point-set-topology

[surreal lantern]

well

your idea of continuity is probably epsilon delta i suppose

but that uses the concept of distance since R is a normed (hence metric) space

but generally we don't need distance to define continuity on topological spaces

we just need neighbourhoods of points

which is necessary since there exist spaces that aren't metrizable (there's no metric that induces our topology)

and then when you get back to metric or normed stuff you'll see that they are special cases of the general definition of continuity on topological spaces by using the topology induced be the metric/norm

[odd flame]

it's weird man, im doing ind. study w a prof and he starts w topology but idk what a metric space is lol

and the only topology he's taught me has been a whirlwind of a phone call that was 10 minutes long lol

he did say that a map between topological spaces is cont if the preimage of an open set is an open set

but having asked what i ddid before about open sets, that seems a lot simpler now ig

[odd flame]

wdym metrizable

[heady grove]

How do i show something is a metric

[gritty widget]

show that it satisfies the properties of a metric

[heady grove]

teacher never gave us the definition

ima look it up

[odd flame]

can someone give me an example of a discontinuous map between topological spaces

im trying to understand the def. that a cont. map is one such that the preimage of an open set is an open set

[lunar yoke]

take X = Y = {0,1} with topologies T_X = {{},{0,1}} and T_Y = {{},{0},{0,1}}. Then f : X -> Y, x -> x is not continuous because f^{-1}({0}) = {0} is not in T_X, even though it is in T_Y, i.e. we found an open set {0} in Y whose preimage under f is not open anymore

[odd flame]

this map isn't onto though

oh ig nobody said it has to be

[lunar yoke]

it is onto though

lmao

its just the identity mapping

[odd flame]

oh

it's been a long day

but why does {0} even have a preimage

or is that the point you're making

[lunar yoke]

wdym

every set always has a preimage

[odd flame]

f^{-1}({0}) = {0}

[lunar yoke]

if f : X -> Y is any function, then for any subset S of Y i can consider f^{-1}(S) = {x in X | f(x) in S}

[odd flame]

ok yeah

[lunar yoke]

and in the above case f is just the identity

[odd flame]

i mean {0} in particular, no element in T_X should map to it right...?

[lunar yoke]

0 -> 0

its the identity mapping

f(x) = x

its just different topologies on domain and codomain

same underlying set

[odd flame]

OH the map is between the X and Y not the topologies themselves

[lunar yoke]

yes

[odd flame]

im being dense

[lunar yoke]

being continuous is a property of a function between topological spaces

the function itself us just a set-function between the underlying sets of the topological spaces

and you use the topologies of the spaces to check that the function has the property of being continuous

[odd flame]

ajsdblksahdbfadsbh ive been overthinking this

[lunar yoke]

happens

i mean the continuity you know from functions between the reals is just a special case of this

would be a good exercise to check that the definitions are equivalent

[odd flame]

you're assuming i paid attention in analysis

jk ik what you mean lol

or at least i will after a quick look through my notebook

[lunar yoke]

do you know the topological definition of being continuous in a single point?

[odd flame]

i do not

[odd flame]

.

[lunar yoke]

do you know what a neighborhood of a point is

[lunar yoke]

oof

well in contrast to topological spaces i'd say that working with metric spaces is pretty straightforward if you're comfortable with R

just whenever you say |x-y| < eps replace that with d(x,y) < eps

d is just a distance function

[odd flame]

what i mean is im kinda groping around atm lol

just tryna get a better idea of shit while i meet w prof again

is a homeomorphism a map between the actual topologies or the topological spaces again

[lunar yoke]

again the spaces

you dont really look at maps between the topologies

[odd flame]

,w topological space

[lunar yoke]

the topologies are just extra data so that you know which sets are open

[gentle osprey]

Results provided by WolframAlpha

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[odd flame]

so should i think of the "topological space" as just the set on which the topology is being built (like R for example) or as the set and the set of subsets together

[lunar yoke]

well you need to consider them together of course because otherwise you would just be looking at an ordinary set

[odd flame]

butt

[lunar yoke]

im not really sure why you're having such a hard time with this

[odd flame]

no i get it im just being unnecessarily exhaustive probably

[lunar yoke]

if you consider the reals with their natural ordering, you dont suddenly ask whether order-preserving maps are defined on the reals or in the relation-set of the order

[surreal lantern]

[fair idol]

Given a map between two topological spaces, there is an induces homomorphism between the fundamental groups. Is this induced homomorphism an example of a categorical pushout

[plain raven]

no, fajitas

i guess you're already discussing this in #category-theory but yeah no, it's a functor

[icy schooner]

why is it phi_i+ compose phi_i-

phi_i^- -1 maps B^n back to U_i^-

which isn’t even the domain for phi_i^+

how do you compose them

nvm it’s in the errata

[limber ravine]

f is continuous if any open U in X is open in A (since we're dealing with inclusions, it is the identity map)

Let X be the discrete topology on {a,b}

A be the induced topology of {b} on X

So, A = {{},{b}}

We claim in : {{},{b}} -> {{},{a},{b},{a,b}} is continuous. But {a} is open in X but not in A

Where am I wrong?

[vocal wharf]

this implies A having the subspace topology

[limber ravine]

but

Ok I will review, there is something not making sense

[vocal wharf]

i mean you are not wrong

but the question assumes A having subspace topology

[limber ravine]

That's what's not clicking

what is it having a subspace topology, does it mean, open sets in X are open in the subspace? Closed sets in X are closed in the subspace? I mean, how can it be that {a} is open in A (example above) if it isn't an element of A

[vocal wharf]

ah, the open sets in A are open sets in X intersect A

so you start with a topology on X and endow A with the so called subspace topology

(there is also a definition in terms of the inclusion map, but that makes this exercise trivial)

[vocal wharf]

also i assumed in your example above you had A=X, if {a} isnt even in A, its preimage is the empty set, which is open

[limber ravine]

if {a} isnt even in A, its preimage is the empty set

oh oh oh

wut

[vocal wharf]

i mean

say X = {a,b,c} and A = {c} both with discrete topology

then the preimage of {a} under inclusion is {}

[limber ravine]

why is that

[vocal wharf]

by the definition of preimage

[limber ravine]

let me check

[vocal wharf]

the preimage includes all elements of A that get mapped to a

but c gets mapped to c, so c cant be in it

and there arent any other elements

[limber ravine]

oh I see lol f^{-1}({a}) = {x in A : f(x) = {a}} which is indeed the empty set

[vocal wharf]

your counterexample still works for A=X, so the word "subspace" implies "endowed with subspace topology"

[limber ravine]

now I am not seeing how the counter example works

[vocal wharf]

i mean just have A = X, X discrete top and A indiscrete/trivial top

then the inclusion will not be continuous

[limber ravine]

ok, A = X = {1,2}

{1} open in X but not in A

no wut

the preimage of {1} in A is open indeed

How is the inclusion not continuous?

[vocal wharf]

hm?

{1} is open in X but not in A

and the inclusion is just the identity

[limber ravine]

but the preimage of {1} is the emptyset, and indeed the emptyset is open in A. Why doesn't it make f continuous?

[vocal wharf]

in this case the preimage of {1} is {1}

[limber ravine]

the condition for continuity is open in X its preimage must be open in A ?

[limber ravine]

since {1} notin indiscrete topology of {1,2} that makes inclusion not continuous

[limber ravine]

so the map in : A -> X (A endowed with subspace topology) generally corresponds the mapping $A \cap U \mapsto U$ where $U$ is open in X

[gentle osprey]

mns

[vocal wharf]

the map is just the identity restricted to A

and the open sets in A are of the form $A \cap U$ with $U \subseteq X$ open

[gentle osprey]

Lochverstärker

[limber ravine]

exaclty, I see now, it is continuous because the inverse of the mapping will be open since for any open U in X the restriction to A is U cap A which is open in A

[vocal wharf]

yeah

[limber ravine]

I think I will leave the counterexample to rest a little bit and comeback to it later today

[vocal wharf]

the thing with that counterexample is

when you talk about subspaces in topology you want them to have subspace topology generally and not just any topology

because subspace means more than just subset

you want at the very least the inclusion to be a "good" (in this case continuous) map

otherwise it isnt really a subspace, because subspaces have to play nice with the maps we care about (in this case continuous maps)

[limber ravine]

oh I see

Hum, I will think about it

thanks!

exactly

we can't find any coarsest topology in A such that the mapping in : A -> X is continuous with respect to the topology of X

Because letting A be the trivial topology in : A -> X won't be continuous

because in the example above, {1} is open in X but is not open in the indscrete topology of A = {{},{1,2}} thus making in discontinuous

[vocal wharf]

what is the issue? the subspace topology is the coarest topology of those that make the inclusion continuous

i.e. the subspace topology has enough open sets to make the inclusion continuous, but not more

[limber ravine]

Thanks!!

[fair idol]

Is every continuous injective map a local homeomorphism?

[empty grove]

Identity map from discrete topology on a set to the indiscrete topology on the same set

[fair idol]

Rats

[fair idol]

Is there a criteria for an open map?

[gritty widget]

whats the free homotopy class

there's a statement in my book which states that every free homotopy class admits X (X irrelevant here)

homotopy class but you don't fix a basepoint

[gritty widget]

There is no nice statement like you are looking for

what structure do these form?

[marsh forge]

They are in general a set, but that is about it

[gritty widget]

you need a fixed base point to create the group, are the free ones some kind of a groupoid?

[marsh forge]

Well, you need more than just a basepoint for a group structure on homotopy classes of maps in general

[gritty widget]

how so?

[marsh forge]

[X,Y]_* is not a group in general unless we choose X,Y specifically

[gritty widget]

I mean we're looking at maps I to X

because I care about paths

[gritty widget]

It probably would help if you gave the

(Irrelevant here info)

[marsh forge]

The fundamental groupoid of a space is the correct thing to think about here. One takes as objects the points of X and as morphisms that paths from a point x to a point y.

One might quotient the morphisms by based homotopy

Or one could try to remember this structure (obtaining some sort of higher fundamental groupoid)

[gritty widget]

"every free homotopy class is represented by a length minimising curve" is the full statement

[marsh forge]

That seems reasonable enough

[gritty widget]

and for each real r there are only a finite number of homotopy classes represented by curves of length less than r

[marsh forge]

I would not know how to prove such a statement but it does not seem nonsensical

[gritty widget]

So you are looking at riemannian manifolds

Ascoli + the fact that homotopy classes are open subsets of C^0(S^1, X)

compact path metric spaces

apparently both statements also true when the base point is taken to be x, but we can't expect the curves to be smooth at x anymore

alright, thanks for the help, I'm gonna go study

[limber ravine]

a submap of a continuous map is continuous
X,Y topologies with f : X -> Y continuous
A subset X and B subset Y
we must show ab(f) : A -> B is continuous
Take any open set U in B. We must show ab(f)^{-1}(U) is open in A

[fading vale]

What is ab

[limber ravine]

ab(f) is abbreviation of f to A, B

[fading vale]

So its the restriction of f to A and you're assuming that the image lies in B?

[limber ravine]

oh yeah, that's an assumption in the definition, for A subset X and B subset Y, then for every f : X -> Y such that f(A) subset B there's a mapping ab(f) : A -> B defined by the formula x |-> f(x) called abbreviation of f to A,B

[marsh forge]

What have you tried

[limber ravine]

yeah, so, since U in B is open, we have U is open in Y. Since f is continuous we have f^{-1}(U) open in X

now I have many questions

what exactly is A? 1) subset 2) subspace 3) a random topology on a subset A

x open in X means x is open in A?

[marsh forge]

A and B should both be subsets of X,Y with their respective subspace topologies

for this statement to make any sense

[limber ravine]

ok, so its done

[marsh forge]

Not quite

[limber ravine]

A cap f^{-1}(U) is open in A

[marsh forge]

U open in B does not imply U open in Y

[limber ravine]

I see

[tough imp]

Hi max

:)

[tight agate]

max

[marsh forge]

hello!

I am sorry I missed ur talk chm

I rejoined the server like 15 min after it started lol

[fading vale]

Max

[tough imp]

It’s okay

[fading vale]

Hi

[limber ravine]

should I take an open set in f(A) ?

no

it seems so wrong

[marsh forge]

You need to take an open set in B and prove that it has open preimage

[swift fjord]

Max

[marsh forge]

The idea behind your proof is reasonable, it just needs more thinking

(in particular, you have not at all used the definition of the subspace topology)

[limber ravine]

U = B cap V for some V open in Y

f^{-1}(U) = f^{-1}(B cap V) = f^{-1}(B) cap f^{-1}(V). Note f^{-1}(V) is open in X

[marsh forge]

Much better

Do you see why this last set is open in A?

[limber ravine]

I am trying with the f^{-1}(B)

[marsh forge]

huh?

[limber ravine]

oh, A subset f^{-1}(B)

because f(A) subset B ?

no

[marsh forge]

this is true

[limber ravine]

oh yes

but why does it help

but then, f^{-1}(U) is open in A if f^{-1}(B) = A

and we already have A subset f^{-1}(B)

If x in f^{-1}(B) then there's some x in X with f(x) in B

note f(A) subset B, so the image of every element of A is an element of B

but it can be the case that the x we take in f^{-1}(B) is not an element of A (right?)

in fact, it can be the case that f(x) in B and x notin A

[marsh forge]

This is a valid concern

However, there is a solution

[limber ravine]

[marsh forge]

(remember that we want to be intersecting with A, eventually)

[limber ravine]

eventually

[marsh forge]

So maybe f^{-1}(B) cap f^{-1}(V) is not good enough, but can we find a set with the same A-intersection that is more obviously open in A?

[odd flame]

what is meant by point set in "point set topology"

im assuming it has to do with a literal set of points but what does it mean

[gritty widget]

Point set topology is a way of doing topology by describing your space as a set of points and having another family of subsets which describes how the points are connected

[marsh forge]

It also tends to mean you are studyung something that emphasizes like, the nitty gritty study of spaces that can be gross or have weird properties

[atomic parrot]

If I'm not mistaken, point set topology is a synonim for general topology, as opposed to, say, algebraic topology or topology of manifolds.

[marsh forge]

Like point set topology is used to describe the study of topology that does not really vibe with how future topological subjects go

Point set is more specific than general, I would say

At least as a Concept With Attitude(tm)

[limber ravine]

hum

[odd flame]

[odd flame]

im gonna assume the latter half of this is the def of topology that ive been taught so far

set of subsets that contains nullset, entire set, closed under inf union and finite intersection

[tough imp]

Yes they’re just saying you’re giving neighborhoods of a point

[odd flame]

now im just wondering how the def of a topology is connected to describing how points in a topological space are connected

as in the 3-rule def i regurgitated justnow

[marsh forge]

Do you know what a connected open set is

[odd flame]

i do not

,w connected open set

[marsh forge]

Ah, well you can define these

and then the picture will start to emerge

arguably, the actual connectedness properties should be thought of in terms of how certain "test spaces" continuously map into your space

For example, maps from the unit interval [0,1] with its standard topology into your space tell you which points can be connected by paths

[odd flame]

studying this makes me feel so cool but at the same time adfhasbdhbfsdhbfrea

[limber ravine]

@marsh forge Not getting it

we're going to need something like A cap W where W is open in X

I will sleep under this exercise, tomorrow I let you know tho. Thank you!

[odd flame]

is there any reason we say homeomorphic and not bijective

am i missing some difference in their meaning or is it just to promote the idea of a "physical" map

[odd flame]

what did you mean by these

i want the picture

like connected open set?

i dont wanna just jot down def's and not understand why

[hollow harbor]

the picture is that a connected open set is an open set that is connected haha

[marsh forge]

lol

[hollow harbor]

but the definition is that you can't split it up into two separated pieces

[odd flame]

[marsh forge]

ryc is thinking like a real topologist

[hollow harbor]

(i.e. pieces whose closures intersect)

hello max! it's good to see you back

[marsh forge]

hiya

[hollow harbor]

these people are slowly making me more algtop/alggeo brained

[odd flame]

moldi has left his cavern in #groups-rings-fields to watch me suffer here how fun

[marsh forge]

moldi is gonna learn all about the adams spectral sequence with toki

right moldi

[empty grove]

Yessir

I suppose I am

[hollow harbor]

bijective could mean you cut things apart or glue things together (like the function which sends [0, 1] union (2, 3] to [0, 2] by shifting the second interval to the left is a bijection)

[odd flame]

this channel has scary words

[hollow harbor]

homeomorphisms cannot glue things together or tear them apart

can't tear apart = continuous

[odd flame]

ohhhh that makes sense

[hollow harbor]

can't glue together = inverse is continuous

(gluing things together is just tearing them apart in reverse)

[marsh forge]

Keep in mind that the definition of a bijective function makes no reference to the topology

so there is absolutely no way a bijection can say anything meaningful about the topology

[odd flame]

chicken or the egg

[hollow harbor]

well, i guess my perspective is that's it's very clear to me why tearing things apart should be bad

but gluing things together looks continuous to me, so what's the problem with that?

and in a lot of fields of math, structure preserving bijections are automatically also structure preserving in reverse

(for example, if you have a linear transformation, and if it's invertible, then the inverse is automatically also linear)

[marsh forge]

Gluing things together is fine, but it is also destructive

[hollow harbor]

but in topology this doesn't happen: a continuous bijection can also glue stuff together. so you need to be really careful.

[marsh forge]

like, it is continuous, just not invertible

what is the point set condition for a continuous bijection to be a homeomorphism

[hollow harbor]

i don't actually know if this property of "inverse of a bijective homomorphism is a homomorphism" is given a name / is discussed somehow

[marsh forge]

i feel like there is one

[hollow harbor]

well, one is that the domain is compact and the codomain is hausdorff

maybe you can weaken the former to locally compact?

[marsh forge]

thats what i had in mind yeah

[odd flame]

is this why we say a continuous map is one where the preimage of open sets are open sets?

[marsh forge]

maybe

[odd flame]

or am i mixing things

[hollow harbor]

or sigma compact surely

idk

[marsh forge]

I would say that the strict definition of continuous is the way it is "because it works"

[marsh forge]

It does capture the intuition of "sends nearby points to nearby points" whenever one can make sense of the latter

[hollow harbor]

not quite. this seems like it's a weird definition until you start thinking about what it means to tear things apart - when you tear things apart, you create new open sets against the boundary of tearing that then would preimage to non-open stuff

for example, when i tear [0, 2] into [0, 1] and (2, 3], i get a new open set (1/2, 1] in this space

but that has a preimage of (1/2, 1], which is not open in [0, 2].

so the "preimage of open sets is open" condition really does capture the idea of "continuous maps can't tear stuff apart"

i guess another way to look at it is from the perspective of the definition of continuity on metric spaces (the epsilon delta one)

[odd flame]

oh wow i need a moment for that to sink in

[marsh forge]

(Proving things like image of compact is compact, image of connected is connected, etc also helps with this intuition a lot)

continuous functions really are deep down functions that look continuous though lol

[odd flame]

need a topology version of this bc i forgot how hard step 1 is lol

i have no resource to give me problems tho so if yall wanna give me homework that could be cool 👉 👈

[marsh forge]

1. yes, metric spaces make sense
2. what the fuck is a hawaiin earring
3. okay metric spaces make sense

[hollow harbor]

lmao

[odd flame]

No one can be told what the Manifold is. You have to see it for yourself.

[marsh forge]

manifolds make sense as long as one does not corrupt them

[odd flame]

i think i get those

at a high level at least

my prof explained them as crumpled up pieces of paper glued together and that made sense

[hollow harbor]

accurate

[marsh forge]

I would go with like, any material other than paper

but sure

[hollow harbor]

smooth structure

[odd flame]

im assuming he just meant it as R^2 for its simplicity

[hollow harbor]

i guess the problem is flatness

[marsh forge]

sure i just feel like cloth might be better

silk

manifolds are very silky in my mind

[hollow harbor]

hmm

i think manifolds are made out of some kind of plastic in my mind

flexible acrylic sheets

[marsh forge]

im thinking more rubbery

latex

[hollow harbor]

yeah latex was coming to mind initially

[odd flame]

LaTeX

[hollow harbor]

i'd go for the acrylic for a riemannian manifold

symplectic manifolds are made out of bent wood, but i'm not even sure how people do that

[gritty widget]

Often you steam the wood

[hollow harbor]

i see

that fits in with the vibe of symplectic manifolds

[gritty widget]

what's a poisson manifold under this?

what do you get when you stick together a bunch of pieces of steam bent wood

You add a fish stock to your steamer

Just for the person who was asking about pointset

I personally don’t think of all point set topological spaces as being “spaces”

Point set topology

Is a very general way to talk about spaces

But

You only really use pointset topology to touch base and connect different fields

Like in algebraic topology we work with simplicial sets or CW complexes

And we prove results about these

But you can translate this back into pointset topological theorems

And there people in other fields can make use of what you prove

So don’t spend too much time trying to worry about what is this intuitively in general point set settings

Just get loads of examples

And get an intuition about nice spaces

[marsh forge]

algebraic topologists dont even work with spaces anymore

[hollow harbor]

nah poisson manifolds are just a bunch of fish in a pile

i wish this were ironic

something in my brain is broken

[gritty widget]

this is fine

[hollow harbor]

each fish is a symplectic manifold

they're all together in like

a school of fish

[gritty widget]

some fish may be bigger than others but they're all doing working together

to do.... whatever fish do

[hollow harbor]

yeah

idk it's like something in finding nemo

and then a contact manifold is also bent wood but there's a car driving around on it

it's got a little pointy tangent vector arrow sticking out the front. it's red and looks like the car from the game Rush Hour

[odd flame]

simple question

what is a simplex

i understand low dim examples ofc, should i worry about a "rigorous definition"

[lunar yoke]

what kind of simplex are we takling about

[marsh forge]

A simplex is just the convex hull of the unit vectors

i.e., a dot, a line, a triangle, a tetrahedron, and so on

and so on = "i have no geometric intuition for these after this point"

[plain raven]

yo tell me if this makes sense to you

[marsh forge]

few things do

[plain raven]

I think CW complexes are like, made of this kind of dark void material which you can spread with a trowel kind of like it's umm something a plasterer or bricklayer would use. It's firm and it retains its shape in place, you can shovel it into gaps in the CW complex and patch holes when you want to kill off elements of the higher homotopy groups. Whenever I do CW complex stuff I imagine i'm reaching my trowel into the void of CW source material, lifting it out and scooping it into my complex to patch a hole

[marsh forge]

interesting

i think actually

topological spaces are to me

made of up this white material that like

has no earth analogue

[lunar yoke]

platonic realm gang

[marsh forge]

it is like, the platonic material out of which one makes spaces

yeah lol

[plain raven]

yeah it's the same for me

i'm describing its texture and feel but it's not earthly of course

[marsh forge]

i think it would feel bendy

but also soft

[plain raven]

do you know what the cone of a space is

do you understand like, quotient topologies, product topologies, this stuff

[odd flame]

no

still very early on in this but i can look up definitions

[plain raven]

ok well both of these things are important in the long term learning about quotient topologies or identification topologies as they're sometimes called but

do you know like

linear algebra in an abstract vector space

[hollow harbor]

i agree with max

definitely

it's a really nice sort of doughy stuff

a little firmer than dough though, it sort of returns to its desired shape when i push at it too unless i hold it in place for a few seconds to get it to set

hmm

[lunar yoke]

now i feel weird that i have never thought about what topological spaces or their "material" would feel like

apparently many people did?

i just stuffed it in my platonic realm box

like most mathematical objects

[hollow harbor]

i definitely think this is something people would call me a weirdo for

i'm surprised to find any camaraderie in it

[marsh forge]

you forget that i am also very weird @hollow harbor

[hollow harbor]

lmao

[marsh forge]

i have the weird philosophical position

of being very much not a platonist

[hollow harbor]

i see

[marsh forge]

but none of the objects i study have a real existence

[hollow harbor]

yes, that makes sense

i think my past platonism is being torn from me by the fluids people

[hollow harbor]

i'm not sure if the takeaway is that the platonic realm isn't real, or if it's that it's the only thing that's real

[fading vale]

@marsh forge

[finite forge]

Is every zero set a closed set and if yes, how does one show it?

[empty grove]

Zero set of functions of which kind?

If ℝ^n to ℝ^m under the standard topologies yes

The preimage of a closed set is closed

[finite forge]

My bad for not clarifying, I mean the null set which has measure zero

[empty grove]

Like any set of measure 0?

Or by null set do you mean the empty set

Measure theory goes in #advanced-analysis

But if you just mean empty set

Then yes, the empty set is always closed in any topology

[finite forge]

I mean this one, I'm sorry I'll ask it there then

[pearl holly]

No fucking way

Is that Max?

Yoooo Max is back

[empty grove]

You say that as if you haven't talked to him in months

[empty grove]

Anyway I have been told to learn this with you

How do I start

[pearl holly]

Yo fr let’s go

But ye how do we start

[empty grove]

Idk what these are so you gotta guide me

[pearl holly]

Lmao I will try to look up some nice sources later

Or just ask Max where to read about these stuff

[empty grove]

I wonder what the easier way is 😌

[pearl holly]

oh I forgot to ping lmao

[pearl holly]

@marsh forge 👉 👈

[azure heath]

topology alg top

[limber ravine]

@marsh forge

[limber ravine]

shit

I think I did it

Recall f : X -> Y continuous

A subset X and B subset Y induced subspaces

ab(f) : A -> B such that f(A) subset B

We want to prove ab(f) is continuous

Take U open in B. So U = B cap V for some open V in Y

Note B = B \ f(A) U f(A)

So U = [B \ f(A) U f(A)] cap V

Take the preimage of both

f^{-1}(U) = f^{-1}([B \ f(A) U f(A)] cap f^{-1}(V) (notice the skipped step)

Now we can make f^{-1}([B \ f(A) U f(A)] = f^{-1}(B) \ A U A

And we have [f^{-1}(B) \ A U A] cap f^{-1}(V)

take the distributive

[ f^{-1}(B)\A cap f^{-1}(V) ] U [ A \cap f^{-1}(V) ]

Now notice that f^{-1}(B) \ A cap f^{-1}(V) must be empty. Here's why:

When we took V open in Y. Either V subset f(A) or not. Indeed, if V subset f(A), then clearly we have the left side equal the empty set. If V wasn't a subset of f(A), then f^{-1}(B) \ A cap f^{-1}(V) is not empty in X but it is in A since it won't belong to such set A

this last statement is so badly written sorry.

Something wrong?

[marsh forge]

blue book!

you two should just read the blue book

[limber ravine]

what's blue book o:

[marsh forge]

Adams "Stable Homotopy and Generalizes Homology"

[pearl holly]

oh ye lol true

@empty grove you hear this?

[empty grove]

Yessir 🙉

[limber ravine]

oh that's not for me. What do you guys think of my answer?

[pearl holly]

honestly I think my cat theory is too weak for the blue book

I will have to work my cat theory up if I want to read it properly

[marsh forge]

I think you don't need a ton of cat theory

unless im misremembering

like nothing past the first quarter of riehl or whatever

You should 100% read riehl if you havent

at least up until some point

[pearl holly]

there were some like, representable functors and stuff and I don't know those yet

[marsh forge]

Right

[pearl holly]

ye I will have to read riehl lmao

[marsh forge]

Representable functors are just functors that are of the form Hom(-,X)

[limber ravine]

[marsh forge]

i.e if I have a functor F: C-->D and F is naturally ismorphic to Hom(-,X) where the Hom lives in D then we say X represents F

woke

[pearl holly]

hmm okay I see

[marsh forge]

It is difficult to read this without proper latex

But I think there is still an error

Or maybe not?

Sorry the last statement is confusing and my coffee is still kicking in

[limber ravine]

+1

[marsh forge]

this should be false, so the proof of this statement must be false

but i am too tired to figure out why yours is wrong lol

[limber ravine]

fair enough xd

[marsh forge]

"If V wasn't a subset of f(A), then f^{-1}(B) \ A cap f^{-1}(V) is not empty in X but it is in A"

I think the part after "then" is on the right track, but that it does not prove what you want.

Like at the end of the day, you can kind of ignore what's going on outside of A, but you need to formalize this.

[limber ravine]

yeah that's what I wanted to say

[lunar yoke]

the typesetting in that book

im spoiled by latex

[marsh forge]

Here's a hint. Say I want to prove that the preimage of f in A is of the form U cap A for an open set U in X. Then the set you're currently working with is kind of gross, but if I can throw out some part of it so that it looks like A cap U then we are done

Like my recollection is that you currently have something like Z cap U where U is obviously open but Z is not obviously open and also Z can contain points outside of A

But what if we could just ignore the bad parts of Z

[marsh forge]

Adams typesetting is bad but he makes up for it by being a truly excellent writer

[limber ravine]

could I write that?

[marsh forge]

Unlike rudin

[marsh forge]

if you were a grad student you could write that and people would assume you knew what you were doing, but when you are first learning you have a greater burden

but you can rewrite what i said with simple set stuff

[limber ravine]

hum ok interesting

thanks!

finally I got the right mind set on this proof

it was indeed a good exercise

[marsh forge]

my suggestion would be to like

draw some "worst case scenario" style pictures

like start maybe drawing a nice picture where this is all easy to see

and then ask how it could go wrong

and see if you can figure out how to fix it

[limber ravine]

Like

[marsh forge]

honestly 99% of point set problems can be solved using complicated ven diagrams

[limber ravine]

.rotate

[marsh forge]

Okay I think the right hint to give you is as follows

first, let's review

We take V to be open in B. By definition this means $V=U\cap B$ where $U$ is open in $Y$. Then $f^{-1}(V)=f^{-1}(U)\cap f^{-1}(B)$. We know $f^{-1}(U)$ is open in $X$. Let's call it $Z$ to simplify notation. So now we are looking at $Z\cap f^{-1}(B)$. But this is the preimage for $f$, not for $ab(f)$ (this notation is nonstandard but let's use it for consistency). Do you know how to relate the preimage of $f$ and $ab(f)$?

[gentle osprey]

MaxJ

[marsh forge]

@limber ravine

[limber ravine]

I know that f|A = f o in

with in : A -> X

and that this map is continuous

but I think that would be the next exercise

because

[marsh forge]

Well

I meant a simpler relationship

the preimage f^-1(C) is all those points in X that map to C

but the domain isn't X for ab(f)

it's A

So ab(f)^{-1}(C)=f^{-1}(C) \cap (fill in the blank)

[limber ravine]

A?

ab(f)^{-1}(C)=f^{-1}(C) \cap A

So, taking where you were:
$f^{-1}(V) = Z \cap (ab(f)^{-1}(B) \cap A)$ ?

[gentle osprey]

mns

[limber ravine]

@marsh forge

[marsh forge]

replace ab with just f but yes

Can you simplify $f^{-1}(B)\cap A$

[gentle osprey]

MaxJ

[limber ravine]

That is the set of all the points in A which the image under f is in B

[marsh forge]

Which is....

[limber ravine]

Ab(f)^{-1}(B)?

[marsh forge]

Okay so we agree that f(A) is a subset of B, yes?

[limber ravine]

Yes

[marsh forge]

so A is a subset of f^{-1}(B)

yes?

[limber ravine]

Yes

[marsh forge]

So if M is a subset of N

what is M cap N

[limber ravine]

Oh lol

M

[marsh forge]

perfect

[limber ravine]

A

[marsh forge]

Can you simplify $A\cap f^{-1}(B)\cap Z$?

[gentle osprey]

MaxJ

[limber ravine]

A cap Z

[marsh forge]

And Z is....

[limber ravine]

https://tenor.com/view/party-lets-party-dj-cat-dance-gif-9305331

Open in X!

[marsh forge]

Perfect

[limber ravine]

Awesome that was much simpler

Thank you very much

[marsh forge]

no problem!

[gritty widget]

Is there any way to do a sphere eversion in real life? Like could we use bubbles to allow the possibility of self-intersecting surfaces? Or could we make a wire mesh sphere with gaps in it to allow parts of the sphere to pass through?

[marsh forge]

Did someone ping me

[gritty widget]

Did you just delete this so you could ask it again?

[limber ravine]

I did

but I noticed something wrong

[marsh forge]

Ah

No worries

[gritty widget]

Opps sorry I

[marsh forge]

Just making sure I’m not senile

[gritty widget]

I thought person deleted their question to ask it again

[limber ravine]

next time I will say if I pinged someone, sorry!

[marsh forge]

Hahaha no worries at all

[gritty widget]

In any reasonable way no

Could you contrive something

Probably

[limber ravine]

HOLY SHIT

I just noticed how to much the prove so much much easier

[gritty widget]

Which way do you think would be more feasible, bubbles/soap films or wire meshes?

None of these

I mean

Contrived contrived

Doing it with bubbles

Will be even more unreasonable

[gentle osprey]

mns

[gritty widget]

You need your animation to go through minimal surfaces

When I say contrived I mean

Extremely contrived like

Get enough drones of a small enough size

And stand far enough away so that’s they look like pixels

And make them perform a 3D animation

And really I only say maybe you could contrive something because

I’m not going to argue why this is clearly impossible

Sorry I don’t mean to say clearly in the sense this is obvious

I mean clearly in the sense

I’m not sure how to finish my assignment so I write clearly

[fair idol]

If I wanna demonstrate a map p:R+ ->S^1 doesn't have the path lifting property is it sufficient to show p is not a covering map?

[empty grove]

No

[limpid kernel]

https://tenor.com/view/cute-anime-girl-blushing-happy-gif-18663876

[novel acorn]

It is nice to see the topology channel still has it's anime roots

[cursive spade]

What is geometric topology

[gritty widget]

As far as I know its a more detailed look at manifolds

[mint sage]

E^5 is Euclidean space

[gritty widget]

look at df(x,y)

[mint sage]

What does it do

I thought this was a purely topological problem

[gritty widget]

oh

uh

oh

in general topology wikipedia says find a homeomorphism from E^2 onto f(E^2)

not sure where to start besides trying to understand what image of E^2 is in simpler terms

is topology on E^5 standard?

[mint sage]

I think so

[marsh forge]

So cursed

[gritty widget]

What is this from?

Im genuinely interested in trying to solve this problem

[mint sage]

Basic Topology written by Mark Anthony Armstrong

[gritty widget]

wow

i have that book

[mint sage]

in Chp 4

[gritty widget]

,rotate

[mint sage]

That's 11

[gritty widget]

I guess the part im trying to wrap my head around are open sets in the image

I dont really see how the image is the same as [0,2pi]^2

i guess the idea is to change of variables the image

if we transform all the cos into sqrt(1-sin^2)

nah i cant really think of what to do

oh wait

everything with y can be strictly positive

so maybe you can only transform those

nvm those 2y’s

[gritty widget]

oh i see something

i think the idea is to use angle identities

yeah

i think you can write each coordinate as either sin (u+v) or cos(u+v)

and then maybe something obv shows up

because f(x,y) = cos x, cos(y+y),sin(y+y),sin(x+y)-cos xcosy,cos(x+y)-

yeah idk lol

oh

oh bruh

got to. use identification mapsnfor this

[gritty widget]

yeah

i think i figured it out but i wont do details

you use fact that klein bottle is given by opposite identifications

and then you prove that your map f is injective wrt identification

so f:[0,2pi] x [0,pi]->E^5, g:[0,2pi] x [0,pi] -> K=Klien Bottle. And then you check that f is injective on the boundary

so interpret image of f if it were on one of the identified edges

we know that the interiors are homeomorphic because they both look like squres

[mint sage]

excellent, thanks

[sweet walrus]

Hey, had a doubt in fundamental groups. Suppose A and B are closed sets in X, such that X is A union B, and A intersection B is non empty and path connected. Is the fundamental group of X generated by fundamental groups of A and B?

In case of open subsets, this is true by Van Kampen Theorem but the same proof cant be adapted to closed subsets

[pearl holly]

I think that for closed set in van kampen, you want an additional criteria that A intersection B is a neighborhood deformation retract in A and B, i.e there exists open sets V in A and U in B containing A intersection B such that V and B deformation retracts onto A intersection B. I think you can then use van kampen on the pair (A union U, B union V) and show that the "amalgamation product" you get is equal to the one you get when you apply van kampen to (A, B)

or something like that I'm not entirely sure

[sweet walrus]

Thanks, yeah i was thinking along these lines to generate a counterexample for the case where we just replace open by closed, but couldnt come up with any

[marsh forge]

I would add that for most decent spaces, and reasonable closed sets we can without issue slight grow or shrink them to open sets

[cursive spade]

say r is a continuous retraction from D to S, is it the case that the fundamental group of S is a subgroup of that of D

[marsh forge]

Yeah

this is not so hard to prove

[cursive spade]

ok I see, thank you
also is it true if you have a continuous map from D^2 to D^2 , it is then homotopic to a continuous map that maps its boundary S^1 onto itself

[marsh forge]

This is true in the most trivial way possible

Every map D^2->D^2 is homotopic to the identity

[cursive spade]

how do I prove that

that's what came to mind but I was doubting the continuity of the in-between maps

[marsh forge]

You can also prove that every map is nullhomotopic

which is easier maybe

easier to believe at least

[cursive spade]

nullhomotopic?

[marsh forge]

homotopic to a constant map

[cursive spade]

yeah I just did that ,lol

thanks @wooden falcon @marsh forge
I was asking these questions to get a better sense of the Brouwers fixed point theorem.
feel free it mention anything that might be useful to think about

[bleak path]

People use htpy as a short form of homotopy, is there a short form for homotopy equivalent or is it just htpy equivalent

[lunar yoke]

i tend to use eqv for equivalent/equivalence, but not sure if thats common

[bleak path]

I'm sure it's understood at the very least, was just wondering if there was a standard out there

[marsh forge]

$\simeq$

[gentle osprey]

MaxJ

[marsh forge]

I wouldn't use htpy in formal writing and in informal writing you can use whatever you want

[bleak path]

I use simeq as well, but I dislike that you have to read into the context to distinguish when it's being used for homotopies and when it's being used for homotopy equivalences. Maybe that's just my inexperience showing though.

[marsh forge]

it should be fairly clear from the objects on either side

like if you write simeq between functions its clearly a homotopy and if you write it between spaces it means htpy eq

the bigger issue is distinguishing between weak and strong homotopy eqs

but again this is normally clear from context or you can restrict to a situation where they are the same

[bleak path]

I've not learnt much about those, do people use the simeq notation for both strong and weak homotopies?

[marsh forge]

Yes

The meaning rarely causes any actual confusion, in my experience

And people are normally careful when the difference could matter

[gritty widget]

I use $\cong$ for homeomorphism, which is similar

[gentle osprey]

Blitz

[marsh forge]

\cong is good for any isomorphism

I would avoid it for homotopies

[bleak path]

one of the (IMO) weak points in math is how people use the same symbol /definitions in related contexts :/

thanks for the short discussion though!

[marsh forge]

It might feel awkward at first but I promise it’s better than having to learn a million symbols

[bleak path]

Hahah I wish there were like some fancy H for homotopy, and then you could just add an equal sign somewhere to make it homotopy equivalent

[marsh forge]

keep in mind that like

a lot of these notations

[gritty widget]

Like $=_\mathcal{H}$?

[marsh forge]

are really only for situations where you meaning is totally obvious

[gentle osprey]

Blitz

[marsh forge]

in real mathematical writing

you should just write out what you mean in words

[bleak path]

maybe! Or possibly even the reverse

[marsh forge]

i think about symbols like pronouns. You want to state your meaning explicitly first and then you can use symbols after the meaning is established

For example

if I was writing a paper i might write

[bleak path]

$\mathcal{H}$ and $\mathcal{H}_{=}$

[gentle osprey]

wOne

[marsh forge]

let $X\simeq Y$ be homotopy equivalent spaces, then ...

[gentle osprey]

MaxJ

[marsh forge]

then later I can use \simeq freely

and people know what i mean

[marsh forge]

this type of stuff gets very unwieldy very fast

you want to write for people, not a compiler

[bleak path]

I do that as well, but when I find my literature gets too long, I'm always worried people will forget and then the meaning becomes lost :/

[marsh forge]

i have rarely thought that an explanation was too long

and certainly not from the addition of extra clarification

[bleak path]

It's more of when it detracts a bit far from the original point, you know? When you explain a bit more and more, etc.

But i"m sure it's something that resolves itself as I do this more and more

[marsh forge]

idk I think you get used to the like

sweet spot

for mathematical writing

symbolic notation should play a minima role

only for things that really are best explained in symbols

[bleak path]

It's mostly for assignments now, when the lecturer says to target it so that your peers will understand, but when I actually discuss the homework with them, a lot of gaps are present here and there and so sometimes it goes a bit off tangent

[gentle osprey]

Blitz

[bleak path]

Is meager set something topoligical? Haven't heard of tha tbefore

[marsh forge]

(I would avoid defining ones own nonstandard notation if you want your TAs to like you)

[gritty widget]

Yes, it comes up when studying Baire spaces

[marsh forge]

theres a big difference between holes in notation and holes in the argument

[bleak path]

Definitely don't, which is why I started by asking if there was any conventional short form for homotopy equivalent

[marsh forge]

I would just write homotopy equivalent

I also wouldn't really write htpy in homework

but thats a personal thing

[bleak path]

Interesting, don't think those are in my syllabus though, so I might read up more

[marsh forge]

they aren't central objects in AT

if ur in an AT class

[bleak path]

I am

(in an AT class)

right after GT, which is from what I gather the usual order

Either GT and AT "together", or AT after GT, never really before

[marsh forge]

But as a parting remark I would say that your readers and graders will rarely be upset that you write out what you mean explicitly in english, even if it increases the length a little. I'd rather read 2 well written pages than half a page of condensed gibberish 🙂

[bleak path]

While I agree with you, I find that when I reread my own work there's much that can be cut down or shortened in future passes

But maybe that's just me cringing at my own work, something something imposter syndrome

[marsh forge]

That’s just learning better writing, you’ll improve in time

My writing now is a lot better than when I first took AT hahaha

[bleak path]

Mine's continually improving

[gritty widget]

Well, it was just an example that you can put * in your equality/equivalence symbol somewhere anyway, Baire spaces are just a side note

[bleak path]

Not that there was much to "write" in earlier classes, topology has had the most explanations and least computations by a long shot in the classes I've taken so far

[marsh forge]

It’s probably a lot of peoples first experience with proofs that involve a lot of arguments that can’t be expressed notationally

[bleak path]

Definitely, and it's often easier to write a sentence or two in latex rather than come up with a diagram, though it's often the case that a diagram helps more than a sentence can

[long coyote]

this may be a dump question, i couldn't figure out Hom(Hn(X;Z),Z)=?

where Hn(X;Z) is free

[marsh forge]

Do you understand homomorphisms from Z^n to Z?

[long coyote]

i got it, Hom(Hn(X;Z),Z)= prod of generators

[coral pawn]

Can someone explain the construction of a genus 2 surface using cell complexes?

We take a point, attach 4 1-cells to it using the constant maps

Then we attach a 2-cell to this

But I don't know what the attaching map is

[marsh forge]

aba^-1b^-1

i.e. glue around one circle, then along the other, then back around in the opposite direction, and then back around in the opposite again

[storm nova]

Does knot theory fall into this room?

My question is "Can rings (of the same size) link in any combination in 3d space?" I put it into stackexchange but am getting no traction... https://math.stackexchange.com/questions/4338935/can-identical-rings-intersect-in-any-combinations-in-3d-space

[rancid umbra]

an open, injective map should preserve hausdorff-ness right?

[empty grove]

The points outside the image could do whatever they want

[gritty widget]

at least onto its image, but that's not a very interesting statement

[empty grove]

[gritty widget]

sniped me

[rancid umbra]

yea, i just meant onto its image. my bad

hmm. i’m just trying to think of what maps preserve hausdorff-ness. seems like continuity isn’t even needed

[gritty widget]

all spaces are hausdorff

so all maps preserve it

[gritty widget]

At the end of Engelking's "General Topology" there is a table with what kind of properties are preserved by what maps etc.

It says there that image and preimage of a Hausdorff space by a perfect map is Hausdorff

[lunar yoke]

today i also learned that nlab has unexpectedly much content on pointset topology

like under what circumstances which separation axioms are preserved by pushouts, coproducts,etc

neat proofs that every cw complex is normal, or even paracompact

[gentle osprey]

Blitz

[gritty widget]

So the "Hausdorff preimages via perfect mappings are Hausdorff" doesn't tell us anything actually

suppose that i have a pushout square (maybe a homotopy pushout)

A B
C D

Is H^*(D,B) iso to H^*(C,A)

[lunar yoke]

for homology it works if A -> C is a cofibration

other than that, if you know that A -> C is a cofibration and A -> B is a homotopy equivalence, then you get a pair homotopy equivalence (C,A) -> (D,B)

i dont know about homotopy pushouts sadly

[gritty widget]

yeah if we have a homotopy pushout the A to B to D is a cofiber sequence

i’m basically thing of the situation for CW complexes

B and D bring the Xn-1 and Xn

[gritty widget]

but i want to know in how much more generality this holds

[lunar yoke]

yeah also the S^n-1 -> D^n is a cofibration and coproducts preserve that

so in homology the situation is pretty good

but im not sure about cohomology

maybe ill know that in 2 days though when im done revising cohomology

[gritty widget]

yeah for homology and actually the case of CW complexes this is fine

[lunar yoke]

this is not #chill

[pearl holly]

yes. Plz don't shitpost the anime channel

[gritty widget]

i think i might be wrong

[marsh forge]

I think homotopy pushouts give you a Mayer-Vietoris like statement but I don’t think it’s quite this strong

[gritty widget]

In the definition of local contractibility on wikipedia, does neighbourhood mean open neighbourhood?

Seems so

Not necessarily

[marsh forge]

Local always means open

[gritty widget]

Look at the alternative definition of local path-connectedness: it says that every point has to have a neighbourhood basis consisting of path connected neighbourhoods, but not necessarily open

[marsh forge]

The alternative definition may indeed be alternative

But if you are reading something and it says locally P

It means P on open sets around points

[gritty widget]

What about locally compact

I don't think there's a rule, one just has to be careful

[gritty widget]

what are some good topology lectures/yt series?

[quartz edge]

does seifert van kampen theorem give you any actual way to determine precisely what the fundamental group of some space is?

[gentle osprey]

grist bundle

grist bundle

[quartz edge]

i see now that this is the case indeed

very cool

i do wonder now what the difference between wedge and connected sums is

topologically speaking

[empty grove]

They should have different homotopy and homology groups

Connected sum of 2 spheres is just 1 sphere

Not so for wedge

So 2nd homotopy and homology are both different

ouch

[quartz edge]

o_o

oh yeah because the connection is a hollow cylinder

i was thinking you meant S^1 by sphere

[empty grove]

Is connected sum of 2 S¹s not S¹?

I have never actually looked up the definition of connected sum I just know what people here have told me

Ye the definition on wikipedia looks like it makes S¹ # S¹ = S¹

[quartz edge]

ohhh

i was thinking it was a line

i have a Q

there can't be a homotopy from a dim k manifold to a dim k+1 manifold can there

hm

yeah it seems there cant be

[empty grove]

Do you mean homeomorphism

[quartz edge]

no

i can't see a continuous map from R^2 to R^3

[empty grove]

Do you mean surjective

[quartz edge]

doesn't the afterimage of a homotopy need to be surjective?

[empty grove]

Because there is an obvious embedding there

[empty grove]

no

[quartz edge]

huh

[empty grove]

homotopies can be constant

[quartz edge]

i dont mean codomain

wait

wat

[empty grove]

Define H((x,y), t) = 0 for all (x,y) in R^2 and t in [0,1]

This is a homotopy from R^2 to R^3

[quartz edge]

that's not continuous, firstly

so it can't be a homotopy

[empty grove]

All constant maps are continuous

[quartz edge]

ah, then it cannot be a homotopy as the H((x, y), 0) != (x, y)

iirc a constant homotopy needs to be H((x, y), t) = (x, y)

[empty grove]

(x,y) is not in R^3 so that doesn't make sense

[quartz edge]

well

some embedding then

ah

yeah

lol

now i see

[empty grove]

[quartz edge]

still, not what i meant

clearly the image is at most a 2-manifold

no homotopy can elevate the dimension of its preimage

it can just embed it in a higher dim codomain

[empty grove]

Have you seen space filling curves

[quartz edge]

yeah, but let me think for a sec about a trivial example

oh shit

that actually works in homotopy

nice

wow that really does do the trick

wow ok

[empty grove]

[quartz edge]

i'm just so used to that not being feasible

[empty grove]

I have no clue what you are referring to lol

[quartz edge]

like those dont work via homeomorphisms

i kind of ignored them when starting to read ab homotopy

[empty grove]

I'm still lost lol leave it

[quartz edge]

R not homeomorphic to R^2

[empty grove]

Oh

[quartz edge]

i'm unsure if this is right but here goes

i'm trying to compute the fundamental group of S^2 punctured in n antipodal pairs of points

[gentle osprey]

grist bundle

[quartz edge]

i.e. the free product on 2n-1 generators

is this right? i had also tried to use van kampen but i always get a cylindrical intersection in my decomposition and am not sure how to recover exactly what the fundamental group of the whole punctured sphere is

[empty grove]

Yes that is right

Another way to look at this would be to use the homeomorphism between a once punctured n-sphere and R^n

[quartz edge]

what does that tell us?

[empty grove]

k-punctured n-sphere is k-1-punctured R^n

[quartz edge]

ah

yeah, that's right.

just pick a puncture and flatten that sucker

[empty grove]

Yep

[quartz edge]

well

there's really no way to use van kampen eh

[empty grove]

There is

How would you show that k-1-punctured R² has pi_1 = free prod of k-1 copies of ℤ?

One way to do that is to induct on k and use van kampen for the induction step

You could probably do something similar on the sphere directly

[quartz edge]

yeah

[empty grove]

Just that you'd have to do this standard argument again

[quartz edge]

van kampen makes this quite weird

i always end up with a cylinder intersection, which makes the kernel of the map from the free product to the fund group of the whole punctured sphere nontrivial

unless you mean doing something else first

[empty grove]

You can use a second hole to make the intersection a band that goes around the sphere but pinches off at exactly that hole

So that it's not actually a cylinder

This is probably where you'd get a k-1 instead of k

[quartz edge]

oh shit

that's crazy i didn't even think about that

[empty grove]

[quartz edge]

damn we can do all kinds of funky stuff here

thanks for the guidance

one more thing if you don't mind

is there a standard notation for the free product on n copies of Z?

[empty grove]

I would imagine it would be this

You could also use the upside down capital π since free product is the coproduct in the category of groups

[quartz edge]

oh shit you're right, it's even called \amalg

,,\amalg

[gentle osprey]

grist bundle

[empty grove]

Nice

[quartz edge]

"disjoint union" also

same idea ig

[empty grove]

Ye that's the coproduct in the category of sets

[gritty widget]

All contractible spaces are homotopic

[quartz edge]

sure, but for some reason i was thinking that because you only needed continuity (not bicontinuity) that it could be deleterious

of course homotopy classes would mean nothing then, though

[cyan halo]

why is the orientation of 0 mfd +and-?

the orientation of 0-dim simpleces is only one.
what is the difference between them?

[marsh forge]

It should be a choice of generator effectively

in H_0

i.e. unreduced H_0 will be Z

there are two "fundamental classes"

1 and -1

[cyan halo]

0-degree homology group?

[marsh forge]

yes

which is also the top degree

[bleak path]

Hello, I was wondering if I could get some feedback on my attempted proof?

The goal is to show that homotopy equivalence is an equivalcen relation

It feels like I'm in the right direction, but also something feels missing, if anyone is willing to help me that would be great

[marsh forge]

This is all fine

Depends on the level of course you are in

they might expect you to be more explicit about the homotopies between the maps

The main ideas here are probably like

Reflexivity: maps that are actually equal are trivially homotopic

Symmetry: The definition is inherently symmetric anyway

Transitivity: Homotopies compose

(In fact, one has two different ways to compose homotopies, usually referred to as "vertical" and "horizontal")

[bleak path]

I'm in a first homotopy Algebraic Topology course, but that is exactly what I think can be improved, I would like to be more explicit about the homotopies between maps. Could you elaborate more on that?

It's interesting to hear about vertical and horizontal compositions, could you share more as well?

And also I'm wondering what the exact difference is between homotopy and homotopy equivalence, I haven't been able to find a distinction that I understand yet

[marsh forge]

I'll start with the last question

a homotopy is a relationship between functions

a homotopy equivalence is a relationship between spaces

One way to think about homotopy equivalence is that if you take homotopy classes of maps (homotopy between maps is an equivalence relationship) then the two spaces are homotopy equivalent if and only if there are maps f:X->Y and g:Y->X such that [f] and [g] are inverses of eachother, where brackets are homotopy equivalence classes

Vertical composition says the following: If H is a homotopy between f and g and H' is a homotopy between g and h then I can compose H with H' to get a homotopy between f and h

Horizontal says that if I have a homotopy between f and f', both functions X->Y and similarly g,g':Y->Z

then I get a homotopy between the compositions gf and g'f'

The top left here is a vertical composition

and bottom left is a horizontal

[bleak path]

Okay, perhaps this is a good place to pause before you talk about the first question I asked, if you don't mind

What does the down arrow notation mean here?

[marsh forge]

For this one you would name the given homotopies and explicitly construct new ones that prove your claims

[marsh forge]

c_1,c_2 are homotopies

between the black single line arrow maps

(they are arrows because sometimes one studies non-invertible versions)