point-set-topology

[lament needle]

Do you care about bijecting or only a surjection

[pseudo coral]

ive shown 1-1 already

bijection

[lament needle]

1-1 is injective

[pseudo coral]

yep

if $x \neq y$ then $2x-1 \neq 2y-1$

[gentle osprey]

MyMathYourMath

[lunar yoke]

well ok so for r = 2 your solutions are x = (1+4)/(4+2) = 5/6 and x = -1/(4-2) = -1/2. One of those is in (0,1), so you know 2 is in the image

[pseudo coral]

so its certainly injective

[lunar yoke]

now do this for general r instead of 2

[lament needle]

well this map divides also right its like $\frac{2x-1}{1-|2x-1|}$

[gritty widget]

it'd be easier to not use this map and use arctan instead

[gentle osprey]

matthew

[pseudo coral]

yep

[pseudo coral]

i was really trying to avoid that map lol

[lament needle]

so you are strugglng with the surjective part

[pseudo coral]

yes

[lament needle]

what happens when x gets close to 1 or 0

[pseudo coral]

you get positive and negative values

as outputs

[lament needle]

are you unable to prove that this is a surjection explicitly

do you know how to do it explicitly

[pseudo coral]

yes, but I get stuck since setting r equal to the function and solving for x gives me two values

im having a tough time showing only one of the two values is inside of (0,1)

[lament needle]

ok so i see your problem You can prove there is an inverse just the inverse has two real values

[pseudo coral]

yes

[lament needle]

i think you actually constructed a bijection on your interval but in general the inverse of a continous function need not be a function

[gentle osprey]

MyMathYourMath

[lament needle]

like the map f(x) = 0

this surjects onto the set {0}

to prove surjectivity you just need to show that there exist some x' such that f(x') = 0

this map just isnt injective

[pseudo coral]

yes since everything hits 0

[lament needle]

so you only need to prove that one of the values mapping onto some real r is in the interval (0,1)

[pseudo coral]

precisely

[lament needle]

right so your function f: (0,1) -> R is a bijection, However it naturally extends to a function F:R(removng 0,1) -> R but this function is not injective

[pseudo coral]

yeah that guys not 1-1

[lament needle]

so for every real r there are two reals x,y such that F(x) = F(y) = r

[pseudo coral]

namely,

[pseudo coral]

these values

can I use the breaking points to show one lands in (0,1)

by that I mean the two points that make the denominator zero

i.e., 1,-1

[lament needle]

yeah

as long as one of these values is always in (0,1) you are good

[pseudo coral]

considering points between -1 and 1 and outside of -1 1

showing one lands inside of (0,1)

oh i see it now i think !

[lament needle]

yeah its a little weird

[pseudo coral]

lol right?!

[lament needle]

but we are just proving existence

not that the thing that exist is nice

In general math is mildly mean.

[pseudo coral]

tell me about it lol

[lament needle]

But we try to make it nice

[pseudo coral]

i think i got it, I need to consider cases for r i think

to show only 1 lands in open unit interval

[lament needle]

so you can show that if r is on a certain interval you can find A inverse(not a unique inverse) in (0,1) and if r is on another interval you can find a inversem and if this works forall reals then you have surjected

[pseudo coral]

yep

[lament needle]

you dont actually need to show only 1 lands in the unit interval

[pseudo coral]

thats what i was thinking

[lament needle]

because you claim to have already shown injectivity

[pseudo coral]

i think i do need to show one of them lands inside of (0,1)

if both land in there then its not 1-1

if $r \in [1,\infty)$ then the first one is in the interval

but not the second

[gentle osprey]

MyMathYourMath

[lament needle]

yeah but you can show injectivity and surjectivity independtly

[pseudo coral]

if $r \in (-\infty,-1]$ then the second is in (0,1)

[gentle osprey]

MyMathYourMath

[lament needle]

you skipped over the inteval (-1,1)

[pseudo coral]

i know im working my way to it

lol

what if r = 0

theyre both in (0,1) 😦

[lament needle]

I think your inverse function is wrong

so here i plugged your inverse function in when r = .4, both your inverse functions gave me diffrent values in (0,1). When i plug in one of these values it spits back out the desired value of .4 but for the other one it doesnt work

so in other words you have two inverse functions $g^{-1}$and $h^{-1}$ when i do

$f(g^{-1}(r)) = r$
however
$f(h^{-1}(r)) \neq r$

[gentle osprey]

matthew

[lament needle]

for the function to be an inverse it needs to behave like g^-1

[pseudo coral]

so only one of them is the correct inverse?

[lament needle]

... i think so

[pseudo coral]

hhmmm

[lament needle]

we should expect your inverse functions to agree

[pseudo coral]

i checked on wolfram just now and it got the same thing as me

[lament needle]

on the 0,1 interva

i could be bad at math show me your wolfram

[pseudo coral]

here

[lament needle]

yeah so see the $\frac{2r+1}{r+1}$ breaks over the interval that i put

[gentle osprey]

matthew

[lament needle]

wait a second

[pseudo coral]

you mean 2r+1/2r+2?

[lament needle]

wait my bad thats the one that worked t was the -1/2a-2 that broke

so the $-1/(2r-2)$ is not an inverse on the earlier interval.

[gentle osprey]

matthew

[pseudo coral]

OH

[lament needle]

or less than -1

[pseudo coral]

if r=0 then you get 1/2 for both values!

omg

how did we miss that

so its the same element of (0,1)

!!!!

ayyyyyyyyyyyyyyyyyy

it works !

🙂

[lament needle]

wait uhhh i think the interval where it breaks is (0,1)

uhhh the point is to be careful on what interval you use what inverse

[pseudo coral]

yep

ur right

[lament needle]

because if you think about your function the absolute value is flipping stuff around depending on wether x>=.5

right so this is the really wonky value

in terms of (0,1)

[pseudo coral]

yeah 1/2 is the break point i think and -1/2

i think..

wait now im stuck again, at least we figured out the r=0 case

and from -\infty to -1 and 1 to positiv einfinity

[lament needle]

ok i see whats going on. These are your inverse functions

you want their y values to stay between 0,1 because that is where you are mapping back to. They are both defined on most of the reals besides between1, -1

They do not always agree. You always want them to return an inverse in 0,1

Beyond this interval it should be clear what function should be your inverse

[pseudo coral]

so far

[lament needle]

If you look on wolfram alpha the blue function which is -1/2x-2 does not define your inverse on the interval

[pseudo coral]

$r \in (-\infty, -1]$ the second is in (0,1) the first isnt

[lament needle]

(0,1)

[gentle osprey]

MyMathYourMath

[pseudo coral]

$r \in [1,\infty)$ then the second one isnt in (0,1)

[gentle osprey]

MyMathYourMath

[pseudo coral]

if r = 0 theyre both 1/2

so just concerned now with (-1,0) and (0,1)

[lament needle]

if you look at -1,0 the green function veers off towards negative infinity and the blue function behaves nicely

the opposite is true on 0,1

and this lines up with what wolfram alpha is saying

wolfram alpha has told you where these functions behave like inverse

[pseudo coral]

so on one of those intervals one is in (0,1) and on the other the other is in (0,1)?

[lament needle]

yeah

and the give eachother a little smooch at r=0

[pseudo coral]

lemme do some calculating

makes sense

[lament needle]

or high five whatever you prefer

[pseudo coral]

lol

[lament needle]

ok you can take it from here

sorry i got lost in the sauce

we did tell you we dont know how to calculate

[pseudo coral]

lol all good

thanks so much !

so if $r \in (0,\infty)$ the first one is in (0,1) but not the second

[gentle osprey]

MyMathYourMath

[lament needle]

uh that is what it seems

Check this though

[pseudo coral]

how come as r=1/4 both are in (0,1)?

[lament needle]

this is fine but they are not both inverses in this region

[clear jackal]

If $F = \operatorname{Hom}_{Top}(-, X)$ is a representable functor from Topological spaces to Groups, is it necessarily (naturally isomorphic to the functor) induced from a topological group structure on X?

[gentle osprey]

datorangeguy

[clear jackal]

I think I just need that F(singleton) is a topological group structure on X but I don't think that's guaranteed

sorry if this is too algebra for the topology chat but I hope this is understandable and interesting to ATers

[coarse night]

if you mean to ask that if the functor takes values in groups then is it necessary that X is a topological group, then yes

you can recover the group structure on X by taking F(X)=Hom(X,X).

• f is a function from X to X and F(X) is a group, so there is a inverse function f^(-1), just use this to construct the identity and inverse of each element of X.

I believe it is in general true that if a functor
$Hom(-, X): \mcal{C}^{op} \to Set$
factors through the category Grp, then $X$ is necessarily a group object in category $\mcal{C}$.

[coarse night]

but someone can correct me if I am wrong

[clear jackal]

wow ok cool! That sounds right I might've heard that in the context of hopf algebras but forgor

[devout sorrel]

Is it possible for A \cap (A^c)' to be nonempty when A is a closed set?

' representing the set of all limit points

[wise ruin]

For sure! Let the interval [0,1] be the space in question. Then A=[0,0.5] is an example. A^c = (0.5, 1], so (A^c)'=[0.5, 1]. So, A cap (A^c)'={0.5}

In fact, as long as A^c isn't clopen, A cap (A^c)' is nonempty.

[devout sorrel]

Hmm i see

[candid hedge]

How to show that C without a point is connexe

i took 2 points a,b. Now if the point we removed isnt on the segment

ta+(1-t)b

is a continuous path and were done

if the point is on the segment, should i take a point c

[vocal wharf]

does this argument continue?

[gritty widget]

You can prove it's image of a product of (0, infinity) x S^1

And S^1 is also an image of interval

So it's an image of continuous sets

Alternatively, decompose it into sets you know are connected

Or just show it's path connected

You could show that R^2 without a countable amount of points is connected by considering segments like you just did too

You have a continuum amount of paths from a to b which are only not disjoint at common point a, b

This means that removing any set of points < continuum results in a connected set

You do this by considering an intermediate points sitting at a line perpendicular to your interval

[gritty widget]

Same for R^n with n > 1

[small hemlock]

What is a good book for introductory category theory?

[lunar yoke]

Riehl's Category Theory in Context

[gritty widget]

Not sure why you're asking here. #book-recommendations or #category-theory would be a better place imo.
There's also book by Mac Lane

[small hemlock]

No I totally forgot about those channels, that's my bad. Thank you though

[coarse night]

Let's say G is an abelian topological group and X be arbitrary. Let $N$ be the connected component of $G^X$ ($G^X$ is a topological group). I need to show $G^X / N \simeq [X, G]$.

[lunar yoke]

Finally I can learn about group theory

[pliant hornet]

I'm a little confused about this definition of connectedness.

Is it the case that, say {0, 1} is considered a connected subset of C, despite it not seeming like it should be connected, intuitively?

[lunar yoke]

No, {0,1} is not connected in C

maybe you're missing that the G_1 and G_2 only have to be open in G, not in C

so we can take G_1 = {0} and G_2 = {1}

unless you're working with some fucky spaces connectedness usually means what you'd expect it to mean

[pliant hornet]

I see. The text I'm working from only defined being open in C. This is an analysis text rather than a topology one, so perhaps it's an oversight. Thanks.

[gritty widget]

Yeah. They're talking about G being connected as in the subspace topology

that is, open sets in G are intersections of open sets in C and G

[gritty widget]

If X is a CW complex, and A a subset

is (X,A) always a relative CW complex

or do we need A to be a closed (if A is closed then it is a subcomplex and certainly (X,A) is a relative CW complex)

[lunar yoke]

this seems wrong

take R^2 and A the hawaiian earring

latter is compact hence closed in R^2

R^2 is CW

but i think hawaiian earring is not a cw complex

[gritty widget]

yeah you are right

[unreal stratus]

ye you need to be a union of cells (and closed) to be a subcomplex innit

[gritty widget]

yeah you are both completely right

thank you

[gritty widget]

when you say a union of cells

this should be equivalent to

for any a in A

if a lies in e^n

then all of e^n lies in a

where e^n is the image of some D^n under the characterstic map

[steel glen]

a surface is smooth on the complement of the support of it's gradient by definition correct

oh i had it backwards

i'm trying to find a motivation for the support of a function and i figured smoothness could be one

what i was going for is
if F:R^3 -> R, F=0 is a surface, we can say F=0 is smooth on supp(gradF)
but that's not necessarily true since {(x,y,z) | gradF(x,y,z) =/= 0} \subseteq supp(gradF)

also i guess supp(gradF) could have points such that F=/=0. so it’s just a really bad example lol

[gritty widget]

What can we do about this?

[woven sundial]

for part 1, note that if $A \cup B = U \cup V$ for $U,V$ disjoint, then $A = (U \cap A) \cup (V \cap A)$ and likewise $B$. but you need to show $U \cap A$ and $V \cap A$ are both nonempty, which is why you need to $A \cap B$ to be disconnected.

[gentle osprey]

twiceshy

[gritty widget]

Ahh.

I can't figure out what is claim in your argument

Can you please explain it in more simple way

[gritty widget]

If intersection of connected sets is non-empty then their union is connected

And if its empty then its connected

For 2) use rational numbers

[gritty widget]

does the neighborhood of a point contain the point itself?

[unreal stratus]

What is your definition of a neighbourhood of a point?

[gritty widget]

the set of all points in a set that’s distance from a main point is less than a r. O wait I just realized lmao

[unreal stratus]

Yeah aha

Although neighbourhoods are often defined in a more general way

I guess you would call that an r-neighbourhood or something

This looks like rudin chapter 2 or something though

[gritty widget]

yes, always

in analysis you consider "deleted neighbourhoods" where you remove it

[coarse night]

can anyone give an explicit description of the boundary map in the Mayer-Vitoris cohomology sequence

only the $H^0(U \cap V) \to H^1(X)$

[coarse night]

I do not understand how a map from U\capV to Z can give a map from X to S^1

[plain raven]

Let f be a 0-cocycle defined on U \cap V, so a (perhaps discontinuous) function from U \cap V to Z.
I define a function g which sends cycles in X (kind of Eulerian paths) to elements of Z.
If sigma is a path in X, decompose sigma into a sequence of paths sigma1, sigma2, .... all concatenated from head to tail, and where each of sigma_i lives in either U or V. Let us write this as a pair (\sum sigma_i, \sum tau_j), where the sigma_i are in U and the tau_j are in V. The boundary of the combined sum \sum sigma_i + \sum tau_j is zero by assumption that \sigma is a cycle, but it is not necessarily the case that the boundaries of \sum sigma_i and \sum \tau_j are both zero, only that they are equal and opposite formal linear combinations of points.

All the points in this formal linear combination \sum d \sigma_i = -\sum d\tau_j must live in the intersection U \cap V, because \sum sigma_i is in U and so \sum d sigma_i is in U, and \sum tau_j is in V, so \sum d\tau_j is in V.

Now we have a formal linear combination of points in U \cap V, and we can apply f.

Does that make sense

[coarse night]

actually we haven't been introduced to cycles/cocycles stuffs yet. Our instructor has asked us to prove this bare hand.

[plain raven]

if you haven't been introduced to cycles i need you to give me your definition of mayer vietoris because the definition i learned was in terms o fcycles

actually what is cohomology

let's start there

i assumed you meant singular cohomology but that was probably a faulty assumption

[coarse night]

we aren't even introduced to cohomology, he said that given G a top group, [X, G] becomes a group. Hence define H^0(X) = [X, Z] and H^1(X) = [X, S^1]. Then we are asked to show the sequence is exact

,tex \begin{tikzcd}
0 \ar[r] & H^0(X) \ar[r]&H^0(U) \oplus H^0(V) \ar[r]&H^0(U \cap V) \ar[r, "\delta"]& H^1(X) \cdots\
\end{tikzcd}

[coarse night]

he has also given the map f: X -> S^1 then (f|U, f|V), then (f, g) -> f-g

[plain raven]

so he didn't give you the definition of delta you're supposed to figure it out on your own?

[unreal stratus]

Is this like cohomotopy cohomology theory

Interesting pedagogy

[coarse night]

yes

[plain raven]

can you tell me what spaces you're considering

is X an arbitrary topological space or a CW complex or what

[coarse night]

he didn't say anything but I assume it's LCH

[plain raven]

where did LCH come up prior to this

do you expect some fact about LCH spaces to be relevant

[coarse night]

to give a topology on S^1 ^X?

he did ask us to prove (S^1)^X/connected comp of the identity is == [X, S^1]

[unreal stratus]

i mean you can still define compact-open topology anyway

but ig being lch is nice for evaluation map and currying ofc

[coarse night]

yes

gives as that homotopic maps are path connected and those stuffs

[little hemlock]

what would be an example of singular simplex that misses a subspace A but its boundary does not?

[plain raven]

take X = the n-simplex

take A = its boundary

then the identity map \Delta^n -> X misses the subspace A, but its boundary does not.

[unreal stratus]

Is there any nice place to look for an intro to the Serre spectral sequence?

I guess there's hatcher but i generally don't like the style

[plain raven]

The book by Switzer has a good chapter on SS'es

[unreal stratus]

Noice thankss

[little hemlock]

in what sense is this true? The identity \Delta^n -> X is surjective. I'm not sure I understand how it can miss A

[plain raven]

The image has to be contained in the subspace

X is not a subset of A

[little hemlock]

by "miss A", do we not mean the image is disjoint from A?

[plain raven]

i'm looking at the screenoshot - the singular n simplices whose image is not contained in A.

If that's not what you mean i'm confused

[little hemlock]

yeah... i see i misunderstood 🤦‍♂️

[wise ruin]

In the Van Kampen theorem, what is the intuition behind the requirement that all 3-fold intersections are path connected? Hatcher provides an example where the lack of that condition causes an incorrect calculation of the fundamental group - namely, he considers the suspension of three points and the open cover is the complements of each point.
I understand that this fails, but I don't quite understand what's going wrong.

[candid hedge]

why are finite dimention spaces closed?

By the linear stability of vector spaces

?

[gritty widget]

Because they are complete

[broken nacelle]

Does there always exist a continuous map between topological spaces?

[swift fjord]

A constant map is always continuous

[broken nacelle]

Right

Lol

[unreal stratus]

Prove it

[swift fjord]

Let f:X->Y be constant at some y\in Y, and let U be open in Y. Either y\in U in which case f^-1(U)=X or y\notin U in which case f^-1(U)=\varnothing and in both cases is open, so f is continuous

[unreal stratus]

Lol you actually could be bothered

[gaunt linden]

Because any two norms on a finite-dimensional space are topologically equivalent -- in particular they're all equivalent to the Euclidean norm on R^n which we know is complete. The proof of this hinges on the fact that if you take the norms of all the vectors in a basis for the space, they will be bounded above and below by some fixed positive numbers -- which is not necessarily true if the basis is infinite.

[gritty widget]

And linear isomorphism preserves completeness

I think it's easier to just say "any linear map between finite dimensional spaces is continuous"

[tough imp]

Is there a formal sense in which the fiber and cofiber sequence are dual?

By this I mean, if one shows that say, the cofiber sequence exists, can one use some fact about loop space and suspension being adjoint, along with cones and mapping fibers being adjoint in order to deduce the existence of the fiber sequence

Without just proving it exists and essentially writing down a dual argument to that uses to demonstrate the existence of the cofiber sequence

[bitter smelt]

Here is a proof with fewer words and more pictures, and I think the last paragraphs explain explicitly why the triple intersection is needed for it to work

https://www.google.com/url?q=https://jlmartin.ku.edu/courses/math821-S14/VanKampen.pdf&sa=U&ved=2ahUKEwjTjZTYnrX6AhVDIn0KHUVqA8gQFnoECAkQAg&usg=AOvVaw1CPJvIw1iZv9UP7it30aEf

I'm not sure if this gives the intuitive idea you were looking for, but this is the resource I used when I couldn't get Hatchers VK statement and proof (I mean seriously, who doesn't include the diagram?)

[obtuse meteor]

yeah I mean the trick here is that when you nudge rectangles around the best you can hope for is to get a max of three things intersecting at any place

[bitter smelt]

The triple intersections allow your factorization to simplify, otherwise things couldn't be homotoped nicely

It's a really pretty proof. Van kampen was a clever guy

[candid hedge]

Is a space closed if and only if its complete?

!!!!!

and how do we show that Q isnt complete using the sequence xn+1=xn/2+1/xn

Or at least how do we show that this sequence is cauchy

[gritty widget]

No

Completeness doesn't even make sense for most spaces

Also wdym by closed space

[candid hedge]

for the spaces where completeness makes sense, is closed and completeness equivalent

[candid hedge]

sequentially closed

I mean set

[gritty widget]

That only makes sense for subsets

Also, it's still no. You need your space to be complete at least

[candid hedge]

okay

[candid hedge]

and this

[gritty widget]

A subset of a complete metric space is complete (with the same metric) iff closed

If we can switch metrics then we're talking about being completely metrizable, and that's equivalent to being G_delta instead

[gritty widget]

Show that it's convergent to sqrt(2) (I think)

[candid hedge]

yeah but how

x=x/2+1/x and then i dounf that its limit is sqrt(2) if it was convergent

but why is it convergent

[gritty widget]

That's something you do when justifying Dedekind cuts in construction of R

Iirc

Not really topology anyway

Q isn't complete because it's not closed in R

It's not even G_delta, so it can't be given a complete metric either

Sad space

[unreal stratus]

Sad!

[lone merlin]

I can't find the english word for the property of an \Omega where every point in \Omega can be connected

[gritty widget]

can you write the property in mathematical notation

i think you're talking about path connectedness. but just to be sure

[lone merlin]

yess it is that

thank you

does anyone know a link to a proof of

if a vectorfield $\vec F \in C^1$ and irrotational in an open simple connected area $\Omega$, then is $\vec F$ conservative in $\Omega$

[gentle osprey]

ɥɾʞʞɥʌzd

[lone merlin]

because i don't understand why it can't just be a path-connected area

maybe if anyone would know a counterexample

[wise ruin]

The vector field <-y,x>/(x^2+y^2) on the (path-connected but not simple) punctured plane R^2-{0} is irrotational and not conservative

[gritty widget]

irrotational
vector field on R^2

[lone merlin]

owh god that was an example in the lesson 😭

i guess there is no proof needed

[gritty widget]

the procedure for defining a potential function f typically goes as follows:
-pick a base point p in \Omega
-for x in \Omega, you want to define f(x) to be the line integral of F over a curve in \Omega from p to x
-but you can pick any curve, so why should this be independent of the choice?
-pick two curves from p to x. if the region is simply connected, you can find a surface S in \Omega whose boundary is the concatenation of the first curve and the reverse of the second
-since the vector field is irrotational, the integral of the curl over S is zero. but by stokes' theorem this is the line integral over the first one minus the line integral over the second. this is zero, so f(x) is well-defined

if \Omega is not simply connected, you can't do the fourth step of picking a surface

something along those lines. details omitted for the sake of giving the general idea

[lone merlin]

I guess I'll wait until I see line-integrals

i'll save this comment

thank you ❤️

[gritty widget]

i don't like doing this but i should remark on alex's example

"irrotational" refers to curl zero, and curl is only defined for vector fields on three dimensional spaces (typically). alex's example is defined on two dimensional space

but it's still a good example of this general phenomenon, and you can do something very similar to get a counterexample in three dimensions

[lone merlin]

you can extend it to three dimensions

but i think you know that

[cedar pebble]

it's not that you can it's that you have to for this to make sense

but the basic idea is still there

[lone merlin]

can you do curls in 7 dimensions?

[cedar pebble]

good question

[lone merlin]

because I know crossproduct has also a kind of definition in 7 dimensions

don't know how it looks

[cedar pebble]

essentially yes, since that's the other dimension where cross products exist

it's more complicated and doesn't enjoy exactly all the properties that the cross product and curl do in 3 dimensions

[lone merlin]

is this even the best channel for nabla operators?

[cedar pebble]

it's fine

anyways one big problem (or feature depending on how you look at it) is like

curl has something to do with rotations, and indeed the cross product on R^3 is invariant under the group SO(3)

cross products on R^7 aren't quite invariant under SO(7)

it's only G_2 invariant

[gritty widget]

i fucking KNEW the cross product on R^7 sucked omg

i didn't have a reason but now i do

[lone merlin]

this is exciting👀

[cedar pebble]

so somehow this is telling you something about "special" rotations, namely those that live in G_2

this is definitely really weird!

[lone merlin]

owh god I don't know Lie groups

any sources that have a kurzgesagt document?

[cedar pebble]

G_2 is the automorphism group of the octonions. It's the smallest exceptional Lie group

the cross product on R^7 comes from the octonions too, so G_2 is just the group coming from symmetries of this

[lone merlin]

owh I guess that makes sense as well

[cedar pebble]

yeah idk if this is useful at all haha

[lone merlin]

automorphism is a permutation of the elements right?

in a group

or is a permutation an example

[cedar pebble]

sure yeah, in a way that respects the group operations

the other slick way to say it is like

[gritty widget]

(ab)^2 = a^2b^2 in groups isnt always true, right?

[cedar pebble]

the group of rotations SO(7) of R^7 has something called the spinor representation, this is an 8-dimensional representation

pick any vector in this spinor representation

G_2 is the subgroup of SO(7) that preserves this vector

so you can think of elements of G_2 as "special" rotations

it's just the way you distinguish which rotations are "special" is weird

[lone merlin]

so a spinor is only existent within the octonions

[lone merlin]

so SO(7) aren't that special anymore

[lone merlin]

didn't that have to do with the determinant being 1?

or are we talking of SO in a more general sense?

[cedar pebble]

the S in SO means orientation preserving

so determinant 1

O has determinant \pm 1

[lone merlin]

so where does the crossproduct become variant since it's just developing the rows?

[cedar pebble]

yeah I guess it's kinda hard to see why this cross product is only G_2-invariant from the definition

I guess what you can do is like

pick an element of SO(7) that is not in G_2 and apply it to some elements and see what happens if you want to cook up explicit counterexamples to SO(7) invariance

but idk how enlightening this will be

[lone merlin]

i am only second year in my engineering degree so probably not much

but thank you for the effort

are there any good video's on octonions or lie groups you would recommend?

to give a general idea

[cedar pebble]

to be fair I study stuff where the octonions actually come up, and idk if I've ever used the 7-dimensional cross product in my life lmao

[lone merlin]

documents are also welcome :))

[cedar pebble]

John Baez has a fantastic expository article about the octonions

[lone merlin]

yeah my professor also said it just exists because it can

[cedar pebble]

and the exceptional Lie groups

https://math.ucr.edu/home/baez/octonions/

[lone merlin]

do they work with octonions in physics?

searched it up apparently a lot

damn

[cedar pebble]

Not really. There are some attempts at deriving parts of the standard model (e.g. finding U(1)xSU(2)xSU(3) inside something coming from the octonions) and these are sorta successful but I'm not sure this actually gives any additional insight

the main utility of the octonions is in constructing some of these exceptional Lie groups. G_2 arises as the automorphisms, but the other exceptionals F_4, E_6, E_7, E_8 arise in some more complicated ways from the octonions as well

[lone merlin]

allright i'll look into it this weekend thanks a lot

[lone merlin]

btw any requirements before starting it besides some group theory?

[cedar pebble]

if you're comfortable with group theory and linear algebra you can definitely start learning some Lie theory if you want to

also knowing smooth manifolds helps for Lie groups but you can get away with studying Lie groups and their Lie algebras just in terms of matrices so it's linear algebra again

[lone merlin]

can not thank you enough

[gritty widget]

does ab = ba in groups?

[gaunt linden]

Not in general, but that question belongs in #groups-rings-fields, rather than here.

[gritty widget]

ok mb

[sleek thicket]

Probably the right context is something infinity categorical

[tough imp]

well, okay

I'm out

hahahaha

[sleek thicket]

like the defintion of stable oo category is kind of

"these coincide"

Okay okay actually

Thinking

Yeah so

Still model categorical but it kind of has to be to phrase your question

take f : X -> Y

Alex how do you feel about homotopy limits in model categories

[tough imp]

[sleek thicket]

I think the infinity picture is genuinely simpler

[tough imp]

I don't know

[sleek thicket]

And you will not hate it

[tough imp]

what they are

I mean sure

idk just say words

and I'll go "okay" and maybe they pop up again later

[sleek thicket]

So the kernel of a morphism f in an abelian category

[tough imp]

yee

[sleek thicket]

f : A -> B

Is the pullback of f with pt -> B

pt is the zero object here

Yeah?

[tough imp]

uhh

yes

[unreal stratus]

uhh

[sleek thicket]

Okay and the cokernel

[tough imp]

is the pushout

[sleek thicket]

Is the pushout with B -> pt

yeah

[tough imp]

yeah

[sleek thicket]

These are the defintions of the homotopy fiber and cofiber in an infinity category

this is the fiber

cofiber is the pushout

[tough imp]

uh

[sleek thicket]

Note the underlined the

[tough imp]

lmfao

[sleek thicket]

Anyways

the point is

[tough imp]

let me think like

[sleek thicket]

This makes them completely formally dual

[tough imp]

isn't this also like

the normal way to define

the mapping fiber?

Is it not as a pullback via * -> B

ah it is not

[sleek thicket]

Yeah that's only if f is a fibration

Right?

[tough imp]

it's the pullback along FY -> Y

[sleek thicket]

And this is something model categorical

[tough imp]

right

I see I see

so if you take like fibrant replacements or whatever

[sleek thicket]

Exactly

[tough imp]

and cofibrant I guess for the dual one

then you can describe them in a totally dual manner

[sleek thicket]

So yeah the simplest duality is really "pullback vs pushout in infinity category"

[tough imp]

or like, actually formally dual

I see

[sleek thicket]

There is a story for model categories

[tough imp]

okay

[sleek thicket]

And it'll be similar

[tough imp]

gotcha

cool

[sleek thicket]

But idk this is less complicated imo

[tough imp]

[unreal stratus]

fibre bundel

[sleek thicket]

I've been thinking a lot about this lkke

[tough imp]

potato bundle

[unreal stratus]

yes!

[sleek thicket]

fiber = cofiber = kernel = cokernel

Bc that's what the point of triangulated categories is

[tough imp]

chmonkas

[tough imp]

Call that a Macca's large fries

[tidal cedar]

oh yeah

[gentle osprey]

potette

[tidal cedar]

i mean it's less kind of and more of exactly, triangle is fiber sequence iff it is cofiber sequence

[sleek thicket]

ywah

[tidal cedar]

when is a vegetable not compactly generated that sounds fun to eat

[sleek thicket]

that's what I said

[tough imp]

he said "kind of" to not scare the Chmonkey

[tidal cedar]

I'm saying it's not kind of it's just they do on the nose

fair

[unreal stratus]

These are secrets that would crash the global economy

[tidal cedar]

I've been trying to tackle these and triangulation still fucks my tiny brain

[unreal stratus]

But it's important to work in that category so that products of vegetables work nicely

[sleek thicket]

I've been doing a lot of triangulated stuff recently

It's not as bad as I thought

[tidal cedar]

the TR4 seems nonsensical

[sleek thicket]

Octahedral axiom?

[tidal cedar]

tbh most of it does I have no idea what it's trying to do

[tough imp]

boy oh boy does Shamrock have thoughts about that

[tidal cedar]

yeah

[sleek thicket]

I thought so too

[tough imp]

I mean the other ones seem sensible?

IMO

[tidal cedar]

can I just intuit it

[tough imp]

It's second iso

[tidal cedar]

"the shift functor is basically ..."

[sleek thicket]

yeah so basically what it says is that the mapping cone of M(f) -> M(g°f) is M(g)

M here denote mapping cone

[tidal cedar]

[sleek thicket]

But mapping cones aren't defined by universal property

there's very little coherence

They're unique up to nonunique iso etc etc

[tough imp]

lmao

what a shit object

[sleek thicket]

So it says this is true and you get coherence

So really you start with the first two triangles

And you can choose the third one but you can also just complete vu to an exact triangle

[tidal cedar]

ughhhh

[sleek thicket]

And this tells you the mapping cones Z', Y', X' fit into an exact triangle

With nice maps

[tidal cedar]

so like our motivation here is

shifting a chain complex

right

[sleek thicket]

The exact equations are really saying "all diagrams commute"

yeah

[tidal cedar]

hmm

this stil feels so alien

the mapping cone then is the mapping complex?

[sleek thicket]

So I guess exercise if you want, show that the mapping cone of g is quasi isomorphic to the mapping cone of g°f

No the mapping complex is internal Hom

[tidal cedar]

[sleek thicket]

The mapping cone is like

Well it's the homotopy cofiber lmfao

Sorry

you take f : C -> D

you get a complex cone(f) which is easy to write down

It comes with two maps C -> D -> cone(f) and D -> cone(f) -> C[1]

[tidal cedar]

hmm

ok

so those are those two maps in whatever axiom of the triangulated category?

[sleek thicket]

yep

[tidal cedar]

and we say a third one is also exact

[sleek thicket]

it's all very explicit

[tidal cedar]

wait what is that third one then

[sleek thicket]

You should look at weibel

What third one?

Ah the third axiom sorry

It's a weak version of functionality

*functorality

So for triangulated categories we do not have a functor Arr(T) -> T taking the mapping cone

[tidal cedar]

yeah I thought there was one that's like almost permutations of the exact triangles

[sleek thicket]

oh that's different

That's just like

Kind of the point imo

That triangles aren't oriented

[tidal cedar]

o

does it make sense if I draw these as actual triangles and then like rotate by 120°

[sleek thicket]

Sure

That's doing the shift operation

people draw it like that, usually with a +1 for the map C -> A

Also, lmao, this was in a book I was looking at recently

[tidal cedar]

Jesus

[sleek thicket]

Lol

[tidal cedar]

I may just need to wait for more motivation of these to work through this

[sleek thicket]

Yeah

The point of them is sort of similar to model categories

Extra structure on a 1-category (a homotopy category) that lets you do a lot of things that infinityy or homotopyy but still tractable and computational

That's my take at least

Like you can define homotopy limits and colimits using the triangulated structure

[tidal cedar]

oh

that is cool!!

[bitter smelt]

I wrote the solution to a quadratic and the prof circled it and wrote "details?"

I hate my life

[unreal stratus]

Ouchies

[bitter smelt]

Grad students are dumb it's true but maybe we can solve quadratics? Idk?

This was most certainly not topology related lmao

[gritty widget]

[long hornet]

Source?

[sleek thicket]

Something about tensor triangulated geometry

Tensor triangular geometry by Balmer

[coral pawn]

What is the smash product in the category of simplicial sets?

[lunar yoke]

You can define it as usual. Given pointed simplicial sets X,Y you have a map X u Y → X x Y. Taking the cofiber gives the smash product.

[lunar yoke]

One can then check that pointed simplicial sets are closed symmetric monoidal with this

[coral pawn]

Gotcha

[lunar yoke]

A nice reference with most of the details on this is Riehls Categorical Homotopy Theory, chapter 3.3

[coral pawn]

By S_s^1, do they mean (constant sheaf of S^1) \coprod {*}?

[coral pawn]

Similar question here: what is the morphism Spec(k) ---> A^1 - {0}?

The morphism I'm thinking of is the one given by evaluating at 1, but I'm not sure if its the same

k[x,y]/(xy-1) ---> k given by f(x,y) ---> f(1,1)

[wide egret]

hello all, i have a question regarding knot theory
why aren't 90 degree rotations included in the set of valid tangle mutations, that is, why isn't it all the transformations in D_4, but just D_2?

[sand hill]

Why are T4 spaces called normal spaces

strange choice of terminology no?

[pearl holly]

I'm sorry if this is dumb but what does Omega^infty Sigma^infty mean in the start of the second page in this pic? Like is it colim Omega^n Sigma^n?

[gritty widget]

It's a space with nice properties. If you want to learn more about origins of its name, maybe you should take a look in some book about history of topology? I doubt it'll give you much different explanation - they are nice, hence, normal

[pearl holly]

this is from May's "geometry of iterated loop spaces" which I'm reading to hopefully learn some operad stuff, but tbh I have like no background in loop space stuff so maybe I should learn more about them before going in to this lmao. So I also want to ask if there are any nice intro's to loop space stuff with motivation etc.

or maybe I should be reading something completely different if I want to learn operad stuff, I'm not sure. Any suggestions?

[sand hill]

I see

Thanks

[opaque cloud]

hausdorff spaces are very nice 😌

[pearl holly]

Okay nice, thanks!

[bitter smelt]

Is a crab pincer with a hole homotpic to torus wedge circle

I wanna say no

But then what is it 🤔

Oh is it just a pinched torus

It feels like it shouldnt be since it started as a genus 2 surface…

[little hemlock]

if by pinched torus, you mean a torus with 2 points glued together, then yes i believe so. ive used this on hatcher exercise anyway

[bitter smelt]

Yeah I'm currently doing maybe the same exercise

2.1.17

I just don't immediately see the homotopy to the pinched torus. I'll play with it a bit more

[echo dove]

Imagine the crab claws being a bit slimy (as topological crab claws so often are). If you pull them apart, you'll end up with a thin string of goo connecting the tip of one claw to the other. Keep going, and you can deform the claws all the way back to the torus, leaving a loop of goo attached to the side.

[little hemlock]

yea, same exercise. I didn't explicitly construct a homotopy or anything tbh

[bitter smelt]

I did some quick sketches of this and ended up with torus wedge circle. Ill

I'll draw more carefully tomorrow and see if I don't get that again

Obviously I'm not gonna go through the process of showing the goopy strip and the claws form a CW pair and so blah blah blah, but just getting the pictures down right

[rancid umbra]

the crabs joints break back right?

[echo dove]

Not sure what you mean, but topological crabs are always in a just-molted state. The shell is surprisingly flexible.

[candid hedge]

is there some signficance to this sequence?

where (xn) is a bounded sequence

Not an important question, just wondering

[flint sinew]

is the standard topology in some way analogous to base-10 number system in that it's not inherently special in any way, but we're just using that as the default by convention?

[hidden crag]

what is the standard topology for you here

[unreal stratus]

I guess I would turn this on its head and say that topology was meant to generalise stuff like this tbh lol

[hidden crag]

yes i agree

[unreal stratus]

I mean it is also the most natural topology associated with the order on R for example

[hidden crag]

i mean "standard" is just not a thing for general topological spaces

I might be misunderstanding the question but idk

[gritty widget]

it is for R^n

[crisp helm]

Different topologies correspond to different things

Unlike how if you change the base of your number system you're still referring to the same underlying concept

[gritty widget]

I'd say the standard topology on R is pretty special. It's the unique topology which gives it a (Hausdorff) tvs structure for example

The question would maybe be more worthy of discussing if it were asking about, say, metrics/norms on R

well... the norm is also pretty unique here. Assuming we know that 1 is of norm 1

up to a constant it's the only metric induced by a norm

[unreal stratus]

I think I'm being silly, lol. If $M$ is some (connected, not compact - though doubt it's relevant at all) n-manifold, $x,y \in M$ and there's a path from $x$ to $y$ with image $L$, why is $H_n(M, M \setminus x) \cong H_n(M, M \setminus L)$?

[gentle osprey]

potette

[unreal stratus]

I imagine the point is you can use excision to argue that we can replace M with some open subset of R^n? But it still seems messy in the case that L has some self-intersections or something (maybe we can just pick a nice enough path though)

[cursive vigil]

What does this notation mean, I keep seeing it

[dry jolt]

It usually refers to the identity map from X to itself

[cursive vigil]

Thx😎

[bitter smelt]

In this context it is indeed what it means 👍

I've never seen it mean something else, actually!

[unreal stratus]

Well, in analysis I suppose it could be an indicator function of X, but here it's not gonna be that

[gritty widget]

I wish I could help you potato but I haven't learned any algebraic topology

[unreal stratus]

Dw Blitz aha appreciate it though

[bitter smelt]

Sorry potette I'm about to head out actually but the case of n=0 should be obvious (assuming L isn't separating if x is not) and for n>0 try passing to the quotient and using homotopy invariance. May not work but worth a shot

If that doesn't (morally I feel like it should) I can VC some time during this 5 hour drive B)

But you are right in that we are missing some sort of hypotheses. Like if x and y are the same point and L bounds a disk

[unreal stratus]

Yeah

[unreal stratus]

(thanks btw)

[steel glen]

my book is defining a fuzzy topology for a set X as $\delta\subseteq I^X$ satisfying:\
$\forall \alpha \in I,\alpha \in\delta (?)$\
$\forall A,B\in\delta\implies A\land B \in\delta$\
$\forall(A_j){j\in J}\in\delta\implies\lor{j\in J}A_j\in\delta$

[gentle osprey]

maximo

[steel glen]

i don’t understand that first condition

how can a family of fuzzy sets have real numbers in it

[gritty widget]

@steel glen If we think of elements of delta as functions, then the first condition probably means that all constant functions are in delta?

[steel glen]

ok that makes sense to me as well, but i thought that would be way too strong of a condition

most other texts say the 0 and 1 constant functions need to be in delta

maybe my author is just weird

[gritty widget]

@steel glen check out paper Concerning the Constants in Fuzzy topology by Lowen and Wuyts
It explains that if we include 0 and 1 that definition is by their terms called quasi-fuzzy topology.
It was original definition and people still use it, without explanation for why.
And what they call fuzzy topology is which includes all constants

[steel glen]

ok good to know. thanks for taking the time to search it up

[gritty widget]

👍

[tidal cedar]

Are there any spaces that are connected and nowhere path connected

[gritty widget]

Wdym by nowhere path connected?

There exist neighbourhood which doesn't contain a neighbourhood which is path connected?

If yes then any solenoid is an example.
It's even a compact group, and as a topological space embeddable in R^3

[tidal cedar]

Mhm, thanks!

that's wild

[feral copper]

Hey there! Probably a dumb question, but I'll try:
If I have two embedded (closed oriented) surfaces F and G in CP², respectively homologous to a[CP¹] and b[CP¹], then they have algebraic intersection number a×b. But can I make the actual number of points of intersection arbitrarily large?

I'd say yes, otherwise the Bézout theorem has nothing to do with algebra and I'm quitting maths x')

Yeah okay, asking the question can sometimes help answering it x')
I can take b=0 (i.e. G is nulhomologous), and move it so that it does intersect F, and I get that the algebraic intersection number is zero but the number of points of intersection is at least one, so this breaks Bézout x')
Sorryyyyyy!

[graceful sequoia]

Hello. I was asked if R is homeomorphic to (a,b) or not. I'm thinking it is. First I showed it for (-pi/2, pi/2). Let R be equipped with the standard topology, (-pi/2, pi/2) with the subspace topology. Let f: (-pi/2, pi/2) --> R s.t f(x) = tan(x). Take any open set U in R. U is either (c,d) or (c,d) U (e,f).

1-) f^-1 (U) = f^-1 ((c,d)) = (arctan(c), arctan(d)) which is in the subspace topology,
2-) f^-1 (U) = f^-1 ((c,d) U (e,f)) = (arctan(c), arctan(d)) U (arctan(e), arctan(f)) which is in the subspace topology and so our function is continuous.

Actually now it just came to my mind that any open set U in R is either (c,d) or (c,d) U (e,f), (c,d) U (e,f) U (g,h) and so on.

Is this correct or should I do it some other way?

[nimble pebble]

have u done anything related to bases of topologies

[graceful sequoia]

Not much but yes we have studied bases

[nimble pebble]

those should be able to get u the result for general open subsets instead of the last part ure trying to do

[gritty widget]

Why do you take U to be union of two intervals?

[graceful sequoia]

I realized later that U can be an open interval or a union of open intervals

[gritty widget]

Of?

[graceful sequoia]

Sorry

[gritty widget]

That's true but the union can be arbitrary

[graceful sequoia]

Yes

Is there any way I can do this without including bases?

[nimble pebble]

this is basically what a basis is

so i wouldnt think so

epdel just uses that epsilon balls are a basis of standard topo

[dapper dagger]

If you need to avoid bases, then how the standard topology was introduced in your case? Usually it’s introduced using base.

[graceful sequoia]

Bases scare me.

I'm trying, hold on

[graceful sequoia]

This is what I came up with

I'm unsure as to how I can justify line 3, that is f^-1 (union of B_alphas) = union of f^-1 (B_alphas)

Other than that tho, is this correct?

the function f here is tan(x), from (R, T_s) to ((-pi/2,pi/2), T_SB)

[nimble pebble]

line 3 is a set theoretic argument

its a good exercise if uve never tried it urself

[graceful sequoia]

Is my proof correct?

[nimble pebble]

ye as long as u attach the paragraphs that justify "arctan maps open to open" and "f^-1(cupU)=cupf^-1(U)" i dont see any problems

[dapper dagger]

It is not true that any open set in R is a union of intervals. There are dense open sets.

It is true that R is homeomorphic to (a,b)

and indeed you can shift tan(x) to do the job

what you need is to know that tan(x) and arctan(x) are continuous

because the real topology is generated by intervals it suffices to check that the preimage of an open interval is open

so your proof works

[graceful sequoia]

I'm not sure how to justify "arctan maps open intervals to open intervals" w/o looking at the graph

[dapper dagger]

Is this a homework question

[graceful sequoia]

Which?

No

[candid hedge]

Intuitively, why are all norms equivalent on a finite dimensional space?
or maybe why they are not on an infinite dimensional one

[lunar yoke]

well i think intuitively you only have to "measure the difference" between two given norms on the finitely many basis vectors, and can thus find a bound on it by taking the maximum, which gives you that the two norms are equivalent

if you have infinitely many basis vectors you may not be able to take the maximum in this way

[candid hedge]

What goes wrong when we're working on the polynomial space for example

[lunar yoke]

this

i mean there just are counterexamples

what the actual space looks like doesnt really matter, vector spaces are uniquely chracterised by their dimension

as long as the dimension is infinite you can find nonequivalent norms on it

[candid hedge]

Why? :p

i still dont understand exactly whats the intuition behind this

[lunar yoke]

take your favourite infinite dimensional space with two different norms

then any infinite dimensional space of the same dimension is isomorphic to it

well idk ofc i could write a proof but you ask for intuition

[lunar yoke]

for me this is the intuition

[candid hedge]

You mean finite dimension right?

[lunar yoke]

no

i mean it works for finite as well ofc

any cardinality

[candid hedge]

Wdym same dimension in infinite dimensions?

are polynomials isomorphic to functions space?

[lunar yoke]

if V,W are K-vector spaces and dim V = dim W then V and W are isomorphic as K-vector spaces

[lunar yoke]

well K[x] has countable dimension over K, and i think most function spaces have uncountable dimension over K

[candid hedge]

And where does this lead?

[lunar yoke]

given two non-equivalent norms on some space they transfer to any isomorphic space

[swift fjord]

I think for me the best intuition is that one proof of equivalence relies on the extreme value theorem, which only works when the closed unit ball is compact, which is true iff the space is finite-dimensional

And it's maybe intuitively clear why the unit ball is noncompact in infinite-dimensional spaces (with ANY norm)

There's 'too much space'

[candid hedge]

i wish :p

why is it clear xd

Why closed and bounded isnt enough for compactness in infinite dimensions

[lunar yoke]

vector spaces are metric spaces so it suffices to check its not sequentially compact, and you can easily find a sequence in the infinite dimensional ball that does not have any convergent subsequence

[candid hedge]

ohhhh

[lunar yoke]

e.g. by enumerating normalized basis vectors

[candid hedge]

yeah

Okay this is incredible thank you

[coarse night]

just want to confirm that the mapping fiber $$Mf \coloneqq { (x, \omega) | x \in X, \omega \in Y^I, f(x) = \omega(1) }$$ is the pullback of the diagram, given $X \xrightarrow{f} Y$ \
\begin{tikzcd}
Mf \ar[r, "\pi_1"] \ar[d, "\pi_2"] & X \ar[d, "f"] \
Y^I \ar[r, "ev_1"] & Y
\end{tikzcd}

[coarse night]

X, Y, f are pointed

[graceful sequoia]

Is this proof okay?

Proving that composition of continuous functions is continuous.

[unreal stratus]

Yeah nice

[graceful sequoia]

Alright, thanks!

[candid hedge]

a hyperspace is always closed in a finite dimensional space? and if its infinite dimensions then its dense?

[gritty widget]

In what topology?

[candid hedge]

Normed vector spaces

[gritty widget]

Yeah but which norm

[candid hedge]

theyre all equivalent in a finite dimension

[gritty widget]

Not in infinite dimensions

[candid hedge]

None is specified

i think density probably doesnt depend on norms in infinite dimensions

The proof is very simple that if H isnt closed then its dense

if H isnt closed then there is a in H_ and not in H

then E=H+Ka which is in H_

so H_=E

I dont see how we needed to specify a topology for this

[candid hedge]

.

[gritty widget]

No

Take any discontinuous linear functional f and f^-1(0) is an example

[west spindle]

can anyone help me see how the red and blue loops are homotopic

cause i'm having some trouble visualizing it

or even whether they're homotopic

[graceful sequoia]

Is it okay to look at the graphs of f(x) = tan(x) and f(x) = arctan(x) to justify that they map open intervals to open intervals? Need this result to conclude that the functions are continuous.

[candid hedge]

derivative doesnt work

[woven sundial]

if you can't take as given that tan or arctan are continuous, i think appealing to the graphs would be insufficient.

you might as well conclude that they are continuous directly from the graphs.

i think you have a good idea, but you should try constructing a function which you can more easily show to be continuous that has a similar graph to tan, i.e. is a bijection from an open interval to R.

all injective continuous maps from R -> R are open (this is worth trying to prove itself imo)

[unreal stratus]

invariance of domain

😩

[woven sundial]

hint: try constructing ||a rational function which fits this bill||

[gritty widget]

Wdym?

[candid hedge]

(0)^-1 is the constants

all real numbers

[gritty widget]

[woven sundial]

it's a lot easier to prove for R than R^n unless im being v silly

[unreal stratus]

yup yeah it is

I mean being injective and continuous place a lot of restrictions when it's R -> R

[wide egret]

Is this proof incomplete or missing any cases?

perhaps i should have said "knot projection" instead of "knot" in a few places but outside of that

[uncut surge]

i know nothing about knot theory but i'm convinced

since the circle is connected, every knot is connected, so the two crossings must be connected by paths, and then your argument follows; i guess "crossings" only make sense when considering knot projections, so yeah arguably every "knot" should be "knot projection" but it seems clear enough

and the simple twist that undoes the first crossing is reidemister move type 1 so that's also good

[wide egret]

thank you

i ask bc i lost points for it missing cases or smth, but i don't see how it could be otherwise

[uncut surge]

tho once again i've never done anything with knot theory, i've only heard that reidemeister moves exist and that you can colour things once in a talk

[wide egret]

the lack of rigor is because this was before formal tools such as reidemeister moves were introduced in the book

hmm fair

[graceful sequoia]

What do you mean by all ..... maps are open?

[hidden crag]

they map open sets to open sets

[wide egret]

i will ask the grader abt it ig; there was another one abt decomposing into reidemeister moves where it said I missed some or something, but i'm convinced i didn't
i figure nobody wants to check that though lol

[graceful sequoia]

Oh okay.

[woven sundial]

what timo said

[graceful sequoia]

Thanks for the hint I'll try

[uncut surge]

ye might be good to ask directly, i don't think your proof is missing anything but maybe they had a specific way of proving this in mind

maybe they just wanted you to draw things explicitly

[wide egret]

perhaps indeed

i don't like drawing sadge

[uncut surge]

i guess for knot theory you have to like drawing a little bit

but that's just a guess

[coarse night]

in Rotman, the definition of exactness in hTop* is given by X' →X →X" is exact if
[Z, X'] →[Z, X] →[Z, X"]
is exact in Set* for all Z. Why is this way of defining exactness is eqv to how we define exactness in other categories?

[graceful sequoia]

@woven sundial The thing is, I'm trying to prove continuity, and for that I need the result that my proposed function maps open intervals to open intervals

[woven sundial]

yes, i understand that you want your function to map open intervals to open intervals so that the inverse will be continuous. one way to sidestep this requirement would be to use a rational function instead of tan so that you know the inverse is continuous immediately.

[graceful sequoia]

How do we know that the rational function is continuous immediately?

[woven sundial]

okay, well not immediately. but it's easier to show continuity for than tan/arctan

[graceful sequoia]

Okay so lets say I take f(x) = x/(1-x^2)

and I want to show (-1,1) is homeomorphic to R.

assuming R is equipped with the standard topology, (-1,1) with the subspace topology

How should I go about proving that the function is continuous?

[gritty widget]

It's rational and the denominator doesn't vanish

[graceful sequoia]

Uh yes but shouldn't we do it by showing that for any set U in the standard topology, f^-1 (U) is in the subspace topology?

Or for any open set U in R, f^-1 (U) is open in (-1,1)?

[gritty widget]

That's equivalent

[graceful sequoia]

Oh, how?

[gritty widget]

Use epsilon Delta and the definition of limits of sequences

[gritty widget]

No, $f^{-1}(0)$

[gentle osprey]

Blitz

[gritty widget]

The inverse is on f

[woven sundial]

every continuous function is still continuous when restricted to an open subset of its domain

[unreal stratus]

Well any subset

[celest loom]

Is there an easy proof of the other way? Where theres a way to extend domain of a continuous function to the whole space and still have it be continuous? I did the easy case in R in rudin but I didn't do the harder one for a general metric space

[gritty widget]

First you might want to prove Urysohn lemma for metric spaces @celest loom

The proof is easier than the general case

[graceful sequoia]

If f: (X,d) --> (Y,d) (where both are metric spaces) is continuous, then is f: (X,T_1) --> (Y,T_2) continuous, where T_1, T_2 are topologies induced by the metric d? I did it this way: Take U in T_2, it will be open in (Y,d). Since f is continuous, f^-1 (U) is open in (X,d) and also in (X,T_1) and so f^-1 (U) is in T_1.

[gritty widget]

Yes. If and only if

[graceful sequoia]

Ah okay, thanks.

So lets say I'm able to prove that f: ((-1,1), E) --> (R, E) is continuous (by epsilon-delta), where f(x) = x/(1-x^2), then I can conclude that f: ((-1,1), T_SB) --> (R, T_U) is continuous? E is the Euclidean metric, T_U and T_SB are the usual and subspace topologies.

[gritty widget]

Yes

[graceful sequoia]

Ah finally! 🙂

Now I just have to do the epsilon-delta thing

Thanks!

[gritty widget]

You have to have an assumption on the domain of your function of course. Tietze extension theorem demands it to be closed

[celest loom]

oh yea my bad, I forgot about that part

thanks

[wide egret]

is this proof i found online not incomplete? (proof that type III reidemeister move preserves tricolorability)

it forgets about the case in which none of the crossings have strands with the same color

[hard wind]

Can you draw a picture of the case you're describing?

[wide egret]

this sort of thing

[hard wind]

Is that not the top left?

Just reversed

[wide egret]

wait a min

ah i got mixed up considering the ones where the moved string is on the bottom as well

thanks

[hard wind]

No problem

[wide egret]

^ but on that subject, shouldn't one prove it for those cases too, where the bit being moved is below the crossing?

what's the rationale for saying those cases are taken care of? it can't be flipping the knot over right since we haven't proven it's an invariant yet

[hard wind]

I believe that case is identical to the ones above

You don't need to flip it over or anything

But moving the top string and moving the bottom string give you the same pictures, unless I'm mistaken

Worth noting at least

[wide egret]

like this vs the image above

ig i'm struggling to see how they are equivalent

[wide egret]

like i don't think you can construe this move as some upper string being moved

[hard wind]

Posting them next to each other for convenience

No I definitely think you can, let me see if I can graphically explain how I am visualizing it

[wide egret]

hmm i'm starting to see smth now but i think it's like a mirror image thing

[hard wind]

You can choose to view this transformation as sliding the bottom string northwest

[wide egret]

oh

lol

damn why the book gotta introduce it as two different kinds

that makes sense tho thanks

[hard wind]

Yeah no problem

If only I had physical color changing string...

[wide egret]

lol that would be pretty sick

come to think of it i had one other question - is there a definitive way to prove nontricolorability? and/or if i'm asked to "determine which of the 7-crossing knots are tricolorable", need I prove it for each?

[hard wind]

I'm not much of a knot theory guy, so I'm not really sure

Generally you want to run into some sort of contradiction; you choose some starting colors that force you to do other colors and you run into an error somewhere

[wide egret]

yea ig i'm just worried about how to be sure to show all cases

it seems there could be a ton of cases for 7-crossing knots and while i bet a bunch of them are the same, i don't know which or how to show it, and i'm not sure how many cases that'd get me down to

plus there are 7 of them and this is just one exercise of 15

[wise ruin]

In class, we were asked to show that the projection of quasi-circle (closure of sin 1/x with an arc making it path connected) onto S^1 does not lift through the covering map R->S^1. Is there anything pedagogically interesting about this exercise? It seems to me that the proof is exactly the same process as showing the identity map on S^1 doesn't lift through R->S^1. Perhaps, it could be demonstrating that a certain property of the quasi-circle does not change that fact, which might be counterintuitive. Though, if it is counterintuitive, I'm not sure why.

[plain raven]

I think that it's interesting because S^1 is locally path connected, but the quasi-circle is not. So it might feel like you could "break it apart" at the discontinuity at zero and lift it that way.

There are at least two homotopically distinct paths between two points in S^1, but there is only one path between two points on the quasi-circle.

So if your intuitive conception of the reason S^1 fails to lift is that it would "break a path", this reasoning no longer works to explain why the quasi-circle doesn't lift.

[wise ruin]

Ah, that concretizes the thought quite well. My initial thought was "why would a local discontinuity intuitively imply that the space could now provide a lift," but thinking one could "break apart" the space at that discontinuity seems like a reasonable intuition. Now I understand why this exercise is interesting 😁

[wide egret]

in knot theory tabulation, is it ever the case that a knot and its mirror image are both given names (like a_n and a_m)? or is each name supposed to represent both knots?

i have a 7-crossing knot that I know is the mirror image of 7_4, and I want to make sure it's not any of the other 7-crossing knots

for the same problem, another question i guess; how can you tell if two dowker-thistlethwaite codes represent the same knot?

[candid hedge]

Can someone explain the infinity notion in a normed vector space

I am not sure what to make of it

does it exist?

is 1/x for example a function that goes to infinity in a normed vector space when x goes to 0?

Can f(x) goes to infinity?

so for example if i take an orthonormal base, e1....en, can f(e1) go to infinity?

[gritty widget]

what does 1/x mean in a normed vector space

[candid hedge]

Idk thats what i am asking

I am a bit lost

[gritty widget]

alright. Is it a part of a bigger problem

[candid hedge]

i found this proposition in my textbook

[candid hedge]

Tell me

[gritty widget]

3?

[candid hedge]

E

its full of mistakes like that

[gritty widget]

alright, what are you confused about
you don't know what to make of the infinity symbol there?

[candid hedge]

this

Is there an infinity notion in a normed vector space?

[gritty widget]

it doesn't mean anything. Division only makes sense after we give it sense

[candid hedge]

Are applications a group of elements?

[gritty widget]

There could be

[candid hedge]

so i could say lim f(x)=+inf?

[gritty widget]

Just like how you take a one-point compactification of R^n say, you could try the same thing with a normed space.
But the resulting space will be almost never Hausdorff

This is because infinite-dimensional normed space is never locally compact

[candid hedge]

i dont understand what youre saying

many times we wrote in proofs for example : f(x) < the sum of f(ei)*xi if f linear

how can we be sure that f(ei) isnt infinite

[candid hedge]

To deduce a certain propert

like there exists a k such that f(x)<kx

[swift fjord]

this seems to imply f is valued in real numbers

so it makes sense to talk about when f tends to infinity

the book states $f(x)\geq M$, this does not make sense if we don't have an order

[gentle osprey]

ShiN

[gritty widget]

Just replace $f(x)$ with $|f(x)|$

[gentle osprey]

Blitz

[swift fjord]

yea, perhaps the author means $\lim_{x\to a}\abs{\abs{f(x)}}=\infty$

[candid hedge]

Yeah in my textbook they seem to be talking about any normed vector space

[gentle osprey]

ShiN

[candid hedge]

and in any case i dont see why R would be different than other normed vector spaces

[candid hedge]

Do you know?

[swift fjord]

because R is ordered

[gritty widget]

You could write $\lim_{x\to a} f(x) = \infty$ even if $f$ takes values in a normed space imo

[swift fjord]

yes, but then it doesn't make sense to write that the limit is infinity

[gentle osprey]

Blitz

[swift fjord]

how would you make sense of this

[gritty widget]

why not? In C we often write this

[candid hedge]

Norm goes to infinity

Not coordinates

[gritty widget]

Just like in C, the norm goes to infinity, but we write the limit without norm

[swift fjord]

idk if I like this

but sure I guess

[candid hedge]

Any clue?

[swift fjord]

because infinity isn't an element of your space

and the norm of any given element is finite, by definition

[candid hedge]

This is exactly what i want to understand

[candid hedge]

So there isnt infinity in a normed vector space?

[swift fjord]

no

[candid hedge]

what about 1/x when x goes to 0

[swift fjord]

it was explained to you that 1/x doesn't make sense in a general vector space

since that isn't defined

[candid hedge]

Because we cant divide?

[swift fjord]

and that writing that the limit is infinity is just symbolic of the norm tending to infinity

we don't even know what multiplication is a priori in a vector space

[candid hedge]

So The norm of any element is finite, but a sequence can go to infinity?

[candid hedge]

What kind of function then are defined on a vector space?

Only addition and multiplication by a scalar?

Are all functions linear

then>

[swift fjord]

the sequence goes to infinity in the sense that the norm goes to infinity as the sequence progresses yes

you can certainly define nonlinear functions on vector spaces

they're just usually not as useful

[candid hedge]

like what

[swift fjord]

choose a nonzero vector v and send all elements to v

this is nonlinear because 0 is not mapped to 0

[gritty widget]

bilinear functions

[candid hedge]

Ah fuck ok

[candid hedge]

Lol