subset U$$
Consider partition of unity subordinate to ${U(r)}$ , $f_r$
\
Define $f(x)=\sum_{r\in X}h(r)f_r(x)~(\forall x\in X)$ this is continuous because for any neighborhood of $x$ in $X$ only finitely many $f_r$ are positive in this neighborhood now take a convergent net in this neighborhood.

[gentle osprey]

CityHunter

[gritty widget]

and for the last part

if y ≤ f(x)

only finitely many f_r(x) are non zero

take that r for which h(r) is maximum

say r'

y ≤ h(r')

but f_r'(x) > 0

Hence (x,y) lies in U(r')×[0,h(r')] subset U

[gritty widget]

no

Ok

I see how

now

[gritty widget]

Bcuz thats what i literally used here

But yeah

We can run from semicontinuity

when taking your partition of unity, you're using paracompactness of which space?

My space is hausdorff and paracompact , X

U(r) is open cover of X..

okay, I see

omw to homotopy theory

[cosmic beacon]

My algtop course is going through H. Miller's notes on the subject (http://math.mit.edu/~hrm/papers/lectures-905-906.pdf), which defines the n^th cohomology boundary map like this, with a sign change if n is even:

Yet hatcher just defines it like this

Without that sign change

Are those actually different maps? Or am I not seeing it correctly?

[marsh forge]

is the dsigma defn identical

the sign change in millers notes is indeed mysterious to me

[cosmic beacon]

Miller defines the homology differential as:

where d_i is the defined like so

And d^i is

So I think it's the same?

Miller has a habit of defining things in stages where the pieces are scattered all over the notes, and I'm really not a fan lol

[marsh forge]

its possible these are the same up to chain homotopy but I am not able to think about it right now

[cosmic beacon]

Later Miller does the following calculation

I'm pretty sure that's not what d sigma is, no? It should be sigma(e1) - sigma(e0)

[marsh forge]

Maybe yeah

That is rather inconsequential for that remark I think

Its possible miller was not super careful about this

[warped rover]

If I have a space X and all it's isolated points are given by I, shouldn't X\I not have any isolated points?

Oh I see, so each of the 1/n is isolated but 0 is not, once you remove each 1/n you get just 0, which is isolated.

Is there a simple example of a space with height of at least \omega_0?

Ohhh ok that makes sense.

Thanks

I will watch after class

instead of typing up computability homework

[jolly garden]

Can someone give a concrete example of filter convergence? I get the concept, but I'd like to see it in action.

[gritty widget]

Take $\mathcal{F}(x) = {F\subseteq X : x\in F}$, then $\mathcal{F}(x)\to x$

[gentle osprey]

Blitz

[gritty widget]

@jolly garden here you go, trivial but an example

you can think of filters, well, more precisely about ultrafilters, as limits of sorts
and this will give you Stone-Cech compactification

[jolly garden]

Yep, I see. Easy example, thank you. @gritty widget

The name "filter" suits the concept well. It's like a continually tightening set system and thus predestined to describe convergence.

[golden gust]

why is this?

[fading vale]

Honestly not totally sure but Id guess they mean something like CW approximation and then thinking about attaching maps as maps S^n -> S^m

But that seems like a bit of a stretch

[marsh forge]

Homotopy groups preserve homotopy colimits and the spheres generate the homotopy category

So once you have a recipe for X in terms of sphere colimits you can in principle just compute from pi_*S^n

@fading vale should see this 2

[fading vale]

Oh that makes sense

Well not the generating part but the homotopy colimits bit

Does spheres generate the homotopy colimit mean every CW complex can be written as the homotopy colimit of spheres

[marsh forge]

Every homotopy type is a homotopy colimit of spheres

[marsh forge]

Up to homotopy

I mean also just on the nose

[fading vale]

Yeah

Well

things stronger than homotopy are too scary

[marsh forge]

They are iterated pushouts

Bc everything is cofibrant in cw land this is the same as homotopy pushout

[fading vale]

I see

I should learn about these constructions

Ive only really seen the homotopy pushout

[marsh forge]

Like the attaching map construction

Is just an obvious pushout diagram

“Obvious”

the n skeleton is the pushout parameter used by gluing disks along spheres

From the n-1 sell

Skel

[fading vale]

[plain raven]

I have no idea what's going on here

[fading vale]

Max where should i learn like

[plain raven]

but i'm curious

[fading vale]

This kind of homotopy theory

[marsh forge]

Lurie

[fading vale]

I feel like it so hard to find places to see it

unironically?

[marsh forge]

Almost yeah lol

[plain raven]

oh brother

ok

[fading vale]

Man

HTT right

[marsh forge]

If you wanna do model cats you can do like blue book stuff

Htt and ha

I am drives one sec

[fading vale]

np

Drive safely

[marsh forge]

Yeah the literature around here is hard

[plain raven]

i just wanna learn what a homotopy colimit is exactly

[marsh forge]

A colimit in an infinity category

[plain raven]

i get it intuitively

Ok