point-set-topology

[lunar yoke]

and you wont have to compute the homology like we just did for simplicial homology

[fading vale]

I feel like this is super normal for AT

[lunar yoke]

well i didnt really have to worry about orientations and the like anymore after we went from simplicial to singular

[fading vale]

Orientations and degree arguments r still important

[lunar yoke]

i actually feel like i understood singular homology quicker than simplicial

[lunar yoke]

yeah

but not when i compute the homology of say CP^n

[fading vale]

They will be if you use cellular

Which is the bestest way

[lunar yoke]

idk about that

getting the boundary maps right is a pain

but for CP^n you can just see that is Z in even dimensions

since you dont get cells in the odd ones

so you can even skip computing boundary maps there

[fading vale]

Well, if the boundary maps are trivial then yes but uh

Generally speaking

[lunar yoke]

i was referring to CP^n, not generally of course

i also know how cellular homology works

[fading vale]

Oh sure I thought you were saying you prefer singular to cellular

Well

Idk I guess its kind of silly to prefer one method inherently since theyre just tools but I do think that cellular is often more convenient

[lunar yoke]

i mean i try to avoid computing boundary maps of cellular homolgy whenever posible

cause its a pain

but also, i dont have to comput homolgy of spaces anymore really

[fading vale]

Freedom

[lunar yoke]

yeah

who needs computations when instead you can devote 1 month of lectures to proving the hurewicz theorem

[fading vale]

Dude why do so many classes have like, loooooong hurewicz proofs

[lunar yoke]

ask schwede

[fading vale]

the one i saw was very nice

[lunar yoke]

do you have a source?

[fading vale]

I learned it out of homotopical algebra by fomenko/fuchs but it was done through recourse to nice things about weak equivalence and pi_n of CW complexes

[lunar yoke]

sounds good, ill look at it sometime

tbf at the time we did it we didnt have cw approximations yet

but the whole order of things could have been different

[fading vale]

Yeah I feel like doing pi_n things before homology is kinda the way to go

But i can understand why people do it in the order they do

[lunar yoke]

yeah i guess both have their pros and cons

[fading vale]

(all you really need to know is that if X is n-1 connected you can assume it has one vertex and no cells of degree between 1 and n-1. then doing cellular homology H_n(X) is C_n(X)/Im Delta_n and pi_n(X) has generators the n-cells and relations the n+1 cells given by the boundary maps so these are just definitionally the same lol)

this assumes n > 1

[lunar yoke]

oh yeah that makes sense

we had that lemma later on in the course

[fading vale]

Its a nice result

[lunar yoke]

i first thought you can just take X/X_n-1

but that was complete bullshit

[fading vale]

to compute pi_n?

[lunar yoke]

no to get a htpy eqv cw complex with trivial n-1 skeleton

[fading vale]

Oh yeah

[lunar yoke]

for n-1 connected

anyway it was a very nice lecture course overall

i learned alot

continuation with spectra and stable htpy theory starts on 4th

[fading vale]

the construction is so wacky, i remember thinking it was weird until I saw it applied to eilenberg maclane spaces and was like

"Ok i guess it makes sense that its not that nice"

[lunar yoke]

hmm i dont remember it being that weird

i think we showed that you get a htpy equivalence from X_n to a wedge of spheres

and then you glue X to the wedge of spheres along X_n-1

and the map from X to that is a htpy equivalence by some general lemma involving cofibrations and htpy equivalences in a pushout

[fading vale]

The one i saw was given by attaching the right number of cells of each dimension up to n and deformation retracting back to X

[lunar yoke]

this was our proof

ah

[fading vale]

This is how i saw eilenberg maclane spaces constructed

[lunar yoke]

but that uses hurewicz

[fading vale]

Oh yeah the proof i saw didnt make any recourse to homology or anything

it took a bouquet of n spheres so that the homotopy groups for i < n would be trivial

so you get a free group of whatever for pi_n

and then attached n+1 cells to give the right relations

then n+2 cells to kill pi_n+1

n+3 to kill pi_n+2

so on

[lunar yoke]

yeah we had some lemma that we can kill homotopy groups from smoe point

[lunar yoke]

to show this we used cellular homolgy and then hurewicz

i.e. that this gives you the correct relations

[fading vale]

capping and killing

[lunar yoke]

i guess its also possible without that

[fading vale]

I should learn more of this stuff I only know some model category stuff

Have not explored spectral sequences or stable stuff or any of the more advanced computational tools

[lunar yoke]

im trying to learn model cats rn

[fading vale]

Im using riehls paper

[lunar yoke]

me neither, but startin soon xd

[lunar yoke]

im using the dwyer spalinski paper

got that recommended by my prof

i think i sent it in here some time ago

when max was recommending stuff

[fading vale]

Lmfao

did not mean to send that

https://arxiv.org/pdf/1904.00886.pdf this

[lunar yoke]

yeah i also have that in my to-read folder

[fading vale]

homotopical derived functors are really nice

[lunar yoke]

oof that reminds me i also really need to learn some proper homological algebra

i still havent every dealt with derived functors

so many things to read

so little time

[fading vale]

you can define them as kan extensions along the localization into the homotopy category

[lunar yoke]

im also still at the start of ch6 in riehl on kan extensions

[fading vale]

Then the derived category is the homotopy category of the category of chain complexes with weak equivalences the chain homotopies

i also actually dont know that much homological algebra

I am learn sheaf cohomology to fix this

[lunar yoke]

if i hadnt dropped algebraic geometry i would also learn this soon

[fading vale]

always more time to pick it up later if u want

[lunar yoke]

true

[gritty widget]

closed sets in a metric space are G_delta

this is simple to prove, you just consider their neighbourhoods D(F, r) = {x : d(x, F) < r} and you can prove that the intersection of those is F (here we use that F is closed)

[feral copper]

Wait, I'm having an existential crisis now... Milnor says that any two Riemannian metrics µ1 and µ2 are joined by a smooth one-parameter family (1-t)µ1+tµ2
This basically boils down to saying that SO(n) is convex, right? But now, the thing is: is it true that SO(n) is convex? x')

Like, what's a dumb argument for that?

Wait, no, SO(2) is the circle, it is not convex :\

[gritty widget]

I need to show that if I have $X$ a topological space, $A\subset \bR^n$ a star domain, and $f: X \to A$ a continuous mapping, then $f$ is null-homotopic. I don't know where to start though

[gentle osprey]

Entelechy

[lunar yoke]

you can contract the star-domain onto its middle point

[gritty widget]

cause I suppose the point a0 for which any segment {ta + (1-t)a0} is contained in A, is the middle point

[lunar yoke]

yes

[gritty widget]

It's like Phil says. Wlog assume a0 = 0, and for the contraction just multiply the whole thing by a constant

0 <= t <= 1

I mean like the time

Uh. $H_t = t\cdot f$

[gentle osprey]

Blitz

[gritty widget]

Like that

but I think I really struggle understanding what homotopic means. From what I have, $f$ and $\bar{f}$ are homotopic if there exists a continuous map $F: X \times I \to A$ such that $F(x,0) = f(x)$ and $F(x,1) = \bar{f}(x)$ for all $x \in X$. So here $\bar{f}$ must be constant ?

[gentle osprey]

Entelechy

[lunar yoke]

correct

nullhomotopic means homotopic to a constant map

[gritty widget]

okay I understand a bit more now, but intuitively, what does the constant map do to, say, a curve ? Caue what I have in mind is the identity which preserves the curve but that is of course not constant

I guess it's what you said, it brings the whole thing down to a single point

[lunar yoke]

yeah

and at least for me, the way I imagine "bringing something down to a point" is already as a homotopy

since in my head i would gradually contract the curve, rather than go from the whole cruve to a single point in an instant

[gritty widget]

yea, hence the "time" dimension

from the unit interval

[lunar yoke]

yep

but what im saying is that when you try to picture the constant map, you often actually picture the homotopy instead, which may make things confusing

but in your case you're interested in the homotopy anyway, so all good

[gritty widget]

alright, it's still a bit hard to visualise

from the formal definition

[lunar yoke]

i mean you can always start with an easier example

and then try to generalise

for example lets pick X = A = D^2 and f the identity

can you construct the homotopy in this case?

D^2 is the 2-dimensional unit disk in R^2, centered at 0, although the dimension really doesnt matter here

[gritty widget]

D^2 is a star domain itself so I can see the link, if my f is continuous then it's null-homotopic
I suppose I can arbitrarily define H(x,t) = (1-t)f(x), so
H(x,0) = f(x) = x
H(x,1) = 0 = the 0 map

[lunar yoke]

yes, exactly

nice

so yeah the point is that you can basically "first do f, then contract everything"

i.e. there will be a homotopy H : A x I -> A starting at identity and ending at a constant map, which in you case was just H(x,t) = (1-t)x

and then you can consider the homotopy H' : X x I -> A, (x,t) -> H(f(x),t)

often times one writes Hf for this homotopy H'

this is called "whiskering of homotopies", and can also be done from the other direction in the sense that given a homotopy H : X x I -> Y and a map f : Y -> Z, you can consider fH : X x I -> Z, (x,t) -> f(H(x,t))

this comes up pretty often

and often helps deconstruct problems into simpler ones

in you case, we went from "find a nullhomotopy for f" to "find a homotopy contracting A to a point"

[gritty widget]

wait, so here we have the "trivial" homotopy (A to A), and from there we want to define another homotopy H' (X to A)

[lunar yoke]

well you can also just write down the homotopy from X x I -> A immediately

i just like to do it that way personally

to first find H : A x I -> A, and then consider Hf

[gritty widget]

ah, but H : A x I -> A will not always be null-homotopic

[lunar yoke]

yes, this only works because A is contracible

this is basically the definition of being contractibe

that you find a homotopy A x I -> A from id to constant map

and star-shaped things in R^n are a special case of that 🙂

[gritty widget]

okay, I understand now why "homotopic to a constant map" is denoted as null-homotopic, as it all contracts to a single point

so H' : X x I -> A, (x,t) -> H(f(x),t) will simply be H(x,t) since f(x) = x ? I don't quite get how to explicitly express a homotopy for X x I -> A considering X is arbitrary

[lunar yoke]

no above we just discussed this one example

generally f is not the identity of course, since X is arbitrary as you said

The homotopy you look for will still be of the form H' = Hf though

you just have to write down the homotopy H : A x I -> A that contracts A onto the middle point

if you assume that the middle point is 0, then this is just H(x,t) = (1-t)x as before

otherwise you have to do some convex combinations, but it will basically look the same

[gritty widget]

okay, I guess it's all equivalent up to a translation anyways, so I can assume the middle point is 0 to avoid overcomplicating things

[lunar yoke]

yep

i mean the whole point of all these messages was basically that the difficulty of X being abitrary is kinda fake, since you only need to find the contracting homotopy H : A x I -> A and then form Hf

and since A is star shaped in R^n you can actually write down this H

without too much pain

[gritty widget]

yea I see, so it's way simpler than it looks

alright, that was very helpful, thanks for taking time to explain 🙂

[lunar yoke]

np

[limber ravine]

Hey guys
So I have an argumento to check

I want to show that a normal space is regular. That is, a normal space satisfies T_3

Let me remind you what T_3 means. It a space such that for any C subset X closed and any x in X\C, there is open sets U,V with C subset U and x in V such that U cap V is empty

So, take C in X closed and x in X\C which is open. But then, there's an open set U subset X\C such that x in U subset X\C. Now consider the closure of U. If U intersects C then consider another open U_1 subset U with x as an element. Consider the closure of U_1. With this iterative process we can find two disjoint closed sets. By T_4 there's an open set for each one of them.

Note x would be an element of the open set of U_1 and the closed set C would be a subset of the another open set

[gritty widget]

I don't get it

What's your definition of normal space

[limber ravine]

satisfies T_1 and T_4

oh, I could take a singleton of x

by T_1

I forgot this property lol

[gritty widget]

If you have T_4, then why go through some elaborate argument? Simply say {x} and C are disjoint closed

[limber ravine]

yeah indeed

thanks

[steel glen]

i’m given that f_i: X_i -> Y_i, i = 1,2 are each continuous.
we then define f1xf2 : X1 x X2 -> Y1 x Y2 as f1 x f2 (x1, x2) = (f1(x1), f2(x2))
how do we show continuity?

i’m confused since we usually show continuity with some associated space/s

this is from my notes right before my class discusses topological groups, so maybe that added context can shine a light on what to do

[river granite]

in this case you should assume that the products have the product topology

[steel glen]

ah of course, thanks

[winged viper]

Is there a common way of proving that maps from a product topology are continuous? Like, for maps to a product topology f : Z -> X x Y, you just need to check that f composed with the projection maps is continuous. Is there a similar criterion or way of thinking about continuous maps from a product topology?

[gritty widget]

No

But there is one from disjoint unions

Think about if there is some way of telling when a function f:R^2 to R is continuous

We could consider g(x)(y) = f(x, y)

Obtaining a function g:R to R^R

Then you can prove under some mild assumptions like local compactness that this should be continuous iff f is continuous

R^R equipped with compact-open topology

Other than that, I see no way of phrasing continuity of f differently

@winged viper hope that exhausts your question

R^R meaning continuous functions from R to R

[gritty widget]

If you want to know more google things like, exponential law for topological spaces, evaluation map etc

[winged viper]

Ah I see okay, this validates my fear of product topologies

[marsh forge]

This is secretly a category theory question. Essentially products are limits and mapping into limits is easy. Disjoint unions (coproduct) are colimits and mapping out of colimits is easy. In general the reverse is not true for either.

[vast estuary]

Just trying to understand what goes wrong if F is not compact

In that case, we may not be able to find a finite subcover as done above

0.9 doesn't require neighborhoods to be open though, so what goes wrong?

[gentle osprey]

Hausdorff

[river granite]

$\bigcap_p U_p$ might have an empty interior. Consider $E={0}$ and $F=(0,1)$. For each $p\in F$ there's some $V_p=(p-\varepsilon,, p+\varepsilon)$ that doesn't contain zero since $F$ is open, and if $U_p=(-\varepsilon, , p-\varepsilon)\ni 0$ then $U_p\cap V_p=\varnothing$ for all $p$, but $\bigcap_p U_p$ has an empty interior and thus isn't a nhood of $0$.

[gentle osprey]

derivada.schwarziana

[river granite]

the idea is that compactness here guarantees that the condition on disjoint nhoods implies a positive distance between E and F (if the ambient space is metrizable)

[river granite]

this also serves to show that the hypothesis on compactness can't be dropped since obviously {0} and (0,1) have no disjoint nhoods in R

[vast estuary]

Gotcha, thanks!

Also, in the case where F is compact, we are just using that int(A) \cap int(B) = int(A \cap B), right?

so the interior of a finite intersection is just the finite intersection of interiors, I hope that makes sense

This doesn't work for arbitrary intersections, as you said

[river granite]

int(A \cap B) is a subset of int(A) \cap int(B), generally strict even in the finite case nevermind, I was thinking about unions. You're right

however an infinite intersection of open sets isn't necessarily open

that's what fails

[hot night]

don‘t know where else to ask: looking for cool math background image for my ipad

any ideas?

[lunar yoke]

since you're asking in this channel, look up images of the hopf fibration

[hot night]

hopf fibration looks cool but it seems that there are nk images suited for a background.

actually i like your profile picture. i‘m looking for smth in that direction, i prefer smth decent

[lunar yoke]

you can find the image somewhere in Amann & Eschers Analysis I

[fading vale]

so the homotopy product in a simplicially enriched category is being defined as an object m equipped with a weak equivalence Hom(x, m) -> prod Hom(x, m_i) for all x

since coproducts are supposed to be dual the weak equivalence would go from coprod Hom(m_i, x) -> Hom(m, x) right

[crude vector]

dear #point-set-topology, could someone look at this fella for me real quick?

I did it for an arbitrary separable metric space instead of R^k

and you should replace "is a neighborhood of" with "is an open set containing"

rudin terminology, idk why he calls it neighborhood

is everything looking yay here?

[gritty widget]

Yeah, this is true in this case also

[crude vector]

Proof works here and for R^k?

or just the statement?

[gritty widget]

Sure

[crude vector]

Sure to this?

[gritty widget]

Yes

If your space is additionally complete, we can even prove that P has size continuum if non-empty (Cantor set embeds into it)

And that it's Borel isomorphic to real numbers

All perfect Polish spaces are Borel isomorphic

Perfect is essential here, like, suppose your space looks like {0} and {1/n : n in N}

It's a Polish space, but it's countable

And it has no perfect subset

[stone cipher]

what does it mean to a sequence be precompact?

I know that $A\subset X$ is precompact if its closure is compact. So, if $A=\left{x_n::n\in\mathbb{N}\right}$ then we say that the sequence is precompact if $\overline{A}$ is compact?

[gentle osprey]

RaD0N

[stone cipher]

Ok. Forget my question xd

[wise walrus]

Can someone here help me understand the amalgamated free product? I'm working with "repairing" the punctured torus with Van-kampen theorem, and the notation and how the amalgamated free product is used confuses me:

here I have the fundamental group of the punctured torus deformation retracts to a boquet on 2 circles, the fundamental group of the disk is trivial, and the overlap is a circle. I get:

$\pi_1(\mathbb{T}) = \langle a, b \rangle *_{\mathbb{Z}} \{1\}$ (via the mapping $[a,b] \mapsto 1$)

$\pi_1(\mathbb{T}) = \langle a, b : [a,b] \rangle = \mathbb{Z} \oplus \mathbb{Z}$

[gentle osprey]

darkninja175

[wise walrus]

But I don't really understand how the mapping of the commutators arises, like where it comes from?

Is it chosen arbitrarily? I guess this mapping is somehow the kernal of some map I'm not displaying here?

I guess partially what I'm asking is how the commutators arise within the presentation of the fundamental group of the torus as well

[umbral finch]

what is the tightest bound on each coordinate of the n-torus T^n (say embedded in R^(2n)).
is it 2?

[silk ember]

Is R² with a small line segment removed, say from (0,0) to (0,1) a star domain? I personally thought it wasn't

[umbral finch]

this should be isomorphic to an annulus, which is not a star domain.

[silk ember]

Thank you. I already assumed so but just wanted a confirmation haha

[still vessel]

if i have a subset S which is not open or closed, how can i argue that it has to have a boundary?

[dull zinc]

Use bd(S) = cl(S) \ int(S)

Since it is not closed, S ≠ cl(S)

Since it is not open, S ≠ int(S)

See if you can find a nonempty set contained in bd(S) in this way.

[still vessel]

hmm, i tried it like that, but i just feel like i am stuck in a logical loop

what i am trying to do is to prove that if the boundary is an empty set, then S is clopen. But i can not get rid of the possibility that S is neither open or closed

so if boundary is empty set then int(S) = cl(S), but how can i than say that S = int(S) = cl(S) and not that S is neither open or closed

[dull zinc]

I'm confused as to your confusion. int(S) is contained in S and that is contained in cl(S), so that triple equality follows.

S = int(S) gives openness and S = cl(S) gives closedness.

Hence if bd(S) is empty, S is clopen - which, if that's your ultimate goal, you're done.

[still vessel]

yeah, i just realised i can do it very simply with $int(S) \subseteq S \subseteq cl(S)$

[gentle osprey]

FireLi0n

[still vessel]

ok, thanks for patience

[gritty widget]

Being a star-domain isn't a homeomorphism invariant, no?

[gritty widget]

It's not, since if you pick any point from your domain, and any point on the segment, then you consider the ray obtained this way

Well, I guess what AdrainV said also works but not directly

If it was a star domain then it would be contractible, but it's not

[umbral finch]

indeed!

[umbral finch]

thanks for pointing it out!

[fading vale]

The free product with amalgamation here is given by quotienting out by the normalizer of f(x)g(x)^-1 where x is a loop in the intersection and f and g are the induced maps of pi_1 of the intersection into pi_1 of each element in the cover

So the point here is that your generator for your intersections image in <a, b> will look like aba^-1b^-1 because you have to wrap around the entire boundary of the polygon. So f(x) = aba^-1b^-1. And it's image in the interior of the square is trivial because that has trivial pi_1. So we have to quotient our free product out by the normal subgroup generated by f(x)g(x)^-1 = [a, b], ie we send [a, b] in the free product to 1

[fierce rock]

Hello! How ccan I prove this?

[gritty widget]

It's been 13 minutes. Do you really have a question

[hollow harbor]

I think they posted it and deleted it, it was $\partial(A \times B) \subset (\partial A \times B) \cup (A \times \partial B)$

[gentle osprey]

ryc

[hollow harbor]

I'm not sure why they deleted it

Also I just realized that the boundary operator satisfies a leibniz rule... holy shit

[cedar pebble]

@frigid patrol let f:C^{n+1}->C be a regular or analytic map with 0 a critical value, let X_0=f^{-1}(0) be the special singular fiber, let X_s=f^{-1}(s) be a generic smooth fiber. For a point x in X_0 take a small ball B in C^{n+1} centered at x with boundary (2n+1)-sphere S. Then you can study the topology of the hypersurface singularity (X_0,x) as follows:

B\cap X_0 is contractable and is homeomorphic to the cone on K_x=S\cap X_0 the real link of x in X_0

the real link K_x is (n-2)-connected

the map f/|f|:S-K_x->S^1 is a topologically locally trivial fibration; this is the Milnor fibration of the hypersurface singularity

If the germ of the critical set of X_0 at x has complex dimension r, then the fiber F_x of the Milnor fibration is (n-r-1)-connected. In particular if x is an isolated singularity then F_x is (n-1)-connected; this is the Milnor fiber of f at x.

the point of the theory is that the Milnor fiber F_x says a lot about the geometry of the hypersurface singularity you're interested in

For example, if (X_0,x) is an isolated hypersurface singularity, then F_x has the homotopy type of a bouquet of n-spheres; the number of n-spheres is the Milnor number of f at x, which is dim_C C{x_0,...,x_n}/(∂f/∂x_0,...,∂f/dx_n) and this is a ubiquitous invariant in singularity theory

one of the main ways this comes up in algebraic geometry is the monodromy of Milnor fibers and their relation to nearby/vanishing cycles. You get a monodromy map h_x:F_x->F_x given by going once around the circle in the Milnor fibration f/|f|:S-K_x->S^1

it's a fundamental theorem of Grothendieck and others that the induced monodromy morphism h*_x:H^i(F_x,C)->H^i(F_x,C) is quasi-unipotent, that is its eigenvalues are all roots of unity

this implies among other things that the Lefschetz number L(h_x)=0 if x is a singular point for f, so this gives a cohomological criterion for detecting singularities

the other main applications include construction of exotic spheres and homology spheres

another thing is that if you assemble the cohomology of Milnor fibers into a single sheaf, this is the nearby cycles sheaf that is ubiquitous in the theory of perverse sheaves. In the case of isolated singularities, the circles in the bouquet are your vanishing cycles, and the specialization map from nearby to vanishing cycles has an obvious topological interpretation

/end chatspam

[marsh forge]

This is literally called the leibniz rule by a lot of people haha

[hollow harbor]

Really?!

[marsh forge]

Yeah

[hollow harbor]

Wow that makes me very happy

[marsh forge]

Me for example

There’s also a generalized leibniz time

Rule

For massey products

[hollow harbor]

Ah

Huh

Is there a sense in which taking boundaries is actually some kind of derivation

on a module or something

uhhh

Ok, you can say something like this with an identification from stokes theorem i guess

[marsh forge]

Stokes theorem can be thought of as exactly the statement that integration is a chain map from deRham to singular chains @hollow harbor

[hollow harbor]

That makes a lot of sense

[marsh forge]

Makes it sound a lot less cool

Lol

[cedar pebble]

And the de rham theorem is simply a statement that the de rham complex is a resolution of the constant sheaf

A great way to make the subject completely lose all of its magic in two statements

[plain raven]

yes, differential forms were such a quaint way to resolve the constant sheaf. Good to mention on day 3 of a geometry class, along with Alexander-Whitney cohomology etc before saying "Of course, any resolution by injective sheaves will do" and then move onto nonsensical juggling of H^k(X;F) for the rest of the semester

[cedar pebble]

lmfao

algebraic geometers: just write down an injective resolution
algebraic geometers when asked to write down an explicit injective resolution:

[old violet]

is the span of a countable set always separable?

[old violet]

Thanks! That makes a lot of sense.

[crude vector]

Any hints on how to approach this problem, ladies and gents?

[gritty widget]

We've gone through this already

[crude vector]

All I've got is I should use the fact that R is separable somehow, so worrying about Q, and taking open balls centered at rationals, but im not sure where to go from here. Blitz mentioned something about connectedness, but I still don't see how to make use of that

[gritty widget]

Note that R is locally connected

This means that components of open subsets of R are open

[crude vector]

I don't even know what that means, if we're supposed to be using only ideas covered in the book

Okay, okay, I see

Hmm

[gritty widget]

Write open U as union of its connected components

Since R is separable, each component contains an element from a countable dense set

So there is at most countable amount of components

Finally, open connected subsets of R are open intervals

[crude vector]

When you put it like that, it's pretty simple, but are you aware of any other way to prove this? I find it hard to believe the book would expect me to use an idea it's never introduced

[unreal stratus]

I have another way that is a bit lower tech but is ultimately the same

Consider the relation ~ on U by saying x~y iff the closed interval between x, y is contained in U

It's an equivalence relation, the classes are open, and you can pick distinct rationals in each class to show there are countably many classes

[crude vector]

The classes are open?

[unreal stratus]

Yeah

That's where we use that U is open

[gritty widget]

Yes, those are basically path-components and they are the same as components in this case

[unreal stratus]

exactly yeah ^

Just has the very slight advantage that you don't have to prove the components are the intervals

But ultimately the same

[crude vector]

Goootcha, gotcha

I was going about this the wrong way then, good thing I didn't commit

Still kinda weirded out by the fact that rudin never mentions components, but at least now I know

Many thanks fellas

his section on connectedness is less than a page long, so I'm well equipped

[unreal stratus]

Well this bit of rudin is not really that general iirc, mostly just gives yoy stuff you need for the rest of the book

[crude vector]

You're right yeah, doesn't cover more than necessary

that's fine I guess, planning on going through munkres eventually

[midnight echo]

The product topology of Hausdorff spaces is Hausdorff. Does the converse hold? If a product topology is Hausdorff it is a product of Hausdorff spaces?

[fading vale]

Yes if X is the product of non-empty hausdorff spaces X_i then the X_i are homeomorphic to subspaces of X, hence hausdorff

[midnight echo]

brilliant thanks

[crude vector]

@gritty widget I ended up looking up the solution just to see if people did it the same way, and turns out you can do it without bringing up connectedness

wanted to only read the first line or something then attempt the rest on my own, but I ended up reading almost the entire thing...

oh well, least it's done

just wanted to check with you if the equivalence relation part works though? since that's the part I wrote myself

[gritty widget]

can someone explain the difference between the direct sum and product of groups

they agree on finite products?

[lunar yoke]

this is probably better suited for #groups-rings-fields, but no finite products and coproducts do not agree for groups

for example Z x Z is still abelian, but the coproduct of Z with Z isnt

unfortunately the coproduct of groups is often called the free product

for Z with Z it would be the free group on 2 generators

[gritty widget]

by the coproduct of Z with Z

is this the same as the direct sum Z\oplus Z

[lunar yoke]

no that would be the coproduct in the category of abelian groups

and there finite products and coproducts agree

[gritty widget]

I've noticed you post a lot of questions from abstract algebra and category theory here

I know it connects with algebraic topology, but it doesn't have any topology in it tbh

maybe this will make it more suitable

I would have thought this should be a direct sum

[lunar yoke]

that would be for homology

cohomology is contravariant so you want to switch to products instead

[gritty widget]

ah thank you

[hasty pasture]

Hey im trying to understand hatchers proof of Borsuk-Ulam and im failing to understand why this is implied

The last line says that the restriction is nullhomotopic, but why does the restriction being nullhomotopic imply the theorem?

is this supposed to be a contradiction?>

because it is nullhomotopic, it cant be odd

and also one more thing, why does the restriction of f to a hemisphere being null homotopic imply that f restricted to the equator is nullhomotopic?

[marsh forge]

A map can't have odd degree and be nullhomotopic

[weak narwhal]

I still dont see whats wrong with my proof on this question can someone take a look

ig the grader wants me to show topoogical equivalence a different way, but I couldve sworn I've seen it done the way I did many times

for any open U in tau_1, and any x in U, there exists V in tau_2 such that V\subseteq U and x\in V and vice versa

[little hemlock]

if you're talking about the note they left, it looks to me like they just repeated exactly what you wrote, but in a different way.

[weak narwhal]

I think the graders saying show that a basis elt in T_2 is the union of basis elements in T_1 and vice versa, which is slightly different from my method no?

but should be equivalent

[little hemlock]

once you show what you wrote, what the grader wrote is pretty clear. Its just one little extra step

[weak narwhal]

ah true i guess i can just show it explicitly

thats so dumb

[little hemlock]

they took off points for that?

[weak narwhal]

correct

[little hemlock]

[weak narwhal]

this is the method I see every person use ever

[weak narwhal]

$\exists U_x\in \tau_2$ such that $x\in U_x$ and $U_x\subseteq B$ for each $x\in B$ so $\bigcup_{x\in B} U_x=B$

[gentle osprey]

Dpao

[little hemlock]

yea

[weak narwhal]

alright fuckin a

I literally did this style of proof one other time

and the grader took off for it, but I wasnt as explicitly clear what i was doing

got points back

now it happened again 💀 ig they already forgot that this is sufficient

@little hemlock the grader took off points cause when i wrote sine has zeroes at x=n\pi, i didnt specify that n was a fucking integer

[little hemlock]

ah okay, damn. imho that's slightly more fair, but i still would've given benefit of the doubt for something like that

[weak narwhal]

i guess but using n to denote integers, esp in the context of roots of a trig function, seems like it should be obvious

i will say this class allows u not just to request regrades but to do redos so i cant complain too much

[little hemlock]

yea i don't necessarily disagree, but in general uninstantiated variables make people feel uncomfy.

[weak narwhal]

yea i do usually specify

usually to make it clear N, Z, etc

[stone pine]

can two knots with the same knot group have different crossing numbers?

[coral pawn]

How does one realize a delta complex structure on A such that this structure is a substructure of the delta complex structure on S^2 indicated in 7?

[coral pawn]

I figured it out

Didn't need to put a delta complex structure on it

[gritty widget]

I need to explicitly write a homotopy between $f\cdot e_q$ and $f$, such that $f\cdot e_q \simeq f$ where $e_q$ is the constant path at point $q$ and $f$ is a path between $p$ and $q$. This seems extremely trivial, yet I'm not familiar enough with homotopy to do it, if anyone could help me understand

[gentle osprey]

Entelechy

[gritty widget]

nevermind I got it

[cursive vigil]

Homotopy is cool lol

[haughty wave]

Well you can sort of squeeze the first half of f into a point and stretch the 2nd half of f out.

So for example, f(x, t) = f(max(x - t, 0)) could accomplish the squeezing for t < 1/2.

[mint sage]

How to do this? And how to deal with a topological group generally?

[lunar yoke]

generally its very helpful to note that multiplication by a fixed element is a homeomorphism G -> G

[fresh mirage]

hey guys i couldnt find this on google for some reason

how is the metric d_2 defined on C([0,1])?

i know for $$x,y \in \mathbb{R}^n$$ it's just $$ d_2(x,y) = ||x- y||$$

[fresh mirage]

not sure if d_2 defined on C([0,1]) is just a notation thing or if its related to the above definition

oh i found it never mind

ignore the above guys

$$ d_2(f,g) = \sqrt{\int_0^1 \left( f(x) -g(x) \right)^2 \ dx $$

[gentle osprey]

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

[fresh mirage]

just thought i'd write it out because latex is fun :_)

[bright acorn]

So, I am trying to prove the following:

If $A \subset X$ is a contractible subspace of a topological space $X$ and $i : A \rightarrow X$ is a cofibration, then the quotient map $\pi : X \rightarrow X/A$ is a homotopy equivalence. I was able to prove this in the case where $i$ is the inclusion map, by taking any homotopy $H : A \times [0,1] \rightarrow A$ between the identity of $A$ and the constant map wrt a given point $a_{0} \in A$, then composing this with the inclusion to get a homotopy between the inclusion of A into X, applying the homotopy extension property wrt the identity of $X$ and working from there to find a homotopy inverse to $\pi$ via standard arguments related to the universal property of the quotient map.
\
\
The fact that $i$ is the inclusion is then crucial in this argument I just presented and I don't really know how I could generalize this to a general cofibration between $A$ and $X$, any ideas?

[gentle osprey]

MISTERSYSTEM

[gritty widget]

rate

https://media.discordapp.net/attachments/359052581022203914/961736207576997928/image0.jpg

[empty grove]

Cofibrations are always inclusions, but I don't think you need the map to be an inclusion for this to work. A homotopy H: A × I → A (rel a_0) gives a homotopy iH: A × I → X (rel i(a_0)) which collapses everything to i(a_0)

[bright acorn]

wait, are you using inclusion in the sense of being an embedding (i.e being injective and a homeomorphism between its image) or in the sense of really being the inclusion map of a subspace into the underling space?

Because yeah, I am aware of cofibrations being embeddings, but idk if every cofibration is in fact given by an inclusion.

[lunar yoke]

no you could just compose with some homeomorphism

oh wow i finally reached "very active"

anyway your statement and the proof can be found in hatcher, 0.17

[bright acorn]

Yeah I am aware of that, but the point is that I was trying to prove this by myself

[lunar yoke]

fair enough

[bright acorn]

Anyways, I think know how to prove the more general statement.

Thanks!

[lunar yoke]

if you're interested in cofibrations, the above statement follows immediately from some other very useful lemmas

first, cofibrations are stable under pushouts

this means that if you have a pushout
A -> X
| |
B -> Y
where the top map is a cofibration, then so is the bottom map

second, in the above case, if the left map is a homotopy equivalence, then so is the right one

now you can apply this to the pushout
A -> X
| |

• -> X/A

and you're done

[bright acorn]

Tbh, I am still not so used to category-theoretic arguments.

Which is a really bad thing

[lunar yoke]

they're the best kind of arguments

[bright acorn]

Because homotopy theory uses a lot of category theory.

[lunar yoke]

true

[bright acorn]

And I am having kind of a hard time following May's concise course because of this.

[bright acorn]

Ok, here's a slightly crazier question.

Suppose we are given a group G. Is there a way to define a topology on G in such a way that for any topological group H, the continuous maps from G to H are precisely the homomorphims from G to H?

I.e, for any topological group H, we want to define a topology on G such that f : G -> H is continuous iff f is a homomorphism.

[gritty widget]

I rate this T_2/paracompactness

[coral pawn]

(S^2, A) where A is a finite set of points is a good pair right?

[gritty widget]

i'd hope

[coral pawn]

Here's the definition. If by neighborhood they mean an open set, then it is pretty easy to see that it is a good pair

If they require neighborhoods to be connected, then this is not true

The reason why I'm questioning this is because we were given this problem

And if its a good pair, then b follows trivially from a

As H_n(X,A) = reduced(H_n(X/A))

[gritty widget]

i see no reason to require connectedness

[coral pawn]

Okay cool

Thanks

[coral pawn]

So we can use this right?

[gritty widget]

yeah

i think that's a proposition a little later in hatcher

[coral pawn]

Oh you know what I think this problem set was assigned before we did that

That would explain why its listed as two separate parts

Doesn't mean I'm not gonna use the proposition though

[gritty widget]

based

[gritty widget]

thanks

[gritty widget]

discrete maybev

[gritty widget]

oh

like the cts maps from G to H are nothing more than homomorphisms

this might not exist for every group H im feeling

[gritty widget]

like a baby counter example im thinking

is having f a trivial homomorphism

if H is a hausdorff space then we are fkd i think

or wait

nvm

[golden gust]

I don't think so, because the identity G --> G is a group hom but it's not continuous if the codomain group has a finer topology than the domain group. so the only candidate is the discrete topology, but then every map is continuous without necessarily being a group hom

[cursive vigil]

@coral pawn would you be opposed to infinitely many bananas

[limber ravine]

So, suppose X is the discrete topology of {a,b}

An open cover would be ${\varnothing, {a}, {b}}$?

[gentle osprey]

mns

[limber ravine]

Do I need to put the empty set in every cover to make it an open cover?

I mean

[gritty widget]

you could, but it's not like you need to. no elements are gonna be in there any time soon

[limber ravine]

so

when I do the union of such a cover, the result must be {a,b} or the space {varnothing, {a}, {b}, {a,b}}?

oh

nvm

I just need to have X (the space) as a subset of that union

so {a,b} with the discrete topology is compact since {a,b} is an open cover and {a},{b} is another such cover contained in the former

{{a,b}} and {{a},{b}}

[gritty widget]

are you trying to show this space is compact by writing down all the open covers?

[limber ravine]

Yeah, I read the definition and I am trying to make sense of it

[gritty widget]

there are easier ways to prove spaces are compact than writing down all of their covers

for example, finite spaces, no matter the topology, are compact

[limber ravine]

that's the first exercise

Moreover, the subcover must be open as well?

like with the indiscrete topology

[gritty widget]

yes

[limber ravine]

of {a,b}. An open cover would be {{a,b}} but another subcover is {{a},{b}} but it isn't open

[gritty widget]

because you started with an open cover, if you pick away some of its sets, those will still be open

[limber ravine]

uh

[gritty widget]

so?

this isn't even a subcover

{a} and {b} are not in {{a, b}}

[limber ravine]

I didn't understand

So for the indiscrete topology of {a,b}, i.e., {{}, {a,b}}. An open cover would be {{a,b},{a,b}}. Then a subcover is {{a,b}} which is open. Therefore {a,b} with the indiscrete topology is compact ?

[gritty widget]

your argument is fine because the only open cover in this case is {{a, b}} (note {{a, b}, {a, b}} and {{a, b}} are the same thing), but you can't just show a space is compact by showing one open covering has a finite subcover

[limber ravine]

I see

[gritty widget]

if you could, every space would be compact

[limber ravine]

ikr

that's why I need to study the countability axioms

[cursive vigil]

Hi μωμ

[coral pawn]

Yes

[cursive vigil]

Nooooo

[vast estuary]

Hey

What's an example of a pseudocompact space which is not countably compact?

How should I go about thinking of one such example?

[gritty widget]

@vast estuary Let X = N and U is open iff U is empty or 0 is in U

[cursive vigil]

Uh

[limber ravine]

Hey guys, just want to confirm if the arrow and R_T^1 are not compact

[limber ravine]

hum, I think R_T^1 is compact.

Let T be an open cover for R_T^1. Take A in T. So A contains finitely many points. Since T covers R, consider B such that B contains all those finitely many points. But then A U B = R and {A,B} is a finite subcover

[gritty widget]

you need to tell us what "the arrow" and "R_T^1" are

[woven sundial]

what do people think of sutherland's intro to metric and topological spaces as a first topology text?

[limber ravine]

Arrow: is the topology from [0,infty] of all intervals of the type (a,infty) with a >= 0

R_T^1 is the topology on R of all complements of finite sets

[golden gust]

the complement of any open (a,infty) is [0,a], try to prove that these are compact

then you can use the open cover definition

[frank bolt]

The proof seems to be that it's true because it is. Is there some nuance I'm missing?

Oops my purple note is wrong, disregard. Should say simple arc

[gritty widget]

What do you mean?

Here's (imo) a simpler proof

There exist a simple chain of open balls contained in Omega connecting x and y

Say B_1, ..., B_n

Take x_i in the intersection of B_i and B_(i+1)

And let x_0 = x, x_n = y

Connect x_i with x_(i+1) by a segment

Since those are always contained in an open ball, which is convex, we are allowed to do that

Since the open balls form a simple chain, we need the polygonal arc just defined to be simple

[gritty widget]

You can prove that the set of points which can be connected by a simple chain with x (with respect to the open balls contained in Omega) is clopen in Omega, hence equal to Omega

[weak narwhal]

@little hemlock i think the grader is at it again

Compact + possibliy infinite collection has empty intersection => contrapositive, theres finite subcollection not satisfying fct
not compact=> contrapositive, there is a collection of closeds with null intersection that satisfies FCT

did i screw something up or is that right

[little hemlock]

yea, you got it right

@weak narwhal

[weak narwhal]

this is getting ridiculous at this point

i would gladly email the prof that the grader needs to be more careful--its just that the grader may very well be the prof

and saying "you keep grading wrong" is gonna get a negative response no matter how i phrase it

[frank bolt]

I mean that the proof hardly seems like a proof. "Suppose the supremum is not 1. This is a contradiction because the supremum must be 1."

I think I'm just not understanding it

[gentle osprey]

AstroCode

[vast estuary]

Interesting. Thank you!

[gritty widget]

I think in some cases you just have a space and think about its properties and see that it has this but not this property
For counter-examples like that it's useful to check out pi-base

You know, instead of trying to study if particular property follows the other, you study particular space

[gritty widget]

They clearly want you to take some open ball around p(t_0)

Honestly, their proof is kind of sloppy

I just don't like it

[gritty widget]

Here's the result I used before, the proof is wrong but you can fix it

(second part of it)

[vast estuary]

I'm trying to see why this is pseudocompact, firstly

We need continuous functions to be bounded

[gentle osprey]

Hausdorff

[gritty widget]

Suppose that f(x) =/= f(0) for some x

[vast estuary]

Okay

[gritty widget]

Take open U containing f(x) but not f(0)

You see now?

[vast estuary]

Well then the pre-image of U under f is empty, right? No other option

[gritty widget]

But it contains x by assumption

[vast estuary]

which is a contradiction

So f(x) = f(0) for all x

[gritty widget]

Yeah, so f is constant

[vast estuary]

Thanks!

[gritty widget]

Np

[gentle osprey]

Hausdorff

[vast estuary]

Yeah clearly this has no finite subcover

[gritty widget]

Yep

[frank bolt]

Thank you!

[gentle osprey]

DarQ

[broken nacelle]

wait, no

[rain ether]

O is a subset of the power set of X

[broken nacelle]

hmm

it can't be any subset of the power set

[rain ether]

Wdym?

[broken nacelle]

[pearl holly]

it has to satisfy the axioms

[broken nacelle]

I was wrong

[broken nacelle]

for example take $\mathcal{O}={\emptyset, {1}, {2,3}, {1,2,3,4}}$ and $X={1,2,3,4}$

[gentle osprey]

DarQ

[broken nacelle]

$\mathcal{O}$ is obviously a subset of the powerset of $X$

[gentle osprey]

DarQ

[rain ether]

[broken nacelle]

yeah, I get that now

[rain ether]

Did you want to say something else with your example?

[broken nacelle]

I was explaining what I meant here

[broken nacelle]

this is quite confusing

can we say a set is open in some topology when it isn't actually open?

[unreal stratus]

I don't see what you mean by "actually" open

A set isn't inherently open, only relative to some topology

given one set X there will in general be various different topologies you can place on it, so a given subset of X may be open in one topology and not in the other

(I mean that's just exactly what it means for the topologies to not be the same)

[bright acorn]

Suppose that we have a topological space $X$ and subspaces $A,B \subseteq X$ which are homeomorphic, then is it true that the quotient spaces $X/A$ and $X/B$ have the same homotopy type?

[gentle osprey]

MISTERSYSTEM

[bright acorn]

There is one case where it is easy to see this is true, if $A$ and $B$ are homeomorphic and moreover, we have find a homeomorphism $f : X \rightarrow X$ such that $f(A) = B$.
\
\
Because in this case, we can define a map $\tilde{f} : X \rightarrow X/B$ which takes a point x to the equivalence class of $f(x)$ modulo $B$, we can then use the universal property of the quotient by $A$ to induce a homeomorphism between $X/A$ and $X/B$. (so in particular, they have the same homotopy type).

[gentle osprey]

MISTERSYSTEM

[bright acorn]

But is this true in general?

[swift fjord]

This seems wrong. Maybe smth like this? Consider two disjoint intervals, say [0,1] \coprod [2,3]. Consider the subspaces {1,2} and {0,1}. Clearly these are homeomorphic as finite discrete subspaces, but one quotient is contractible while the other isn't (and is not even connected)

[bright acorn]

Ah, [0,1] coprod [2,3] quotient out by {0,1} would be the disjoint union of S^1 and [2,3], right? While the quotient of [0,1] coprod [2,3] by {1,2} is homeomorphic to [0,3], which is contractible.

That's neat

[swift fjord]

Yea

Exactly

[bright acorn]

That's a nice counterexample

Thanks

[swift fjord]

Np

[bright acorn]

I had something else in mind

[swift fjord]

Fwiw I kept trying to come up with examples on connected spaces with no luck

So maybe with some extra conditions (maybe u need cw) this could be true?

[bright acorn]

What if we take the plane R^2 and the open disk (let's call it D^2) in R^2. Then the open disk is homeomorphic to R^2. I was thinking about taking X=R^2, A=R^2 and B=D^2. Then, we know that R^2/R^2 is just a point. What I was having trouble with, was trying to specify R^2/D^2 up to homotopy type.

And I was trying to use this as a counter example

I mean, the quotient of R^2 by the open disk is not even T_1.

[swift fjord]

Yea i thought of something like that too, but I don't really know what that quotient would look like

Might still be contractible

[bright acorn]

Yeah, and I spent so much time thinking about this example with no luck.

This also made me think about another question

In general, how hard is it to compute the fundamental group of the quotient of a space?

I know some stuff related to taking quotients of (nice) spaces by nice group actions

but that's about it

But are there more general results?

Knowing about how singular homology and cohomology behave under certain quotients would be nice too.

Btw, I am aware that this might not be such a well posed question.

[magic geyser]

what if you take the big space to be D^2 wedge S^1. then take A= the S^1. then take B = the S^1 thats the boundary of the D^2 .

[supple locust]

$CP^{n-1} \hookrightarrow CP^n$

[gentle osprey]

bert

[supple locust]

as [a0,...,an-1] -> [a0,...,an-1,1] right?

why does this induce isomorphism on H^i for i <2n-2

[supple locust]

(I am reading example 3.40 in Hatcher)

[swift fjord]

Oh that's a nice example

[swift fjord]

For singular homology I know there's some conditions for say CW complexes where the homology of the quotient coincides with relative homology

[gentle osprey]

Hausdorff

[vast estuary]

Note that a zero-set is just the collection of zeros of some real-valued continuous function in this case

[gentle osprey]

Blitz

Blitz

[vast estuary]

Why is that true, i.e., why are the f_i's continuous

[gritty widget]

@vast estuary because X is normal

In a normal space there exists a Urysohn function separating any two disjoint closed sets

[fading vale]

There's an LES of relative homotopy groups for nice spaces (for example CW complexes and subcomplexes)

[shadow charm]

i was thinking about directed graphs earlier and wondering if there was a use in studying "continuous versions" of them. By that i mean that it feels like there must be useful ways to encode a weighted direction to a neighborhood (simplest one would be to consider an implicit hypersurface in R^n and to take directional derivatives) and im wondering if there are ways to associate digraphs to such structures to get information on them

particularly to simplify computations on very large digraphs

[magic geyser]

for a manifold example take S^1 x R^2. A = S^1 x p, B = p x S^1

[bright acorn]

This example seems a bit harder to picture, how does the quotient of R^2 by S^1 look like? But yeah, we could also use a fundamental group argument to see that since S^1×R^2/S^1×{p} ≈ R^{2}, then its fundamental group is trivial, but S^1×R^2/{p}×S^{1} ≈ S^1 × (R^2/S^1) has non trivial fundamental group and these spaces can't be homotopy equivalent.

[magic geyser]

R^2/S^1 looks like S^2 wedge R^2

[gritty widget]

Hey, does anyone know what finite space has a fundamental group of Z2?

That is, {0, 1}.

[jaunty pumice]

hey im in an undergrad topology class and working on a presentation on higher homotopy groups. the topic itself seems pretty broad and mostly above my level. are there any interesting results they give that are appropriate for an undergrad topology class?

[cedar pebble]

I mean you can try to give some kind of an overview of what homotopy groups of spheres look like, but certainly not with any proofs

higher homotopy groups are insanely hard

[jaunty pumice]

i've realized haha, really got the short end of the stick with this one

[cedar pebble]

cool so you can try to survey what is known about homotopy groups of spheres and their stabilization

in order of difficulty it's like
π_k(S^n)=0 for k<n
π_n(S^n)=Z
Freudenthal suspension theorem and stabilization
stable homotopy groups of spheres (a lot more is known about this)
π_k(S^n) for k>n (very little is known in this range)

probably very little you can say about the unstable range since the techniques there are insane

very little you can say about how you actually compute things in the stable range, just that things stabilize

[jaunty pumice]

ooh interesting. i haven't heard of the freudenthal suspension theorem, so this does seem like something interesting to mention

[cedar pebble]

yeah so this is the theorem that tells you that the homotopy groups of spheres stabilize in some range

(or more generally for other spaces)

so this is the theorem that is the starting point for stable homotopy theory

[jaunty pumice]

alright, i'll see if i can work with this thanks. thankfully i dont have to go in depth or anything

[cedar pebble]

yeah

you can maybe mention at the end why stable homotopy groups of spheres matter, other than the obvious motivation that they tell you a lot about the homotopy groups of spheres in general

e.g. the stable homotopy groups of spheres classify framed cobordism classes of manifolds

[jaunty pumice]

that might be hard to explain to the class in the alotted time, cobordism isn't something we have covered, but i'll try to get that in there if i can.

[gritty widget]

It'd be weird if you did cover cobordism

[shadow charm]

Just putting this out here again in case someone smarter than me knows if examples

[shadow charm]

Not exactly but interesting to know there’s another field for « directed spaces »

Thanks

[fading vale]

if you guys have talked about H spaces maybe you can do the puppe sequence

I can elaborate on it a bit if you want

or the long exact sequence of a fibration

[jaunty pumice]

sure! i've mainly been going over homotopy groups of spheres, but i am interested

[fading vale]

A bit late lol but if you have seen an H-space you know that it is a space X equipped with a map h: X x X -> X that obeys nice homotopical properties. A fairly direct consequence of this is that for any other space Y, [Y, X] obtains a group structure: given two classes f: Y -> X and g: Y -> X, we ave a map fg: Y -> X x X, and composing with h you get another map in [Y, X].

Similarly there is a notion of co H-space, which is a space X equipped with a map h: X -> X wedge X that obeys nice homotopical properties, and in this case for any space Y, [X, Y] obtains a nice group structure. The most important examples of H and co H-spaces are loop spaces and suspensions respectively (with suspensions the map X -> X wedge X is given by quotienting out the middle to a point for example).

So lets say i have a map of CW complexes f: A -> B. Then if C(f) is the mapping cone of f (take the mapping cylinder of f and then quotient out the top to a point so it looks like a cone) I have a natural map B -> C(f) given by sending B to the copy of B in the bottom of the cone. So i have a sequence A -> B -> C(f).

[gentle osprey]

Moth

Moth

[fading vale]

this is called the coexact puppe sequence, coexact meaning that if I apply Hom(-, X) to this sequence (recalling that this will reverse the direction of the arrows) we get an exact sequence $\ldots \to [\Sigma A, X] \to [C(f), X] \to [B, X] \to [A, X]$

[gentle osprey]

Moth

Moth

[fading vale]

Dually we have a similar exact sequence using loop spaces and mapping fibers but I chose to go with this one because idk if you know what a mapping fiber is I think the cylinder is taught more commonly in these kinds of classes

but you can read that one online

[gentle osprey]

Moth

Moth

Moth

[cursive vigil]

@fading vale who is simone lol

[fading vale]

Maybe not in this channel cause I was writing something up

anyway this was kind of a lot to throw at you and I have no idea if its actually fit for your class/how much time you have but I think this is a potentially very good choice. Higher homotopy groups are in general very not nice unlike homology. However these sequences (the puppe sequences) are one of the rare instances in where they are nice: from them we can see that generally computing the homotopy type of the total space of a fibration in terms of the base space and the fiber is generally not super bad (covering theory is sort of a nice case of this for n = 1 actually)

Doing the same thing for singular cohomology is a lot harder and requires use of spectral sequences

[vast estuary]

Thanks, didn't know that

[gritty widget]

what is example of weak homotopy equivalence that isnt homotopy equivalence

are the examples very pathological

nvm

found examples

i wonder what the motivating example was for weak homotopy equivalence

if not example why does it exist as a notion

[hidden crag]

@jaunty pumice I’m actually in the the exact same situation as you, I also have higher htpy groups as a presentation topic with it technically being way above my level :D feel free to ping me if you find something nice to present and I can do the same for you if you want me to

Currently reading what all the others already suggested

[gritty widget]

ping@me also

im interested in topic

[vast estuary]

Additionally, you need to argue by Weierstrass M-test and uniform convergence that f is continuous

Then, the proof is complete

[pearl holly]

That was very nice moth

[lunar yoke]

Nice writeup, thanks

[lunar yoke]

oof sniped

[pearl holly]

Lmao

[gritty widget]

That was implicit but yes

[gritty widget]

how can i see that

suspension spectra are connective?

[broken nacelle]

nvm

I answered myself while typing

[vast estuary]

I don't understand the remark in yellow, could someone help?

[gritty widget]

@vast estuary what does completely seperated mean

[vast estuary]

Well there's an Urysohn function

Basically if A, B are completely separated in X

There is a continuous function from X to [0,1], which takes 0 on A, and 1 on B

[gritty widget]

Can it be equal to 0 outside of A?

[vast estuary]

Yes

[gritty widget]

A is contained in f^-1(0)

B is contained in f^-1(1)

Both are closed in S

So closed in X

@vast estuary

[vast estuary]

Ah! Right, that's very neat and much simpler compared to the proof I came up with

[vast estuary]

Also see the first line here

Suppose you have a metric space X, and two disjoint closed sets A, B. A,B are zero sets of some continuous functions f, g on X. I want to show that A, B are completely separated. Taking h = |f| - |g| works, right?

h is negative on A, and positive on B. Then we can modify h appropriately to get an Urysohn function

[gritty widget]

Hmm... not really

[unreal stratus]

There is fairly canonical example of a Urysohn function

consider distance from the sets A,B and modify appropriately

[vast estuary]

Why?

[gritty widget]

Why would it work

[vast estuary]

Oh you're right it doesn't work

[vast estuary]

Well that's how you'd show that every closed set is a zero set in the first place, right?

The distance functions do it

[unreal stratus]

I didn't even think of that but fair enough

[gritty widget]

Distance functions also "completely separate" closed sets in a metric space

[unreal stratus]

ye

[vast estuary]

Okay wait I've a brilliant idea

[gentle osprey]

Hausdorff

[unreal stratus]

that's exactly what i mean

yes

[gritty widget]

They also serve as a kind of, decomposition of unity

[vast estuary]

Amazing! Yay

[gritty widget]

So it's useful in paracompactness shenanigans

Similar

[vast estuary]

The author says, "every closed set is a zero set. SO THAT any two disjoint closed sets are completely separated"

but in our proof, we directly produced an Urysohn function

What is the author's intention behind noting that closed sets are zero sets? How does that help?

[unreal stratus]

ig when you have two zero sets you can always do the f/(f+g) type construction?

Or perhaps |f|/(|f|+|g|) to be more careful (or equivalent)

[gritty widget]

True

[jaunty pumice]

super late but thanks for the writeup @fading vale. was really clear and interesting

[jaunty pumice]

so far my presentation focuses on homotopy groups of spheres and briefly touches on stability. my presentation only has to be 15 minutes so i think touching on those would be enough.

[fading vale]

https://math.stackexchange.com/questions/2876463/why-is-it-that-homotopy-is-better-described-by-weak-equivalences-than-by-homotop Here is an MSE post that might be helpful

[gritty widget]

thanks

[cold vine]

CW Complexes (& Whitehead): What does the first sentence here mean? This is a proof used in Whitehead's Thm in Dieck. (8.4.1 is the compression lemma: f:(X,A) -> (Y,B) is homotopic rel A to a map X-> B with dim constraints)

I guess it's to do with f: X -> Y being equal to f:X -> M_f -> Y with first map being an inclusion and the second a deformation retraction and thus a homotopy equivalence, but it's a bit too shorthand

Also for a second point: There's a theorem in Spanier which is basically the exact same theorem, but the map h is n-equivalence instead of n-connected. What is the connection here?

[limber ravine]

which definition for compactness do you use the most?

[gritty widget]

Why do you ask

I know 3 differet ones equivalent for metric spaces, all of which are useful

1. open covers have finite subcovers
2. sequences have convergent subsequences
3. no closed subspace homeomorphic to natural numbers

[golden gust]

that third one is cool

I guess in practice you test it by testing (2) so it's not very useful? but still I've never really thought about it

[limber ravine]

wondering

[unreal stratus]

There's also in terms of finite intersection property which is basically dual to open cover way (and works in any top space)

a collection of closed subsets has nonempty intersection iff the interaction of every finite (non trivial) subcollection is nonempty

[fresh mirage]

nerd

[unreal stratus]

Lol hi

[fresh mirage]

xddd

[unreal stratus]

long time no see

[fresh mirage]

yh idek why this channel was open but it turned out it has u

❤️ ❤️ ❤️ ❤️

[unreal stratus]

❤️

[limber ravine]

yeah that's the one I was studying

[unreal stratus]

Noice

[tired quail]

Hello there and good evening! I am currently looking for recommendations on books on general topology, and i mean only general topology, i plan studying algebraic and differential topology later, i already know some basic definitions of topology like certain properties of topological spaces, homeomorphisms and embeddings but i would like to go deeper with my studies, i have no problem with lengthy and terse books, in fact i would much rather prefer those, i like books that keep me busy and are challenging to my brain, so if anyone has any recommendations on such a book i would love to hear it. Thank you very much in advance.

And i am very sorry if my grammar is incorrect, i am not from an English-speaking country. I also apologize if my way of writing seems pretentious, i am trying my best to write in a formal manner due to the fact that i'd like to seem polite, i have recently joined this serve so i don't want any bad first impressions of me.

[hollow harbor]

the book that is recommended most often is Munkres' topology book. it's very good.

i also love seebach and steen's "counterexamples in topology" cause it's fun, though it's not a proper book on topology.

[tired quail]

Hm i'll definitely look into it then! Thank you very much, if anyone else has any other recommendations i'd love to hear them but i'll check this one out first.

Thanks!

[hollow harbor]

It's pretty easy to find free PDFs online, so definitely give any book a test run like that even if you plan to buy a physical copy.

[river granite]

I'm not sure which's your native language but Munkres' topology book has a pretty good Spanish edition as well

I like Willard's book over Munkres but it's a tad bit more abstract (though not lacking in exercises and examples); it also introduces nets and filters in the main text which can be useful to prove some things

[tired quail]

ah don't worry i can read english books pretty well so it's okay if it's only available in english.

hm i'll certainly check willard's too but currently Munkres' does seem interesting

[gritty widget]

lee's introduction to topological manifolds has some nice general topology content, and then about halfway through it becomes a more algebraic-topological kind of book

[bright acorn]

Suppose that we have a topological space $X$ and a group $G$ acting on $X$ via homoeomorphisms, i.e we have a morphism of groups $\psi : G \rightarrow \text{Homeo}(X)$. Then, there's naturally an induced group action of $G^n$ on $X^n$ given by:
\begin{align*}
\psi^{n} : & G^{n} \rightarrow \text{Homeo}(X^{n}) \
g = (g_{1}, \cdots, g_{n}) & \mapsto \psi^{n}{g} \in \text{Homeo}(X^{n})
\end{align*}
Where $\forall x = (x{1}, \cdots, x_{n})$ we have:
$$
\psi^{n}{g}(x) = (\psi(g{1})(x_{1}), \cdots, \psi(g_{n})(x_{n}))
$$
Then, is it true in general that the spaces $(X^{n}/G^{n})$ and $(X/G)^{n}$ are homeomorphic ?

[gentle osprey]

MISTERSYSTEM

[shut moat]

Why wouldn't this be true? The orbits of psi^n all factor into products of the orbits of psi

[bright acorn]

The main problem would be stablishing that the map that takes an orbit of psi^n into the product of orbits of psi is in fact a homeomorphism.

And not only a bijection

[shut moat]

a set is open in (X^n/G^n) if its preimage under the quotient map is open. If you take the preimage of such a set by your bijection, then you can argue that that that also has an open preimage because they are essentially the same quotients

This is a bit informal but hopefully you get what i mean

jesus
https://math.stackexchange.com/questions/228687/products-of-quotient-topology-same-as-quotient-of-product-topology

[sharp frost]

Hi I'm having trouble reconciling an easy result with an explicit example. If we have pointed topological spaces $X$ and $Y$ with an inclusion map $i:X\hookrightarrow Y$ and a retraction map $r:Y\to X$ with $r\circ i=id_X$ then the functorality of $\pi_1$ implies that we have a short exact sequence $$N\longrightarrow \pi_1(Y)\stackrel{r_#}{\longrightarrow} \pi(X)$$ where $N=\ker{r_#}$. If we take the explicit example of a torus being retracted into the figure 8 (https://www.youtube.com/watch?v=j2HxBUaoaPU) then $\pi_1(Y)=\mathbb{Z}\times\mathbb{Z}$ and $\pi_1(X)=\mathbb{Z}\mathbb{Z}$ but is it even possible to have a surjective homomorphism from $\mathbb{Z}\times\mathbb{Z}$ to $\mathbb{Z}\mathbb{Z}$? this feels impossible

[gritty widget]

it says punctured torus

[sharp frost]

o

whats the fundamental group of a punctured torus

[honest narwhal]

Punctured torus is homotopy equivalent to wedge of two circles @sharp frost

[sharp frost]

oh that would make sense ty

damn it so this is a bad example then because X and Y are homotopy equivalent

[cold vine]

I'm gonna second that Munkres is definetely the clearest approach for general topology! I used a different book on my courses, and every time something was confusing I knew I could check out if Munkres would have the same thing and be certain it would help.

[gritty widget]

It's old and his approach to simplexes is horrible, but I like Dugundji a lot

His books make for a nice read

[empty grove]

I heard about this book from some of my juniors too, who said it's way better than Munkres

[radiant glen]

hi @midnight echo,
i'm reply in response to your question about compactness in a metric space posted on a math help channel which is now closed.
#help-15 message
you don't need surjectivity of $f$ for the conclusion of part (2) to hold.

[gentle osprey]

vin100

[mint sage]

X a connected totally ordered set, does its section S connected？

[gritty widget]

Is there some nice way to prove that a union of a simple chain of n-cells is an n-cell?

This is not true actually

I think it's true if I assume the n-cells are convex and embedded in R^n

[bright acorn]

Yeah, the category of topological spaces is weird

I am slowly getting more and more convinced that I should care a lot more about the category of locally compact hausdorff spaces, the category of compacly generated spaces or the category of weakly hausdorff spaces.

rather than working with weirdly behaved spaces

[still vessel]

Can someone help me with trying to understand this definition of a compact space?

A space X is said to be compact if every open covering of X contains finite subfamily that also covers X.

So my understanding of compact spaces is that it is a space which includes all of its limit points ( i dont know if that is correct); but i cant really understand how this definition means the same as the one i intuitively get.

[hollow harbor]

That understanding isn't really correct. R contains all its limit points, but it's not compact. "Containing all your limit points" is something that only makes sense in the context of a bigger space (where those limit points might come from). So this isn't really a workable definition.

For metric spaces, there are two ways of defining compactness. The one you gave is the one that applies to general topological spaces too.
The other one is about limits. It says that a space is compact if every sequence has a convergent subsequence.

In other words, every sequence has a limit point.

[still vessel]

i thought infinity can is also a limit point (while not being a point exactly), which if you include in R it becomes compact, no?

[hollow harbor]

Again, that totally depends on the broader space you're looking at.

The best way to understand this definition is that a compact space can be understood / dealt with by just looking in finitely many spots.

For example, on R, we can take the intervals (n, n + 2) for each n. These intervals cover R, but there's no way to just look at finitely many of them and understand all of R.

[gritty widget]

I think you mean, every sequence has a limit point

And if you were to define compactness in this way, then it's correct for, say, metric spaces

But for general spaces we need to use filters instead of sequences

[hollow harbor]

But in a compact space, this definition tells you that you can "see" your whole space (given some collection of open sets that you want to understand: these could represent sets where some function varies by a certain amount, or open connected components, or plenty of other things) by just looking at finitely many of your open sets.

[still vessel]

hmm, ok so these finitely many subsets are representative of the whole space

which is kind of explained by the "covers all of X"

i dont know, this definition is not intuitive at all, the one with sequences is much more clear what it means

[still vessel]

also, these finite sets cant be open, right? Becouse than you would have an inifite loop of those

[gritty widget]

We can't use sequences in general topological spaces

[hollow harbor]

You're right that it's not intuitive, it takes a while to learn and internalize.

[gritty widget]

You can have equivalent definition with filters, which are a way to generalize sequences, but you need a new definition for that

[hollow harbor]

It's also hard to prove things are compact using this definition.

The benefit is that it's a very very useful definition for doing proofs about compact things.

[gritty widget]

It's a simple definition

[hollow harbor]

And a very general definiton

Yes

[gritty widget]

I think that's the main advantage

[hollow harbor]

I would say that this definition becomes more and more intuitive and useful the more you use it in other proofs. You see why it makes sense.

But it's not like... super easy to prove that in metric spaces the sequence definition and the open cover definition are equivalent.

[gritty widget]

I wouldn't say that "it's not easy"

Just that it's not obvious at all

[hollow harbor]

Yeah. It's not so hard. But the ideas feel weird when you see them for the first time.

(turns out those ideas are super ubiquitous and important, but they're not clear at first)

[gritty widget]

I didn't think about it much when I first took topology, so I didn't have this problem

[hollow harbor]

i'm dodging actually typing them out cause i don't want to remember the details. lol

[still vessel]

i will trust you on this, as i do have some proofs coming up with compact spaces

[gritty widget]

Speaking about compactness, what I think sucks a little is that the construction in the Čech-Stone compactification is almost never given more generally

You can use pretty much the same construction but with different filters and obtain different compactifications

[hollow harbor]

You might be interested to look into the construction of compactifications of locally compact spaces using a correspondence with C* algebras. The stone-cech compactification corresponds to the spectrum of the algebra of bounded continuous functions on your space. Then you can look at subalgebras in between the bounded functions and the functions which vanish at infinity, and see these as different compactifications.

https://en.wikipedia.org/wiki/Stone–Čech_compactification#Construction_using_C*-algebras

then this generalizes to the noncommutative setting, you can look at "multiplier algebras" of nonunital C* algebras (for the case of continuous functions vanishing at infinity on a locally compact space, this corresponds exactly to the stone-cech compactification) https://en.wikipedia.org/wiki/Multiplier_algebra

[gritty widget]

@hollow harbor do you have any, say, paper or monograph about this instead, I hardly like to browse wikipedia

[hollow harbor]

I will take a look

Here are some papers which you could look at, which I also should look at:
https://core.ac.uk/download/pdf/82567939.pdf
https://www.uni-muenster.de/FB10/mjm/vol_4/mjm_vol_4_05.pdf
the intros / background sections of these papers might be helpful.

I don't know if this is recorded in a book anywhere.

Yeah. That's what I'm looking at now and it's just ok

Actually this isn't bad

[shut moat]

oh wow!

[hollow basin]

this is a weird question but - I'm in an algebraic topology course and I need to give a presentation on some topic in or related to the field. Any suggestions for subjects?

[golden gust]

if you like data analysis, maybe topological data analysis? it uses something called persistent homology to analyse the shape of datasets

[hidden crag]

@lunar yoke another one LOL

how about higher homotopy groups

[still vessel]

can someone clear this up for me, i am getting different facts from different sources.

If we say that some family of subsets F covers a subspace A of a topological space, then does this mean that a union of this sets equals A or does it mean that A is a subset of union of F?

[old girder]

A is a subset

[wise sigil]

I think thats the standard interpretation

i know an author that means A is equal to the union though

its best to check the authors convension

[gritty widget]

Honestly, just figure it out from context

[still vessel]

ok, i think that A should be subset so i will go with that

[weak narwhal]

Anyone know how to demonstrate a countable basis around 0 for the K-topology

Not regular at origin since K closed wrt K-topology, so showing separability unfortunately isn’t sufficient

Maybe it’s easy and there’s some reason just taking intervals w rational endpoints and removing K is necessarily countable

[wide kayak]

@weak narwhal intervals with rational endpoints should be countable, even before removing K

you can identify it with Q x Q

and since Q is countable, so is Q^2

it's actually a strict subset of Q^2

because a < b in (a,b)

but I don't quite see why rational intervals (a,b) centered around 0 in R\K forms a basis for the K-topology

I see it's a basis for a topology

[weak narwhal]

every other point but the origin contains only finitely many elements of K, so other points inherit a countable base from the standard topology

so I'm thinking i need only show that 0 has a countable local base and union it in

[wide kayak]

ah

[weak narwhal]

hmm could you say a little bit more about this

[wide kayak]

I meant to say you can identify it with a subset of Q x Q

because an element of Q x Q is a point (p, q), both points rational

an interval centered around 0 and with rational endpoints is also of the form (p, q) but with p < q

so there's a one-to-one function from your set of intervals into Q x Q

[weak narwhal]

ah i see

[wide kayak]

actually you can say more

your interval is of the form (-p, p)

where p > 0 is rational

but again you can realize all of those as a subset of Q x Q

[weak narwhal]

u could prob construct it explicitly by ${(1/n,1/(n+1))\cup {0},{(-1/(n+1),-1/n)\cup {0}\mid n\in \mathbb N}$ right

[gentle osprey]

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

[wide kayak]

I'm confused, that looks like a totally different set than we were just talking about

[weak narwhal]

omfg idk what i was thinking

@wide kayak i already knew that the intervals with rational endpoints is countable for the reason u stated, idk why it wasnt immedialtey obvious that removing points ofc doesnt take away countablility

[wide kayak]

oh, yeah. It could only make it go down

but it doesn't actually since countable with finitely many points removed is still countable

[weak narwhal]

i was thinking about taking intervals between N and unioning it with 0 to get a local base, but this actually may be finer on second look

[wide kayak]

oh, so the ones centered around 0, together with the ones like (1/n, 1/n+1) ?

and the negative ones

I don't think you want to include {0}

[weak narwhal]

the singleton 0 together with the opens between elts in K

its not even a base

the intersection of any two nonequal elements in that set i proposed is {0} which isnt in the set

so picking x=0 we dont have any subset of the intersection containing x

[wide kayak]

I think I need to hear the problem again

finding a countable basis for the K-topology that's centered around 0?

[weak narwhal]

yea, we found one its just $\mathcal B={(a,b)\setminus K\mid a,b\in \mathbb Q}$

[weak narwhal]

i just wanted to point out that this was demonstrably wrong

[gentle osprey]

Dpao

[weak narwhal]

so taking this together with the usual basis of opens w rational endpoints gives a countable basis for the total space

[cold vine]

Hi! This is a part of the proof in Dieck for Whitehead theorem for CW complexes. What does the first sentence "h is an inclusion" exactly mean here?

I know it has something to do with the map h being deconstructed to maps h: B -i- > M_h -r-> Y where i is an inclusion to the mapping cylinder and r is deformation retraction and thus a homotopy equivalence

[wide kayak]

does it mean B can be identified with a subspace of Y?

[cold vine]

That would make sense, dont exactly see why though. Maybe it is just to mean up to homotopy, since it is definetely a composition of an inclusion and homotopy equivalence. Weird way to shorthand that though.

(Is your name a reference to Kayo Dot perchance? :D)

[wide kayak]

yes!!

[cold vine]

Damn! Love them so much! One of my absolute favourites

[wide kayak]

fantastic

I kind of fell out of keeping up with their newer stuff, have a few albums to catch up on

[cold vine]

Yup they definetely arent all consistently as good. The newest album is a bit more like the older stuff. Love Plastic House and Blue Lambency tho. maudlin stuff still some of the best

[wide kayak]

yea BLD is great. Dowsing is probably my favorite. I saw a really positive review of their newest one, should check it out

[gritty widget]

#discussion #serious-discussion or dms tbh

[cold vine]

Ya just a quick chat nw ^^

[weak narwhal]

this is a good one, consider what b\mapsto (b,0) looks like in the mapping cylinder of h

oh i see thats not the issue here

[weak narwhal]

i think its just saying we have a homotopy equivalent inclusion M_h, ill have to check

strange way to phrase it though

[cold vine]

Yes I agree. I think this is right since as we see the homotopy equivalence I think we can regard h as the inclusion of a relative CW complex (M_h, B) and now we can use the compression lemma 8.4.1 to a map (X,Z)->(M_h, B) to gain a homotopic map X -> B rel Z. That's kinda confusing. Still to realize why this gives surjectivity though when Z={0}.

[obsidian socket]

Is there a type of topological space where every closed neighborhood of a set contains an open neighborhood of that set?

[unreal stratus]

Well any such space is either discrete or not even T1 if you consider singletons

[obsidian socket]

ahh that's a good point

[unreal stratus]

Well it's also imply any closed set is open more generally

[gritty widget]

Huh? A neighbourhood by definition contains an open neighbourhood

[gritty widget]

Every such space? Since it's a neighbourhood, it needs to contain an open neighbourhood by definition

Is there a name for a retraction which maps the complement of our retract into a specific subset of our space?

[unreal stratus]

Sorry i was sleepy idk why i said that lmao, ig i assumed to avoid triviality they meant smth slightly different by closed nbhd

[gritty widget]

how can i see that suspension spectra are connective?

im confused

we want \pi_k(E)=0 for k<0

if E is the suspension spectra of a space X then

$\pi_k(E)=\operatorname{colim}n(\pi{n+k}(\Sigma^n X))$

[gentle osprey]

lime_soup

[gritty widget]

i think the arguement should be something like, for large $n$ $\Sigma^nX$ is $q$ connected, so $\pi_{n+k}(\Sigma^n X)=0$ for $n+k<q$

[gentle osprey]

lime_soup

[gritty widget]

but I don't see why the colimit is then 0

I'm a little confused

i thought this theorem had restrictions on n>1

[gentle osprey]

Nobody

[gritty widget]

Okay so

$\Sigma^n X$ is at least $n$-connected

[gentle osprey]

lime_soup