point-set-topology

[grave fossil]

SO(n) is compact, so if you embed it it's gonna be a little weird I think?

[gritty widget]

Why weird?

I mean topological embedding btw

Can we use something like $\exp(iA)$ where $A$ is diagonal for this?

[gentle osprey]

Blitz

[gritty widget]

Well maybe not this but I think exponential map might be useful somehow

[gritty widget]

Because we can embedd R into S^1 = SO(2) using exponential function

[vast estuary]

Yep, works, thank you

[gritty widget]

I think I want to embedd R^n into U(n) instead

That's easy though

We can embedd R^n into (S^1)^n into U(n)

I feel kinda stupid now

[fathom shale]

Doing more reading this morning in Ozsvath and Szabo. Wish me luck!

[gritty widget]

No idea what that is but good luck

[shy moss]

you mean the function $f(x)=e^{ix}$? this is not injective

[gentle osprey]

Or x1

[gritty widget]

Generalization of it

Besides, I don't need it to be injective on all of R

If we were able to embedd R^n into SO(n+1) then that'd be pretty strong I think, but U(n) is alright too

I'm basically embedding R^n into (S^1)^n

[marsh forge]

I think or x1 was objecting to the use of embed here

Since the map is not injective

[gritty widget]

I know, like I said, that's not the point

[marsh forge]

No I mean

You keep using the word embed hahaha

[gritty widget]

R and (0, 1) is the same thing

I can use that word

I never said I'm using exponential as an embedding

[marsh forge]

?

[gritty widget]

Just that I use it to form an embedding

using

[marsh forge]

Using

[gritty widget]

This is a crucial detail you missed

[marsh forge]

Using

[gritty widget]

Whatever

If we all know what is meant then this discussion is pointless

[marsh forge]

I genuinely did not know what you mean

[gritty widget]

Oh

[marsh forge]

When we think of exponential as a map R to S^1 I usually think of the universal covering space

Which is certainly not objective

Injective*

[gritty widget]

I meant that I'm mapping R to (0, 2pi) homeomorphically and then use exponential to map it to S^1

[marsh forge]

Ah, sure

[gritty widget]

Yeah, sorry, I thought you were just being picky

Idk if picky is a good word

Bothered with details

[marsh forge]

No I think both me and or x1 had the covering map in mind

Exponential is the way most topology books define it

[gritty widget]

Anyway, in the same way I can embedd R^n into (S^1)^n which then embeds into U(n)

[honest narwhal]

Why are you using the word "embed" for R^n to (S^1)^n? Do you mean cover?

[coral pivot]

he is including R into S^1

[honest narwhal]

Why don't you include yourself into my foot?

[coral pivot]

thinking of S^1 as basically one point compactification

thats not how it works dami

infact thats the opposite

[honest narwhal]

But that wouldn't really be an embedding I don't think

If we're assuming you're not trolling me here

Or wait

No nvm that's not a problem

[gritty widget]

But I didn't say any maps

[honest narwhal]

I was thinking of embedding as topological groups

You're just thinking as spaces

[gritty widget]

Yes, top spaces

[coral pivot]

u said exponential which will make everyone think of covering map

lol

[gritty widget]

I was thinking what kind of spaces have compactifications which are topological groups

So I was trying to see if I can come up with one for R^n

[coral pivot]

just any compactification?

[honest narwhal]

Yeah so, obv every n-manifold contains a chart

And in principle that's something

[gritty widget]

Yes, any compactification which admits a group structure

[coral pivot]

well a compactification requires like, R^n to be dense in it dami

[honest narwhal]

I see, I've never seen the term "compactification" on its own used actually

Usually either one point or one time I saw Stone-Cech lol

[coral pivot]

yeah its like, a compact space Xtilde with an inclusion of X in it

st the image is dense

I wonder if the C* perspective could be helpful to this

[gritty widget]

There's something called Bohr compactification for top groups but it's not an actual compactification

Well actually the question is:
What top spaces/top groups/other are such that some comp./any comp./Alexandroff extension/Stone-Čech comp. admits a top group structure

It'd be cool if any of this could be moved

[warped rover]

So at least with regards to the Stone-Cech compactification, all X not pseudocompact are such that beta X are non-homogenous and hence, do not admit a group structure.

[gritty widget]

I see. So if we go further to metric spaces (say), then only Stone-Čech compactifications of compact spaces can admit a topological group structure

That's pretty cool

So iff such space is a compact top group

[flint cove]

Does anybody kow common notation for the specialization preorder? something like SpP(X)? or SP(X)

[gritty widget]

I am given a covering map $p: S^1 \to S^1, z \mapsto z^4$ and a continuous map $f: S^1 \to S^1, z \mapsto z^5$ and I am asked to determine if there exists a lift $\hat{f}:S^1 \to S^1$ of f

[gentle osprey]

Entelechy

[gritty widget]

but I am a bit struggling on this, I don't know where to start

[marsh forge]

Do you know the lifting lemma

If not, you can try to compute what such a lift would have to be at the level of the fundamental group

[weak narwhal]

can someone sanity check my proof of sierpinski space is contractible

[bitter smelt]

What is quotient space of square either opposite sides identified negatively

Like the mobius band square but with the other edges also identified in the same way

Klein bottle but reverse one of the double arrows

[weak narwhal]

I think its the real projective plane

[bitter smelt]

I can see how this might be simular to antipodal identification on disk

[weak narwhal]

@bitter smelt round it out, connect two opposite sides to get something that looks like wedge of two circles, then glue across the circumference after spinning one of them 180deg

[gritty widget]

what is the free action of S^1 on S^\infty?

I know there is such an action since S^\infty is EU(1)

[gritty widget]

What's S^infinity? Hilbert cube?

[gentle osprey]

Blitz

[marsh forge]

Actually I’m not sure that even makes sense

But it’s a colimit that’s well defined

I am speaking nonsense sorry

[marsh forge]

Sorry why do we know that it is EU(1)?

[cursive flume]

does this seem right?

[gritty widget]

pls try to distinguish your union Us and set Us

[marsh forge]

lol

yeah the handwriting is good modulo that

looks good otherwise

only other comment is to avoid using things like \exists randomly in a paragraph

[cursive flume]

I am unsure about oen step tho

is this surely true?

[marsh forge]

Prove it 🙂

[cursive flume]

if $U_x \subseteq U, \forall x \in U \implies \cup_{x \in U} U_x \subseteq U$

[gentle osprey]

ProphetX

[marsh forge]

It is straightforward to prove this with elements

[cursive flume]

let $x_1,\dots,x_n \in U$. Then $U_{x_1} \subseteq U, \dots, U_{x_n} \subseteq U$

[gentle osprey]

ProphetX

[cursive flume]

therefore $\cup_{n \in N} U_{x_n} \subseteq U$

right?

[gentle osprey]

ProphetX

[cursive flume]

i could generalize this for abritrary index set,but just for the sake of idea

[cursive flume]

I actually omitted a step: $\cup_{n \in N} U_{x_n} \subseteq \cup_{n \in N} U$

[gentle osprey]

ProphetX

[cursive flume]

but $\cup_{n \in N} U=U$, so this finishes it, right?

[marsh forge]

uh

[gentle osprey]

ProphetX

[marsh forge]

I don't think you've really done anything

except re-index the union

?

What I had in mind was to like

[cursive flume]

I mean idk how to formally prove it

[marsh forge]

How do you prove two sets are equal

or sorry

[cursive flume]

a=b iff a subseteq b and b subseteq a

[marsh forge]

how do you prove A is a subset of B

[cursive flume]

let a in A. Then a in B

[marsh forge]

okay

[marsh forge]

do that with the union

i.e.

Let $x\in \cup_xU_x$ and show $x\in U$

[gentle osprey]

i_hate_printers

[gritty widget]

overloaded x

[marsh forge]

oh

oops

[cursive flume]

let $a \in \cup_{x \in U} U_x$. Then $a \in U$

this is what we want

[marsh forge]

Yes

[gentle osprey]

ProphetX

[cursive flume]

right?

hm

i don't see how to do it. I get back to the previous point

it seems ciruclar

[gritty widget]

you can definitely do this

[cursive flume]

I mean $x \in U_x$, therefore $\cup_{x \in U} U_x \subseteq \cup_{x \in U} x =U$

but....

[gentle osprey]

ProphetX

[cursive flume]

this seems literally the proof I did before

i don't see why this helps

[marsh forge]

What does it mean for x to be in a union

Like if x is in A union B

[cursive flume]

x is either in U_1, or U_2, or....

[marsh forge]

what does that mean

okay

great

now use that haha

Suppose $a\in \cup_{x\in U} U_x$. Then what do we know about $a$?

[gentle osprey]

i_hate_printers

[cursive flume]

a is either in U_1,U_2,...

[marsh forge]

Well, $a$ is in $U_x$ for some $x\in U$ right?

[gentle osprey]

i_hate_printers

[marsh forge]

Not necessarily naturally indexed

[cursive flume]

yes

[marsh forge]

Okay

then what do we know about $U_x$

[gentle osprey]

i_hate_printers

[cursive flume]

U_x subset U

[marsh forge]

And finally?

[cursive flume]

a is in U_x for some x, but U_x subset U, therefore a is in U

lel

[marsh forge]

great

[cursive flume]

thanks for help!

[marsh forge]

np

[still vessel]

One of the proofs i am following says this: "$\forall n \in N \exists x_n \in A$ such that $d(x_o, x_n)< 1/n$", where A is a subset of a metric space and $x_0 \in \overline{A}$. Where does this come from? The proof just sort of says it like its obvious.

[gentle osprey]

FireLi0n

[gritty widget]

@still vessel $x_0\in \overline{A}$ means that for any neighbourhood $U$ of $x$, $U\cap A \neq \emptyset$

[gentle osprey]

Blitz

[gritty widget]

Pick an appropiate neighbourhood U for each n

[still vessel]

i understand that, but how do you get to distance being 1/n

is it just, there is always at least one that is exactly that distance?

nothing more than that

[gritty widget]

What are the most basic neighbourhoods of a point in a metric space that you can get?

[gritty widget]

You never said you want them to be exactly 1/n distance apart

[still vessel]

okay yes, less than that

but the point is that you can put any distance there as an upper limit, right?

there will always be a point like that

[gritty widget]

Upper bound

As long as it's positive, yes

[still vessel]

yeah, bound not limit. Okay i guess that makes sense, thanks

[coarse night]

there's compact spaces and para compact spaces. Is there something as para normal space?

[lunar yoke]

every paracompact space is normal

or maybe im not understanding what you are asking

[coarse night]

It's a joke

[lunar yoke]

oof

[coarse night]

not a serious q

[unreal stratus]

Lol

[gritty widget]

paranormal spaces are spaces that are not hausdorff

[coarse night]

non T0, non first countable

0'th countable

[wise walrus]

I have a question about finding normal closures under the free group on 2 elements. How do I find the normal closure of the subset {a^2, b^2, (ab)^3} of the free group <a,b>? This is to subsequently find a presentation for this subgroup and create a corresponding 2-orientable graph.

[marsh forge]

The idea is most likely to avoid actually finding the normal closure

and instead realize that if you force a^2,b^2,(ab)^3 to be 0 you automatically force that on the normal closure

Oh sorry, you want the subgroup generated by those

Then the idea is to make those loops the nonzero ones

i.e. "a" should not be a loop in your graph but a^2 should be

and so on

[wise walrus]

yeah so I can just directly take those set elements directly to constructing the graph?

Well with those elements in particular I feel a bit stuck. the a^2 and b^2 elements feel like they "lock" me into a very rigid structure and I can't find a place to allow the last element

I feel quite locked into this structure, not understanding how to alter it to accommodate for the last element being cubed

[marsh forge]

The way I normally do this by trial in error is a draw a point

and then i draw an "a"

and then I try to draw another "a" and a "b" starting at the end point of the first "a"

and whenever the subgroup generators tell me i need to have a loop

i force it back to the starting point

[wise walrus]

yeah, so I guess just some hardcore trial and error

[marsh forge]

so for the one in the picture

i would draw a

and then see a^2 so my next a has to go to the start

then the ab also forces the b to go back to the start

and you're done

now you do the first b

and you repeat

until youve exhausted everything

its fairly algorithmic (and might actually be decidable?)

[wise walrus]

hmm yeah that does make sense, I guess I just have trouble visualizing everything

[wise walrus]

This seems somewhat like it, but it induces commutativity of the ab within the (ab)^3... which i don't think works. like it would give <a^2, b^2, (ab)^3, (ba)^3> i think

@marsh forge I hope a ping is okay

[marsh forge]

ababab=1 implies babab = a implies bababa=1

[wise walrus]

Oooohhhhh the a^2=1 allows for the inverse to be removed, amazing ! I didnt think of that. Thanks for all the help 😄

[marsh forge]

no worries

you could even think of your diagram as a "proof" of that extra relation

since you did everything right and still ended up with something extra

just be careful to actually check that the algebra works out

[still vessel]

so there is a theorem that says that in an inner product space $(X, <,>)$ for any $U \subseteq X$, the orthogonal complement $U^{\bot}$ is always a closed subspace of X. My question is doesn't that mean that U is also closed as its also a orthogonal complement of some other subset of X?

[gentle osprey]

FireLi0n

[rain ether]

The orthogonal complement of the orthogonal complement is not always the original subspace

It's the closure of the original subspace

[gritty widget]

It doesn't even have to be a linear subspace, just a subset

[rain ether]

Yeah, in any case, the original subset does not necessarily have to be the orthogonal complement of some other subset

[gritty widget]

Hi!

I'm wondering how this is true

For example, take the closed real interval [0, 1]

There is no way 1/2 has a neighbourhood that intersects [0, 1] in 1/2 alone

Alright understood thanks

[weak narwhal]

fwiw accumulation point means the same but sometimes sounds better

[fair idol]

Is $RP^2$ homotopy equivalent to a closed disk?

[gentle osprey]

limbostar

[wise walrus]

I know that RP^2 minus a point/disk is equivalent to S^1

I don't know if that helps lol

I dont... think it's equivalent?

[fair idol]

Yeah that's true, just by imagining removing the 2-cell away from a common CW structure for RP^2. Then keeping the 2 cell should give the interior of the S^1 I think. So yeah

[wise walrus]

the real projective plane has fundamental group Z/2Z, and any closed disk has trivial fundamental group. Ignoring base points because both spaces are path-connected.

[fair idol]

Ahh that's true. That is a problem

I think it must be the disk after identifying a pair of antipodal points along the boundary of the disk and identifying the remaining arcs of the boundary

[wise walrus]

well RP^2 itself is a sphere with all antipodal points identified, so yeah you can't just "collapse" the sphere down to a closed disk because you would be forgetting about some of the antipodal points and subsequently forgetting about some of the structure

@fair idol if you're comfortable with Van Kampen then this is a pretty approachable link that helped me a little while ago with understanding RP^2: https://www.math3ma.com/blog/the-fundamental-group-of-the-real-projective-plane

[gritty widget]

what does it mean that the top and bottom halves can be separated by a left-right mirror reflection?

[snow cobalt]

Hello
What is the usual topology on R² ?

I'm new in topology

Is it the set of unions of open discs ?

[uncut surge]

Yeah; that's the same as the topology induced by seeing it as a metric space with the Euclidean metric

Or any other metric on R^2 since they're all equivalent

[snow cobalt]

Okay so I can consider rectangles instead of discs ?

[uncut surge]

Sure, that's what you get with the infinity-norm

[snow cobalt]

for ||.||_infinite

[gaunt linden]

It doesn't say that; you're trying to continue reading after the end of one line at the beginning of the line you just finished ...

[snow cobalt]

okay thank you

[uncut surge]

Yeah, the norms are different but the topologies are all the same

you're welcome!

[gaunt linden]

It says "top and bottom halves can be separated from each other by a two-sided Möbius strip".

The important point here is, I believe, that if the ambient space is not orientable, then a non-orientable surface can have two distinct sides.

[gritty widget]

If I could explain it in my own words, would the left-right mirror reflection of the cube be the non-orientable ambient space since going through through the mirror would mean you would end up in the flipped surface?

[gaunt linden]

The left-right mirror reflection in the example is not a reflection of the entire cube; it's only there to describe which point on the front face each point on the back face is glued to.

So if you're standing inside the cube and step out through the back side a little left of center, you'll arrive into the cube through the front side, with your head still pointing up but a little right of center. (And you yourself become mirrored by stepping through the glued face, so your subjective experience would be of having stepped into right left side of a different cube that's the mirror image of the one you came from. But "really" according to the mathematical model there's only a single cube that you're now experiencing in a different way).

[gritty widget]

Yes so like this. But Im still confused about how the mobius strip is two sided. Would the mobius strip be 2 sided from the gluing of the two sides of the cube?

[gaunt linden]

The Möbius strip is flat and horizontal, stretching from the front side of the cube to the back. Thus it has a top side and a bottom side.

[gritty widget]

Ah I see thank you!

[gaunt linden]

Alternatively you could stretch a pipe horizontally through the cube; it would then match up across the glued face to become a Klein bottle with a definite interior and exterior.

[solemn temple]

If Y is a Hausdorff space then any two maps g, h : X -> Y are equal if they are equal on any dense subset of X. Is the converse also true? That is, if every two maps g, h : X -> Y that agree on a dense subset of X are equal, is Y Hausdorff?

[lunar yoke]

If we set X = {0,1} with the indiscrete topology, then all functions out of X are continuous. Given two points y,y' in Y, we have maps f,g : X -> Y with f constant on y, and g(0) = y, g(1) = y'. Note that f and g agree on {0}, which is dense in X. Hence f = g. But that means y = y'.

This proves that every such Y is already a one-point space

thus in particular hausdorff

if you mean for a fixed and not arbitrary X, then the answer would be no since e.g. we can take X to be a one-point space where trivially all maps out of it that agree on a dense subset agree on X (the only dense subset is X itself)

[solemn temple]

Wait doesn't X have to be the discrete topology for all functions out of X to be continuous?

[lunar yoke]

oof yeah you're right

been a while since ive done these pointset things

[solemn temple]

An algebraic topology person told me this is an iff and I am becoming increasingly skeptical

[lunar yoke]

i mean it still feels like a really strong condition

[solemn temple]

There's a standard exercise in Munkres along the lines of: let Y be a Hausdorff space and g h : X -> Y arbitrary maps into Y that agree on a dense subset of X. Show that g = h
I was told the converse is also true but struggling to prove it

My current idea is: suppose Y is not a Hausdorff space, so there exist x != y such that every pair of opens around x and y overlap. Then it suffices to find two maps g, h : Y -> Y that agree on a dense subset yet aren't equal. Note that at least one of Y\{x}, Y\{y} must be a dense subset (if they were both not dense, then {x} and {y} would be disjoint opens around x and y). wlog say Y\{y} is the dense one. If the closure of y is more than just {y} (I can't get rid of this condition) then the map
h(z) := if z = y then some arbitrary element of Cl({y}) other than y else z
is continuous and distinct from id : Y -> Y, but h and id agree on the dense subset Y\{y}

[grave maple]

You're having problems proving that if g = h, then they agree on a dense subset?

[gritty widget]

lol

i don't think that's the converse they had in mind

[solemn temple]

I'm having problems proving the converse of (Y hausdorff) implies (for all X and maps g, h : X -> Y, g and h agree on a dense subset of X implies g = h)

[grave maple]

Ah, OK.

[gaunt linden]

g and h need to be continuous rather than completely arbitrary, I trust?

[solemn temple]

Yes

[solemn temple]

This person sent me a proof, turns out it's much easier to show (for all X and g h, ...) implies the diagonal Delta(Y) is closed

[gentle osprey]

Blitz

[gritty widget]

@solemn temple like this?

[gaunt linden]

Cool!

[solemn temple]

Yep

[hollow ermine]

https://proofwiki.org/wiki/User:Dfeuer/Definition:Usual_Topology

[gritty widget]

I never thought of this as a real definition, but something informal

[sturdy notch]

Hello

Does anyone here know of an explicit homotopy between $C(\vee_A S^1)$ and $\vee_A e^2$ which fixes the copies of $\vee_A S^1$?

[gentle osprey]

dadaurs

[sturdy notch]

Basically, I need to show that the pushouts obtained from $C(\vee_A S^1) \leftarrow \vee_A S^1 \rightarrow \vee_I S^1$ and $ \vee_A e^2 \leftarrow \vee_A S^1 \rightarrow \vee_I S^1$ are homotopic which is equivalent to showing the above, but the parametrization of the homotopy seems non-trivial and when I asked my prof he didn't know it either

[gentle osprey]

dadaurs

[sturdy notch]

where the right to left maps are always the trivial inclusions and the left to right maps are the same in both pushouts

[snow cobalt]

Thx

[lament geode]

for those who have studied degree theory

could u give me some insight

on how to get started

i plan to go through math 327 utoronto topology course (is it a good place to start with?)

bit about me: sophomore math student, taken real analysis courses, lin alg, odes and intro abstract algebra courses

[hollow ermine]

i'm sure that point-set topology is not enough for topological degree theory

math327 is an introductory course on point-set topology

[lament geode]

oh

sorry for the trouble

but do u know any course/ online resource/text to get started

[hollow ermine]

actually the origins of degree theory lies in algebraic topology but some people do degree theory in analytic viewpoint with point-set topology only.

a common introductory textbook for topology is munkres topology which covers both point-set topology and intro algebraic topology.

[lament geode]

Thanks a lot!!!!!

have a great day!

[hollow ermine]

I believe you are speaking of finite-dimensional topological degree theory since you don't know linear functional analysis.

You can go really deep into degree theory but for the basics you don't need algebraic topology actually

you should also mention why you want to learn degree theory.

[lament geode]

umm

i am doing a project under a prof

for my odes class

which requires proving schauders fixed point theorem

[hollow ermine]

you would only need some basic concepts in point-set topology

[lament geode]

oh!

thanks for the information!

[gritty widget]

mat327 won't teach you enough topology (neither algebraic or differential) to be able to jump into degree stuff. you'd want to look at 1301 or 367/1300

source: took all these courses

1301 is algebraic topology, 1300 is differential topology, and 367 is the undergraduate version of 1300

[hollow ermine]

what he's looking for is applications of fixed point theorems to ODEs but not something deep in degree theory. He certainly only need the basic topology

[lament geode]

Thank you guys! i see what u mean @hollow ermine

thank you for your time

[lament geode]

ooo are u a grad student at uoft?

[gritty widget]

undergrad

i just finished my 4th year (doing 5)

[lament geode]

oh damn!

u took all those grad courses in ur ug?

m only planning to do mostly 2, a grad ode n func analysis course

p scared about them

i am planning to apply to uoft/ uwaterloo n some unis in eu for my masters as well 😊

[gritty widget]

i did 367 instead of 1300

1300 really is differential topology, whereas 367 is like "here are manifolds, vector fields, etc etc, stokes"

general advice for taking graduate courses is: if it's a "core" course then it's going to be intense (exceptions apply, 1301 this semester was super easy), and if it's a seminar or topics course, it's not going to be very difficult workload-wise

ie the workload for topics courses is whatever you make it

[lament geode]

oh damn

yea the ode one is a core course

thanks for the advice

[gritty widget]

what's the course code? i know there's a PDEs one

is it the new one they added?

[lament geode]

oh m a student at uoft

m a student at ualberta

[gritty widget]

ah

[lament geode]

yeeee

[gritty widget]

assumed you were talking about uoft courses mb

[lament geode]

nw nw

my ode prof here taught at uoft

[gritty widget]

the advice on grad courses still holds though

[lament geode]

and he often refers to those courses

[lament geode]

yep thanks!!

[bright acorn]

Suppose that we have a pair $(X,A)$, where $A \subset X$ is closed and the inclusion $\iota : A \rightarrow X$ is a cofibration. And let's assume that we have a self-homeomorphism $f : X \rightarrow X$ that restricts to a self homeomorphism $\tilde{f} : X \setminus A \rightarrow X \setminus A$ of the space $X \setminus A$.
\
\
If we know the fundamental group of the quotient space of $X$ obtained via the orbits of $f$ and we also know the fundamental group of $A$ (possibly of $X \setminus A$ too if such a hypothesis is necessary), can we recover the fundamental group of the the quotient of $X \setminus A$ by obtained from the orbits of $\tilde{f}$.

[marsh forge]

You are using setminus, do you mean quotient or set difference?

[gentle osprey]

MISTERSYSTEM

[bright acorn]

Set difference

The question is basically like

[marsh forge]

I see, including the last statement?

you say quotient

Oh

I see now

sorry I was misreading

[bright acorn]

Yeah, I was also having a hard timing writing this down for some reason haha.

I hope this is not so confusing.

[marsh forge]

Hm

This seems tricky

Any time set differences show up in topology things get funky

at least algebraic topology

[bright acorn]

This might sound silly, but I thought about this more general question when trying to solve a really elementary problem.

Suppose that we have a linear automorphism f : R^2 -> R^2 given by (x,y) |-> (2x,y/2). Now, we know that this restricts to a self homeomorphism of R^2 \ {0} given by:

f' : R^2 \ {0} -> R^2 \ {0}, (x,y) |-> (2x,y/2)

One of my problem assignment questions was to prove that the quotient of R^2{0} by the action induced by f' is in fact a covering space map and also to calculate the fundamental group of the quotient of R^2 \ {0} by the action of f'.

And my idea to solve this problem was, precisely, to extend f' to a linear automorphism of R^2 (the f we had previously defined) and calculate the quotient of R^2 by this extended linear automorphism (which is in fact an easier problem).

And see what kind of information we can recover.

I actually could solve my assignment problem using this idea.

And I was thinking if like, there was a way to generalize this.

Idk, just giving some motivation for my question.

[cerulean oriole]

But @Phil showed Y has to singleton (or empty), which is much stronger, anyway.

[cerulean oriole]

Although this do be very neat

[gritty widget]

All Hausdorff spaces are singletons?

[gritty widget]

Phil was wrong, he even corrected himself later. It holds for any Hausdorff space Y

[cerulean oriole]

Oh right, my bad

[dawn sphinx]

on what channel, do i ask doubts on lattice and linear algebra

[gritty widget]

Definitely not here if it's not topology related

[mystic reef]

I'm having trouble with the following proof: Let $\mathcal{T}_1,\mathcal{T}_2}$ be two topologies on a set $X$. Suppose $\mathcal{S}_1$ is a subbasis for $\mathcal{T}_1$ and $\mathcal{S}_2$ a subbasis for $\mathcal{T}_2$. Show that $\mathcal{T}_1$ is finer than $\mathcal{T}_2$ if and only if for every $U\in\mathcal{S}2$ and every $x\in U$ there exists a finite collection ${V_1,\dots,V_n}\subset \mathcal{S}1$ such that $x\in\bigcap{i=1}^n Vi$ and $\bigcap{i=1}^n V_i \subset U$. I know that I need to show one implication and then the other, but I really dont know where to start..

[gentle osprey]

Ursus1234
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[gritty widget]

the intersection is a member of the basis generated by this subbasis

U needs to be union of such sets, so naturally, x is contained in some set of this form

now conversely this shows that any such U is open in the topology T1

so finite intersections of elements of S2 are also contained in T1

so arbitrary unions of finite intersections are

and this is precisely the topology generated by S2, that is T2

so T2 is contained in T1

[mystic reef]

thanks, I'll try wrapping my head around it. Still quite confusing to me

[sharp frost]

Hi, my lecturer was showing the our very first example of computing homologies straight from the axioms when the lecture recording cut off. The example was to compute the homology groups $H_*(S^n,\text{pt})$ for the $n$-sphere. She started off by choosing the upper hemisphere of the $n$-sphere and using the homotopy and excision axioms to write this isomorphism down

[sharp frost]

and then she used the long exact sequence axiom to write down this sequence

and then the recording stopped. I can see that the homology groups of the lower hemisphere are going to be isomorphic to the homology groups of a point, but how do we use this sequence to obtain an expression for the object in the top right?

[gritty widget]

probably induction, so you can assume something about the homology of the intersection of the hemispheres (the guy on the left)

that's typically how the computation of the homology of S^n goes. note that the intersection of the two hemispheres is homotopy equivalent to S^{n-1}

[sharp frost]

Ah yep, so the point is to get this kind of diagram?

these sequences still look like squiggly diagrams to me atm, the only cat theory I've learnt is the stuff I've accidentally inhaled from taking this algebraic topology course

so does this give us some sort of expression for $H_(S^n)$ in terms of $H_{-1}(S^{n-1})$

[gritty widget]

yeah, something like that. you should write down the homology of the point. in positive degrees it'll vanish, and you can use exactness to say something about the homology of S^n

[sharp frost]

Okay I think I got it, thanks for the help!

[arctic relic]

If I want to work out the set of bordism classes of dimension 10 manifolds, would the only possible choices be 10 copies of dimension 1 manifolds and a dissension 2 manifold Cartesian product a dimension 5 manifold

Or do I just have to enumerate all linear combinations of bordism classes with index i st that the sum of i’s = 10

[arctic relic]

Never mind I got it

[gritty widget]

Hi, I'm having a hard time getting the intuition of why the fundamental group of the real projective plane is isomorphic to Z/2Z. From what I understand, any loop based at x in RP^2 corresponds (via the quotient map) to either a loop based at x (or -x) in S^2 or a path from x to -x. But this is where I'm stuck

[plain raven]

have you drawn a bunch of pictures

[gritty widget]

I have yes

[plain raven]

can you think of an example of a path that you're not super convinced should be homotopic to one of these

[gritty widget]

a path in RP^2 ?

[plain raven]

yeah

[gritty widget]

hmm no not really

[plain raven]

also like, i'm assuming you know the statement of van kampen's theorem but like

my recommendation would be to take the proof of van kampen's theorem and specialize it to the case of this theorem

so don't treat it as a black box

but see how it works in this situation

[gritty widget]

I have heard of this theorem though it's not in my course

[plain raven]

oh

ok

let p be like

the simplest nontrivial path

so it leaves the basepoint at p, leaves one side and comes back at the other

and joins the basepoint

do you understand intuitively why p^2 = 0?

[gritty widget]

let me think about it

[plain raven]

remember that when you multiply two paths, the new path becomes detached from the basepoint at the midpoint and we can move it freely near the midpoint

[gritty widget]

so f*g is essentially just ignoring the point y

[plain raven]

yeah, x and y should be glued together at the basepoint but y is no longer required to be attached to the basepoint when we homotope f * g

[gritty widget]

alright

how does it tell me that p^2 = 0 ?

[plain raven]

that doesn't tell you itself, i'm just saying it's a helpful reminder lol

here's how i draw RP^2

i draw a circle with a dot in the middle

the dot is the basepoint

whenever a line goes out of the circle at one side it comes back on the other 180 degrees removed

so like

[gritty widget]

yea the boundary points are identified or something

[plain raven]

draw p so that it starts at the midpoint of the circle, goes up to 12 o clock, comes up from the bottom at 6 o clock and rejoins

[gritty widget]

okay, this is a bit different from representing RP^2 as a sphere and imagining all antipodal points are the same

[plain raven]

yeah it's easier to draw for me

[gritty widget]

well, it's equivalent in the end

[plain raven]

so p^2 can be drawn like:

• starts at the basepoint
• goes up to 12 o clock
• comes in the bottom at 6
• comes up but veers off a bit to the left instead of rejoining at the basepoint
• keeps going up and hits 11 o clock
• comes in at the bottom at 5 o clock
• rejoins at the basepoint

[gritty widget]

okay so the "veers off" part comes from what you just told me before

[plain raven]

yeah

i'm detaching the "midpoint" of the path from the basepoint

then like, what i'd do is homotope the point at 11 o clock/5 o clock in the counterclockwise direction

and you can eventually just pull the whole loop on the left hand side through the wall of the circle

idk if this makes any sense

[gritty widget]

okay

https://math.stackexchange.com/questions/629595/why-is-the-fundamental-group-of-the-projective-plane-c-2?noredirect=1&lq=1

that's what I just saw there earlier

but now I actually understand it aha

okay so now that p^2 = 0, it says that any non trivial loop is of order 2 (well, their class rather) so the group structure is the same as Z/2Z

[plain raven]

idk if that's a complete proof but basically like, i can't think of any simple nontrivial group elt that's not homotopic to p

and any more complicated element, imo intuitively it should be easy to break this down into the product of copies of p

[gritty widget]

alright I see, thank you for your time 🙂

[mystic reef]

I'm trying to solve following problem: Let $X$ and $Y$ be topological spaces and equip $X\times Y$ with the product topology. Define $\pi_X:X\times Y\rightarrow X$ by $\pi_X((x,y))=x$, and $\pi_Y:X\times Y \rightarrow Y$ by $\pi_Y((x,y))=y.$ Show that for every open set $U\subseteq X\times Y, \pi_X(U)$ is open in $X$ and $\pi_Y(U)$ is open in $Y$.

[gentle osprey]

Ursus1234

[mystic reef]

I have a proposition that says something about if I have a subset of X that is open , then every open set in the subset is also open in X, but I dont know if that is usable here

[sturdy notch]

recall how you defined the product topology and notice you only need to show this result for a basis

basis of opens on X that is

[mystic reef]

so the basis that generates the product topology is $\mathcal{B}={U\times V|U\in\mathcal{T}_1,V\in\mathcal{T}_2}$ and since $U$ is an open subset of $X$ then $\pi_X(U)$ is also open in $X$?

[gentle osprey]

Ursus1234

[mystic reef]

is that correct or didn't I understand it correctly

$\mathcal{T}_1$ is the topology on $X$ and $\mathcal{T}_2$ is the topology on $Y$

[gentle osprey]

Ursus1234

[sturdy notch]

Seems right, you do have to show it is sufficient to show this for a basis

[mystic reef]

Im not quite sure why that is true, but I would think it has something to do with the fact that a basis generates a topology, which means that the elements of a topology are unions and intersections and so on of the elements of the basis...

I dont know whether you would always just be able to show it for a basis, or its possible in this case because of how we have defined $\pi_X$ and $\pi_Y$...

[gentle osprey]

Ursus1234

[sturdy notch]

well you checked it for a basis of open sets, now to check it for any open set, you have to write this open set as a union of basis elements

[gritty widget]

$f(\bigcup_i U_i) = \bigcup_i f(U_i)$

[gentle osprey]

Blitz

[gritty widget]

It's just this fact applied

[sturdy notch]

was trying to hint at this, but yes, that's it

[cosmic beacon]

Is it possible for two spaces to have pairwise isomorphic fundamental homotopy groups but different cohomology rings?

[lunar yoke]

by pairwise isomorphic fundamental groups, do you really mean just pi_1, or actually that pi_n(X) is isomorphic to pi_n(Y) for all n?

if you only need pi_1, then this is pretty easy; just take S^2 and S^3. They're both simply connected, but H^2(S^2;Z) = Z and H^2(S^3;Z) = 0

if you want that all homotopy groups are isomorphic this will be a lot harder, im not sure if its possible actually.

[cosmic beacon]

the latter

[lunar yoke]

The main point is that in you statement, the isomorphisms between the homotopy groups need not be induced by a map f : X -> Y on the level of spaces (such a map would be called a weak homotopy equivalence if it works for arbitrary basepoints)

if you had such a weak equivalence, then they also have the same cohomology groups

but this map need not exist if these groups are just abstractly isomorphic

[cosmic beacon]

I am trying to check: I think S^3 x RP^2 and S^2 x RP^3 would be a counterexample to this?

The cohomology ring of S^n with coefficients in Z/2 is Z/2[a]/(a^2) right?

[lunar yoke]

i would be really surprised if something as easy as this works

[lunar yoke]

well yeah if you say a is degree n

[cosmic beacon]

Yea

[lunar yoke]

but i dont think these two spaces have the same cohomology ring, do they?

[cosmic beacon]

They don't

Unless I'm messing up my calculations

That's what I'm trying to find

[lunar yoke]

oh yeah that way around

[cosmic beacon]

two spaces with same hty groups but different cohomology rings

[lunar yoke]

how are you going to compute the higher homotopy groups of your space though

this is generally so hard that its still open, even for spheres

[cosmic beacon]

so then by the K"unneth isomorphism:
[H^(\bR P^2\times S^3)\cong \bZ_2[\omega]/(\omega^3)\otimes\bZ_2[\alpha]/(\alpha^2)]
but
[H^(\bR P^3\times S^2)\cong\bZ_2[\omega]/(\omega^4)\otimes\bZ_2[\alpha]/(\alpha^2)]
those are different, right?

[gentle osprey]

@cosmic beacon

[cosmic beacon]

For n >= 2

[lunar yoke]

oh

you just dont

neat

[cosmic beacon]

lol yea

Do these calculations look correct?

[lunar yoke]

well pi_1(RP^n) is Z_2, so your thing only works for i > 1

[cosmic beacon]

yea sorry

but for i=1 it's easy

[lunar yoke]

but yeah the example works anyway

[cosmic beacon]

thx!

[lunar yoke]

interesting

i woudl have never guessed you could find such an easy example

[cosmic beacon]

lol "easy" is an overstatement, I had somehow convinced myself this was impossible, but then got confused when working with RP^2 x S^n and S^2 x RP^n, realizing it contradicted that

So I just wanted to make sure I didn't make any mistakes calculating

[lunar yoke]

easy in the sense that you can basically write and prove it in a few lines

[cosmic beacon]

But I didn't go looking for a counterexample, just one popped up

[solemn temple]

What's the expected solution here? I have a solution but it seems overkill. (If T' finer than T then id : (X, T') -> (X, T) bijective and continuous; T' compact and T Hausdorff so id closed; hence id is a homeomorphism)

[empty grove]

That it

That is it*

Probably

[solemn temple]

OK

[gritty widget]

Is there just one way to talk about cohomology here?

[lunar yoke]

i think we were talking about singular cohomology with Z_2 coefficients

[cosmic beacon]

Is every Eilenberg MacLane space necessarily homotopy equivalent to a CW complex?

I know that any two CW complexes which are both K(G,n)'s are hty. equivalent

Nvm I don’t think that’s true

[shy moss]

Consider the Hawaiian earring

[sly mural]

I'm trying to show that if X is homotopy equivalent to the one-point space {a} then X is contractible. By definition we have maps f: X -> {a} and g: {a} -> X such that their compositions are the respective id maps. I feel like this should be obvious but i just can't quite get it 😭

is there some way i can show the homotopy between id_X and the constant map on a using the transitivity of homotopy equivalence or do i need to specifically construct a homotopy

[gritty widget]

what's your definition of contractible

[sly mural]

X is contractible if id$_X$ is homotopic to a constant map

[gentle osprey]

rustyjpeg

[sly mural]

iff*

[gritty widget]

think about what it means for the compositions of f and g to be homotopic to certain identity maps

maybe the one you want will pop out

(and by "think about" i mean write out every assumption you've made in this problem)

[sly mural]

my assumptions are that if $f: X \to {a}$ and $g: {a} \to X$ then $f \circ g \simeq id_{{a}}$ and $g \circ f \simeq id_X$. i'm also considering the fact that since $f: X \to {a}$ then we must have $f$ is the constant map to $a$

[gentle osprey]

rustyjpeg

[cosmic beacon]

If M is a compact manifold with Euler characteristic 0, does its universal covering space also have Euler characteristic 0?

If the cover is of finite degree, yes, but is this also true for covers of infinite degree?

[gritty widget]

you have a homotopy between id_X and something. what can you say about that something?

you are almost there

[sly mural]

are you referring to g \circ f?

im sorry im probably missing the obvious 😭 my brain is a little fried

[gritty widget]

i am

what is g \circ f?

you know id_X is homotopic to it. you want to say id_X is homotopic to...

[sly mural]

f is the constant map to a and then i think g is the inclusion of a into X?

[gritty widget]

so overall as a map from X to X, g \circ f is...

[sly mural]

the constant map to a!

[gritty widget]

yes!

and what does that mean?

[sly mural]

my only question is couldn't g send a to some other member of X - must it be the inclusion since its homotopic to the identity on {a}

[gritty widget]

ok well the claim about g \circ f being constant is still true

you caught me, i have lied

[sly mural]

hahahahahha

[gritty widget]

my brain may be a little fried too

[sly mural]

i think it still might be right that g circ f is homoptic to the constant map on a

ok no well actually

[gentle osprey]

rustyjpeg

rustyjpeg

rustyjpeg

[gritty widget]

and that's exactly what you want

[sly mural]

I was caught up thinking it had to be homotopic to the same constant map, but it just has to be any constant map

thanks for the help @gritty widget ❤️

[cosmic beacon]

I have an embedding $f : S^1 \times \mathrm{int}(D^2) \to S^3$, is $S^3 - \mathrm{im},(f)$ necessarily homotopy equivalent to $S^3 - \overline{\mathrm{im},f}$?

[gentle osprey]

@cosmic beacon

[cursive flume]

here f is continuous bij. map

how can I conclude that all open sets in X are of the type f^{-1}(U)?

[gritty widget]

f^-1 is continuous iff (f^-1)^-1(U) = f(U) is open for any open U

And this is precisely what it means to be an open map

You don't need for f to be continuous for this

[cursive flume]

I am on the direction. Let f^{-1} continuous. then f is open

[cursive flume]

this is true

but in my argument i don't see how this helps

[gritty widget]

What you wrote is overly complicated

[cursive flume]

I mean I agree with what you write

but i'd like to fill in that. detail in my profo

[gritty widget]

This is straight from the definitions

There's no details untouched there, really

[cursive flume]

in your profo not

in mine yres

[cursive flume]

here i'd still need to show that all open in X are preimages of opens in Y

[gritty widget]

Your proof is overly complicated and I don't see reason to work with it

So I guess I won't help you, sorry

[cursive flume]

nvm got it

[buoyant oasis]

How can I get good at topology?

[gritty widget]

do topology

[river granite]

work through a ton of examples and exercises, see how the general theory ties in with those

[buoyant oasis]

What chapters of math do I need to know before I get started with topology?

[river granite]

some real analysis is good to know

really you don't need a ton more than set theory and familiarity with reading and writing proofs

[river granite]

but many interesting examples come from this and other things you might've already seen in math

[buoyant oasis]

Real analysis? What school years is that done in?

[river granite]

depends on each math program

sometimes freshman calculus is taught as an analysis course right off the bat

and in some other programs there's a "real analysis" course in third year or something

[buoyant oasis]

Then it’s a long way for me

[mystic reef]

hmm for our class tomorrow, our little assignment is to try to find an example of sets $X$ and $Y$ and subset $Z\subseteq X\times Y$ where 1: $\pi_X(Z)$ is open in $X$, 2: $\pi_Y(Z)$ is open in $Y$ but 3: $Z$ is not open in $X\times Y$ with respect to the product topology. I've been thinking about it, it confuses me how the 1,2 are open, but 3 is not...

[gentle osprey]

Ursus1234

[mystic reef]

$\pi$ is defined as: $\pi_X:X\times Y\rightarrow X$, $\pi_Y:X\times Y\rightarrow Y$

[gentle osprey]

Ursus1234

[mystic reef]

would appreciate some suggestions

[plain raven]

maybe narrow down the search space by looking for an example where X = Y

and look for a geometric space

one that's easy to visualize

[mystic reef]

ahh yea, thank you

[mystic reef]

hmm I've been thinking about it and I still cant see of $\pi(Z)$ would be open but $Z\subseteq X\times Y$ is closed...

[gentle osprey]

Ursus1234

[mystic reef]

can you maybe enlighten me some more?

I was thinking about letting $X=Y=\mathbb{R}$ and $Z={(x,y)|x,y\in\mathbb{Z}}$ because the integers is a closed set in $\mathbb{R}$ but then $\pi(Z)$ would not be open either...

[gentle osprey]

Ursus1234

[mystic reef]

Im a little confused

[gritty widget]

@mystic reef good thinking but instead of this take Z = {(x, x) : x in U} where U is your favourite non-empty open set

In fact the diagonal is open iff X is discrete

So for all non-discrete spaces you can just take the diagonal Delta(X) = {(x, y) : x = y}

In X x X

[mystic reef]

Delta(X)?, should this be my Z?

[gritty widget]

Yes

It can be

[mystic reef]

ahh yea, and then pi(Z) is open but Z is not open in X x Y

[gritty widget]

Unless X is discrete

[mystic reef]

but the real numbers arent discrete are they?

[gritty widget]

Yeah they aren't

[mystic reef]

thanks a lot @gritty widget

[lament needle]

The discrete topology on a space X is the topology that takes every subset of X. As I understand it.

But I mean would this mean if we took the real numbers in [0,1] and took every subset, we would have a discrete topolgy? Why would we think of this as discrete

[sharp frost]

yeah it's discrete, the intuition is that if you look at any point x then every other point is far enough apart from x that you can draw an open neighbourhood around x that includes no other points

so you're taking [0,1] and yanking it apart so that each point is isolated from every other point

[lament needle]

ohhh so you have all these singletons as open sets

[sharp frost]

yeah

[lament needle]

and thats why it differs from the metric intuition

I guess I just didnt think of real analysis in terms of sets LOL

[sharp frost]

if you like you can even define a metric that's consistent with the discrete topology, and think of it as a metric space

there's a "discrete metric" but I won't ruin your fun trying to find it

hey if X is a topological group and Y is a subgroup of X then does anyone know of a strategy to get the fundamental group of X/Y from the fundamental group of X?

[lament needle]

im familiar with the discrete metric

[cosmic beacon]

For $n\geq1$, does the quotient $q: X\sqcup Y\to X\vee Y$ induce an isomorphism
[H_n(X\sqcup Y)\xrightarrow[]{\sim}H_n(X\vee Y)?]

[gentle osprey]

@cosmic beacon

[cosmic beacon]

I don't believe there is a strategy in general, what are the spaces in question? You should be able to say something if the quotient map X --> X/Y is a covering map

[sharp frost]

I have SO(3), and the subgroup is A_4 embedded into SO(3) as the symmetries of the tetrahedron

since posting the question I think I found a proposition in Hatcher, Prop 1.40, that looks like it should do the job

though I haven't had time to work it out yet

[cosmic beacon]

Yes this is what I was referring to, are you talking about the orbit space of SO(3) under the action of A_4?

[sharp frost]

yeah

or at least that's what I assume, the way the question is worded is a little ambiguous

[cosmic beacon]

Yea, if that action is properly discontinuous and free, then you have A_4 ~= pi_1(SO(3)/A_4) / Z_2

[sharp frost]

yaaay thanks for confirming

and then I can just undo the quotient with a semidir product right?

[cosmic beacon]

hmm I'm not exactly sure what to do from there tbh

[sharp frost]

probably need to argue that there's a short exact sequence $$1\to\bZ_2\to \pi_1(X)\to A_4\to 1$$

[cosmic beacon]

Interesting, how would this help? Also note A_4 isn't an abelian group, FWIW

[sharp frost]

according to my classmates the answer should be $\bZ_2\rtimes A_4$ so my brain is just doing whatever it can to get that lol

[sharp frost]

a fallacy I know

and that short exact sequence would imply the semidirect product (I think)

[empty grove]

If the basepoints have neighbourhoods that deformation retract to those points, then that is an isomorphism for n > 0, and a (split) surjection for n = 0 with kernel ℤ

[vocal anchor]

If I regard the component colour as an important property of links, then these two are not the same, right? I apparently have absolutely no visual thinking capacity, and I'm not sure if the Reshetikhin-Turaev invariant distinguishes them.

[lean cedar]

It seems to me like I can just rotate the red ring 180° and get exactly the right picture

Dont know the invariant, but they seem highly the same

@vocal anchor

[vocal anchor]

oh dang you're right lmao

[gritty widget]

oh wait yeah

[vocal anchor]

I see it, but I can't prove it, because "rotate 180 deg" doesn't translate to Reidemeister moves for me. But I think I have found the proof

[lean cedar]

Like cant you literally write down an isomorphism of knots that leaves the components alone??

[vocal anchor]

I'm dumb, so my proofs are complicated

[lean cedar]

why do you need the second row

isnt the firsy row exactly what you need

ah nvm

I look at it later

will leave the house now

[vocal anchor]

you read the second row from right to left, and then the 1st row from left to right 😄

Don't worry about it 🙂

thanks 👍

[gritty widget]

you don't need those though?

those are two spaces in R^3

[vocal anchor]

Well, I'm not a topologist, so don't get me started with constructing explicit homeomorphisms

[gritty widget]

like why don't you just go like this

[vocal anchor]

Like, for me "rotate 180 deg" is not a proof, it's a topologist's proof.

[gritty widget]

rotate it a little bit not even really changing it in place

[vocal anchor]

I understand that, but I don't have a theorem that tells me this operation is good, and to be one the safer side of things, I'd rather use Reidemeister moves: I know they work, and I don't have to rely on intuition (in my case, topology intuition is usual quite wrong, so)

[lean cedar]

I mean link you fingers like that

and rotate both hands in the same way 180°

Its like literally a linear rotation of the whole R^3

More iso you wont get anywhere

[vocal anchor]

No, you only rotate one hand though, not the whole R^3

[lean cedar]

Oh I see

well yeah ok, in that case its hatder to write down the ambient iso

No actually

I will write it down when I'm home

[gritty widget]

why would you want to write it explicitly

[vocal anchor]

I'm an algebraist, I need to detail every step, and put equation numbers over equalities, when I use previously known facts.

[gritty widget]

if you say so

[thin girder]

For the following subsets U ⊆ R 2, determine the closure, interior, and boundary. I couldnt understand what is meant with for example sin(x1)<x2
All examples I found had a set with points ... and a topology on the set, tau and a subset

I am new to topology so excuse me if my questions are to easy

[gritty widget]

It means $\sin(x_1)<x_2$

[gentle osprey]

Blitz

[gritty widget]

I don't know what else to tell you

[thin girder]

wow 🙂

how would you do the question then

[gritty widget]

Did you try graphing those sets

[thin girder]

no do you have an advice for an online graph?

[gritty widget]

You can just use a piece of paper

[thin girder]

without understanding the given how can I do that

[gritty widget]

what do you not understand?

[thin girder]

what is meant by sin(x1)<x2

ofc I understand that that means

[gritty widget]

it means that sin(x_1) is less than x_2

[thin girder]

but in this question

[gritty widget]

how's it different in this question from its usual meaning

[thin girder]

ofc its not but what am I suppose to do with that

[gritty widget]

If you had x_2 = sin(x_1) instead, can you identify what U would be

[thin girder]

no

[gritty widget]

do you know what sin(x) means

[thin girder]

yes

[gritty widget]

okay, that's something

[thin girder]

-__-

[gritty widget]

{(x_1, x_2) : x_2 = sin(x_1)} is just the graph of sin(x)

it's a set of points (x, sin(x)) on the plane

[thin girder]

okay

[lean cedar]

@thin girder do you know what this set builder notation means?

[thin girder]

no

[gritty widget]

so now {(x_1, x_2) : sin(x_1) < x_2 } would be everything above that graph

{x : p(x)} where p(x) is some logical statement means the set of elements x such that p(x) holds

here it's meant that (x_1, x_2) is an element of R^2, which is the standard Euclidean plane

[lean cedar]

{x in A : phi(x) } is the set of all elements of A which statisfy phi. So {(x,y):x^2=y} is the set of all 2D points that satisfy x^2=y (ie the graph of a parabola)

[gritty widget]

yeah. Like A = R^2 in this case but the author didn't write it, which is a slight abuse of notation but it's forgivable

${x\in A : \varphi(x)}$

[gentle osprey]

Blitz

[gritty widget]

just so you don't get confused by what we mean when writing it in text

[thin girder]

okay

[gritty widget]

it reads "the set of elements x belonging to A such that phi(x) holds"

[lean cedar]

Maybe try {(x,y): x^2>y} as an easy example first

Before going to your actual exercise

[thin girder]

alright

[gritty widget]

@thin girder are you trying to learn topology on your own?

[thin girder]

kinda

[gritty widget]

did you try learning some basic set theory first

it'll be a problem if you don't understand the notation

[thin girder]

I dont know the name of the topics but I mightve

understand

[gritty widget]

yeah, union, intersection, inverse images, images etc.

notation for a function

properties of functions

[lean cedar]

Yeah like topology is literally impossible otherwise

[gritty widget]

Well, that's used all over mathematics, so it should be learned

[thin girder]

so this would be the the area left on the parabola?

[thin girder]

ofc

[gritty widget]

it's points below the parabola

[thin girder]

:/

[gritty widget]

{(x, y) : y < f(x) } are things below graph of function f, {(x, y) : f(x) < y } are things above it

[thin girder]

oh yeah

[gritty widget]

here f(x) = x^2 is a parabola

yeah, you just have to think about it for a little while

[thin girder]

thats what I thoughtt but I guess I saw the > wrong

ok but is x1 =x and x2=y necessarily?

[gritty widget]

it's irrelevant what we name those

in the set notation those are just dummy variables

[lean cedar]

Nope

[thin girder]

then how should I image the graph

[gritty widget]

{(a, b) : f(a) = b} = {(x, y) : f(x) = y } = {(x_1, x_2) : f(x_1) = x_2}

those are all the same

[thin girder]

okay

[gritty widget]

just we named those variables, in the set notation, differently

[thin girder]

yeah but since for example a is before b (a,b) it is f(a)=b

[gritty widget]

note that those aren't explicit numbers, it's more like a description for a method for obtaining an element of this set

[thin girder]

okay

[lean cedar]

{a:a>1} is the same set as {z:z>1}

[gritty widget]

I can help you with this question but you should read about set notation from a book asap

and then continue to read about topology

[thin girder]

okay

is there a book you would advice in general

for topology

[gritty widget]

Oh, for topology? I have a book that I have sentiment for since I've learned from it but I don't know if it'll be suitable for you

Dugundji is the author

[thin girder]

from 1978?

: D

[gritty widget]

most of the book should be understandable but he omits a lot of details when it comes to simplicial approximations

[gritty widget]

the math didn't change

[thin girder]

I guess

[gritty widget]

it just expanded

[thin girder]

okay then I will check it

but just to stop my suffering could you explain the question

just an example

[gritty widget]

Okay, maybe I will answer this question for c)

just to show you the logic behind it

[thin girder]

okay

[gritty widget]

https://www.desmos.com/calculator/2pc0larkqc

okay so U is the intersection of the red and blue areas

[thin girder]

yep

[gritty widget]

so to take the closure you think like this

which elements of the plane can you get by taking a sequence of elements from this area and then their limit

https://www.desmos.com/calculator/udr1kheet6

and this will be intersection of those two areas

note how the top left part of the "boundary" of this area is now filled in

before it wasn't in this set

[thin girder]

yeah

[gritty widget]

https://www.desmos.com/calculator/uyvx20diet

the interior is this

you have no boundary at all this time

[thin girder]

so literally interior is the inside

[gritty widget]

yeah, something like that

[thin girder]

so one question

[gritty widget]

and now the boundary is the closure without the interior

[thin girder]

is closure just the boundaries or boundaries plus intersection

[thin girder]

hmm so again literally bounds

[gritty widget]

closure is this whole thing

you have your set and then some more

[thin girder]

okay

so I understand their meaning

do they have a mathematical writing

[gritty widget]

yes

you can write them in terms of closed and open sets for example

or differently, there's a couple equivalent ways you could define them

(in this context)

[thin girder]

okay for example how would boundary look like

[gritty widget]

$\partial U = \overline{U}\setminus U^\circ$

[gentle osprey]

Blitz

[gritty widget]

you could define it like this, closure without interior

[gentle osprey]

Blitz

Blitz

[gritty widget]

this is the closure

[gritty widget]

this is the interior

[thin girder]

I suppose these are in general , is there a way to add values given?

[gritty widget]

what do you mean exactly

[thin girder]

these are notations right

in general

what would these look like for our example

not all of the situations will be the same

what separates them

[gritty widget]

oh, like do you want me to write explicit formulas for boundary etc.

[thin girder]

dont know the name but might be what I mean

[gentle osprey]

Blitz

[gritty widget]

so here the only thing changing is that one inequality is changed

[thin girder]

ah alright

[gritty widget]

this isn't really how it works in general though

just a coincidence

[gentle osprey]

Blitz

[thin girder]

what do you mean

[gritty widget]

well I am explaining mainly based on intuition so if you don't understand then that's ok

[thin girder]

okay

[gritty widget]

and now the boundary is the hardest to write explicitly

[thin girder]

I understood the logic only thing I need to work on is the notations I believe

[gritty widget]

yeah

[thin girder]

so I can understand what a question asks : D

[gritty widget]

visit #book-recommendations and ask for some basic set theory book

[thin girder]

okay

[thin girder]

if thats not so hard how would it look like

or any example would be fine

doesnt have to be about our question

[gritty widget]

${(x, y) : x^2+y^2 = 16, x-y\leq 3}\cup {(x, y) : x^2+y^2\leq 16, x-y = 3}$

[gentle osprey]

Blitz

[gritty widget]

in general I already wrote how to obtain it here

[thin girder]

yeah

but I assume there is one last writing

which is the hardest?

[gritty widget]

one last writing?

[thin girder]

is there a way to write this without the union

the cup symbol you used in latex

${(x, y) : x^2+y^2 = 16, x-y = 3}$ maybe like this?

[gentle osprey]

Ege █▀█ █▄█ ▀█▀

[gritty widget]

no, this is just two points

[thin girder]

okay

[gritty widget]

${(x, y) : x^2+y^2 = 16, x-y\leq 3\text{ or } x^2+y^2\leq 16, x-y = 3}$

[gentle osprey]

Blitz

[gritty widget]

you can write it like this if you prefer

[thin girder]

okay

well thanks a lot

helped me to understand it a lot

[gritty widget]

np
come back when you learn some set theory

[thin girder]

yeah

I hope its not to late

: D

anyways thanks again

[gritty widget]

Can $\mathbb{N} \cross {0, 1}$ have a one-point compactification?

[gentle osprey]

kaz

[gritty widget]

every space has a one-point compactification, just it's not necessarily Hausdorff

😮

but here it is

My attempt: no, as limits of the sequences x_n = (n, 0) and y_n = (n, 1) are both not in closure of this set

the one point compactification of this space is homeomorphic to ${0}\cup {1/n : n\in\mathbb{N}}$

[gentle osprey]

Blitz

[gritty widget]

Wow yeah definitely

$\mathbb{N}\times {0, 1}\cong \mathbb{N}$

[gentle osprey]

Blitz

[gritty widget]

those spaces are homeomorphic

any bijection is a homeomorphism

Okay, could you tell me what's wrong with my approach?

you say "limits of the sequences" but they don't converge

even if they did converged, this doesn't show anything

So what I aimed for, if there was an one point compactification

Since N X {0, 1} would be metrizable

it is metrizable

Then it being compact is equivalent for it being sequentially compact right (compactification of N X {0, 1])

you don't know the compactification is metrizable a priori

ah I see, thanks

but here it is metrizable

[lean cedar]

Well first of all consider what a knot diagram actually is: A suitably nice projection of a knot onto a plane. Consider in $\mathbb R^3$ the two circles $S^1={x\in\mathbb R^3\colon (x_1-1/2)^2+x_2^2=2^2, x_3=0}$ and $\tilde S^1={x\in\mathbb R^3\colon (x_1+1/2)^2+x_3^2=2^2, x_2=0}$. Consider the projection as if you were at $\lambda\cdot (0, 1, 1)^\top$ for $\lambda\to \infty$. Then this is precisely your first diagram. If you view from $\lambda\cdot (0, -1, 1)^\top$ instead (or from the original viewpoint and rotating the whole $\mathbb R^3$ with the knot) you get your second diagram.

[gentle osprey]

MrMonday

[lean cedar]

@vocal anchor

[lean cedar]

Or this

lol, knot theory is fun

literallt drawing a bunch of doodles

[rugged rock]

Let $p: E \xrightarrow{} B$ be a covering map. Let $e \in E$ and $b = p(e)$. Let $A \subset B$ such that $b \in Cl(A)$. Show that $e \in Cl(p^{-1}(A))$.

[gentle osprey]

Évariste Galois

[rugged rock]

while I understand the question, I'm not sure how to approach it

I could use a few hints

[lean cedar]

\xrightarrow{}???? My guy, \to???

[rugged rock]

lmao my bad

[lean cedar]

Ok, now that I've complained let me look at the question lol

[rugged rock]

Constructive criticism 🙂

[lean cedar]

Is Cl(A) the closure?

[rugged rock]

Yup

[rugged rock]

Any updates?

[lean cedar]

By $p$ being a covering and $b\in B$ there is a neighborhood $U$ of $b$, such that $p^{-1}(U)$ is the disjoint union of some open sets $U_{E,i}$ all homeomorphic to $U$ via $p$. That is we have homeomorphisms $p\vert_{U_{E,i}}\colon U_{E,i}\to U$. Now from "closure in the relative topology" we have that the closure of $A\cap U$ in $U$ is $\overline{A\cap U}^U=\overline{A}^B\cap U$. But $b \in U$ and $b\in \overline A^B$ so $b\in \overline{A\cap U}^U$. Now let $e \in E$ such that $p(e)=b$. In particular $e\in p^{-1}(U)=\coprod_i U_{E,i}$. Thus we have a unique $j$ such that $e\in U_{E,j}$ and $e$ corresponds via the heomemorphism $U_{E,j}\cong U$ to $b$. Thus walking backwards through the homemorphisms we obtain from $b\in \overline {A\cap U}^U$ that $e\in \overline{p^{-1}(A)\cap U_{E,j}}^{U_{E,j}}$. By the same "closure in relative topology" we again have $\overline{p^{-1}(A)\cap U_{E,j}}^{U_{E,j}}=\overline{p^{-1}(A)}^E\cap U_{E,j}$. In particular we have $e\in \overline{p^{-1}(A)}^E$. I guess that should be it

[gentle osprey]

MrMonday

[lean cedar]

@rugged rock

took a little while lol

Was a bit rusty

[rugged rock]

That’s awesome. Clean argument. Thanks a lot 🙂

[lean cedar]

(Everybody please fact check this proof lol)

[cursive flume]

isn't there a typo?

shouldn't g: Y,S->Y,R?

how does f o g make sense like this?

[lean cedar]

I mean whats f? I think its just a weird rephrase of a universal property. Is there anything above missing?

Otherwise its just written very sloppily

[woven haven]

If A is subspace of X ( both path connected) then does the inclusion map induce identity map on the first homology groups?

[lean cedar]

Does this help: https://en.m.wikipedia.org/wiki/Relative_homology?

I guess you are talking about singular homology

[woven haven]

yeah singular

[woven haven]

i am sorry i dont get it

[lean cedar]

I mean I haven't solved your question yet, this was just a heads up

Thig is you get a long exact sequence $\cdots \to H^0(X)\to H^0(A)\to H^0(X,A)\to H^1(A) \to H^1(X)\to H^1(X,A)\to H^2(X)\to \cdots$. So if $H^0(X,A)=0$ and $H^2(X)=0$ then you have a short exact sequence with the first map your map in question, which then will be injective.

[gentle osprey]

MrMonday

[lean cedar]

But I'd guess this holds much more often. But not sure about the general case

Eh

Oops, homology

make everything _n instead of ^n, let degrees decrease to the right and adapt the statements I guess lol

[cursive flume]

it's the universal property yes

it's the universal property of induced topology

but isn't there a typo?

f is a map (Y,R)->(X,O_X)

it's the proof of uniqueness of induced topology

[lean cedar]

Well of f:Y->X and g:Z->Y then f°g:Z->X?

Why not?

[cursive flume]

because the topologies don't match

[lean cedar]

Oh you mean the later bullet points

give me a second

[cursive flume]

yeah

after he writes consider g=...

g should be id_y: Y,S->Y,R

[lean cedar]

Yes, g should have switched domain and codomain

[cursive flume]

[cursive flume]

and thus f o g:Y,S->X,O, right?

[lean cedar]

yeah

[cursive flume]

how to show that the induced topology satisfies this universal property?

I only showed uniqueness

didn't show that the induced topo satisfies it

[lean cedar]

remind me tomorrow