point-set-topology

[hollow harbor]

it doesn't mean "T is an endomorphism of V" since T may be defined on a larger space.

but all of this is just like

notational right?

in practice, these are very very similar things.

(also this probably isn't a topology question, and might make more sense in #groups-rings-fields unless it came up in a topology/geometry context)

[zealous glen]

Ok I get it then. Didn't know endormorphism referred to the whole space.

Well, It came from a dynamical systems/number-theory context. But yeah you are probably right. Thanks 😄

[marsh forge]

Endomorphism refers to the domain and codomain being the same but people are often sketchy with ignoring restrictions

Well not sketchy

Its all accurate

[zealous glen]

I havn't taken any classes in dynamical systems though so I figured topology/geometry was close enough.

[hollow harbor]

ah

that's fine

i mean this is important in topological dynamics

and yeah dynamicists almost always use the "invariant" vocabulary instead of talking about restricted endomorphisms

because the invariant vocabulary offers some nice extensions

like "A is an almost T-invariant subset of X" could easily be taken to mean that T(A)\A is measure zero.

whereas it's basically meaningless to try to define an "almost endomorphism"

and things like this are super important in dynamics

[zealous glen]

Ah neat. So far I only experienced a quite narrow part of dynamics. But it is actually quite exciting.

[marsh forge]

Don’t people study like

[hollow harbor]

it's very nice

[marsh forge]

Almost everywhere defined maps

[hollow harbor]

yeah

you just quotient by almost everywhere equivalence

but this doesn't really help you afaik here

[polar whale]

Can someone explain how every nonempty subset of X has a limit point?

X is 'lim point compact' if every infinite subset of X has a limit point

[gritty widget]

think about what kinds of sets are open in X

[polar whale]

Hmm i was thinking the topology of Z+ is power set of Z+ right?

I mean the discrete topology?

[gritty widget]

if that were the case then no nonempty subset of X would have a limit point (reason: if X is a discrete space, {x} is an neighborhood of x which contains no points other than x)

[polar whale]

The problem i m facing is this:
Suppose Y={a,b}. Then Q=(Z_+) × {a} is a subset of X.Let's assume x=(1,a) be a limit point of Q. Then Q intersection {(1,a)} \ {(1,a)} = phi .... so (1,a) is not a limit point of Q.

[polar whale]

Yes exactly! Where am i wrong i can't figure out

So what is the topology on Z_+ then?

[gritty widget]

hint: the only non-empty open set in Y is Y itself, so what can you say about a (basic) open set in X?

[polar whale]

Must be U × Y where U is any open set in Z_+

[gritty widget]

yeah

[polar whale]

Yes this much i got...but exactly how do the open sets of Z_+ look...

For here no specific topology is mentioned

Sorry to bother if these were too trivial questions T-T
Thanks for the hints I'll try to take it from here :)

[gritty widget]

it's the product topology

[polar whale]

Yeah...thanks

[fading vale]

So for context phi is a map between connected schemes and phi* is the induced map on etale pi_1

at the end of the second part they must mean that phi* and not phi is injective right

Its not in the errata but like... lol

[swift fjord]

I don't understand what you're asking

No matter the metric an interior point is defined the same

@gritty widget

[gritty widget]

Yes

[swift fjord]

For a subset $U \subseteq X$ of a metric space we say $a \in U$ is an Interior Point of $U$ if there exists some $r>0$ such that $$B(a,r)\subseteq U$$, where of course
$$B(a,r)={x \in X: d(a,x)<r}$$
Is the open ball centered at a of radius r

[gentle osprey]

ShiN

[gritty widget]

Can we say for what r

We know that there exists a r

But can we define that r

Life for r =(a-b)/2

[gritty widget]

This won't work

[swift fjord]

Well it depends on the point and the set

[gritty widget]

But as an example

[swift fjord]

Sometimes finding the proper r constructively is just really difficult

[gritty widget]

Set is x^2+y^2<1

[swift fjord]

It's really dependent on context

[marsh forge]

you can often find a largest possible r

or at least define one as a supremum

but actually computing it is in general impossible, basically

esp. if you want to compute it like, generically

[gritty widget]

Yes That's what i was saying

[marsh forge]

basically just write

$R:=\operatorname{sup}_{r>0} B_r(x)\subset X$

oh

rename the first r

to like

R

lol

[gritty widget]

Yea

[gentle osprey]

MaxJ #MiuArmy

[marsh forge]

but finding what R is probably requires like

ad hoc arguments based on X

[gritty widget]

Yea but I'm in final year of My Undergrad And I haven't seen Thua sup thing

[marsh forge]

uh

are you a math major

[gritty widget]

Can you tell me briefly what isbit

[marsh forge]

sup just means like

[gritty widget]

Yee

[marsh forge]

'largest possible choice'

note that R need not exist here

by the way

like i said this only works if you have like

[swift fjord]

How are you in your final year and don't know what a supremum is

[marsh forge]

a "hard border"

[swift fjord]

Or do you just mean you haven't seen the notation cuz I think it's pretty internationally standard

[marsh forge]

Yeah SHiN kinda said it harshly but I agree I have no idea how you are getting a math degree without seeing supremums

[swift fjord]

It wasn't meant to he harsh

I just find it weird

[gritty widget]

I mean final year started like week ago

It's not final semester

[marsh forge]

Oh no, my construction is more generic i was being silly, i think it should work in most cases

[gritty widget]

It's 5th semester

[marsh forge]

yeah but like

idk this is like

1st or 2nd year material

[gritty widget]

So basically we can't generalize the value of r

[marsh forge]

(out of 4)

generalize?

[swift fjord]

For reference supremum was the 3rd definition we learned in my calc 1 course

Where are you from @gritty widget

[gritty widget]

I mean for like relation of r to the point we want to be interior

[gritty widget]

India🙂🙂

[gritty widget]

Something Like This

[swift fjord]

That's impossible to answer for a general set

As max said

[gritty widget]

Yes

Makes sense

[grave maple]

Anyone familiar with the concept of a topological universe?

[swift fjord]

I feel like i'm being dumb here. Suppose I have a homotopy $F:f_0 \simeq f_1$ where $f_i$ is the constant map for some point $x_i \in X$. I want to find a continuous path between $x_0$ and $x_1$. Now, I define $$\gamma: [0,1] \rightarrow X$$ by $$\gamma(t)=F(x_0,t)$$. I'm getting stuck on proving that $\gamma$ is continuous though. Should be simple unless i'm missing something, I just can't seem to see what the inverse image of an open set would be. (Taking $x_0$ for convenience but I think it could be any point in $X$ in the definition of $\gamma$)

[gentle osprey]

ShiN

[marsh forge]

sorry, all the f_i are constant?

You're missing something

the extra condition in the defn of a homotopy requires that $t\mapsto f_t$ is continuous

[gentle osprey]

MaxJ #MiuArmy

[marsh forge]

so your conclusion is immediate

[swift fjord]

Only $f_0,f_1$

[gentle osprey]

ShiN

[swift fjord]

should've clarified, $i=1,2$

[gentle osprey]

ShiN

[swift fjord]

doesn't it only required that $F(\cdot,t)$ is continuous for all fixed $t$, and that $F(x,t)$ is itself continuous?

[gentle osprey]

ShiN

[swift fjord]

We're varying over $t$ in this case

[gentle osprey]

ShiN

[marsh forge]

It is continuous in both variables

[swift fjord]

why is that implied by continuity?

I guess that's the question

[marsh forge]

If AxB->C is continuous then the induced maps A->Maps(B,C) and B->Maps(A,C) are both continuous. In the case of H:Xx[0,1]->Y we have that X->Paths(Y) is continuous and that the inclusion {x_0}->X->Paths(Y) is continuous

Wait this might be one of those things that is not true in general for Top one second

Here's a better way to see it sorry

Let x be the desired point of X. Then we have a map $[0,1]\cong {x}\times [0,1]\hookrightarrow X\times [0,1]\xrightarrow{H} Y$ and this composition is continuous

[gentle osprey]

MaxJ #MiuArmy

[marsh forge]

on a point-set level this map is

[swift fjord]

oh yea I see it now

[marsh forge]

yeah

[swift fjord]

Yea ok now that you state it like that it's glaringly obvious

from the continuity of the inclusion

[marsh forge]

this is maybe more obvious from hatcher's preferred defn of a homotopy

[swift fjord]

what is that?

Just to make sure, this kind of continuity in each component is necessary but not sufficient

[marsh forge]

Yes

hatcher says a homotopy is a family $f_t$ of continuous functions with $(x,t)\mapsto f_t(x)$ continuous

[gentle osprey]

MaxJ #MiuArmy

[swift fjord]

ah ok

yea rotman points out that that's equivalent

[marsh forge]

The defn we used is the more common POV in modern alg top

[swift fjord]

I reckon it's slightly more easy to work with in most contexts

[marsh forge]

it is also generalizable

cylinder objects end up existing in more contexts that don't have nice point set tools

[swift fjord]

Ok so just so I make sure I have this 100 percent, if we fix some $t$, then we have the sequence $$X \stackrel{i}{\cong}X \times {t} \stackrel{j}{\hookrightarrow} X \times [0,1] \stackrel{H}{\rightarrow}Y$$, then $$f_t = H \circ i\circ j$$ and therefore continuous

[gentle osprey]

ShiN

[marsh forge]

No

You want to fix in X

[swift fjord]

this is for why the $f_t$ are continuous

[gentle osprey]

ShiN

[marsh forge]

You want a path right

[swift fjord]

This is just in general now

[marsh forge]

Oh yes

[swift fjord]

It's the same process for the path tho

[marsh forge]

That proves f_t continuous

[swift fjord]

ye ok

I just tend to get caught up on small points like this

so I gotta make sure I understand it 100% before I continue

alright thanks a lot

[fading vale]

sorry for the all the images i needed to give proper context lol

but basically i follow the argument up until "they are isomorphic over the generic point"

after that it starts to lose me

I assume that means their fibers are isomorphic but I dont see how this follows from having the same function field?

nor do i really understand the dense open subset argument

@cedar pebble if ur free and dont mind explaining this

[cedar pebble]

yea so like in the world of schemes

so long as everything is locally like spec of an integral domain (as would be the case in this situation), then you have a generic point

(it's the prime ideal (0) in Spec of an integral domain)

what it means for something to be defined on the generic point means for something to be defined on some dense open subset

[fading vale]

[cedar pebble]

This is in some sense what function fields are doing. Think about the case of a curve, where dense open subsets are complements of finitely many points

for a function to be defined on the generic point this means it's defined on some dense open subset, that is away from some finite set of points

those are the poles of the function

[fading vale]

Oh so the follow up statement about being isomorphic over the product w/ a dense open subset is literally the definition of what it means to be isomorphic over the generic point?

[cedar pebble]

mhm!

[fading vale]

I see

[cedar pebble]

again this is why schemes are so cool, you have generic points that organize all this information for you

[fading vale]

Okay right so we have an isomorphism of the function fields

How does that imply the follow up statement

like how does that imply theyre isomorphic over the generic point

[cedar pebble]

I mean that's just what it means

the function field is the stalk of the structure sheaf over the generic point

so this sentence just says that the stalks of the respective structure sheaves are isomorphic over this point

[fading vale]

Uhh why does an isomorphism of stalks imply an isomorphism of X_L times U and Y times U

oh this is fiber product

i cannot read

[cedar pebble]

yea

[fading vale]

tfw illiteracy

[cedar pebble]

Szamuely moment

[fading vale]

Uhh hmm ok so why does this iso of the stalks give an iso of these fiber products

[flint cove]

Does anybody know how one could define the degree of a map S^n→S^n in an elementary way without reference to top-degree homology?
Because I kind of don't want to use homology to define the boundary map of CW-homology…

[bold canopy]

you can look at how many leaves you have in the covering

(i.e., the degree-m map S^n -> S^n has m-leaves generically, you are supposed to count the preimages orientedly)

idk if im making any sense

https://en.wikipedia.org/wiki/Degree_of_a_continuous_mapping#Differential_topology
i mean this

[flint cove]

oh, nice
Not sure if I can assure that the map is smooth tho, need to look at it in more detail

[bold canopy]

https://math.stackexchange.com/questions/176399/continuous-maps-between-compact-manifolds-are-homotopic-to-smooth-ones
there is some thm saying you can homotope continuous maps to smooth ones
but if you really dont want to touch smooth stuff, probably just going with homology is the best way to do things? it feels like the cleanest

[flint cove]

I mean… none of this must be smooth

nothing has an intrinsic smooth structure 😅

But yeah, I will probably need some result of the sort „there is a nice enough representative in each homotopy class“

e.g. if I just want to define it via the degree of the covering

[bleak helm]

@vast estuary any r < epsilon

Probably epsilon/2

Oh, you deleted

[pearl holly]

Okay so I'm reading the proof of this proposition and I can't understand why the set of points where f_1~ and f_2~ agree is both open and closed. How do I know that the set is closed?

So I'm confused by the very last sentence

ohhh okay now I see. Thank you so much!

[pearl holly]

Okay let me just get this straight. If X is homotopy equivalent to Z, is then X v Y homotopy equivalent to Z v Y? (the v stands for the wedge sum). This isn't true, right? Because if we look at the wedge sum S¹ v S¹ in R², then S¹ is contractible. If the above was true, then S¹ would be homotopy equivalent to S¹ v S¹, but this isn't true because they have different fundamental groups

But maybe I can't view S¹ v S¹ as being in R² because I need to consider that to be a whole space in its one

So what if we look at Z, Y, X as being subspaces of some bigger space?

[obtuse meteor]

S^1 is not contractile tho?

I think if everything is CW complex you’re probably fine

And this statement is true

And as we all know those are the only spaces

[pearl holly]

Wait wait, isn’t every loop in R^2 contractible and so the circle is too?

But I guess that you need to view S^1 as being a single space?

[obtuse meteor]

Ah, mess up of language

Yes

The inclusion map S^1 -> R

Is nullhomotopic (or sometimes called contractible)

But S^1 itself is not

[pearl holly]

Oh okay, so you always need to specify whether you should consider a space to be a subspace of something or that it is its own space?

[obtuse meteor]

Yeah essentially. Although most algtop people wouldn’t phrase it as “S^1 is contractible as a sub space of R^2” but rather as “The inclusion map is contractible/nullhomotopic”

[pearl holly]

Okay great thank you so much! So just to make sure, if I replace X, Y, Z to be CW complexes, then the statement is true? Is there an easy proof of this?

If there is, then please don’t spoil it

And I’m referring to my statement above about the wedge sum thing

[obtuse meteor]

Hmmm

The proof will rely on the homotopy extension property

Dunno an easy proof but shouldn’t be too hard?

[pearl holly]

Oh okay I see. Thank you so much! I will at least give it a shot later and we will see how it goes lmao. I think I have something very cool going on

[thin nova]

this isnt exactly advanced but i dont see a geometry channel in the uni section so ill ask here:

why is it x = coscos, y = cos*sin and z=sin?

i thought in spherical coordinates, it's x = cos sin, y = sin sin and z = cos?

[hollow harbor]

i think either one works

they're just different parameterizations

[thin nova]

ahhh

apparently the one in the image is for a ellipsoid

so more accurate i guess

[summer jolt]

Does anyone understand this definition?

[vast estuary]

Yes, that worked, thanks!

[gritty widget]

I think yes

[summer jolt]

@gritty widget I don't quite get why U is a neighbourhood of e though.

[gritty widget]

Do you know what’s a neighborhood?

Well

I don’t know exactly why, but we’ve an open set throughout point e

[swift fjord]

I'm having trouble concretely understanding the open sets in the quotient topology. I understand how they're defined, but I usually have a hard time understanding what they are. Here's an example i'm struggling with a bit, the cone over some space $X$, which i'm defining as $$\sfrac{X \times [0,1]}{\sim}$$ where $\sim$ is the equivalence closure of the relation $(x,t)\sim (x',t') \iff t'=t=1$. So clearly for sets not containing the point $[x,1]$, they identify with the open sets of $X\times [0,1]$ in the product topology, but what about the open sets containing $[x,1]$? I'm having a hard time understanding what their structure is

[gentle osprey]

ShiN

[plain raven]

so $X\times I$ receives the product topology, right? open sets are of the form $U\times (a,b)$ for $b>a$. or $U\times (a,1]$, etc

[gentle osprey]

diligentClerk

[swift fjord]

Yup

[novel acorn]

Here's a maybe stupid question: if $X \subset \bR^m$ is a convex open set, does that immediately imply that it's simply connected?

[gentle osprey]

Irony Incarnate

[swift fjord]

Basic open sets ofc

[swift fjord]

Yes since it's contractible

I'm pretty sure

[plain raven]

a set in $CX$ is open iff its preimage is open. If $q: X\times I\to CX$ is the quotient map, then $A$ is open iff $q^{-1}(A)$ is open. Assuming $A$ is an open set that contains the distinguished vertex at t=1, then that means $q^{-1}(A)$ contains $X\times 1$. Since $q^{-1}A$ is open, it must contain an open neighborhood of that set in the topology; so if it contains $X\times 1$, then for each $x\in X$ it contains $U \times (1-\varepsilon, 1]$ for some small $\varepsilon$ and $U$ with $x\in U$.

[gentle osprey]

diligentClerk

[swift fjord]

Ok yea that makes sense, so that's about as concrete as you can get right?

I guess always when you're dealing with the producy topology general open sets are annoying

Since they're a union of basic open sets which themselves are not always so simple

Thanks

[plain raven]

yeah i'm trying to think if you could prove it contains $X\times (1-\varepsilon, 1]$ but unless $X$ is compact i don't see why this should hold

[gentle osprey]

diligentClerk

[swift fjord]

Yea I think it wouldn't hold generally

Rotman says Cones can get a bit freaky when your space isn't compact hausdorff

[marsh forge]

Linear interpolation

[lean marten]

Yeah because the cone is a join it gets weird for spaces which aren’t compactly generated

[gentle osprey]

Hausdorff

[vast estuary]

Oh I think I can use sequential continuity directly since X and Z as subspaces of R^2 are metrizable by the euclidean metric

[zealous glen]

Could anyone here recommend a good introduction to topological entropy.

[vast estuary]

Although what is meant by "identifying the subset {0} x Z_+ to a point"

[reef shore]

Quotient by the equvalence relation that relates all of those points to each other

and nothing else

Someone came up with this example a while ago and posted it here lol

Surprising example

[vast estuary]

g is one-one right

so the inverse of singletons are singletons

[reef shore]

No

g^-1 of 0 isn't a singleton

It is {0} x Z+

[vast estuary]

oh shit yeah. that's the only weird guy right

[reef shore]

Yeah

[vast estuary]

i don't understand this still

they seem to say that they're identifying that set with 0, right?

but the quotient space also has other elements

[empty grove]

Identifying something to a point means you treat that whole set as a single thing

Which is exactly what you do when you quotient

[vast estuary]

oh okay so they're saying that we need to identify the set {0} x Z+ to a point and let all other points stay as they are

that makes up the quotient space, right?

[empty grove]

Yes

The equivalence relation is x~y iff they're in the same fibre of g

[vast estuary]

and how do you see that A is closed

i'm thinking in terms of: x is a limit point of A iff every neighborhood of x intersects A in a point different from x

[empty grove]

It is discrete

Discrete subsets are always closed in metric spaces

[vast estuary]

yes i know

that works

but i'm trying to show that there are no limit points for A

[empty grove]

Try to get a neighborhood B_n of x_n which contains no other x_i

[rancid umbra]

it has no convergent subsequences (phrased differently, it’s set of subsequential limits is empty in R^2)

[rancid umbra]

@vast estuary another way to get this, view A as a function N —> X : n —> (1/n,n), this map is continuous because N is a discrete space. since X is hausdorff, then the graph of that map, A, is closed, as every continuous map into a hausdorff space has a closed graph

[vast estuary]

overkill 101 but amazing

[rancid umbra]

lol ikr, was just thinking about it for some reason

[astral light]

Isn't it enough to show that (x,n) —> x and (x,n) —> x/n are continuous?

[summer jolt]

But isn't U a subset of X? However e belongs to E.

[vast estuary]

Is Munkres known to come up with hard examples

Anyway, I'm first trying to see why U_n is open in X x Q. On the X axis it's an open set, (n-1/4,n+1/4), but on Y? (It's the product topology, so, we want a cartesian product of two open sets)

[empty grove]

Open sets in the product aren't just products of open sets

They are unions of products

[vast estuary]

Yeah yeah, products of open sets form the basis

[empty grove]

So take some point in that shaded region

And

Draw a rectangle around it

[vast estuary]

Would it matter if the rectangle's boundary intersects Q or not

[empty grove]

It will be a rectangle without boundary

[vast estuary]

Interesting, so are you suggesting that the rectangle is open in X x Q

[empty grove]

Yeah

Both have the subspace topology from R²

From R

[vast estuary]

Yeah the rectangle is open in R^2. So it is a union of products of open sets

[empty grove]

It is a product

What is it a product of

Intersect those 2 open sets with X and Q

Then take product to get open in X x Q

[vast estuary]

Intervals!

With X nothing changes

With Q it's just all irrational points get deleted

So we get the rectangle in X x Q is open

[empty grove]

Yes

[vast estuary]

Am I right?

So now for every point in that weird shape, we can find an open rectangle around it

The shape is a union of such rectangles

So it's open

[vast estuary]

This is a question too 😛
Wondering what to expect from the book

[empty grove]

Idk munkres is pretty good lol

[gritty widget]

Always

If a point e is in X and we’ve a neighborhood it’ll keep like a subset of X

Take this article of Wikipedia if you’d lime to know more about this

https://en.wikipedia.org/wiki/Neighbourhood_(mathematics)

But also we need a Open set around of point e for we’ve a neighborhood U and also this open set A need to be a subset of the set X

[swift fjord]

Anyone have a graphic for the homotopy equivalence of a 2-dimensional cylinder and the mobius strip?

[summer jolt]

I understand the definiton of neigbourhoods and open sets, but what I'm pointing out is that the defintion says $e \in E$. Thus I think that must be a typo, because otherwise it doesn't make any sense.

[gentle osprey]

snypehype

[swift fjord]

Why does that work but retracting it onto its boundary circle is (iirc) impossible?

[bleak helm]

I'm not familiar with the content lol, so I really can't help. But I would guess they meant s(U) is an open neighbourhood of S

Especially since they continue to say the s(U)'s form a basis

[swift fjord]

I know the proof more or less, I guess i'm asking more visually why that wouldn't work

I guess yea

Mine isn't either

[grave maple]

Anyone familiar with "initial structures" in topology?

[swift fjord]

Which is why i:m trying to think of these simpler examples

[marsh forge]

Going to the center circle is easily visualized

Retracting to the boundary would “tear” the Möbius band

[swift fjord]

Yea I see that now

[summer jolt]

Yes, that's my guess as well. It should be s(U) instead of U

[grave maple]

For example, if X, Y are topological spaces, then the product topology X x Y is the initial topology on X x Y making the projections continuous. But as a product in the categorical sense, X x Y is a terminal object. So I guess it's "initial" in the sense that the projections are from X x Y, but it's "terminal" in the sense that the universal property yields unique morphisms into X x Y.

[plain raven]

huh, that's an intereresting distinction. maybe we can also think of the product topology as a terminal topology in the same way that the product is a terminal object - it's terminal among topologies which make all maps Z -> X x Y continuous if the projections Z -> X and Z -> Y are

wait no because you'd always have the indiscrete topology. i guess subject to the constraint that the projections are also continuous

[grave maple]

Terminal objects are just initial objects in the opposite category so... hmm...

[gentle osprey]

Dr.Marschett

[grave maple]

Aha: https://en.wikipedia.org/wiki/Initial_topology#Categorical_description

So in the end, initial topologies are basically terminal objects.

[gritty widget]

🤔

[grave maple]

The nomeclature is kind of backwards.

[summer jolt]

That cannot be the case though. If s is a section then its domain must a set in X

[gritty widget]

I know

That was a typo

[grave maple]

Yes, that's true. Good point.

[gritty widget]

I’m looking here and in a set draw it doesn’t makes sense @summer jolt

[plain raven]

this is pretentious enough that i should probably post it in #category-theory but there are these things called topological categories

which focus on the thing you said about initial topologies

like, the category of topological spaces has the property that if X is a set and {Y_i} is any family of topological spaces (even a proper class) and we have maps f_i : X -> Y_i, there's a unique initial topology on X making these maps all continuous.

and you can reconstruct some interesting aspects of topology in categories with these properties. technically this is really a property of aa forgetful functor p : E-> B with E = Top, B =Sets.

[grave maple]

You anticipated me. I am doing research involving topological categories.

[plain raven]

i haven't spent too much time with them. they are used in oswald wyler's lectures on quasitopoi

[grave maple]

I am generalizing some results of partial group actions on topological spaces to convergence spaces.

I still don't understand partial group actions though.

[gritty latch]

if for any basis elements B of a given topology and some function f you prove f^{-1}(B) is open

is that enough to prove f is continouous?

i'm almost certain this is true but just in case

[true robin]

Yes

f^-1 is nice in that it preserves both unions and intersections

[gritty latch]

yeah I suppose that's why continuity is defined the way it is in topology

thanks!

[grave maple]

I think it follows naturally as the generalization of the delta-epsilon definition.

[true robin]

i suppose the natural way to express the epsilon delta defn in the general case is to say that the preimage of a basic open set is open, so that is actually a nice tie-in to your question

[swift fjord]

It's also sufficient to prove this on a subbasis

[gritty latch]

oh good

i needed this to prove that $\mathbb{R}^{X}$ with the product topology was homeomorphic to the set $\mathcal{F}$ of all functions $f: X \to \mathbb{R}$ with the topology of pointwise convergence

[gentle osprey]

d/dx

[novel acorn]

This problem seems
Too simple

I feel like something isn't right

In my mind identifying this you just get the same torus so the group is $\bZ\times\bZ$

[gentle osprey]

Irony Incarnate

[true robin]

Identifying this does not give you a torus

[novel acorn]

OK

I got it

You get like a torus on top of another torus

[true robin]

Yeah that’s a nice way to think about it

[novel acorn]

I don't know why I though you could overlay two tori but at least I got the stupidity out of me

[pearl holly]

Okay so it's getting late here so this might be a bad idea but I just don't get how the universal cover of a path connected and locally path connected space $X$ is a covering space of every other path connected covering space. So if I assume that $p: \tilde{X} \to X$ is the universal cover (i.e it is simply connected) and $p: X_2 \to X$ is another covering, then I want to show that $\tilde{X}$ is a covering space for $X_2$. Now I can use the lifting criterion since $\pi_1(\tilde{X})$ is trivial to get a lift $\tilde{p}: \tilde{X} \to X_2$ of $p$. But now I need to show that $\tilde{p}$ is a covering space and I can't find a theorem that shows this. Hatcher also just states this as a fact so I feel like this should be trivial.

[gentle osprey]

Tokidoki ✓

[pearl holly]

(sorry if I interrupted but I typed this in before your question)

[novel acorn]

No I managed to solve that other one

[pearl holly]

Okay great! I thought you had another question incoming or something

What did you get btw?

oh wait, I might have misread that lmao

[novel acorn]

Well the structure what homeomorphic to $S^1\times(S^1\vee S^1)$ So I got that the group was $\bZ \times (\bZ * \bZ)$

[gentle osprey]

Irony Incarnate

[novel acorn]

At least I hope that's correct

[pearl holly]

ahhh okay I see. Yeah that makes sense. My solution was written in presentations lmao

and I don't know if my group presentation "simplified" to Z x (Z*Z)

[novel acorn]

Hm

I assume you got that the 1-skeleton was $S^1\vee S^1 \vee S^1$ and then got the presentation by attaching a 2-cell

[gentle osprey]

Irony Incarnate

[pearl holly]

yeah exactly

I think so at least, I don't really remember what I did back then lmao

[novel acorn]

Lol I only got back to this now
Got lazy since summer holiday

Mmmmmmmm but I need to build up a better intuition for this stuff

[prisma kayak]

This feels like a pretty elementary question on here, and I'm not even sure if it belongs in the topology section, but I can't exactly figure out how to solve this. This is from the Loomis and Sternberg book, and bases haven't been introduced yet, so all I'm working with are product spaces, projection maps, and injection maps. Anyone have any suggestions? Thanks

[rancid umbra]

do you know that if V and W are finite dimensional vector spaces, then

dim(V oplus W) = dim V + dim W?

because this trivializes the proof

[prisma kayak]

Well, "technically" I know that, but it isn't in the toolkit offered by the book yet, so I'm trying to work within the confines of what's been introduced. All that's been introduced is a basic theory on Hom(V,W), Projection maps $\pi_i$, and injection maps $\theta_i$

[gentle osprey]

Nivellan

[gritty widget]

isomorphic as vector spaces over R here?

[prisma kayak]

Yeah

[gritty widget]

you need to find a bijective linear map from R^m x R^n to R^(m + n). think about what elements of each set look like and you might be able to come up with something simple

#linear-algebra next time (for questions like this)

[prisma kayak]

okay thanks

[rancid umbra]

i think that map would be a homeomorphism as well doe

[gritty widget]

that's why i asked for clarification

isomorphic as lie groups

[reef shore]

For nice spaces, connected coverings are unique upto isomorphism once you know how their fundamental groups embed into that of X

This proves that the universal cover covers all connected covers

By nice I mean the hypotheses you need to prove the existence of universal covers

[random sky]

why is a circular sector not a triangle

[true robin]

It has a curvy bottom

[hazy nexus]

"Hatcher just states it as a fact" doesn't mean it's trivial, don't worry, Hatcher's just bad at explaining stuff

IIRC Rotman's Alg Top book explains this well

[sweet wing]

agree

[west brook]

Fun fact: trim’s inflection is a complexified Wythoffian.

[flint cove]

isomorphic as sets

[hollow harbor]

Isomorphic as curvy bottoms 😵‍💫

[swift fjord]

Rotman's algtop book explains this well

[hazy nexus]

There's only one part of Rotman's AT book I remember being disappointed with

IIRC it's a small section in the section on higher homotopy groups

He defines a special kind of homotopy, I think? Idk, I just remember it being dense af and very unclear about what we were doing and why

Until I saw a picture in Hatcher and I understood

Also one of the few times Hatcher's been useful

[swift fjord]

Nobody's perfect I guess

But so far i'm very happy with it

[hazy nexus]

I recommend using Hatcher for the pictures alone

Let Rotman do the rest

[swift fjord]

Good point

[swift fjord]

I understood my mistake moldi

[empty grove]

[swift fjord]

I didn't see that it started from p1 and not p0

[empty grove]

I don't know what notation you're using but seems legit

[marsh forge]

i'd be interested to know what you are referencing if you remember

[swift fjord]

Basically tho it's saying that if you take a set of points in an affine space and then 'center' them back to the origin, they should be linearly independent, right?

where p0 was where the affine space was centered and now is the origin

Like basically if $\mathbb A = p_0 + W$ then we want a set ${p_0,\ldots,p_n}$ in $\mathbb A$ to be independent if the corresponding set in $W$ (without 0) is independent

[gentle osprey]

ShiN

[gritty latch]

I have to show that a path connected space is the continouous image of a path connected space

just wanted to ask if my strategy is going to work

[empty grove]

Yep

Affine independent sets are linearly independent sets translated in a higher dimensional space

hmm I guess that's correctly phrased

[empty grove]

Do you mean the converse

The image of a path connected space is path connected

Because what you said is trivial, every path connected space is the image of itself under identity

[gritty latch]

yes

[empty grove]

So take any 2 points in the image, and try to find a path between them. Did you get stuck somewhere with this approach?

[gritty latch]

okay I think I may or may not have been trying to prove the wrong statement all along

as in, I was trying to prove the trivial converse

[marsh forge]

wait what trivial converse

[gritty latch]

I was trying to construct some continuous function between two path connected spaces X,Y with X != Y

[gritty latch]

.

[marsh forge]

thats not the converse but I see what you mean

[empty grove]

oof

[marsh forge]

(the converse is false)

the statement is that if f is continuous and A is path connected then f(A) is path connected. The converse of this is not true

[empty grove]

Converse depends a lot on what you treat as background set up and what you assume are the hypotheses lol

[marsh forge]

huh

[empty grove]

Like being top spaces is also an assumption but it's part of the set up so we don't involve that in the converse

[marsh forge]

huh????

[empty grove]

I mean silly example

[marsh forge]

i think ur being silly

[empty grove]

[gritty latch]

I'm not exactly sure why I got so confused lmao

[marsh forge]

the converse of what I wrote is clear

[empty grove]

Tis what I do

[gritty latch]

all I had to prove was that the cont. image of a path connected space is path connected

[empty grove]

Yeah I was defending my statement of what the converse is lol

[gritty latch]

this shouldn't be too hard. if I get stuck I'll get back to you

thanks!

[hollow harbor]

the moral converse here is clearly "if the image of any path connected set is path connected, then the function is continuous"

this is false i guess?

[marsh forge]

no

the converse is

if f(A) is path connected then A is

[hollow harbor]

[marsh forge]

which is obviously false

[hollow harbor]

what happened to all the quantifiers

are you taking a converse underneath a bunch of quantifiers

[empty grove]

The statement is "continuous functions preserve path connectedness"

[hollow harbor]

but not underneath others

[empty grove]

So rice is right

[hollow harbor]

wait

[empty grove]

Converse is "functions that preserve path connected are continuous"

[hollow harbor]

what's a counterexample to my statement

[empty grove]

[hollow harbor]

i guess

oh

take topologists sine curve with the y axis

and then cut it apart

[marsh forge]

oh no

[hollow harbor]

wait, so does "functions that preserve connectedness are continuous" work?

i've never thought about this

[marsh forge]

I would guess no but I have no counterexample in mind

[versed pivot]

What about discontinuous derivatives?

[hollow harbor]

yeah you're right

just good old sin(1/x), 0 at 0

i think this is a counterexample

[gritty latch]

so if F is the continuous map X -> Y and g : [0,1] -> X is our path

then just taking h : [0,1] -> Y as h(x) = f(g(x)) works right

[empty grove]

Yeah that's the idea but as a proof that's weirdly written

You should start with 2 points in Y instead of a path in X, unless this was supposed to be extremely informal

[gritty latch]

it was informal

[empty grove]

[gritty latch]

now, off to show R^2 is not homeomorphic to R

probably not going to do the rest of the exercises unless I find one that is super interesting

[empty grove]

It's good

oh nvm I misread

[swift fjord]

that's a cute exercise

I've really struggled with convex combinations and simplices in the past, is this intuition correct? We can think of the barycentric coordinates w.r.t some simplex as being inversely proportional to how close you are to that point of the simplex. Like, if one of the coordinates is 1 then you're at that point, if one of the points is 0 you're opposite that point, and if the simplex is spanned by m points, then if all the coordinates are 1/m then you're at the barycenter. Right?

[plain raven]

yeah you can think of the coefficient as a weight. higher weights on a vertex pull the point towards that vertex. a weight of 1 is as high as possible, so the gravitational force is overwhelming and it pulls the point all the way to the vertex. when the weights are all 1/n, then the tension from each vertex on the point is equal, which positions it in the center

[swift fjord]

yea ok that's a good way to think about it

why does rotman define the standard n simplex in Rn+1?

Can't you think of it in Rn?

[marsh forge]

that is weird

yes its the convex hull of the ordinary basis of R^n

[swift fjord]

But you do need n+1 points to define an n simplex

[marsh forge]

wdym?

Yes

sorry

include the origin

[swift fjord]

yea ok

ok yea that's what I imagined

weird that he defined it like that but I guess it doesn't really matter

[marsh forge]

very weird imo

[swift fjord]

it was just an example tho not an official definition

technically

[marsh forge]

i c

[swift fjord]

[plain raven]

it's more symmetric that way, in $R^{n+1}$ it's just the convex hull of all the basis vectors and you don't have to include the origin I guess?

[gentle osprey]

diligentClerk

[swift fjord]

yes

that's now he defined it

[plain raven]

it's not a big deal, it's just convention, it's the difference between $\sum a_i =1$ and $\sum a_i \leq 1$

[gentle osprey]

diligentClerk

[plain raven]

one thing i'll say is that like, this can be seen as a special case of a technique for giving a geometric realization of any simplicial complex

if you have an abstract simplicial complex $K$ with underlying set of vertices $V$ you can define its geometric realization to be the subset of $\bigoplus_{v\in V} \mathbb{R}_v$ consisting of all finite sums $\sum a_iv_i$, $\sum a_i=1$, where all $v_i$ belong to a common simplex of $K$

[gentle osprey]

diligentClerk

[swift fjord]

wait wdym about this part

[swift fjord]

i c

[plain raven]

i'm saying that in rotman's formalism the convex hull would be the set of all points $\sum a_i v_i$ with $\sum a_i=1$, and if you were to work in $\mathbb{R}^n$ it would be the set of all points $\sum a_i v_i$ with $\sum a_i \leq 1$

[gentle osprey]

diligentClerk

[swift fjord]

Ah you're saying that in this case the barycentric and cartesian coordinates coincide

but that in the Rn formulation they don't

right?

[plain raven]

uh hmm yes i guess that's true

[pearl holly]

bois I think I got it on my buss ride

how do I draw diagrams in here?

[swift fjord]

https://q.uiver.app/

[hollow harbor]

buss ride

[empty grove]

COFFIS CUP

[pearl holly]

% https://q.uiver.app/?q=WzAsMyxbMiwyLCJYIl0sWzIsMCwiXFx0aWxkZXtYfSJdLFswLDIsIlhfMiJdLFsxLDAsInAiXSxbMSwyLCJcXHRpbGRle3B9Il0sWzIsMCwicF8xIiwyXV0=
[\begin{tikzcd}
&& {\tilde{X}} \
\
{X_2} && X
\arrow["p", from=1-3, to=3-3]
\arrow["{\tilde{p}}", from=1-3, to=3-1]
\arrow["{p_1}"', from=3-1, to=3-3]
\end{tikzcd}]

[gentle osprey]

Tokidoki ✓

[pearl holly]

okay so we want to show that $\tilde{p}$ is a covering map (which is a lift of the universal covering, $p$, and I can lift it because of the lifting criterion). So let $x_2 \in X$. Now "project" it $p_1(x_2) \in X$. Let $U_x$ be a neighborhood of $p_1(x_2)$ that is evenly covered by both $p$ and $p_1$ (I can do this because that point has a two neighborhoods, one is evenly covered by $p_1$ and the other one by $p$. Now just take the intersection of those two). Let $V_x$ be a slice in $U_x$ by $p_1$ that contains $x_2$. Now $\tilde{p}^{-1}(V_x) = p^{-1}p_1(V_x) = p^{-1}(U_x)$ (the last equality comes from the fact that $V_x$ is homeomorphic to $U_x$ by restricting $p_1$). Since $U_x$ is evenly covered by $p$, we have shown that $V_x$ is the desired neighborhood of $x_2$. Pick $S$ that is a sheet of $V_x$ under $\tilde{p}$. We want to show that this sheet is homeomorphic to $V_x$ by restricting $\tilde{p}$. We can draw a similar diagram

[gentle osprey]

Tokidoki ✓

[pearl holly]

% https://q.uiver.app/?q=WzAsMyxbMiwyLCJVX3giXSxbMiwwLCJTIl0sWzAsMiwiVl94Il0sWzEsMCwicCJdLFsxLDIsIlxcdGlsZGV7cH0iXSxbMiwwLCJwXzEiLDJdXQ==
[\begin{tikzcd}
&& S \
\
{V_x} && {U_x}
\arrow["p", from=1-3, to=3-3]
\arrow["{\tilde{p}}", from=1-3, to=3-1]
\arrow["{p_1}"', from=3-1, to=3-3]
\end{tikzcd}]

[gentle osprey]

Tokidoki ✓

[pearl holly]

(ignore the morphism names)

Now two of them are homeomorphisms, so the third one must too

[empty grove]

Toki I gave a solutions to this

Check your mentions

Different approach though

You're trying to show that universal cover covers all connected coverings right?

[pearl holly]

yeah path connected

[empty grove]

Yeah same thing

[pearl holly]

okay I read it now but how you know that it covers every other one just because it is unique?

[empty grove]

Well you know one of the copies that it covers

And use the isomorphisms to cover all the other ones with the same group

[pearl holly]

wait I don't really see that

I think that there's something wrong with my proof tho

like how do I know that the restriction from S to U_x is a homeomorphism?

[empty grove]

Tbh I didn't bother reading

Revenge 😈

[pearl holly]

revenge?

[empty grove]

For not reading my answer

[pearl holly]

yoo I wan't active on discord then

When I got back I got super hyped and just typed it in as fast as I could

okay no it is right because S comes from p

[pearl holly]

but I still don't reallyt see this, could you maybe explain?

[empty grove]

Yeah I'll do a quiver diagram I guess

Once I finish dinner

[pearl holly]

yeah I need to eat now as well lmao

[empty grove]

[pearl holly]

🍴 🥘

[hollow harbor]

Shallow pan of food.

[gentle osprey]

Coldilocks ✓

[reef shore]

phi is supposed to be an isomorphism, that's what the squiggle means. They were supposed to be in different lines

[pearl holly]

well I gues just phi u?

[reef shore]

yeah lol

[pearl holly]

oh lmao

[reef shore]

Draw commutative diagrams 😌

[pearl holly]

but how do I know that it is a covering map?

[reef shore]

Exercise

Max taught me this 😌

[pearl holly]

it feels like I already done that tho?

but idk

it feels similar to the thing I did above

[reef shore]

Hint is phi goes both ways

[pearl holly]

wait is phi a covering map or is it just an isomorphism?

[reef shore]

Is the identity map on a space a covering map

[pearl holly]

yes?

[reef shore]

generalize

slightly

🤏

[pearl holly]

yeah okay I see lmao. So phi is a homeomorphism and so a covering map. Then apply my solution thing I guess

[reef shore]

I am not sure what your solution thing is

Also composition of covering maps is not necessarily covering I think

[pearl holly]

well my solution basically says that if p and p_1 here are covering maps, then so is tilde p

[reef shore]

Hmm seems legit

But you can do a simpler argument in this case

since phi is a homeomorphism

Like just find evenly covered neighbourhoods directly

I don't think your proof works. The line where you say "Now p tilde inv(V_x) = p inv p_1 (V_x)"

That doesn't seem write, because V_x is not saturated with respect to p_1

So applying p_1 will increase the preimage

[pearl holly]

Okay so let y be in Y. now let U_x be a neighborhood of phi^(-1). Then u^(-1)(U_x) is the desired neighborhood, right?

[reef shore]

phi(U_x) is the evenly covered neighbourhood, and its preimage is what you said yes

Here your previous proof doesn't break because the V_x you took is saturated

Every subset is saturated with respect to an injective map

[pearl holly]

hmm but $p_1 \tilde{p}(V) = p(V)$, and so $\tilde{p}^{-1}(V) = p^{-1} p_1(V)$

[reef shore]

Pre images don't work like that

p inv is not the inverse map of p

[gentle osprey]

Tokidoki ✓

[reef shore]

Decide what you wanna say

[pearl holly]

okay okay I'm done

It was just small errors

[reef shore]

lol

so you applied p inv on both sides then p tilde inv and then cancelled the p tilde, is that it

[pearl holly]

yeah but I didn't cancel p tilde, I just moved it to the other side

[reef shore]

pls publish the steps

I don't get it

[pearl holly]

okay so $(p_1 \tilde{p})(V) = p(V)$ because the diagram commutes. Now $p_1(V) = (p \tilde{p}^{-1})(V)$ so $(p^{-1} p_1)(V) = \tilde{p}^{-1}(V)$

[gentle osprey]

Tokidoki ✓

[reef shore]

You are applying p on V

So V is in that space you wrote at the top

I am not scrolling for the name

[pearl holly]

Yes

[reef shore]

Then how are you applying p_1 on it

Isn't that the map from the bottomological space

Like you have applied all 3 maps to V somehow

That doesn't make sense because at most 2 of the maps have the same domain

[pearl holly]

wtf

what have I done?

[reef shore]

[marsh forge]

You should always always draw diagrams when working with lots of functions that do or don’t commute

I think you should basically always avoid the “algebra of functions” approach

Like just symbolic manipulation

It obscures too many details

[pearl holly]

Okay roger that

[reef shore]

Also when you said you moved p tilde to the other side, did you do the kind of thing where you have a+b = c and you "move b to the other side" by writing a = c-b?

[pearl holly]

yes exactly lmao

but p tilde inv(V_x) = p inv p_1 (V_x) still makes sense in the diagram, no?

[pearl holly]

I mean in this

[reef shore]

Can you prove that moving b to the other side is a valid thing to do

[reef shore]

it would, if your set V changed

[pearl holly]

maaaan why does Hatcher do this to me

[reef shore]

[reef shore]

This is something you need to do toki

urgently

Because you said something very wrong about this

[pearl holly]

but is that even true? It just becomes weird when I do it, right?

[reef shore]

I am asking you to prove that for numbers now

[pearl holly]

oh for numbers

idk, just add -b to both sides I guess

[reef shore]

Yes, and that gives you a+b-b on the left

How do you get a

[pearl holly]

bro

what is this

[reef shore]

[pearl holly]

well b is the inverse to -b by definition

[reef shore]

exactly

So when I said that p and p inverse aren't actually inverses, so you can't cancel them

[pearl holly]

ah yeah that's true

[reef shore]

"because I didn't cancel them, I just moved it to the other side"

is wrong lol

[pearl holly]

oh yeah lmao. I didn't know what you meant by "cancel" so I just ignored it

[reef shore]

🔫

[pearl holly]

okay well anyway, thank you so much!

[reef shore]

np

[pallid lion]

why exactly is it "obvious" that this diagram is commutative based on the axioms from a homology theory?
X,Y are non-empty topological spaces and H is any homology theory

also f is continuous but that should probably be clear already

[empty grove]

The diagram commutes when you ignore the homology

Then applying homology should preserve commutativity I suppose

[swift fjord]

Functors preserve composition so they preserve commutative diagrams

[pallid lion]

ah yeah thanks that was quite the brainfart lol

[pulsar lynx]

is the set of all positive rational numbers whose square is bigger than 2 open or closed or neither in R

[empty grove]

Neither

[pulsar lynx]

why isn’t it open

I think I’m brain farting here

Doesn’t it satisfy the definition of open set

[true robin]

Can you find a point in it such that an open ball isn’t contained in the set?

[pulsar lynx]

oh wait would the set of all reals with square >2 be open

[true robin]

Yes

[pulsar lynx]

is “rationals” the problem here since they act all weird in R

[true robin]

Yes

[pulsar lynx]

how do you articulate that

[true robin]

Take some point in the set, actually any point will work, so take 2 for example

[pulsar lynx]

Ok

[true robin]

Now given any open ball containing 2, can you think of a point not in the set?

[pulsar lynx]

Ah i see

yeah, can’t you find an irrational arbitrarily close to any rational

[true robin]

Correct

[pulsar lynx]

That’s really cool

Is that related to rationals being dense in R

[true robin]

Its related to the irrationals being dense

[empty grove]

Them and their complement both

[pulsar lynx]

ohh

[empty grove]

Rationals being dense and proper subset makes them not closed

Same for the complement

[true robin]

Ah, I meant my answer only for the not open part of the question

[pulsar lynx]

ye

Kind of unrelated question but what’s a canonical example of a (sequentially) compact space

[true robin]

[0,1]

That seems like the first sort of example at the very least

[pulsar lynx]

Wow

[rancid umbra]

any finite set with the discrete topology

[pulsar lynx]

Ye

[empty grove]

Any finite set too

Topology won't matter

[rancid umbra]

also true

[pulsar lynx]

I think I’ve got some misunderstanding here

Is a_n = 1 for all n a sequence

Would that be like (1,1,1,…)

[empty grove]

Yes

[true robin]

Yes

[pulsar lynx]

Ok, so isn’t that a sequence in [0,1]? And if so

[marble socket]

yep

[pulsar lynx]

What is the convergent subsequence

[rancid umbra]

itself

[pulsar lynx]

Oh I’m an idiot

Oh my god

I was thinking sum converge lol

[rancid umbra]

thats a series

[pulsar lynx]

Thank you

Yeah

[swift fjord]

@pulsar lynx it's a well known result that any compact metric space is sequentially compact. In fact it's iff

I can produce examples of a compact space that's not sequentially compact and vice versa but they're a bit messy

[empty grove]

Compact always implies sequentially compact (compact iff net compact → sequential compact)

If there's a sequence without convergent subsequences, for each point in the space you take a neighborhood of that point containing no points of the sequence other than possibly itself

This is an open cover but each neighborhood contains at most one point of the sequence so can't have a finite subcover

[swift fjord]

Are you sure? We've done an example in class of a compact but not sequentially compact space

I mean that sounds correct

[empty grove]

Yeah I think so

The other one has counterexamples, I think ω₁ with the order topology will be sequentially compact but not compact

Not compact is clear because identity net has no convergent subnet

Or also sets of the form [0, a) form an open cover without finite subcover

Sequentially compact should come from cofinality memes I think, all sequences in this are bounded

[swift fjord]

This is the example we gave in class, these are my notes so there may be small mistakes

Maybe the equivalence holds if you allow some countability axiom?

Or the implication rather

Pibase seems.to agree with me

[flint cove]

That shouldn't be the case because the whole point of nets is that you aren't restricted to ctbl stuff
Interesting to see what side is wrong here, nothing obvious to me

[swift fjord]

Same here, really weird

[flint cove]

Oh you mean for sequential compactness, yeah for that you need first countability I think

[empty grove]

Hmm weird

So net compact doesn't imply sequentially compact?

The only place that could break would be in the definition of subnets of a sequence

So there can be a sequence with convergent subnets but no convergent subsequences?

[swift fjord]

Seems like it

[empty grove]

I think I see what's going wrong here

[flint cove]

Oh the sequence might have a convergent subnet!

That iself is not a sequence

Right?

[empty grove]

Yeah I didn't know that subnets and subsequences could disagree like that for sequences

But it seems to be the case

[flint cove]

I feel betrayed

[empty grove]

I hate nets now

[swift fjord]

This is exactly why we need nets

Sequences can't be trusted

[empty grove]

[flint cove]

I did net expect that

[swift fjord]

If your space is too big sequences misbehave

But nets will never let u down

[flint cove]

This deserves to be pinned imo

[empty grove]

[flint cove]

What's the thing missing between the sum and the comma here tho?

[swift fjord]

I think just ti

Lemme check

Or wait

Gotta find it now

[empty grove]

https://math.stackexchange.com/questions/44907/whats-going-on-with-compact-implies-sequentially-compact

Lmao this guy went through the exact same confusion

With the exact same arguments

[swift fjord]

@flint cove it's supposed to be $$\sum_{i=1}^{\infty} \frac{t_i}{2^{i}}$$

[gentle osprey]

ShiN

[swift fjord]

I forgot the infinity so the $\frac{t_i}{2^i}$ got carried over to the limit of the sum lol

[gentle osprey]

ShiN

[swift fjord]

so just like, binary expansion

[flint cove]

Oh wow 70+ votes on a topology question

[empty grove]

All those other people who went through this 😌

[swift fjord]

Lmao

[eternal nimbus]

Turns out, for some circumstances, I need to learn a two semesters worth of geometry/topology in about... 3 weeks. How throughly fucked am I?

[flint cove]

Depends on what you already know

[eternal nimbus]

Only stuff used in other classes (for example we dived a bit on metric spaces/normes vector spaces/compacts so we could properly define and generalize stuff like continuité/convergence/uniform convergence in terms of compacts, complete spaces stuff like that )

Imma just jump on a train

[abstract pagoda]

I never saw [0,1]^{[0,1]}

[pallid lion]

anyone know a proof or where I can find a proof of this statement? f induces a group homomorphism from H_1(S^1) to H_1(S^1) which we will both identify with the additive group of integers via the same isomorphism. Therefore we have a homomorphism from Z to Z which is of the form n \to d \times n for some number d, which is the degree. That is our lecture definition of degree applied to S^1 specifically here

[marsh forge]

Hint: Hurewicz

I mean the idea is that geometrically $z^k$ wraps the unit circle around itself k times

[gentle osprey]

MaxJ #MiuArmy

[pallid lion]

well I don't think I am specifically allowed to use hurewicz here, since we're supposed to work with any homology theory with coefficient group Z.
Now I know that the degree is independent of the homology theory if the coefficient group is Z, but I think that result only came later in the lecture and without that result I cannot directly use hurewicz here, or at least I do not think so.

[marsh forge]

I am not sure I understand what you mean

[pallid lion]

But if I was allowed to do so I guess I just use the fact that H_1(S^1) and the fundamental group are "naturally" isomorphic here since they're both abelian and then just show that the generator (simple loop) is mapped to k times the generator concatenated.

[marsh forge]

yes

naturality is important here btw

not just a vague word

You can prove this for the various equivalent formulations of singular homology but I think you will end up needing to reference the specific one unless you use hurewicz

[pallid lion]

well hurewicz works with p-chains if I remember correctly, but we have not specified our homology theory in that exercise.

[marsh forge]

there is a theory-agnostic construction of Hurewicz

but again if you already know all the theories are the same

there is no reason to tie your hands behind your back

[pallid lion]

well I just use that the degree is independent of the homology theory in that case anyways

[marsh forge]

Oh, then just choose a convenient one then

[pallid lion]

well the one with p-chains does the job, since we have only looked at that one so far and proved hurewicz for it

anyways thanks for the help

[hazy nexus]

I went back to check, and I think it was this part

[marsh forge]

Oh my god that typography

Gouge my eyes out

Okay I’ll read on my flight

[hazy nexus]

Yeah, I had to use an old scan on my phone, too

[swift fjord]

it's an older book

[swift fjord]

Let $M=[p_0,\ldots,p_m]$ be the m-simplex spanned by the independent points $p_i$, Let $C=C[p_0,\ldots,\hat{p_j},\ldots,p_m]$ be the cone over the face opposite $p_j$. I want to show there is a homeomorphism, just checking, the map $f:M\rightarrow C$ defined by
$$f\left(\sum_i t_ip_i\right)=[\sum_{i\neq j} t_ip_i,t_j]$$
Would suffice right?

I can also construct the inverse

[gentle osprey]

ShiN

[swift fjord]

The height represents the missing barycentric coordinate in this case, so I think this was the construction they were shooting fore

tho what's the easiest argument for continuity here?

I mean I could define the map $T:[p_0,\ldots,p_m]\rightarrow [p_0,\ldots,\hat{p_j},\ldots,p_m]\times [0,1]$ as
$$T\left(\sum_i t_ipi\right)=(\sum_{i\neq j} t_ip_i,t_j)$$
Then clearly it is continuous in the first component as an affine map (I think it would be affine, but i'm not sure actually what $p_j$ is sent to in this case, since it can't be 0), but in the first component in fact I need to prove that the barycentric coordinates are continuous

and then I can just compose that with the quotient map to get my desired map

[gentle osprey]

ShiN

[swift fjord]

But how do I prove the barycentric coordinates are continuous without just doing it with epsilonautics

[swift fjord]

actually I don't even think this construction would work because the map in the first component won't be affine

and it won't send convex combinations to convex combinations since i'm deleting a term so it's not even well-defined

[swift fjord]

so what can I do to show continuity here?

ok wait this map won't be well-defined either since it's not sending the first component to an element of the m-1 simplex

how can I remedy this?

maybe sending pj to the barycenter of the m-1 simplex would work

still tho how do I prove that the barycentric coordinates are continuous

[swift fjord]

well still thinking about this ^

but I have a new question

I know that all maps $f,g:Y\rightarrow X$ where $X$ is contractible are (freely) homotopic. I want to show that the fundamental group of a contractible space is trivial. I know I can just do this by showing that a homotopy equivalence induces an isomorphism of fundmental groups, but I want to know if there's a way to do this more constructively

[gentle osprey]

ShiN

[swift fjord]

so I know any 2 loops are freely homotopic, but how do I show there's a relative homotopy preserving the basepoint?

[plain raven]

fix a contraction of X to a point.

show that if f : S^1 -> X is any map, we can use the contraction to extend f to the cone of S^1

then use the fact that the cone of S^1 is homeomorphic to D^2

does that help?

[swift fjord]

I'm working with the definition of loops being from the unit interval to the space. I know both formulations are equivalent, but I wanna show it expilcitly using that formulation.

If using that formulation then yea i've seen that if such a map can be extended to D^2 then it is nullhomotopic rel some point

[plain raven]

a function f : [0, 1] -> X satisfying f(0)=f(1) is equivalent to a function f' : [0,1]/~ -> X. can you tell me why you are not interested in making use of this equivalence?

(where ~ is the equivalence relation generated by 0 ~ 1)

[swift fjord]

If I show that the induced map on the sphere is nullhomotopic, then how can I turn that into a homotopy of the closed path?

[swift fjord]

Honestly just to check my understanding

[plain raven]

gotcha

if you have a map $D^2\to X$, say with the basepoint $(1,0)$ on the boundary getting sent to $x_0$ in $X$

[gentle osprey]

diligentClerk

[plain raven]

then

if you look at the map $f : I \to D^2$ which sends $t$ to $e^{i 2\pi t}$

[gentle osprey]

diligentClerk

[plain raven]

it is obvious that $f$ is homotopic to the constant math at the basepoint, by just shrinking the circle down to the basepoint

[gentle osprey]

diligentClerk

[plain raven]

then if you postcompose this with the map $D^2\to X$, you'll get a homotopy to a point

[gentle osprey]

diligentClerk

[plain raven]

like if you have $f : I\to (X,x_0)$ which sends both $0,1$ to $x_0$, then if you can prove that $f$ extends along the map $I\to D^2$ I just described, then it's clear $f$ is homotopic to the constant map at $x_0$

[gentle osprey]

diligentClerk

[plain raven]

(by a basepoint preserving homotopy)

[swift fjord]

Yea ok I see it now

thanks

Really the formulation of loops being maps from the circle is usually easier to work with

rotman did show it's equivalent

btw @plain raven any way you can help me with the previous problem? To sum it up all that I need to prove at this point is that the barycentric coordinates of a simplex are continuous. Is there a non annoying way to do this?

[plain raven]

a simplex inherits its topology from $\mathbb{R}^{n+1}$, in which it is a subspace

[gentle osprey]

diligentClerk

diligentClerk

[plain raven]

and with this, the barycentric coordinates are the projection functions $\mathbb{R}^{n+1}\to \mathbb{R}$

[gentle osprey]

diligentClerk

[swift fjord]

but the barycentric coordinates don't necessarily coincide with the regular coordinates of the point

[plain raven]

if you work in the $n$-simplex which is the convex hull of the basis vectors of $\mathbb{R}^{n+1}$ they should, right?

[gentle osprey]

diligentClerk

[swift fjord]

oh wait you're right, any affine n-simplex is homeomorphic to the standard simplex

and there barycentric coordinates are just projections

so that's sufficient right?

[plain raven]

yeah i think so

[swift fjord]

eh good enough for me

[swift fjord]

ok, so I have this diagram, where $\beta:X\rightarrow Y$ is a homotopy equivalence with $\alpha:Y\rightarrow X$ being its homotopy inverse. $\psi$ is the isomorphism induced by the homotopy $\alpha\beta \simeq 1_X$. Rotman uses this to prove injectivity of the induced map $\beta_*$ and says a similar diagram can be used to prove surjectivity

[gentle osprey]

ShiN

[swift fjord]

but whenever I try to do a similar diagram, since I don't know if $\alpha$ is surjective, I end up with an induced map with different basepoints

[gentle osprey]

ShiN

[swift fjord]

Like I end up with $\beta_*: \pi_1(X,\alpha\beta(x_0))\rightarrow \pi_1(Y,\beta\alpha\beta(x_0))$ instead of $\pi_1(X,x_0)\rightarrow \pi_1(Y,\beta(x_0))$

[gentle osprey]

ShiN

[swift fjord]

What am I supposed to choose for the basepoint of $Y$ when doing the second diagram

[gentle osprey]

ShiN

[swift fjord]

This is the diagram I forgot to post it

and these aren't based maps otherwise this would be obvious

Oh wait I think I got it. Since it's a homotopy Equivalence then the fundamental groups are isomorphic

Like $\pi_1(X,x_0)$ and $\pi_1(X,\alpha/beta(x_0)$

[gentle osprey]

ShiN

Nobody

[swift fjord]

I got that part

I meant the part where beta* is surjective

Which he says can be done with a similar diagram

The basepoints didn't add up but I think it doesn't matter cuz of the isomorphism

[gentle osprey]

Nobody

[swift fjord]

Ok yea it's surjective onto $\pi_1(X,\alpha\beta(x_0)$ but this means it's surjective onto $\pi_1(X,x_0)$ cuz of the isomorphism right

[gentle osprey]

ShiN

[swift fjord]

Alright yea. The bit that confused me was just that the induced map is not unique is you're working in Top instead of Top*

[gentle osprey]

Nobody

[swift fjord]

Yea alright

Got it

[swift fjord]

But it doesn't matter cuz of the isomorphism

Even though it's technically not the same up (up to composition with an isomorphism)

[gentle osprey]

Nobody

[hard pumice]

hey guys, is there a book or resource that you'd recommend on geometric algebra?

[torpid geode]

Is that valid?

I'd say no, but what mistake happened

[novel acorn]

Hestenes

New foundations of classical mechanics

He basically reformulated everything with geometric algebra

[hard pumice]

Thanks!

[torpid geode]

To be understood my question, δab is the kronecker delta

[drowsy yew]

what is the * here

Ohh thank you 🙂

and ik that S_* is the herre thing right?

kleene star

i mean

[cedar pebble]

It’s not, no

S appears to be referring to the singular simplicial complex, which is nonstandard notation

[gritty widget]

Alright. I’ll prove it for you

First, we’ve $\nabla_a \xi_b + \nabla_b \xi_a =0$

[gentle osprey]

Dr.Marschett

[gritty widget]

I’ll use the manipulation of index with metric tensor, so we get

$g_{ab}\nabla^b g_{ba} \xi^a =-\nabla_b \xi_a$

[gentle osprey]

Dr.Marschett

[gritty widget]

After it we pass the contravariant nabla to the right and make an integral

So

$g_{ab} g_{ba} \xi^a =-\xi_a \Rightarrow \ g_{ab} \xi_b = - \xi_a \Rightarrow \ \xi_a = -\xi_a$ where $2\xi_a =0$, so $\xi_a =0$

[gentle osprey]

Dr.Marschett

[gritty widget]

The fundamental Theorem of Derivative says “If the partial derivative is in relation with a constant, it’ll be 0"

So $\nabla_b \xi_a =0$

[gentle osprey]

Dr.Marschett

[gritty widget]

And with the first equation above, we prove $\nabla_a \xi_b =0$

[gentle osprey]

Dr.Marschett

[gritty widget]

Now, the problem of the Kronecker delta. It only will be useful if and only if $\delta_{ab} =g_{ab}$

[gentle osprey]

Dr.Marschett