point-set-topology

[swift fjord]

path connectedness is a homotopy invariant yes

[unreal stratus]

More generally the set of path components is (pi0(X))

[swift fjord]

it's not

but yea you can show that pi_0 is homotopy invariant

that's much easier

like

taking that approach is easier

[unreal stratus]

Set of path components

[swift fjord]

set of path components

as potato said

you can show that a homotopy equivalence induces a bijection between the sets of path components

this shows path connectedness is a homotopy invariant

[unreal stratus]

I think all the examples kinda show a deformation retract is basically when a space is a "fluffed up" version of a subspace

Like those examples with the letters or things like the circle being a deformation retract of R^2 minus a point

[swift fjord]

This doesn't count as a retract idt

Wait nvm

I'm stupid

I was thinking of compactification

[unreal stratus]

Ah

[swift fjord]

In whicb you add a point instead ot taking it away

[unreal stratus]

Idk tbh of how much interest retracts are

[shadow charm]

Deformation retracts are the useful retracts

[unreal stratus]

Usually it's just that often you can show two spaces are homotopy equivalent as one is a def retract of the other or smth

[gritty widget]

They are of interest

They connect with extension properties

[unreal stratus]

Actually yeah brouwer and stuff

[shadow charm]

Retracts do give you surjections of the fundamental group if im not mistaken

Nice example of a retract that is not a deformation retract is the torus mapping to a circle on it

You can clearly get a continuous map to this circle (retract) but for it to be a deformation retract you’d need to cut up the torus

[unreal stratus]

Ig you always get that for any product of spaces

[shadow charm]

True

[gritty widget]

I feel like it boils down to circle to point example

[shadow charm]

Yeah fair enough

[pallid lion]

^ me every time I ask a prof or supervisor a question

[hushed marlin]

staring an intro to topology course in a few days. Any tips on success in that class?

[gritty widget]

be comfortable with some basic set theory

general point-set topology is easy if your basic proof skills and set theory is solid

the first chapter of munkres' book provides a comprehensive review

[hushed marlin]

awesome I'll check it out, thank you!

[gritty widget]

Depends on how in depth you want to go

But for a course it should be easy. They chop off a lot of material and force feed you it in my experience

Proof of something like, product of continuum separable spaces is separable for example

I wouldn't say it's easy

Or proofs of metrization theorems

fine, "mostly easy with some hard parts"

Proof of Urysohn lemma

It'll be easy if your grasp on topology is good

the context was someone starting a course in the subject

please interpret my messages with this context

Yeah. So they probably won't have intuition for what they're learning (at first at least)

It might make things seem harder

[unreal stratus]

Also btw the most general thing is that pi0 is actually a functor (from Top to Set) sending each space to its set of path components and each map between spaces to the induced map between path components (since continuous maps send path components to path components) If you can prove that then everything else follows

It may seem overkill but ofc the same is true for pi_1 and indeed any pi_n

[polar whale]

In which book can I get a clear idea regarding the concept of a manifold ?

Also, can someone explain it briefly here ?

[shadow charm]

Probably better suited to #diff-geo-diff-top @polar whale . You can get the intuition for a manifold without going through a textbook, but if you want to work with them then the standard introductory texts are lee and tu if I’m not mistaken

With tu being more expository but worst exercises, and lee being somewhat dry but good exercises from what I hear

[gritty widget]

or #book-recommendations

[shadow charm]

The essence of a manifold is a space that looks like Euclidian space at small scales, ie it is locally homeomorphic to R^n (locally flat if you will) This isn’t everything you need for a manifold, you have other assumptions like for the space to be hausdorff etc but that’s the property that captures why they’re interesting the most.

Most nice surfaces you can think of are manifolds for example

[polar whale]

Okay I'll look into both thnx

[polar whale]

Nicee...I think I should delve deeper...thnx for the exposure

[polar whale]

Right!

[chrome ridge]

https://math.stackexchange.com/questions/788097/showing-that-every-finitely-presented-group-has-a-4-manifold-with-it-as-its-fu In the answer here, why is $\pi_1(X \setminus int(N_j)) = \pi_1(X)$ ?

[gentle osprey]

ru0xffian

[brisk cairn]

Is there a name for a topological transformation that creates or closes holes in a surface? (Think sphere to donut or vice versa) Or a sort of transformation that causes two disconnected surfaces to blend into one, or likewise separate into two? (For context I know nothing about topology, I'm writing a paper on implicit surface rendering and these sort of deformations are relevant, but I'm not sure what to call them, for now just "arbitrary topological transformations") Thanks!

[hidden crag]

To stick spaces together you have the wedge sum for example

[gritty widget]

there's gluing

https://en.wikipedia.org/wiki/Adjunction_space

as for "separate into two", hard to say what you mean. Maybe disjoint sum?

[brisk cairn]

I don't mean anything too spooky, for example here's some basic metaball blending

But it could get as trippy as something like this https://youtu.be/-1x-dFD695g?t=667

You're just seeing the zero isosurface of some scalar field that's changing in time

[shadow charm]

Any hint on 5? The second part follows easily from the first since we’ll get that X tilde is homotopic to an infinite wedge of circles, but I’ve absolutely no clue how to produce the required neighborhood. I’ve tried looking at some kind of maximal element of the poset of open sets containing the left edge inversely ordered by inclusion with zorn but I’m not sure how to rule out closed sets that aren’t the left edge itself

[chrome ridge]

Let $X = (S^3 \times S^1) # \cdots # (S^3 \times S^1) $ $n$ times. Then we know that $\pi_1(X) = <a_1, a_2, \cdots, a_n, |>$ i.e. the free group on generators $a_1, \cdots, a_n$. Let $b_j$ be a loop in $X$. Let $N_j$ be a tubular nbhd of $b_j$. Is there a way to figure out $\pi_1(X \setminus int(N_j)) ? $

[gentle osprey]

ru0xffian

[magic geyser]

I think what you wanna do is lift it without the top edge (you can bc it's simply connected without the top edge), then use something like the lebesgue number lemma and induction on the number of delta neighborhoods covering the left edge to show that horizontally adjacent points in a delta nbhd of the left edge have been lifted close to each other. then you can extend to (a piece of) the top edge. does that make any sense? if it doesnt maybe im wrong

[shadow charm]

I’m not sure I understand what you mean by being able to lift without the top edge @magic geyser

[magic geyser]

If you subtract the top of the square you get a “comb” which you can lift into X tilde

[shadow charm]

Sure but what does simple connectedness have to do with that?

[magic geyser]

That’s why that lift exists

[shadow charm]

I’m slightly confused

When does a lift « not exist »

You mean it’s evenly covered?

[magic geyser]

no its not evenly covered

necessarily

lifts dont exist generally

[shadow charm]

By lift we do just mean the inverse image under the covering map right?

I don’t get what you mean by a lift not existing

[magic geyser]

oh so if $p: \tilde{X}\to X$ is the covering map, we say $f: Y\to X$ lifts under $p$ if there exists $\tilde{f}:Y\to \tilde{X}$ st $p\tilde{f} = f$

[gentle osprey]

motivik

[shadow charm]

Right but what’s f?

[magic geyser]

here f is the inclusion of a subset in X

[shadow charm]

Ah i see

Okay I see what you mean

And why it exists by simple connectedness

But won’t that just end up showing that the neighborhood lifts homeomorphically to a subset of X tilde

Not to all of X tilde?

[magic geyser]

yea the comb is simply connected so you can lift it. in general you can lift f if the image of $\pi_1(f)$ is contained in the image of $\pi_1(p)$

[gentle osprey]

motivik

[shadow charm]

Yeah lifting criterion

I just didn’t realize you were originally talking about maps (or that that’s what the exercise meant)

[magic geyser]

not sure i understand ur question

[shadow charm]

The way I understand the exercise it’s asking to show that the preimage of one of the neighborhoods of the left edge is all of X tilde and homeomorphically so

That’s how I interpreted « lifts homeomorphically »

But perhaps I misunderstood the exercise

[magic geyser]

oh no

[shadow charm]

It makes more sense now

[magic geyser]

it means homeomorphically onto the image of the lift

[shadow charm]

Here’s what I’m confused about, isn’t an evenly covered neighborhood always going to satisfy this?

[magic geyser]

yes

[shadow charm]

Then what’s the big deal?

Just take an evenly covered neighborhood of the left edge no?

[magic geyser]

why should there be a nbhd of the left edge thats evenly covered?

[shadow charm]

Ah that’s fair enough

I was gonna say union but that’s dumb

[magic geyser]

yea that doesnt work

[shadow charm]

Okay I’ll give that another think now that I know what the exercise is asking for

And that way there’s actual work to be done for the second part too

Thanks a bunch

[magic geyser]

np

[shadow charm]

@magic geyser doesn’t the argument only work for covering spaces that are path connected and locally path connected?

Since that’s when the lifting criterion applies?

[magic geyser]

oh uhhh maybe

[shadow charm]

I mean it should still work if I only care about the second part I guess since simply connected requires path connected and locally path connected if I’m not mistaken

[bitter smelt]

this is not true iirc hold on

yeah comb space is counter example

[magic geyser]

well the comb is not gonna be locally path connected, but it is simply connected. but if ur saying local path conn is required for all this covering space lifting stuff to go through then i guess im lost

[shadow charm]

Ah yeah local path connectedness isn’t required in the definition of simply connected

[shadow charm]

It’s required of the covering space yeah for lifting criterion to apply

[bitter smelt]

also topologists sine curve

[shadow charm]

Though maybe for this specific case you could show the lift exists without the criterion

[shadow charm]

Yeah one exercise shows a counterexample with quasi-circle

[magic geyser]

maybe im tired but it seems like you should just be able to do this by mimicking the proof that you can lift a path. you know where you just lift one piece at a time under evenly covered neighborhoods.

just start at the bottom of the left edge and work your way up

each time possibly shrinking the width of your neighborhood as needed

[shadow charm]

I’ll give that a shot tomorrow, thanks again for the help

[chrome ridge]

In the footnote, could someone explain how performing the surgery kills off w_j ? I think it is some application of Van Kampen's thm but I couldn't figure out the details

[shadow charm]

https://math.stackexchange.com/questions/1129047/does-there-exist-a-space-w-s-t-x-to-w-y-to-w-are-covering-spaces-when-z/1144239#1144239

Anyone know which graphs with 2 vertices and 3 edges Hatcher refers to here? The only ones that appear to me to have the given covering space Z are the graph with one loop at each vertex and a straight edge between the vertices, and the graph with three straight edges between the two vertices. However the first is a covering space of the second so we don’t get the required solution

[lament needle]

The premeasure her refers to here is $\mathcal{P} = {[a,b):a,b \in \mathbb{R}}$ with $\mu_\alpha([a,b)) = \alpha(b) - \alpha(a)$. I think i was able to figure out which sets are non-measurable and that seems to be any of the sets contaning an interval $(a,0]$ or $[a,0)$. But I wasn't really sure how to show that this is the only class of non-measurable sets. I know for the lebauge measure constructing non measurable sets is not necessairly simple

[gentle osprey]

matthew

[gritty widget]

That's just the dirac delta measure at 0

any subset of R should be measurable

also #advanced-analysis

[bitter smelt]

How is the first a covering space of the second? They each have 2 vertices that are valence 3, so the covering map would have to have sized 1 fibers no?

[shadow charm]

It can be made into a covering space by considering it not as a graph on two vertices but on 4 by introducing a vertex in each of the loops

Which doesn’t change the space in itself

[bitter smelt]

I see

[shadow charm]

Unless I’m mistaken

Huh wait

[bitter smelt]

So what maps to the two vertices in the second graoh

[shadow charm]

I must be mistaken

Since the vertices you include will have degree 2

[bitter smelt]

Because introducing those would just introduce valence 2 points

Yes

[shadow charm]

Yeah okay I’m just an idiot

[bitter smelt]

And if you have a 1-sheeted cover your spaces are just isomorphoc

So it's a trivial covering

[shadow charm]

I was imagining taking the second graph and “unwrapping it” along one edge to the first graph

But yeah that doesn’t work

Alright thanks I’m a fool

[bitter smelt]

[empty grove]

A sufficient condition is for X to be the colimit of a diagram in which the only space that appears is I

Or rather, if the above condition holds for all maps out of X, then that is equivalent to saying that X is the colimit of all the paths in X. Not sure if that's helpful

[ornate berry]

Not really no. Paths aren't really the right way to think about topological spaces

If you take e.g. Q with the usual topology, this isn't even path-connected

In fact its path components are its points

So clearly paths aren't the right way to view an arbitrary topological space.

[swift fjord]

If you look at paths up to homotopy you get the fundamental groupoid

[ornate berry]

Manifolds may behave nicely with regards to this.

[plain raven]

if X is Delta-generated

https://ncatlab.org/nlab/show/Delta-generated+topological+space

this includes any simplicial complex or manifold

[bitter smelt]

is there a classification of all fundamental polygons with four edges somewhere

like what they are, what they look like, are they orientable, etc

[obtuse meteor]

Anything that holds for any convenient category of spaces but not for general topological spaces is true and anyone telling you otherwise is a proponent of BIG TOPOLOGY

[steel glen]

what's the usual topology on C

or what is the usual metric on C

oh google says d(z,w)=|z-w|

[ocean narwhal]

whats nice about spaces like C or R^n is that all norms are equivilent. this means that every norm generates the same toplogy.

hence convergence wrt one norm implies convergence in the other

[steel glen]

right, i just didn’t consider norms and instead sqrt(z^2 - w^2) which is obv not a metric on C

[raw bolt]

Hello. I'm studying paracompactness in topology. What intuition about this property can you give me? Is it some kind of "local compactness" if that makes sense?

[plain raven]

the most important thing about paracompactness is that paracompact Hausdorff spaces admit partitions of unity subordinate to every open cover.

partitions of unity, in turn, can be used to glue a bunch of locally defined things together into a global thing

Two important examples

1. Any (locally trivial) fiber bundle over a paracompact Hausdorff space has the homotopy lifting property. The proof goes by first noting that it's obvious that a trivial bundle has the homotopy lifting property, and then showing that the local homotopy lifts can be glued together into a global homotopy lift using partitions of unity

This theorem is in Spanier's algebraic topology book, section 2.7

second example:

(From Bott, Tu)

The theorem says "compact" but the theorem holds if Y is paracompact and Hausdorff, which all manifolds are (if you require manifolds to be second countable)

the proof again uses partitions of unity

[raw bolt]

uhh, I am trying not to get to manifolds yet

[plain raven]

Great article by Dold

[plain raven]

yep no problem.

[raw bolt]

mmm, I wanted a more pure topological understanding of the property

I think there was an equivalence for compactness, it was something like every sequence has a subsequence that converges

are there something similar for paracompactness?

I am trying to "feel" what paracompactness is, but maybe it is just not posible xD

[plain raven]

The wikipedia article says -

A topological space is metrizable if and only if it is a paracompact and locally metrizable Hausdorff space

so like, if every point has an open neighborhood homeomorphic to a metric space

then you can glue these together into a global metric somehow

using paracompactness

Conversely, every metric space is paracompact

[gritty widget]

yeah but is every metric space locally metrizable???

[plain raven]

so like, it would not surprise me if it's somewhat difficult to explain what paracompactness is using the metric, because it might reduce to something trivial

[plain raven]

Hm, damn, I hadn't thought of that

I will see if i can come up with a proof in my ample spare time

[raw bolt]

that's interesting, so paracompactness allows you to kind of extend the property of local metrizability to the totality of the space

[plain raven]

yeah and this is something that holds more generally for other things besides metrizability. paracompactness is a feature that lets you take things that are true locally - every x has a neighborhood such that this is true - and prove that the statement is true everywhere on the space.

[raw bolt]

sounds cool

but what does it has to do with compactness other than compact -> paracompact?

so, my intuition for compactness is "topological finitude"

[plain raven]

well compactness lets you take an arbitrary cover and replace it with a finite one. paracompactness lets you take an arbitrary cover and replace it with a locally finite one.

So like

For a compact space, you get your finite cover and then you can often proceed to argue by induction on the number of open sets in the cover

For a paracompact space you can do the same thing but like, locally within a neighborhood of each point

[raw bolt]

omg

ok, I kind of see how they are related

so in the end it is indeed like local compactness

in some way

[plain raven]

Yeah.

But there is a different notion of local compactness

which is not the same

[raw bolt]

I see

thanks!

[plain raven]

np

[gritty widget]

Uh what? No matter what neighbourhood you take it's metrizable as a subspace of a metric space.

blitz i think you need to seriously consider the people you're talking to

give some genuine thought as to whether or not diligentclerk or i would actually be tripped up by a near triviality like this

i understand the joking tone may not have been conveyed well through text, but you've surely been around long enough to understand that we might possibly be joking

🙂

[opaque cloud]

you know how in algebra and anywhere you deal with "spaces" you have this notion of a subspace

like subgroup, submanifold, subring whatever

is there a subtopological space?

[gritty widget]

subspace

[opaque cloud]

is that what it's called in topology?

[gritty widget]

yes

[opaque cloud]

the subset of the topological space that has the same topology?

[gritty widget]

see "subspace topology"

[opaque cloud]

hm okay

I feel like I've heard of that

[gritty widget]

you mentioned "submanifold", so probably

[gritty widget]

it's not the "same topology", but the topology on the subset comes from the larger space in a way

[opaque cloud]

it's like derived from it?

like it has similar properties?

[gritty widget]

i think it's best to give you the definition at this point

[opaque cloud]

okay

[gritty widget]

if $(X, \tau)$ is a topological space and $S \subseteq X$, then let $\tau_S$ be the collection of subsets of $S$ of the form $U \cap S$ for $U \in \tau$. prove by yourself as an easy exercise that $(S, \tau_S)$ is a topological space

[gentle osprey]

TTerra

[gritty widget]

the topology on S is called the subspace topology (maybe with extra words to describe the situation, like "subspace topology inherited from X")

and, of course, S with this topology is called a subspace of X

whenever you have a topological space and a subset of it is referred to as a topological space, this is the topology which is being discussed

[opaque cloud]

I see okay, is it like a "canonical" topology or are there infinite choices for a subspace topology?

I was mainly wondering "how big can a neighborhood around a point be? can it be a subspace with a subspace topology?"

[gritty widget]

the subspace topology on S is the smallest topology making the inclusion map S -> X continuous

[opaque cloud]

I see

[gritty widget]

i guess you could say a topology on S is a subspace topology if it makes the inclusion continuous

but we'd rather not have redundant open sets, so we give it the topology which ensures this property with the least amount of sets

[opaque cloud]

ahh okay so there are choices but most of the time we choose the smallest one

[gritty widget]

all of the time

[opaque cloud]

hm yeah makes sense

[opaque cloud]

can it?

[gritty widget]

i should note that "a subspace topology" isn't established terminology. if you say subspace topology everyone will interpret it as the one i defined above (coarsest making inclusion continuous)

[opaque cloud]

okay got it

[gritty widget]

how big can a neighborhood around a point be?
the whole space counts

[opaque cloud]

wow

[gritty widget]

the definition of a topology asks that the whole set be open

typically people ask questions about how large neighborhoods of points or sets are with some restrictions on the set

like

if i have a function on a topological space that satisfies a property locally, i might want to ask just how large i can make the neighborhood before the property fails

[opaque cloud]

ahhh

okay

Thanks!

[gritty widget]

Another thing is that some people assume neighbourhoods are open sets but others do not. I think it's too much of a restriction to assume that they are open, especially in fields like functional analysis.

what is a neighborhood if not an open set?

Neighbourhood of x is a set of which x is in interior of

Upward closed

[opaque cloud]

closed neighborhood, open neighborhood? people could make that distinction

clopen neighborhood

[gritty widget]

I think all 3 are used

[opaque cloud]

[gritty widget]

Clopen sets arise when you're talking about components, especially of compact spaces

Very interesting to think about how they behave

[shadow charm]

I tend to say open neighborhood or just neighborhood when it’s clear from context

[chrome ridge]

In the footnote, could someone explain how performing the surgery kills off w_j ? I think it is some application of Van Kampen's thm but I couldn't figure out the details

[magic geyser]

Do you know what the surgery is?

aren't you are gluing in something simply connected?

namely $D^2 \times S^{n-2}$

[gentle osprey]

motivik

[chrome ridge]

So we cut out a tubular nbhd $N_j$ around the loop that represents $w_j$, then we glue $D^2 \times S^{n-2}$ along the boundary, my question is why the fundamental group after the gluing is the free group with relation $w_j$ ? I know it should be some application of Van Kampen's but i couldn't figure out the details

[gentle osprey]

ru0xffian

[chrome ridge]

Oh yeah I mean that srry

[magic geyser]

yea rigorously it's probably van kampen

but intuitively just think about the fact that after the surgery the loop w_j now lies in something simply connected so it's now null homotopic

[chrome ridge]

Yeah I get that, I am just trying to figure out the details

[magic geyser]

so in van kampen the intersection is something like $S^1\times S^{n-2}$

[gentle osprey]

motivik

[magic geyser]

the generator of $\pi_1$ of which hits $w_j$ when you look at $\pi_1$ of the inclusion

[gentle osprey]

motivik

[magic geyser]

so it kills $w_j$ in the free product

[gentle osprey]

motivik

[chrome ridge]

but what are the open sets we are applying Van-Kampen's on ?

[magic geyser]

how about one is the manifold minus $S^1$

[gentle osprey]

motivik

[magic geyser]

the other is the interior of $D^2\times S^{n-2}$

[gentle osprey]

motivik

[magic geyser]

youre right you annoyingly need open sets in van kampen. which makes this more annoying

id say first figure it out assuming van kampen doesnt need open sets, so you can make the intersection exactly $S^1\times S^{n-2}$. after that it should be clear how to "thicken" the sets to get open sets

[gentle osprey]

motivik

[chrome ridge]

in your first suggestion , isn't the intersection homotopically equiv to the boundary $S^1 \times S^{n-2}$ ?

[gentle osprey]

ru0xffian

[magic geyser]

yes

[chrome ridge]

Yup I see

Okay thanks a lot that helped !

[magic geyser]

sure

[chrome ridge]

Sorry one more question prob trivial, could you clarify why removing $S^1$ from the manifold didn't change the fundamental group ?

[gentle osprey]

ru0xffian

[magic geyser]

good question. definitely we need to use $n\geq 4$. the inclusion $M-S^1\subset M$ is definitely surjective in $\pi_1$. To show it's injective we just have to show that anything that is null homotopic in $M$ stays null homotopic in $M-S^1$. a null homotopy is just a map from $D^2 \to M$. Since $dim(M)\geq 4$, you can use transversality argument to show that you can adjust $D^2 \to M$ to miss $S^1$

[gentle osprey]

motivik

[magic geyser]

is that at all satisfying?

Maybe there is a more elementary argument. im not very good at this stuff maybe someone else knows a simpler argument

[chrome ridge]

Yeah I think this argument might be a bit complicated, I will try to see if i could figure out the details. I would appreciate it if someone else could provide an elementary argument

[magic geyser]

welcome to the hand waving of geometric topology 😄

[magic geyser]

oooh heres a paper that might help by Milnor, the last topologist to ever give a rigorous proof of anything https://www.maths.ed.ac.uk/~v1ranick/papers/milnorsurg.pdf

[high hill]

Hatcher pg 50

https://i.imgur.com/UHbjsCh.png

I don't quite understand (c)

X is an arbritrary path-connected cell complex, and "the 2-skeleton X^2" is...?

[chrome ridge]

I think it's defined in the inductive definition of a CW-complex

[high hill]

this right?

[chrome ridge]

yeah

[high hill]

I won't say I fully understand, but that is to say the n-skeleton is unique up to the starting set!?

I was under the impression it depended on the maps, surely

[chrome ridge]

you start with 0-cells and you get the 0-skeleton. Then you attach 1-cells to the 0-skeleton and you get a 1-skeleton. then you attach 2-cells to the 1-skeleton and you get 2-skeleton and you go on like that

[high hill]

I agree you get a 2-skeleton, surely you can choose how many 2-cells to attach and how to attach them?

[chrome ridge]

I guess so ?

[high hill]

Then which 2-skeleton does this proposition refer to?

Any?

I don't get the reusage of the letter X, either. We start as X^0 = X, build X^1, then build X^2?

In which case I wouldn't understand the inclusion map... surely X is contained in X^2 if so

Ohh ok, I think step (3) is key. X = X^n for some n >= 2 presumably

[chrome ridge]

The paper you sent probably uses a different approach than Van-Kampen's; I didn't quite understand what they're doing. Do you have any other suggestions? could elaborate more on the transversality argument ?

[fading vale]

Any yes

[fading vale]

the inductive process doesnt have to terminate

X can have n-cells for any n

The point of this claim is basically that pi_1 is depending only on the 2-skeleton (meaning the n-cells for n <= 2)

and that adding a 2-cell is equivalent to adding a relation to the group pi_1

[gentle osprey]

c squared

[rancid umbra]

nvm

[gritty widget]

S^1 x R and C are not homeomorphic

if you remove the origin from C you disconnect it, but removing a single point from S^1 x R won't disconnect it

[rancid umbra]

oof

why did i think that

[gritty widget]

even basic algebraic topology tests kill it

one's simply connected (contractible, even) and the other isn't

[rancid umbra]

i was trying to show that M/A is homeomorphic to C where A = S^1 x {0}

and we collapse A to a point

and i wanted to argue by using quotient maps

[gritty widget]

the map q descends to a map S^1 x R / A -> C which might be what you're looking for

(by the universal property of quotient maps)

you should be able to write down a continuous inverse C -> S^1 x R / A to that without too much trouble

[rancid umbra]

i guess my thought was that the quotient map $p:M\to M/A$ and the map $q$ above make the same identifications. if i could show that $q$ was indeed a quotient map then i would be done

[gentle osprey]

c squared

[rancid umbra]

where M = S^1 x R

[bitter smelt]

anyone have a visualization of $S^2 \times S^1$

[gentle osprey]

Migillope

[raw bolt]

I'm trying but my brain crashes

It cannot be embedded in R^3, am I right?

I think so

[high hill]

S1 x S1 you can imagine a circle orbiting around a point to make a torus.
The same for S2, the (surface of the) sphere orbits, but this needs to take place in R^4. You can still 'visualize' the orbit in R^3 if you allow things to go through each other

Another option is to think of S2 as 2 copies of D2 'glued' at the D2 boundaries.
Visualizing the orbit separately for each D2 copy, you get the interiors of 2 tori and the 2 are identified at the boundary of the tori

[sharp frost]

This does Not™️ make any sense because $\partial_{q+1}s_q=\operatorname{id}{Z_q}$ so how are they ending up with $\operatorname{id}{C_q}}$ at the end

[gentle osprey]

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

[high hill]

Can I have a hint on how one formally proves a loop that goes around S1 once is not nullhomotopic. Say f : I -> S1, f(x) = x mod 1

[high hill]

Looking around, I seem to need covering spaces to show this? In which case it seems to me when the concept of homotopy classes is initially defined, it isn't clear (apart from some geometric intuition) there necessarily exists more than 1 class in any space?

[hollow harbor]

Certainly, determining simply connectedness is not easy

I mean

There are proofs that don't use covering spaces

Actually I can't think of one

If you know any diff top, you know that you can define the mod 2 degree of smooth maps and see that it's smooth homotopy invariant, and any homotopy can be smoothened without too much error. This is a way around using covering spaces

[pallid lion]

If you know some complex analysis you can also show it by proving that the path integral of 1/z around 0 is not 0. I don't think the homotopy invariance of holomorphic path integrals relies on any covering space things iirc

[unreal stratus]

Another proof is simplicial approximation, say, which one could prove in this special case specifically if rly needed

But ye idk how long that'd be to do from scratch lol

[swift fjord]

you can also use lifting to define the degree and show that it's homotopy invariant

I think that's the lowest-tech method but it's not very enlightening

or intuitive

like you can do it without 'needing' covering space thoery

[hollow harbor]

Oh that's a good one

[hollow harbor]

I figured that this is what shuri meant by covering spaces. But personally I feel like it's the best way to do it. Not like the lifting lemma here is super hard to prove.

[swift fjord]

yea

it's just that you dont' need the general theory

to do this

[hollow harbor]

ye

idk what else they would have meant by "needing covering spaces" since the lifting method is the standard proof

(and, imo, by far the most intuitive - i guess the diff top one is fine, the complex analysis one is not intuitive at all though from my perspective)

[high hill]

Thanks 👌

[coarse night]

how does removing origin disconnects C?

nvm wrong plot

btw how to show smash product of $S^n$ and $S^m$ is $S^{n+m}$?

[gritty widget]

To avoid further confusion of people, it's not $\mathbb{C}$ but $C$.

[gentle osprey]

Blitz

[gritty widget]

https://math.stackexchange.com/a/2848077/476484

[pearl holly]

I'm very sorry if this is a dumb question but I'm very tired and I can't stop thinking about this. Anyway, why is F_nH_n(E) = H_n(E)?

F_nH_n(E) is the image of H_n(p^(-1)(B^n)) --> H_n(E) where p: E --> B is the fibration and B^n is the n-skeleton of B

[pearl holly]

ok me go to sleeps so I won't be able to respond

[gentle osprey]

bchaotic

[cursive spade]

would the argument here be that since f induces a full functor from the cat of paths in X to the cat of paths.

[obtuse meteor]

that's major overkill

[gritty widget]

oh my goodness

[cursive spade]

the image of a connected space is connected under a cont. function

lol

[gritty widget]

going to revoke your category theory language pass for that one

[cursive spade]

I don't think this is true for path connected

[gritty widget]

have you tried to prove it?

[gritty widget]

Why is it a full functor?

[obtuse meteor]

I believe p^(-1)(B^n) should be the n-skeleton of E

[gritty widget]

pick two points in Y. need a path connecting them? you can find one connecting their preimages, so...

[obtuse meteor]

or at least homotopic to it

which then makes this immediate

[ocean narwhal]

"cube root of 2 is irrational because cube root 2 = p/q => 2q^3=p^3 contradicting Fermat's last theorem"

[obtuse meteor]

or well. actually it's a higher skeleton of E depending on the dimension of F

but either way it's like. Definitely has the same H_n

bc it'll have the same n-skeleton

er well

off by one errors here

but this doesn't matter :)

[obtuse meteor]

it's surjective at least bc of this

[gritty widget]

Wouldn't it being full mean that every path in Y is of the form, f o p where p is a path in X

blitz i don't even know what full means, im trying to tell them the normal way to prove this

(and i do not want to learn what it means either)

Oh. Sorry, I thought you were responding to me

[swift fjord]

Too bad you're going to learn

It means surjective on homsets

[gritty widget]

blocked

🙂

[swift fjord]

He actually blocked me

The madman

[plain raven]

lfmao

get got shin

[pallid lion]

I've known what the projective space is for a good 2 years now

nevertheless my intuition for it is still garbage

the "lines through the origin" analogy is just semi-comforting, will this ever go away

[magic geyser]

wait until the proj construction

[plain raven]

Do you know about the duality principle?

[pallid lion]

most likely not

it doesn't sound like sth that ever came up in one of my courses. at least not those 2 words exactly combined like that

[plain raven]

A lot of the basic theorems of geometry are almost dual in the sense that if you swap "point" and "line" you get something else that's true. Like -

1. Any two points determine a unique line (the line which passes through them)
2. Any two lines in the plane determine a unique point (their point of intersection)

The problem here is that (2.) is not exactly true. Really it's, any two lines determine a unique point *unless they're parallel, in which case they never meet.

That's kind of an annoying counterexample

[hidden crag]

can you kinda trick that counterexample by compactification

[pallid lion]

I mean that's kind of what the proj space does

[plain raven]

Well that's where i'm going.
Projective geometry can be seen as an attempt to rigorously reformulate the basic notions of geometry so that (2.) is true. In the projective plane, for every line in the plane there is an additional point added at "infinity" at the end of the line (which can be approached from both directions

[hidden crag]

brainlag

[plain raven]

Then it's just straight up true that any two distinct lines intersect at a unique point. If they're parallel, they intersect at infinity.

[hidden crag]

that's an interesting perspective on it

haven't heard of that before

[pallid lion]

I feel like this still doesn't quite tell me what happens in higher projective dimensions... well I'm sure there is a good analogon though as bezouts theorem basically tries to tell us about it

[plain raven]

Well, learning about duality in low dimensions will at least give you a better foothold if you want to try and understand n-dimensional projective space!

I think another reason for talking about projective geometry is that even if you start with objects which don't really require projective geometry to formalize, it can be useful to use projective geometry to formulate the general idea of a morphism between objects in your category

and thus to give a notion of when two objects are isomorphic

and "the same" from the pov of your theory

Even if you're really not cat-brained, in order to understand a branch of math you need to know when objects are essentially "the same" from the pov your theory

In projective geometry two things are the same if they differ by a projective transformation

i.e. a change in the pov of the 'observer'

[pallid lion]

ok here things are starting to sound buzzy for me

[pallid lion]

this flies over my head

[plain raven]

You can research projective transformations to learn about more. But basically, think of the affine plane as being embedded into projective space by the map that sends (x, y) to [1 : x : y]

Projective geometry, like any other branch of geometry, has some notion of when two spaces are "the same" - two manifolds are the same if they're diffeomorphic, etc.

Two subsets of the plane are "the same" from the pov of projective geometry if they are the same up to a projective transformation in both directions

If A , B are two sets of the affine plane (regarded as subsets of the projective plane by the inclusion) then A and B are projectively equivalent if there is an invertible linear transformation R^3 -> R^3 sending A bijectively onto B

[pallid lion]

ok this makes more sense. this seems very grassmannianish if I'm not wrong

[plain raven]

Projective n-space is Gr(1, R^n+1), the Grassmannian of lines in R^n+1

[pallid lion]

I meant more with the "being equal" if there is an invertible linear transformation from one to the other. do you mean by "equal" here just isomorphic or actually some equivalence relation

nvm "projectively equivalent"

[plain raven]

"A and B are isomorphic" is an equivalence relation on objects of a category

Two groups are generally considered by group theorists to be the same if they are isomorphic.

All interesting group-theoretic properties are preserved under isomorphism of groups

Projective geomety studies only those properties of objects which are preserved under isomorphic projective transformations

[obtuse meteor]

Strong disagree

[pallid lion]

yeah but no I meant more in a topological sense. if this was an equivalence relation on some top space. which looking at it now isn't what was meant

[obtuse meteor]

Or rather I would be more careful than that lol

[plain raven]

There is a difference between Z and an infinite cyclic group in that one is equipped with a choice of generator and the other isn't, I guess

[obtuse meteor]

Some grp theorists are very interested in particular actions as well which are richer ofc

Or like subgrps contain more structure than just as a grp (more appropriate to view as slice category)

[pallid lion]

I don't quite see how that invalidates what clerk said

anyways, thanks for the help, I gotta go to bed

sweet non-canonical nightmares

[unreal stratus]

The dark side is a path to many abilities some consider non-functorial

[dry jolt]

today I learned a little bit about H-spaces!

but I'm seeing conflicting definitions :( Lecture notes say it's a space equipped with "multiplication" map and "inversion" map while sources online just say "multiplication" map

Or rather, lecture notes call H-space a group object in homotopy category of spaces

[plain raven]

In my experience i've seen the weaker definition (monoid object in the homotopy category) more often

Did you see the eckmann hilton argument yet

that one's so cool

[dry jolt]

I've seen it in a group theory text in the past, think we'll cover it in class tomorrow

that's used to show higher homotopy groups are abelian, right?

[plain raven]

Yeah but the same argument shows that all homotopy groups of an H-space are Abelian.

[dry jolt]

primary example right now are loop spaces (which I'm still trying to internalize) so I'll have to think about that

well I guess suspensions too for cogroups

methinks loop spaces are confusing because my topology notes never covered compact open topology lol

[plain raven]

Actually, this observation lets you see that the theorem "all homotopy groups of an H-space are abelian" implies that all higher homotopy groups of an arbitrary space are abelian.

The argument is as follows: Elements of $\pi_{n+2}(X,x_0)$ are the same as homotopy classes of maps in $[ \Sigma\Sigma (S^n, p) ; (X,x_0)]$. By the exponential correspondence (adjunction) between suspension and loop space (which you can check respects the homotopy equivalence relation) this is the same as $[ \Sigma (S^n,p), \Omega(X,x_0)]$. So $\pi_{n+2}(X,x_0)\cong \pi_{n+1}(\Omega (X,x_0))$. So the former is Abelian iff the latter is.

[gentle osprey]

diligentClerk

[plain raven]

I mean maybe it's not obvious a priori that the bijections are group isomorphisms but this turns out to be true when you check it

[dry jolt]

oh that's a neat way to think about it

[plain raven]

The Eckmann-Hilton argument also shows that if $H$ is an H-space with multiplication $\mu: H\times H\to H$ , the induced multiplication of maps in $[S^n, H]$ given by sending $(f, g)\mapsto \mu \circ (f,g)$ agrees with the multiplication in $\pi_n(H)$, giving two equivalent ways to present the multiplication of the homotopy group

[gentle osprey]

diligentClerk

[swift fjord]

It also shows that homsets in the stable homotopy category are abelian

It's so broken

Like in particular it gives you a group structure

[dry jolt]

I hope we touch on stable stuff sometime

[obtuse meteor]

The compact open topology is just the thing that makes hom adjunction true. Knowing anything else is kinda useless imo

[plain raven]

Walter if the compact open topology seems arbitrary it helps to go back to analysis. the compact open topology on maps R->R is just the same as the topology of uniform convergence on compact subsets

that comes up in like, Arzela-Ascoli or shit like this

[dry jolt]

sorry, could you elaborate a bit on the adjunction here? Is this in relation to the smash product?

But yes, calling back to analysis makes it feel more natural

[plain raven]

This theorem should be equivalent to saying that if you take any subset A \subset C(X) which is equicontinuous and pointwise bounded, then the closure of A is compact in C(X) (with the compact open topology)

I could be fucking up the translation

[gritty widget]

I think it's too weak because we talk only about sequences

[plain raven]

I get your intuition. But I think C(X) should be metrizeable under the given hypotheses. Does that change anything?

[gritty widget]

Yeah it should be.

[plain raven]

For a compact Hausdorff space, C(X;R) should be metrizable under the sup norm. And if X is sigma compact then I think that we should be able to put a metric on C(X) by hacking together the metrics of the compact subspaces.

[obtuse meteor]

Hom(X x Y, Z) = Hom(X, Map(Y,Z))

[gritty widget]

Not necessarily. But if it's LCH then it should admit exhaustion by compact sets

And using it we get topology of uniform convergence on compact sets which is metrizable

Anyway I think Arzela-Ascoli is more general but I'll double check

They use compact-open topology here

Y is Hausdorff

[plain raven]

In what Faye said you need to assume the space Y is locally compact and hausdorff. We can anticipate her response scoffing about pathologies in the category of Top and not working in a convenient category of spaces.

[gritty widget]

I think this is topology of uniform convergence on compact sets when Y is LCH but I might be wrong here.

[plain raven]

Nice.

[hollow harbor]

Yes, that's correct.

[gritty widget]

Thanks. I knew it is for compact Hausdorff Y, but wasn't sure here

[hollow harbor]

Yea

[gentle osprey]

poTato

[swift fjord]

Oh btw @dry jolt I did a talk on the stable homotopy category if you wanna take a look

It was part of MaxJ's seminar

Following afa.s

Adams

[kindred drum]

here's the proof that metric spaces are normal, for the set S let $S_{\varepsilon}^{}={x | d(S,x)<\varepsilon }$, obviously, $S_{\varepsilon}^{}$ is open. Take any closed sets A and B, we will show that for some $\varepsilon$ it holds that $A_{\varepsilon}^{} \cap B_{\varepsilon}^{}$ is nonempty, assume that $x \in A_{\varepsilon}^{} \cap B_{\varepsilon}^{}$, then $d(A,B)\leq d(A,x) + d(B,x) \leq 2 \varepsilon$, for epsilon small enough we get contradiction since distance between closed sets are positive. Is this correct?

[gentle osprey]

poTato

[unreal stratus]

Wdym by common distance here?

[ocean narwhal]

hausdorff distance probably

[ocean narwhal]

thats not a metric (edit: nvm)

[kindred drum]

A and B are subsets of metric space with distance function d, these definitions are just "extension"

[ocean narwhal]

ohh okay

[kindred drum]

you can assume that these are some new functions

triangle inequality holds for such extension, it is easy to prove so assume that it is also true

oh, there's mistake

distance between closed sets could be zero even if they are disjoint

but i think i know how to modify this argument

[dry jolt]

I would in fact be interested in looking at this

[pearl holly]

okay so just to make sure, we're using a CW approximation and I can do this without issue?

[fading vale]

im not convinced this is true...

[pearl holly]

what's not true?

[fading vale]

That the preimage of the n skeleton of B should be the n skeleton of E after homotopying

meaning after doing CW approximation and homotopying your map to a cellular one

i cant think of a counterexample but in general CW complexes are not closed under pullbacks (literal pullbacks i mean)

[pearl holly]

yeah I wasn't entirely convinced either but idk

[pearl holly]

tbh this looks like a typo to me or something

[fading vale]

actually

Okay

[pearl holly]

like not the theorem itself

[fading vale]

I think pullbacks are homotopy equivalent to homotopy pullbacks when you do it along a fibration right?

[pearl holly]

oof I don't know

ye I think so

[fading vale]

Im pretty sure this is true

so p^-1(B_n) is the pullback of the inclusion B_n -> B along the fibration p

hence its homotopy equivalent to the homotopy pullback

which does have the homotopy type of a CW complex

so you can reduce the problem to saying that the homotopy fiber of the n skeleton along a cellular map is the n skeleton, up to homotopy

[pearl holly]

ye right I see

[fading vale]

i dont see an easy proof of this : /

[fading vale]

cant you just take the inclusions E_n -> p^-1(B_n) -> E commuting with the inclusion E_n -> E and use that since the latter induces an isomorphism on H_n, p^-1(B_n) -> E induces a surjection on H_n?

its like 6 am so maybe im like

being extremely fucking stupid

or no not E_n wtf E_n+1

[pearl holly]

oh right lol

sorry it's hard to read while walking

[obtuse meteor]

This is better but I think what I said is true precisely bc it’s a fibration yeah

[pearl holly]

thanks to both you btw!

[obtuse meteor]

What moth does to formally justify what I did is a bit overkill though. You can place a CW structure on the total space of a fibration a priori w/o appealing to CW approximation

From this pov what I said that it’s a higher skeleton is apparent

[obtuse meteor]

Ok as long as the base space is connected which is probably a safe assumption here

[fading vale]

u need to CW approximate for the fiber i think though

bc it doesnt say its a CW space in the problem

but also yeah im dumb if you just CW approximate the fiber then the CW structure on the total space is literally just cells times cells of the fiber

or something

wait is it...

[obtuse meteor]

Ah I didn’t see this

Everything is CW ☺️

[fading vale]

Oh yes right

i see

this is kind of a yass actually

the preimage of any cell under the fibration is cell x F

because the cell is contractible

[obtuse meteor]

Agreed

I thought it must be true and looked it up

[fading vale]

society if moth understood AT

[obtuse meteor]

I was just like “true for coverings…no way it’s not true”

[fading vale]

I think its true for coverings because the fiber is discrete

so its automatically a CW complex

[obtuse meteor]

Ah I mean yeah I assumed fiber + base are cw to get total cw

[coarse night]

Let $X, Y$ be Housdroff topological spaces, then we know that $A \subseteq X$ is a compact subspace and $f : A \to Y$ be continuous, then $X\sqcup_f Y$ is housdroff. My question is, is it true when $A$ is \textbf{not} compact, but only closed? Any CE in case answer is no.

[hidden crag]

what's a closed topological space for you

[swift fjord]

My eyes skipped over the first line

[coarse night]

I meant to write housdroff

Let $X, Y$ be Housdroff topological spaces, then we know that $A \subseteq X$ is a compact subspace and $f : A \to Y$ be continuous, then $X\sqcup_f Y$ is housdroff. My question is, is it true when $A$ is \textbf{not} compact, but only closed? Any CE in case answer is no.

[gritty widget]

Hausdorff

[coarse night]

[dry jolt]

Is the adjunction $[\Sigma X, Y] \cong [X, \Omega Y]$ also an isomorphism of groups?

[gentle osprey]

walter

[bitter smelt]

If I have a direct sum decomposition of the manifold into $n$ pieces, how can I find the orientation of one of the pieces given the other $n-1$? I can give more context if necessary

[gentle osprey]

Migillope

[bitter smelt]

I'd prefer the most generic answer though

[bitter smelt]

Maybe more #diff-geo-diff-top

[gritty widget]

Let Y be a point. Then this is X/A. If X isn't regular, there exists closed set A and point x of X which can't be separated by open neighbourhoods. Then the same is true in X/A so it's not Hausdorff

So let X be a Hausdorff but not regular space

Alexandroff plank is probably the most common

If it's not normal then take Y to be a two-point space and f to map the two closed sets A, B for which it falls onto distinct points of Y

And here you can have stuff like deleted Tychonoff plank

[gritty widget]

So it's not true even if we assume they are regular

Btw. If X, Y are separable metric spaces and A is compact, f:A to Y continuous, then the resulting space is separable metric

But if A is just closed then this can fail to be metrizable as R/Z shows

[dry jolt]

Another fun application brought up in class today is that homotopy groups of Lie groups are abelian 🐱

[spice jungle]

hell ya

[cursive spade]

if I treat the determinant as a function GLn to C. how do I show the the fiber at each point in C\0 is connected.

in am trying to show GLn(C) is both connected and path connected

[magic geyser]

do you know any decompositions for GL or SL?

[cursive spade]

no

@magic geyser

[magic geyser]

the accepted answer here looks good https://math.stackexchange.com/questions/139549/how-to-show-path-connectedness-of-gln-mathbbc

[cursive spade]

ok I believe the problem boils down to showing the set of metrix with det 1 is connected

[cursive spade]

thanks @magic geyser

[elder edge]

How do you visualize topological properties like compactness and completeness?

[gritty widget]

For compactness I pretend I have infinitely many open sets in general that cover my set of interest and then ask myself if theres finitely many of them that will cover it. For completeness you can roughly think of it as saying if I remove bits of these two overlapping open sets or arbitrarily close open sets whose sequence of open subsets get arbitrarily close together as the process continues then they should converge to an element that is still in the space you considered a collection of open sets on. In that sense, whenever you narrow in on something shared by open sets or arbitrarily close open sets it’s not like it’s missing from the original space. All the holes we could try to think of by zooming in on shared parts of overlapping pieces or arbitrarily close bits of arbitrarily close pieces are patched

[elder edge]

Thanks for making it clear

[gritty widget]

Honestly dato i never forced myself to think about completeness until you asked

So thanks for giving me some enlightenment

[coarse night]

I didn't quite get what you are saying. What is not true?

[swift fjord]

Of topological groups in general! No need for smoothness

[gritty widget]

It's not true that it's Hausdorff if we only assume that A is closed

[coarse night]

ok but X,Y normal and A closed does mean that X U_f Y is normal right

[gritty widget]

It's just getting harder and harder

I don't believe it's normal, no

[coarse night]

it's an exercise given to us

I don't believe either but 🗿

[gritty widget]

That's sadistic

[gritty widget]

https://en.wikipedia.org/wiki/Eckmann–Hilton_argument
doesn't even have to be a group either

[gritty widget]

it actually is normal

The proof is 3.4 in Dugundji

My bad, it's not something I remembered

[coarse night]

can you write the full name of the book

[gritty widget]

Chapter 7

Name is "Topology"

easy to find online

[gritty widget]

is something like this true for alexander spanier cohomology

we have a space X

we cover it with closed sets

do this in a sufficently nice way that it makes sense about a set corresponding to a point x

fix some element in the cohomology of X

define a subspace of X as follows

x in X such that f is not in the kernel of the map H(X) to H(Vx)

the map H(X) to H(Vx) is the map induced by the inclusion of the closed set which corresponds to x

[plain raven]

Very cool.

[unreal stratus]

is there any nice or particularly major motivation for configuration spaces? I know about the related physics notion but curious about the motivation to study config spaces in their own right

[shadow rampart]

is a set open iff every element has a nbd contained in the set ?

[unreal stratus]

yes, now try proving it

[spice jungle]

Well, config spaces parametrizes points in manifolds right, for example if you wanna study braid group action

Then you might want to think about n points onR^2

Also they give E_n operads

And there is like this Ran space in geometric Langlands, which is like a complex geometric version of points on Riemann surfaces

[cedar pebble]

the KZ equation and its monodromy is a nice motivation/application of configuration spaces

there's a lot of fun things you can do with the KZ equation depending on how you set it up

[spice jungle]

What is the KZ equation???

Heard it so many time now I’m too afraid to ask

[cedar pebble]

yeah so the definition is kinda complicated and depends on a choice of affine Lie algebra

there's a universal version that is related to this that is somehow a little easier to understand I think

the basic example is the universal KZ equation on Conf_4(CP^1)/PGL_2(C)=CP^1-{0,1,\infty}

it's a Fuchsian differential equation with values in (a completion of) the free Lie algebra on two generators X_0 and X_1, so elements are like formal power series in brackets like [X_0,X_1] and so on

[spice jungle]

Sure you are saying something about the 4 point function right

[cedar pebble]

right

[spice jungle]

Make sense

But KZ is not conformal ward?

[cedar pebble]

I don't think so, since this isn't quite using an affine Lie algebra so you're not getting e.g. an embedding of the Virasoro Lie algebra into your chosen affine Lie algebra by Siegel-Sugawara

but nevertheless the following is illustrative: take the Fuchsian differential equation $\mathrm{d}F=\Big(\frac{\mathrm{d}z}{z}X_0+\frac{\mathrm{d}z}{1-z}X_1\Big)F$

[gentle osprey]

nGroupoid

[cedar pebble]

so it's taking values in the completed universal enveloping algebra of this free Lie algebra, so non-commutative formal power series in two variables X_0 and X_1

[spice jungle]

Interesting

[cedar pebble]

the parallel transport of this equation along paths in P^1-{0,1,\infty} is basically expressed in terms of the monodromy of multiple polylogarithms

[spice jungle]

Do I know you from another server LOL

[cedar pebble]

in particular, if you take the parallel transport of this equation along a straight path from 0 to 1 and suitably regularize it you get multizeta values

yeah maybe lol

[spice jungle]

Ah icic

I’m Leon lol nice to meet ya

[cedar pebble]

yeah we've probably talked before hi

but yeah these 4 point functions from this universal KZ satisfy braid relations

so like Yang Baxter or whatever

you can define the same universal KZ equation for n points in general (and you kinda need to play around with the 5 point functions to prove a pentagon identity for this associator) but as is usually the case in CFT things reduce to the 4 point situation

the usual KZ equation that physicists care about is similar to this, just with affine Lie algebras instead of these free Lie algebras, but it's a similar game

[rich bison]

How much Algebraic Topology does one need to know for Homotopy Theory? Or can I directly jump into homotopy theory?
Also what would be a good introduction to it, videos are more welcome

[lunar yoke]

I dont know of any introductory homotopy theory lectures on youtube, but this is a nice playlist as a first course in algebraic topology https://www.youtube.com/playlist?list=PL2Rb_pWJf9JqgIR6RR3VFF2FwKCyaUUZn

[rich bison]

Hmm... We assume basic knowledge about the fundamental group, categories and CW-complexes throughout the course, these things weren't taught to me in topology (except fundamental group). Should I read about them somewhere or should I check other series which doesn't assume them like - https://www.youtube.com/playlist?list=PLpRLWqLFLVTCL15U6N3o35g4uhMSBVA2b (by Pierre Albin) or go through some topology playlist covering them?

[shadow charm]

Are there situations when computing homotopy/homology groups where the group you get is a topological group in a way that makes sense (ie the topology on the group has something to do with the topology of the original space) and where you can compute the homotopy/homology groups of that to get information on the starting space?

[gaunt linden]

Hmm. I think there are cases where you can get a nontrivial topology on a homotopy group in a natural way -- e.g. the fundamental group of (R\{0})×R union {0}×Q. But the topology would be pretty ugly because "homotopy group" implies collapsing each path-connected component to a single point before you define the group operation. So its homotopy group(oid)s would be trivial.

[unreal stratus]

Oh interesting hadn't heard of this thanks

[mortal iris]

@rich bison I'd say read up on them, they do only say basic knowledge. There's a good intro to categories for AT in Rotman and for CW complexes I'd suggest Chapter 5 of Lee's Topological Manifolds.

[rich bison]

Is that his - An Introduction to the Theory of Groups or Advanced Modern Algebra?

[mortal iris]

It's his Introduction to Algebraic Topology

@rich bison Chapter 0 should have about all you'll need, if it's not enough, there's also Topology: A Categorical Perspective by Bradley, Bryson, & Terilla. But I don't have too much experience with that book

[gritty widget]

Say A is a retract of X and r:X to A is a retraction.
Say H:X x I to X is a homotopy between H_0 = Id_X and H_1 = const.
Consider G = r o H o i where i:A x I to X x I is the inclusion, i(a, t) = (a, t).
Then G_0(a) = r o H(a, 0) = r(a) = a so that G_0 = Id_A and G_1(a) = r o H(a, 1) = r(const.) = const.

@uneven hawk

This proves that G is a contraction from Id_A to a constant function, so A is contractible

[stark creek]

Okay.

[spice jungle]

Okay

[opaque cloud]

how do we define dimension in a topological space?

like are we able to define it in such a way that it naturally generalizes the notions of dimension that we're most familiar with?

for example let's say you think of a finite dim normed vector space as a topological space and ignore the basis

if you looked just at the topology and found out what dimension it would be, would it be the same dimension as looking at the basis set and seeing what dimension it would be?

[tawdry valve]

I don't know anything about it, but I've heard that https://en.wikipedia.org/wiki/Lebesgue_covering_dimension is the thing that you're looking for

so this does meet your criteria---euclidean n dimensional space has lebesgue covering dimension n (same will follow for any finite dim normed space, since all norms are equivalent for finite dimensions). it looks like kinda a weird definition, I don't have any intuition for it yet

[opaque cloud]

thanks

[tawdry valve]

yup

[opaque cloud]

it does seem kinda weird I'm gonna wait until I learn more top

[tawdry valve]

yeah that seems fine lol, I've never seen this in a course, I've only ran into it online a few times

it seems a lil tricky to prove something has lebesgue covering dimension greater than n for some n

[gritty widget]

in algebraic geometry i learned the definition of dimension for topological spaces that is dual to the krull dimension of a ring

the dimension of a topological space is the supremum of the lengths of chains of closed irreducible subsets

so you might call it the krull dimension of a topological space

different requirements necessitate different definitions

worth noting that this dimension is a little goofy if you're not doing algebraic geometry

there's also fractal dimension and hausdorff dimension, but i haven't the faintest fucking clue what those do

[pallid lion]

I thought it was sort of meant to be a thing to see in what dimesion does it "make sense" to give your object a measure is really vaguely what the hausdorff measure does. If you try to measure the "line volume" (dimension 1) of a bended area in 3d space it's obviously infinity, but if you measure it with spheres (dimension 3) it is zero. so you'd have to choose the right method of measure "little pieces of area" (dimension 2) for your paper to have finite volume that is not 0 (sometimes there is none, so you just take the infimum of all dimensions where it is 0). The Hausdorff dimension actually allows for fractal dimensions and it turns out that due to some self similarity arguments (3b1b has a really good video on this), some objects do deserve a fractal dimension (like the sierpinski triangle).

[ornate berry]

Fractal/Hausdorff dimension is specific to metric spaces rather than topological spaces. It requires a notion of distance (specifically, diameter of a subspace).

[gritty widget]

Yeah that's not really... topological

There's small/large inductive dimensions too

https://encyclopediaofmath.org/wiki/Dimension

This all goes under a field of topology called dimension theory

I don't know much about it, but all the standard ways of measuring dimensions should agree for separable metric spaces.

Together with the theory of continua, dimension theory is the oldest branch of general topology. The first concepts and facts predate Hausdorff’s definition in 1914 of general Hausdorff topological spaces and, so, involved only subsets of Euclidean spaces.

https://link.springer.com/chapter/10.1007/978-3-642-61265-7_2

Also see Engelking's "Dimension theory"

[gritty widget]

There's examples of metric spaces where the small end large inductive dimensions don't agree

[stable wing]

do you have a reference for this? i couldn't quite follow from the discussion

[cedar pebble]

this is given in a lot of places, for instance like section 1.5 of this book http://javier.fresan.perso.math.cnrs.fr/mzv.pdf

basically the point is that if the universal KZ equation looks like

$\mathrm{d}F=\Big(\frac{\mathrm{d}z}{z}X_0+\frac{\mathrm{d}z}{1-z}X_1\Big)F$

[gentle osprey]

nGroupoid

[cedar pebble]

then this defines a connection on the trivial vector bundle (with infinite dimensional fibers C<<X_0,X_1>>) with connection 1-form as above:

$\omega=\frac{\mathrm{d}z}{z}X_0+\frac{\mathrm{d}z}{1-z}X_1$

[gentle osprey]

nGroupoid

[cedar pebble]

the parallel transport of this connection is expressed in terms of iterated integrals of this 1-form along a given path

so a typical term will look like uhhhh

$\int_{0\leq t_1\leq\hdots\leq t_n\leq 1}\omega_1(t_1)\hdots\omega_n(t_n)\mathrm{d}t_1\hdots\mathrm{d}t_n$

[gentle osprey]

nGroupoid

[cedar pebble]

where the \omega_i's are either dz/z or dz/(1-z)

but now you have integral expressions like

$\mathrm{Li}n(z)=\sum^\infty{k=1}\frac{z^k}{k^n}=\int^1_0\int^{t_1}0\hdots\int^{t{n-1}}0\frac{\mathrm{d}t_n}{1-t_n}\frac{\mathrm{d}t{n-1}}{t_{n-1}}\hdots\frac{\mathrm{d}t_1}{t_1}$

[gentle osprey]

nGroupoid

[stable wing]

@cedar pebble amazing! thank you so much for explaining. that is really cool! (my background is in physics, so I am used to thinking about KZ but only in the context of Kac-Moody stuff)

[cedar pebble]

haha I am in the opposite camp, I learned this first and at some point tried learning the Kac-Moody stuff

[cedar pebble]

something you'll find kind of amusing... When you study these iterated integrals in this situation, you often want to specialize to the case z=1 so you're integrating along a straight path from 0 to 1. But this requires some regularization, since these differential forms have poles at z=0 and z=1 respectively.

there's a canonical way to regularize this: only integrate from \epsilon to 1-\epsilon, then expand as a power series in log(\epsilon), and take the constant term.

The book refers to this as "the physicists regularization scheme"

[stable wing]

love it

[gaunt linden]

The constant term would be log(epsilon)=0, thus the integral backwards from 1 to 1-1?

[stable wing]

no the meaning is something like: one has an asymptotic series in \epsilon -> 0 of the form a0 log(\epsilon) + a1 + o(\epsilon). the first term is diverging but in the end we're just interested in a1. with some work you can show this is well-defined, if we were to choose a different regulator say

[gaunt linden]

Ah, okay. A "power series in log(epsilon)" sounded like a0 + a1 log(epsilon) + a2 log²(epsilon) + ... to me.

[gritty widget]

Question if my proof is correct: I have to show that given $d_1(f,g) = \int |f-g| dt, d_2(f,g) = sup |f-g|$ on C([0,1]) we have $\tau_1 \subsetneq \tau_2$:

Take $U \in \tau_1, f \in U$. There is $\varepsilon >0 : f \in A_{1, \varepsilon} \subset U$, where $A$ is epsilon nbhood wrt to metric 1. But since $$\int |f-g| \leq sup |f-g|$$ we have that $A_{2, \varepsilon} \subset A_{1, \varepsilon}$, so $U$ is open in $\tau_2$.

[gentle osprey]

Catematician

[gritty widget]

I know that this doesnt show the nonequality of t1 and t2, havent tried that yet.

[abstract saffron]

Wait, what's the question again?

[gritty widget]

@abstract saffron can you troll somewhere else

You can read it again xD

[gritty widget]

well the phrasing of it is arguable but the inequality is the main ingredient here and I think you are thinking of the right thing
just the phrasing is bad

[abstract saffron]

I just didn't see the question, if that's what you meant

Sketchy argument I'd say, but it seems good

[gritty widget]

mainly because A_(2, eps) doesn't contain the information about the center nor it was mentioned anywhere

and the center seems to be f

Oh yeah of course

Didn't really want to write out the definition of it in here.

For me you don't even have to mention open sets, just open balls and that'd be clear too

as for showing they are different, probably the easiest would be to take some sequence which is convergent in one metric but not the other

For the t_2 not a subset of t_1 some argument using unbounded functions that are integrating to 1 should work right>

[hidden crag]

Not sure what you mean by unbounded here

They’re continuous functions on a compact set

[gritty widget]

Okay so yeah that would work to show what you mentioned, but how exactly does that show their topologies are different? I mean, it makes sense because convergence depends on topology, but whats the rigorous arguemnt for that?

[gritty widget]

unbounded set of functions of course (thats still a bad phrasing hah)

Spike getting bigger in say x=1/2 yet integrating to 1 or something like that.

if they are equal then convergent sequences in both are the same, obviously from definitions

I'm not sure what you're asking

convergence is defined using topology

a sequence f_n is convergent to f iff for every neighbourhood U of f, there is N such that f_n is in U for all n >= N

[abstract saffron]

Yes, it should. You can make up a sequence of functions whose integrals are always 1, but nonetheless converging to zero function.

So that a sequence converges in a topology but not another.

[gritty widget]

no that won't work. If it converges pointwise then it converges uniformly

You want to take a sequence of functions which converges to 0 in L^1 but not in the sup-norm

An example should be f(x) = x^n

[abstract saffron]

I'm not sure if I get you right, but convergence in sup-norm (aka L-infinity) is far stronger than in L^1.

Wait, I'm messing myself 😄

[abstract saffron]

It goes the other way, if it's converging uniformly, then it converges pointwise.

[gritty widget]

It does. But here it's equivalent

[abstract saffron]

Take the function defined as f_n(x) = n on [0, 1/n], and 0 everywhere else. The sequence converges pointwise to zero function, but the integeral remains 1.

[gritty widget]

not continuous but yea

[abstract saffron]

With some extra assumption, it will be equivalent, but unless I miss something, I didn't see it mentioned

[gritty widget]

It doesn't converge pointwise to the zero function

In fact a sequence of continuous functions on [0, 1] can't converge pointwise to the zero function

and it's not an example

[abstract saffron]

then you can cook up with affine function, say a line from (0, n) to (1/n, 0). A bit messier to write down, but it goes similarly

[gritty widget]

Yeah but cant converge to 0 pointwise because of f(0)

those won't produce examples of functions convergent in one norm but not the other

Yeah your example x^n works well.

[abstract saffron]

Oh, you need on [0, 1]? I thought (0, 1), my bad

[gritty widget]

(answering now deleted question) that looked like something homeomorphic to R

[odd flame]

i can see how left is a subset of the right but why is the opposite not true?

[gritty widget]

because it's more like cartesian product is multiplication and union is addition

you should have 4 terms on the left, not 2

[odd flame]

i can see that in my head but do u have any counterexampless

[gritty widget]

try writing some yourself

[odd flame]

that's what im trying lol

i dont think nullset messes stuff up

[ornate berry]

The empty set does in fact mess things up

Also, this isn't a topology question. This really should be in #discrete-math or #proofs-and-logic.

[odd flame]

fair, it's ch1 of munkres that's all

[odd flame]

i see this now, oops lol

[gritty widget]

every book usually has some prerequisites but they are still not the main topic of the book

[odd flame]

[swift fjord]

Yea, which is a set theory review

[gritty widget]

Hi, another problem I'm not sure if I got correct. I'm asked to find cts bijection from (0,1] to S^1 and show there is none from S^1 to (0,1]. So for the first part f(x) = (cos2pix, sin2pix) should do the job, since functions on both coordinates are continuous, bijectivity is clear I think. For the second part, if it existed then those two would be homeo, but cannot be since (0,1] isn't compact.

Does that seem reasonable?

[lunar yoke]

It already suffices to say that if we had a cts bijection S^1 -> (0,1] then the latter would be compact, yes

[gritty widget]

Oh true that.

[lunar yoke]

Do you have a homeomorphism if there exist cts bijectiosn X -> Y and Y -> X ?

they dont have to be inverses per se

maybe i just have brainlag

[gritty widget]

For some reason I remember it implying existence of homeo, maybe I'm just misremembering though, cause I can't come up with a proper argument.

[lunar yoke]

maybe one can do some kind of continuous cantor-schröder-bernstein argument

it wouldnt surprise me if there is a counterexample though

arbitrary spaces are wild

[gritty widget]

Hah yeah I just googled something of this sort and yeah uses weird set theory stuff I didnt bother reading.

[lunar yoke]

oh wait i dont think you actually construct an inverse there, you also just get a bijection

[gritty widget]

I'll just stick to compact sets map to compact sets under cts functions lol

[golden gust]

is a continuous bijection from a set to itself necessarily a homeo? feels untrue but for some reason I can't come up with a counterexample

[gritty widget]

with different topologies?

[golden gust]

with the same topology

otherwise yeah of course not

[gritty widget]

No iirc

There should be some example with a subset of R

I think you take a copy of N, a countable amount of (0, 1) and [0, 1) together

[rich bison]

Any topologists here. I have the munkres book and I am thinking of following this UG topology course at University at Buffalo. Are these any good -
https://www.youtube.com/playlist?list=PLoWHl5YajIf6MNTh7Ok024T1JkhZEcYXm

[gritty widget]

like, {-n} for naturals n > 0, [0, 1) and (n, n+1) for n > 1

[rich bison]

They also have a nice website at https://www.mth527.site/ (with notes and homework)

[gritty widget]

you should try mapping -n to -n+1 here, [0, 1) to a part of (1, 2) and so on

the other part of (1, 2) to (0, 1)

I guess you don't need (n, n+1) for n > 1 but just (1, 2)

but yeah, this should be continuous bijection but not homeomorphism

@golden gust visualization of what the map does

oh yeah now I see why we need the intervals (n, n+1) for n > 1

that left part of the interval (1, 2) won't fill itself

so we need something similar to what we're doing with points {-n} on the left side here

like this

[gritty widget]

Back again with another problem: needed to show A in R^n is compact iff every cts function f: A -> R is bounded. => is clear because compact sets map to compact and in R its closed and bounded. Other way: Assume A is not compact, therefore unbounded or not closed. If unbounded then simply f(x)=|x| does the job for contradiction. If it's not closed, then there exists a sequence in A x_n converging to y not in A. Then we can define f(x)=1/|x -y| which is unbounded and cts. Does that work?

If A is not compact then it contains a closed subspace homeomorphic to N

just take f(n) = n and use Tietze extension theorem

Thank you for your idea but I was wondering if my proof is correct.

it's how you prove it in general

yes, your proof works

this property is called pseudocompactness btw

you essentialy shown that pseudocompactness is equivalent to compactness for subsets of R^n

as you already have seen, compactness always implies pseudocompactness

the converse is generally not true

You mean the property of any cts function being bounded?

yes

Never seen that term before

[gritty widget]

Any example of such space off the top of your head?

I'd have to think about it more/research

No worries, thought maybe there is some easy classic example I couldve seen before.

[gritty widget]

[0, omega_1] should work

wdym, like, extended reals?

no wait. It doesn't work

that's a compact space

not closed?

yes. I was going to say this next

there's this book about pseudocompact groups but it's a reference book

https://link.springer.com/book/10.1007/978-3-319-91680-4

should be this one

doesn't feel appropriate for reading but still, good for browsing results I suppose

[white oak]

Hello!

Given continuous maps f, g: A → B, a homotopy is _h: A × 1 → B, h (_, 0) = f, h (_, 1) = g _. In some categories I can transport a function of two arguments h: A × 1 → B into a function into the space of functions curry h: 1 → (A → B). Then a homotopy I should like to be a set of functions indexed by 1 such that an open set in 1 is sent to an open interval in A → B. How should I define a suitable topology on A → B? What do I need to prove to make this work?

Makes sense, A → B being an exponential object.

https://en.wikipedia.org/wiki/Compactly_generated_space ← These k-spaces?

Cool, this is very curious. Is there a book I should check to find out more?

Thanks!

[plain raven]

Charles Rezk has an expository paper on compactly generated spaces here. https://faculty.math.illinois.edu/~rezk/cg-spaces-better.pdf

[lament needle]

What does it mean by on $S^2$

[gentle osprey]

matthew

[lament needle]

does it mean attach 0 cells 1 cells and 2 cells or does it mean construct s^2 via these cells

i feel like the assumption is to construct S^2 since that is the euler characterstic of the sphere

[pallid lion]

it wants you to construct S^2

[lament needle]

Bet. ok i got it. thanks

[bitter smelt]

great question that one

Ive seen it a few times, always a pleasure

[spiral silo]

I'm struggling with Hatcher 1.1 problem 5, (a) -> (b). I think I'm making it harder than it has to be. The extension from S1 -> X to D2 -> X is clearly going to take some homotopy in D2 from S1 to a point (say, transport each point of S1 in D2 linearly along a straight line to a base point at the edge of D2), and correspond it to the homotopy in X from a loop to a point. But I'm having trouble proving that this is continuous.

[bitter smelt]

well post the problem lol

[spiral silo]

Could

I don't really need help, I'm just sharing my problems

[bitter smelt]

uh... ok I guess

cool? I mean I don't understand what any of that is saying without the question but... go you!

get em

[spiral silo]

Okay here

thanks lol

I think it's gonna be something like

Let S1(s) : [0, 1] -> S1 in D2 be defined as:
S1(s) = (cos (2 pi s), sin (2 pi s))

Pick s0 on the boundary of D2.
Let d(t, s) = t * s0 + (1-t) * S1(s)
so that d(t, s) linearly interpolates from S1 to the point s0.

Let f(t, s) : [0, 1] X [0, 1] -> X be the given homotopy in X from the loop to the point, where f(0, s) parameterizes the loop, and f(1, s) is the constant map with image a point.

Let m : D2 -> X be defined
m(p) = let (d(t, s) = p) in f(t, s)

The part I'm stuck on is actually proving m is continuous

I guess if you guys want to help, that would be great

[spiral silo]

Sorry about all that

[rancid umbra]

try using the fact that (S^1 x [0,1])/(S^1 x {1}) is homeomorphic to D^2 @spiral silo

[spiral silo]

Thanks! I'll look at it tomorrow, got to go to sleep.

[rancid umbra]

as an outline for a proof, $p:S^1\times[0,1]\to D^n$ given by $((x,y),t)\mapsto (1-t)(x,y)$ is a quotient map which makes the same identifications as the quotient map $q:(S^1\times[0,1])\to(S^1\times[0,1])/(S^1\times{1})$ and $p:S^1\times[0,1)\to D^n\setminus{(0,0)}$ is a homeomorphism

[gentle osprey]

c squared

[rancid umbra]

then you know that the homotopy factors through $(S^1\times[0,1])/(S^1\times{1})$ which should let you finish the proof

[gentle osprey]

c squared

[rancid umbra]

p restricted to S^1 x [0,1) into D^n\{(0,0)}

[unreal stratus]

(homotopy?)

[gritty widget]

Suppose that the verticle maps are cofibrations

and in cohomology the top map is an injection, and the right verticle map is an isomorphism

can i say anything about if H*(Y) to H*(B) is an isomorphism

[echo dove]

I think no -- should be many counterexamples. Eg. take $X = Y = D^n$ with $B = S^{n-1}$ and $A$ a point.

[gentle osprey]

daveamayombo

[gritty widget]

How do I show there is no cts bijection from (0,1) to S^1? I was thinking that if such function existed, we would have for x_n -> 0 and y_n -> 1 lim f(x_n) = lim f(y_n) = a, but there won't be a point x such that f(x)=a. That's pretty much just intuition, not sure if that approach is correct though.

Couldn't come up with any topological argument, would gladly hear one, although maybe this fits more in analysis channel.

[hidden crag]

S^1 is compact while (0,1) is not

[gritty widget]

Can't noncompact map to compact?

[hidden crag]

so there can't be a continous function from (0,1) with all of S^1 as image

wait

uuh

i didn't read properly

the pre image of a closed set under a continous function is closed

and since you want a bijection the pre image of S^1 would have to be all of (0,1)

[gritty widget]

But like, aren't we talking about the subspace topology for (0,1), in which (0,1) is closed?

So I don't think this argument is valid.

[hidden crag]

yeah i might be trolling

let me think about it a bit more, gimme a sec

[gaunt salmon]

Can anyone think of an example of a continuous map that is not surjective, but hits all the open sets in the target (except the empty set) that is not also an epimorphism?

in the category TOP

nvm

[hidden crag]

@gritty widget would a connectedness argument work here maybe

[gritty widget]

Because if f:(0, 1) to S^1 were a continuous bijection, then so would be f:(0, 1/2) sum (1/2, 1) to S^1\{f(1/2)}

[unreal stratus]

Don't you still have a bit of a gap there?

e.g. we can have a continuous bijection [0,1] cup [2,3] to [0,2]

[ornate berry]

We argue via path connectivity from that point

[hidden crag]

i was thinking something along the lines of removing a point from (0,1) and then restricting our alledged cont. bijection to (0,1) without that point

[gritty widget]

You can argue via "continuous bijection is monotone"

[hidden crag]

but uuh

[gritty widget]

wlog we can assume f is a monotone bijection

now while you approach 1/2, the missing point will provide a contradiction

I did this in this channel before

[hidden crag]

wait what i typed was just doubling what blitz said

mb

[gritty widget]

Contradiction with connectedness right?

contradiction with bijectivity

You take A = sup f on (0, 1/2) for example. A is in (0, 1) but there's no x such that f(x) = A

Oh I see

But also, wouldnt connectedness argument also work? (0,1/2) cup (1/2,1) is disconnected, circle without a point is connected.

Disconnected space can map to a connected space

Not under continuous though?

that doesn't matter

Oh my bad

Wrong direction

So yeah it's more like, open intervals must map to open intervals (under continuous bijections)

And that would contradict that (0, 1) is connected

[candid hedge]

How to prove that any polynome is continuous

[ornate berry]

Certainly the identity function x is continuous, and so are all constant functions. We also know that the product and sum of continuous functions are also continuous. Hence polynomials are continuous, as they are obtained from repeated products and sums of the identity function and constants.

[candid hedge]

what does this proof mean then

[ornate berry]

It's saying that the product and sum of continuous functions are also continuous.

[candid hedge]

okay

[gritty widget]

If you're talking about polynomials of multiple variables then you use projections on each of the coordinate. In the 1-dimensional case it's "the function x"

But other than that the reasoning is the same

[candid hedge]

i still dont get what projection has to do with this. Projecting what on where

[ornate berry]

The projection maps are the functions x, y, z etc.

They are so called because they are like shining a light along a plane onto a line, sending (x, y) to x.