point-set-topology

[lean cedar]

its like 11pm for me :D

[cursive flume]

I think I managed to do it

a little proofcheck though

does this seem to be fine?

proof that induced topology is a topology

[river granite]

all good

[static maple]

hey there, i'm the same tum topology course as you are, but i'm lost, i can't for the life of me conceptualise continuity for topologies... does it simply mean one can approach a point?

[cursive flume]

so as I see it: you know continuity in R^n, right?

[static maple]

i didn't have university level analysis.

[cursive flume]

a topology is the minimal structure, that is needed to be able to talk about a notion of continuity

[cursive flume]

and one can show that the definition of continuity via open sets is equivalent to the definition of continuity on R^n with open balls/epsilon delta

[cursive flume]

did you see the definition of continuity via epsilon delta?

[static maple]

Lemma 1.3?

from 1.1 metric spaces?

[cursive flume]

definition 1.4

[static maple]

oh, sure

[cursive flume]

definition 1.4 is the notion of continuity, which is intuitive

but metric spaces are more structure, than a topological space. We want to be as minimal as possible.

and we ask:what is the minimal structure on a set, so that we can talk about continuity, and in the case of metric space, we regain def 1.4.?

[river granite]

the idea is that for a continuous function f: X -> Y, "nearby" images have "nearby" preimages. This is encoded in the fact that for any open set U in Y, however small, you can guarantee that the preimage f^{-1}(U) is open in X

[cursive flume]

this is the motivation for introducing topology

[river granite]

it's a bit abstract

so that's why knowing the epsilon-delta definition (as an example) helps

[static maple]

i grasp TT and category theory, if anything this is more concrete than i'm used to

[river granite]

in metric spaces (which are in a sense a particular example of topological space) you can characterize continuous functions through this

[river granite]

well, obviously continuous functions are the morphisms in the category of topo spaces

and an interesting characterization of homeomorphisms (continuous fns. with a continuous inverse) is that f: (X, T_X) -> (Y, T_Y) is a homeomorphism iff f(T_X)=T_Y

[static maple]

are there non-continuous functions? or rather are they ever applied?

[river granite]

of course there are, but in topology you generally care about continous functions and properties that are preserved through continuous functions (such as connectedness and compactness)

[cursive flume]

@static maple this might help you, I filled in 80% of proofs/remarks not done in lecture

[static maple]

alright, but how can the intersection of an open ball in X and its complement give a boundary? @river granite

[static maple]

reading, thanks a ton

[river granite]

it might help to show that a simple function such as $f: \bR\to \bR$ given by
[f(x)=\begin{cases}
0, &\text{if $x\leq 0$,} \
1, &\text{if $x>0$,}
\end{cases}]
is NOT continuous, using both the definition by preimages of open sets (Definition 1.12 in ProphetX's notes) and using the epsilon-delta one (Remark 1.7)

[gentle osprey]

derivada.schwarziana

[river granite]

particularly that function is not continuous at 0

[river granite]

not sure if I get your question

are you trying to find the boundary of that open ball?

in that case you need to take the closure of that open ball and then substract the interior

[static maple]

$\delta B = B \cap X / B$ doesn't this apply as well?

[cursive flume]

it's basically this proof more or less, right?

[river granite]

nope, you need to take closures in B and X\B.

[static maple]

right, sorry

[river granite]

the boundary of a set $B\subset X$ is
[\partial B = \overline B \cap \overline{X\setminus B} = \mathrm{Cl}_X(B)\cap \mathrm{Cl}_X(X\setminus B)=\mathrm{Cl}_X(B)\setminus \mathrm{Int}_X(B)]

[gentle osprey]

derivada.schwarziana

[river granite]

there are many equivalent definitions

[cursive flume]

now we take union over interior and we get closure= boundary union interior

and that's th eproof I did

[river granite]

this shows that Cl(A)=Int(A) U Bdry(A)

from that it follows that Cl(A) \ Int(A) is contained in Bdry(A); one needs to show the other inclusion as well

[river granite]

to prove that these defns. are equivalent

[cursive flume]

fair

[static maple]

oh, and the empty set needs be in O because a topology needs be closed under pairwise intersections?

[river granite]

I mean, the empty set is open by definition

any topology must contain it

but I guess it makes sense to contain it since you could take unions with the empty set

and because in all of the motivating examples (e.g. metric spaces) you can show that the empty set is open by a vacuous truth

like, if you define "a set X is open iff for every point x in X there's an open ball, centered in x, contained in X", then the empty set is open precisely because there's no point x in the empty set

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

[static maple]

sure, for every point in Ø there's an open ball

[woven haven]

Any ideas on how to compute singular homology groups of S^1 V S^1 V S^2?

[gritty widget]

the reduced homology of the wedge sum is the direct sum of the reduced homologies, the isomorphism being induced by the inclusions into the wedge

[woven haven]

If I let $a,b $ be the points where the circles are attached to the sphere then $(X,{a,b})$ is a good pair so I have $H_n(X)=H_n(S^1)\oplus H_n(S^1)\oplus H_n(S^2)$. So for $n\ge 3$ we get $H_n(X)=0$, for $H_2(X)=\mathbb{Z}$, $H_1(X)=\mathbb{Z}\oplus \mathbb{Z}$ and $H_0(X)=\mathbb{Z}$. Is this correct?

[gentle osprey]

K零ꓘ

[plain raven]

[sleek thicket]

Let $X$ be (1) the cantor space or (2) the set of rational numbers. What is the group of locally constant functions $X \to \Z$, and in particular in the case $X = \Q$ is it isomorphic to the direct product of countably many copies of $\Z$? (Note that I'm not asking if the evaluation map $\mathrm{Hom}{Top}(\Q, \Z) \to \prod{p\in \Q} \Z$ is an isomorphism. This is false. I'm asking about abstract isos)

[gentle osprey]

Mormon Shamrock

[gritty widget]

@sleek thicket so your functions aren't continuous?

I think in case of 2) your space is isomorphic to something like $\bigoplus_{r\in \mathbb{R}\setminus\mathbb{Q}}^* \mathbb{Z}$ where the star means that we take points from which all but countably many are equal to $0$

[sleek thicket]

the functions are continuous

[gentle osprey]

Blitz

[sleek thicket]

Locally constant = continuous function into space with the discrete topology

Can you explain what the iso you have in mind for that is?

Like what are the irrational number indices representing?

[gritty widget]

Now that I think about it, maybe it's wrong, but the idea was, what kind of sets separate rational numbers? I think it's about have intervals with irrational ends. So maybe for every irrational r we can assign it the value it has on the right of it etc

[sleek thicket]

Right, but there's weird overlap

so like, if you have (a, b+1) and (b-1, c), you can't just take a constant function on both of these and glue to a constant function on (a, c)

So it's not going to be the whole sum and it's also probably not going to be direct no matter how you set it up

But maybe you can write it as a limit/colimit this way? Like Q is the union of (a, b) over all irrational a, b and so maybe you take a limit of the constant functions there, or something?

Ah yeah this is just the sheaf condition for that cover

Still might be useful

Hmm no I think it's like a different limit?

[gritty widget]

Yeah I forgot that it needs to be compatible with the operation

I might think about it later (when I have something to write on) but I probably won't come up with anything

[sharp frost]

Hi does anyone know why exactly H*(i) of this inclusion map would be addition?

[gritty widget]

I think it has to do with that this map needs to be a group homomorphism and additionally defines an operation on Z

And this automatically should imply it's the same as standard addition

[sharp frost]

ah yep I think I see what you mean, but dont maps like $(a,b)\to 2a+2b$ and $(a,b)\to a-b$ also work?

[gritty widget]

Are those group operations on Z?

[sharp frost]

I think the second one is

[gritty widget]

They're not associative

[sharp frost]

oh yeah

and how did you get the requirement that it needs to define an operation on Z

[gritty widget]

If it needs to be addition then it's a group operation on Z

All I'm saying it's enough to prove that

Try showing a group homomorphism which is a group operation is actually the same as the operation on that group

[sharp frost]

that's still assuming that it's not the 0 homomorphism

[gritty widget]

No

[sharp frost]

oh right yeah I misunderstood

[gritty widget]

Why is there a star at the bottom there? Don't you want first homology?

H_n(S^1) = 0 for n>1 I thought

[sharp frost]

yeah * = 0 or 1

[gritty widget]

Ah ok

[rugged rock]

Give an example of a continuous map $f: X \to Y, x \in X, y \in Y$, and $Z \subset Y$ such that $f(x) = y$ and $y \in \overline{Z}$ but $x$ not an element of $\overline{f^{-1}(Z)}$. Is there a particular way to come up with examples?

[gentle osprey]

Évariste Galois

[rugged rock]

I was thinking maybe $p: \mathbb{R} \to S^1$ defined by $p(x) = (\cos 2 \pi x , \sin 2 \pi x)$. But i'm not certain

[gentle osprey]

Évariste Galois

[compact bluff]

take f : {0,1} -> {0,1} the identity with {0,1} having discrete and indiscrete topology. then take Z = {1} . cl(Z)={0,1} but cl(f^{-1}(Z)) = {1}

[rugged rock]

is { = ( or [?

[compact bluff]

no set brackets

two-point set

[rugged rock]

ahh

[compact bluff]

if you wanted to take intervals that would also work

[rugged rock]

How did you come up with it?

That is my main struggle to be honest

[compact bluff]

closures in the indiscrete space give you the whole space, but closures in the discrete topology give you back the same set\

so if you want to keep the set small in the domain when taking the closure but have the set be big when taking the closure in the codomain, these are the toplogies to look at

[rugged rock]

Ahh that is smart

thanks alot 🙂

[compact bluff]

np

[lean cedar]

Doesnt this literally contradict what I wrote you yesterday? Thats a covering, right?

[rugged rock]

Yeah, I didn't catch that to be honest, it was more of a guess as we used this map quite often.

It is a covering

[lean cedar]

:D

[true garden]

Hello. Can you please provide a simple and intuitive proof of the fact that the uncountable product of $\mathbb{R}$ is not normal?

[gentle osprey]

MathPhysics

[plain raven]

oh my GOD

[gritty widget]

What does that mean? Such space is pretty abstract

[uncut surge]

i think it's pretty intuitive that taking an uncountable product is just not a normal thing to do

[woven sundial]

i guess this probably means R^R

idk about simple and intuitive proofs though

[gritty widget]

Uncountable =/= continuum size

[woven sundial]

well sure but it works as an example

[gritty widget]

https://math.stackexchange.com/a/191846/476484

Seems like the best way here is just taking a closed subspace homeomorphic to N^omega_1 and proving it's not normal by definition

this is SO simple and intuitive! thank you!

You're welcome

[sharp frost]

I was trying to prove this thing and I thought it looked kind of similar to a thing I've seen before

I know there's gotta be some connection there using the homotopy equivalence axiom for homology, but I'm not sure what it is

[coral pawn]

Can someone verify my solution for this?

Let $v_0 = (0,0), v_1 = (1,0),$ and $v_2 = (0,1)$ and consider the oriented simplex $[v_0 v_1 v_2]$ . Then the boundary of the simplex is $[v_1 v_2]-[v_0 v_2]+[v_0 v_1]$. We now compute $\int_{\partial c} x dy = \int_{[v_1 v_2]} x dy - \int_{[v_0 v_2]} x dy+\int_{[v_0 v_1]} x dy$

[gentle osprey]

Finitely Many Bananas

[gritty widget]

Did you mean #diff-geo-diff-top

[coral pawn]

Well, its a small tool that we're using for algebraic topology and chains and stuff come up in algebraic topology, so I thought I'd post it here

The first integral becomes (I think) $\int_1^0 (1-y) dy$

[gentle osprey]

Finitely Many Bananas

[coral pawn]

the other two disappear

Because y is constant in one of them and x is zero in the other

So it equals -1/2

Next, we compute $\int_{[v_0v_1v_2]} dx dy$

[gentle osprey]

Finitely Many Bananas

[coral pawn]

This becomes $\int_{0}^1 (\int_0^{1-y}dx)dy$

[gentle osprey]

Finitely Many Bananas

[coral pawn]

Which is 1/2

I'm guessing I oriented things incorrectly, but I don't see where

[gentle osprey]

Nobody

[gritty widget]

@coral pawn you messed up the orientation on the first integral. $[v_1v_2]$ is parametrized by $(x, y) = (1 - t, t)$, $t \in [0, 1]$ (note the orientation!) which gives $$\int_{[v_1v_2]}x,dy = \int_0^1(1 - t) , dt = \frac{1}{2}$$

[gentle osprey]

TTerra

[lean cedar]

@vocal anchor did you read the solution or did it fly past you?

[vocal anchor]

Yeah didn't really check discord. I see, that's just a 90° rotation of the camera. Thanks man :)

Or, well, more like moving along an arc, not Really rotation

[lean cedar]

Yeah

also the knot diagram with rademeister moves :D

[sleek thicket]

If anyone was curious, I figured out how to show these aren't isomorphic as rings. locally constant functions $f : \Q \to \Z$ such that $f^2 = f$ are exactly the (indicator functions of) clopen subsets of $\Q$, and you can encode the subset order by $f \leq g$ iff $gf = f$. Similarly arbitrary functions $f : \Q \to \Z$ such that $f^2 = f$ are (indicator functions of) arbitrary subsets of $\Q$ and the order is the same. Unions of clopen sets might not stay closed, like $\bigcup_{n=1}^\infty (-\infty, -\sqrt{2}/n) \cap \Q = (-\infty, 0) \cap \Q$, and you can turn this into a ring theoretic statement that distinguishes the two rings

[gentle osprey]

Mormon Shamrock

[sleek thicket]

This actually makes me feel even more uncertain about the groupy question though, since this makes heavy use of the order theoretic structure/multiplication

[gritty widget]

this order reminds me of the usual order on idempotents of a semigroup

you define it the same way

[sleek thicket]

for sure

I came it by noticing that the ring of locally constant functions Q -> F2 is a boolean ring, so we're really asking about whether two boolean algebras are isomorphic (and then I noticed that the powerset ring F2^Q has all suprema while the ring of locally constant functions doesn't)

But this is basically what you said, it's just that everything in a boolean is an idempotent

[cerulean oriole]

It's also how you recover the partial order of a lattice from the join/and operation.

[unreal stratus]

Lol I looked at this proof today

V is contained in Y-W. The latter set is closed, so taking closures, cl(V) is contained in Y-W, which is contained in Y-C

[cosmic beacon]

I'm trying to show there's no orientation reversing homeomorphisms $f:\bC P^{2k}\to\bC P^{2k}$. Here's my work so far.

Suppose there did exist such a homeomorphism. The cohomology ring of $\bC P^{2k}$ is given by
[H^(\bC P^{2k})=\bZ[\alpha]/\alpha^{2k+1}, \quad|\alpha|=2.]
Because $f$ is a homeomorphism, it must send the generator of $H^2(\bC P^{2k})\cong\bZ$ to a generator, namely, it must send $\alpha$ to $\pm\alpha$. Then $H^{4k}(\bC P^{2k})\cong\bZ$ is generated by $\alpha^{2k}$. In particular,
[f^(\alpha^{2k})=(f^*(\alpha))^{2k}=(\pm\alpha)^{2k}=\alpha^{2k}.]

Let $[\bC P^{2k}]\in H_{4k}(\bC P^{2k})$ denote the fundamental class, so that $f_([\bC P^{2k}])=-[\bC P^{2k}]$ as $f$ is orientation-reversing. Then we have:
[\langle \alpha^{2k},[\bC P^{2k}]\rangle=\langle f^\alpha^{2k},[\bC P^{2k}]\rangle=\langle \alpha^{2k},f_*[\bC P^{2k}]\rangle][=\langle \alpha^{2k},-[\bC P^{2k}]\rangle=-\langle \alpha^{2k},[\bC P^{2k}]\rangle,]
so that necessarily $\langle \alpha^{2k},[\bC P^{2k}]\rangle=0$.

I feel like I should be able to reach some sort of contradiction from here. Any ideas?

[gentle osprey]

@cosmic beacon

[weak narwhal]

are these two things not the same?

[plain raven]

the first set has one element, the second set has zero elements

like

[weak narwhal]

it seems to me the first set is a set with no elements, aka the empty set

[plain raven]

i said that backward lol

it's late

[weak narwhal]

and the second set is the set containing the empty set

[plain raven]

yes you're right

so the first set has zero elements and the second set has one element

[weak narwhal]

so 2 is definitely a topology on the empty set

[plain raven]

Yeah.

[weak narwhal]

1 is debatably?

idk this seems like a literal semantics question to me

[plain raven]

i'm going to go with no.

[weak narwhal]

yea, i think its supposed to be no

but imo, its bs cause the emtpy set is a subset of every set

therefore that set contains the empty set

therefore its equivalent to the other set right

[plain raven]

no no no

you're conflating two different meanings of "contain"

that' the problem

Sets can contain other sets as elements or they can contain them as subsets and those are very different.

like

It's the difference between $\in$ and $\subseteq$

[gentle osprey]

diligentClerk

[plain raven]

Given two sets $A, B$ we say that $A \subseteq B$ if, for all $x$, if $x\in A$, then $x\in B$

[gentle osprey]

diligentClerk

[plain raven]

it's in this sense that we say the empty set is a subset of every other set, because

for any set $A$

[gentle osprey]

diligentClerk

[weak narwhal]

i guess the issue is that extensionality doesnt apply since the set containing the empty set has an element

[plain raven]

what do you mean it doesn't apply

[weak narwhal]

whereas the empty set does not

[plain raven]

It does apply

[weak narwhal]

cant be used to say they are the same

[plain raven]

ok yeah

but like

do you see what i'm saying here

$\emptyset \in { \emptyset}$

[gentle osprey]

diligentClerk

[weak narwhal]

ok, it applies, it doesnt give that they are the same

[plain raven]

but $\emptyset\notin {}$

[gentle osprey]

diligentClerk

[plain raven]

${} = \emptyset$, so $\emptyset\notin \emptyset$

[gentle osprey]

diligentClerk

[weak narwhal]

yea

not an element, is a subset

of itself

[gritty widget]

The first set is not a topology on any set

A topology always contains an element, such as the empty set

[weak narwhal]

yep clerk and I agree 1 is false 2 is a topology on empty set so true

this one is a bit confusing to me because I thought this was the definition of the closure

do you think it means for me to use the "intersection of all closed sets containing A" defn. and show equivalence?

[plain raven]

sure that's reasonable

[weak narwhal]

if i can use boundary defn. x in A immediately true, so assume $x\in \partial A$; then every neighborhood $N$ of $x$ contains some point in $A$--namely we can choose $N_U$ to be a neighborhood containing $U$

[gentle osprey]

Dpao

[weak narwhal]

but I dont think thats what a teacher would want

[tepid fjord]

I am working on a project that involves algebraic topology and more specifically TDA, I am applying the math to a real world solution and want to write a "proof" proving that TDA can do what i need it to do. Can anyone point me in the direction of learning how to write appllied mathamatics papers? Thanks so much!

[last trout]

Im pretty sure this statement is false

but I cant come up with an example

can anyone help?

nvm got it

its true

[gritty widget]

Yep. It's true because every connected component contains some path-component

So we can construct an injection

[river granite]

not what you asked for but still an interesting example to keep in mind: the topologist's sine curve is connected but not path-connected, in fact it has 2 path-components

[weak narwhal]

is this true?

[gritty widget]

Uh?

[weak narwhal]

it seems very wrong to me

[gritty widget]

1. definitely some assumptions on Y needed
2. I don't think separation is a common term

[weak narwhal]

its not cause its used in a million contexts but afaik from munk it means

a separation is a pair of disjoint nonemtpy open (or both closed) sets that union to the entire space

[gritty widget]

I don't think I ever saw it used

[gritty widget]

Assume Y is connected

[weak narwhal]

in that case, i would agree

[gritty widget]

In general it's obviously wrong, take Y = X

[weak narwhal]

yea

[weak narwhal]

cause I think u could say this a different way: has a nontrivial clopen set

[gritty widget]

I'd rather say nonempty proper clopen subset

[weak narwhal]

yep same idea tho

[gritty widget]

When you say nontrivial it's not really understood what exactly do you mean

But nonempty proper is clear

[weak narwhal]

ya

[mystic reef]

Im sitting with another topology problem, which Im having difficulties with. The description is pretty long so bare with me: Suppose $f$ is continuous and that $\mathcal{U}$ is an open covering for $Y$. Show that $f$ is a homeomorphism if and only if for each $U\in\mathcal{U}$ the function $f_{|f^{-1}(U)}:f^{-1}(U)\rightarrow U, x\mapsto f(x)$, is a homeomorphism (where $f^{-1}(U)\subseteq X$ and $U\subseteq Y$ are equipped with the respective subspace topologies). Im having a difficult time starting this proof. If someone could give me a hint it would be highly appreciated.

[gritty widget]

no $ after mathcal U

first mathcal U has a space

[gentle osprey]

Ursus1234

[mystic reef]

there we go I think

[gritty widget]

It's a continuous bijection by assumption. Show it's open

For this you can write any open V in X as union of intersections of V with f^-1(U), U in fancy U

f(V intersection f^-1(U)) is open in U by assumption, which is open

[cursive flume]

||s||

how to make picture spoiler?

[opaque dew]

can anyone please help me with these? i been trying for a few hours over different days but no result 😦

im new to pure maths and not very good at it

[true robin]

For Q2 consider any y in Cl(B(x)). For any delta, the delta ball around y intersects B(x), say z is in the intersection. Then d(x,y)<=d(y,z)+d(z,x)<=delta+epsilon. Since this is true for all delta, d(x,y)<=epsilon

[gritty widget]

Saketh

[mystic reef]

thanks, I figured it out

[mystic reef]

thanks, I figured it out

[opaque dew]

thank you very much 🙂 i willl try

[bitter smelt]

Is $S^1$ common notation for the space of closed curves on a surface $S$?

[gentle osprey]

Migillope

[bitter smelt]

No, that makes no sense, it's obviously the circle

[gritty widget]

\Omega X is the notation you want, i think

hmmm

though you may want to distinguish between based loops and free loops depending on your purposes

[vast estuary]

hey

do you have any examples of extremally disconnected topological spaces

which are subsets of R^n for some n?

[gaunt linden]

Z^n?

[gritty widget]

A metric space is extremally disconnected iff it's discrete

Which gives us a family of unmotivating examples, all of them are boring

[opaque dew]

does make it sense to write this? feedback?

[gritty widget]

yeah good

[opaque dew]

thank you 🙂

[opaque dew]

can it be by triangle inequality that:
d(x_n,y_n)=<d(x_n,x)+d(x,y)+d(y,y_n)?

i am trying same question with try to use this thing above^

[gritty widget]

yes

[opaque dew]

it is the same question before

[gritty widget]

Looks like an image

but no, I won't click this

[opaque dew]

sorry i send properly

maybe i have to explain more about epsilon

[vast estuary]

Makes sense, yes!

[thin girder]

Let (X, d) be a metric space with the trivial metric

Show that a subset K ⊆ X is compact if and only if it is finite

how does this sequence subsequence thing works?

to show compactness

[golden gust]

what definition of compact are you using

[thin girder]

do you mean sequential and convergent things

if so we use both

[gritty widget]

Only sequence convergent in this space is one that's constant after big enough time

If you have infinite amount of elements, just pick them all distinct

[thin girder]

hmm

[thin girder]

can I do it with the open cover finite subcover

[gritty widget]

Pick cover of singletons

[thin girder]

like what

[gritty widget]

Let me rephrase

Pick the cover by singletons

[thin girder]

I cant, I dont understand

is it 0 and 1?

I really dont understand

[gritty widget]

What is

[thin girder]

what you asked

[golden gust]

the singletons are open because they're the open balls of radius 1/2

[gritty widget]

I didn't ask

[thin girder]

okay okay lets just forget I asked

the whole question

[gritty widget]

I have good memory

I won't forget that easily

[heady skiff]

By projection, do they mean projection in a linear algebra sense?

[gritty widget]

Lol, no

[heady skiff]

what should i review before reading this then lol

[gritty widget]

They mean a function that to each element assigns its equivalence class

[heady skiff]

oh ok thanks

[gritty widget]

If I have a bounded domain $D\subseteq \mathbb{R}^n$, $n>1$, and $U\subset D$ is a non-empty proper open subset, does it follow that $\partial U\cap D$ is infinite?

[true robin]

there is the trivial example of U=empty set.

[gentle osprey]

Blitz

[true robin]

if you want a non trivial example, I have a question first, by domain do you mean that the set is closed or open or something? If not then there are some other trivial counterexamples

[gritty widget]

domain is an open connected set

[true robin]

consider D=open unit ball, U=open unit ball\{(0,0)} in R^2. The only element of D \cap \partial U should be the origin

[gritty widget]

yeah... I need to think about this question further

[unreal stratus]

interesting use of int there

[true robin]

Damn

[mighty stirrup]

Hey, I'm trying to learn about model categories. Does anyone have any recommendations? For reference, I know the basic definitions of category theory and have completed my school's graduate algebraic topology sequence (fundamental group, homology, cohomology, basic homotopy theory)

[bitter smelt]

Does algebraic intersection mumber assume minimal position?

Or, rather, is algebraic intersection number "position" invariant

I would suppose so, that homotoping intersections away will be "accounted for" by cancelation of sign for the bigon that is removed

[fading vale]

Emily riehl has a 60 page short intro to it which is pretty neat

https://arxiv.org/pdf/1904.00886.pdf

[clear storm]

If someone is familiar with homotopy theory can you please have a look at this? I think I am quite close but can’t get the last piece together. https://math.stackexchange.com/questions/4452541/homotopy-fiber-of-of-map-s2-to-mathbbcp-infty-is-homotopy-equivalent-t

[fading vale]

So the map here is given by taking a generator of pi_3(F_f) = Z under the isomorphism of that group with pi_3(S^2), right? then it corresponds under the isomorphism to the hopf fibration, meaning if the map S^3 -> F_f is named, say, g, then g composed with the inclusion has the same homotopy type as the hopf fibration S^3 -> S^2. so then we can apply pi_n to get a commutative diagram. for n > 2 the diagonal and horizontal maps are both isomorphisms so the vertical map is too. for n = 1 or 2 the homotopy groups of the fiber and S^3 are both 0

so this induces isomorphisms on all homotopy groups and by whiteheads theorem you get a homotopy equivalence

[bleak path]

Hello, I have a question or two about deformation retracts

Firstly, is the singleton a deformation retract of any space? What if the singleton is not in the space? (e.g. the singleton {0} with the open interval (1, 2) that has the subspace topology on R)

Any non-empty space*

[gritty widget]

well it has to be in our space

[gritty widget]

if a singleton is a deformation retract of your space then that space is necessarily contractible

[bleak path]

Thanks, have seen almost everywhere assuming it is but nowhere stating this assumption beyond the definition, and was worried about my interpretation of the definition

[gritty widget]

and so every singleton is a deformation retract of such space as well

[bleak path]

Right, then similar to that logic, a connected space with 1 component is only contractible to 1 singleton right, not 2 or more singletons, for example

[empty grove]

All singleton spaces are homeomorphic, so any non empty space contains the singleton

[gritty widget]

a connected space cannot have more than 1 components, that's kind of what it means to be connected

a connected space doesn't have to be contractible

[bleak path]

Thanks for helping my sanity check

I've got a few more questions, if you don't mind

[gritty widget]

what it means to be contractible is that there is a contraction, so a homotopy which sends the identity map to a point

so the space "gets contracted"

[bleak path]

Right, I follow that logic

First is, S^1 (and the family of S^n, in general) is not contractible to the singleton, right?

[empty grove]

ye

[gritty widget]

S^n isn't contractible for any n

[empty grove]

Other than n = infinity

[gritty widget]

I'm not sure what that would be

[bleak path]

Okay, so for finite n I suppose

I don't have an intuition of that either but it's chill

[gritty widget]

sphere in l^2 ?

[empty grove]

Direct limit of the S^n

[bleak path]

Then my other question is a bit more specific

Can projective spaces contract to each other?

I say contract but I meant retract, sorry

For example, I know R3 can retract to R2 or R1, right?

What about RP3 to RP2?

[gritty widget]

what does that mean

[bleak path]

Which part?

[empty grove]

I don't think so, it should follow from D^n not retracting to S^(n-1) for any natural n

[bleak path]

Okay, that makes sense to me, I was confused because my intuition of RPn etc. is not very good so far

[gritty widget]

I mean, I don't think it makes sense to say "X retracts onto Y" unless Y is a subspace of X, since it depends on how Y is located in X

unless you mean some canonical way of treating RP^2 as subspace of RP^3

[bleak path]

Hmmmmm

That's an interesting take

[empty grove]

There is one

For S^2 to S^3 you have equatorial inclusion

That induces an inclusion of RP2 to RP3

Similarly in any dimension

[bleak path]

But that inclusion is not sufficient to admit a retraction, right?

Or am I missing something in my intuition

[empty grove]

Ye

You can prove that there is no retraction using cohomology rings

[bleak path]

Hahah, my course does not cover homology so that's a bit unfortunate

And from my brief reading on category theory, cohomology belongs there too right?

[empty grove]

Ye homology and cohomology are taught together

[gritty widget]

it's homological algebra and it involves some category theory but I wouldn't call it a subfield of category theory

unless I'm wrong of course

[empty grove]

oh ye it's not usually cat theory

[gritty widget]

homology and cohomology is important in topology, theory of modules, algebra in general, algebraic geometry etc I think

[pearl holly]

Moldi 👋

[gritty widget]

it's called resolutions of a module

[empty grove]

Hi toki ✓

[pearl holly]

Long time no see

Or like, long time no write

[empty grove]

Semester ended

And I immediately got sick

[pearl holly]

Bruuuh

I still have like a month left

[empty grove]

oof

[mystic reef]

so $I$ is an infinite index set and for each $i\in I$ I am given a non-empty topological space $X_i$. I need to determine whether the following statements are true when $\prod_{i\in I} X_i$ is equipped with a) the product topology and b) the box topology.
i) If $A_i\subseteq X_i$ are closed subsets for each $i\in I$, then $\prod_{i\in I} A_i$ is closed in $\prod_{i\in I} X_i$
that sthe first one and Im not sure how to do this. I believe that the statements is true for the product topology, maybe as well for the box topology, but Im not sure if that is correct or how can show it

[gentle osprey]

Ursus1234

[mystic reef]

can I not use this tag?

[gritty widget]

it's for the questions channels like the "math help" stuff

not here

[mystic reef]

sorry 😬

[gritty widget]

yes, it's true for the product topology, so it has to be true for box topology as well

[mystic reef]

is that an implication that always holds

[gritty widget]

since all closed sets in the product topology are also closed in the box topology

[mystic reef]

ahh ok

[mystic reef]

I didnt really know how to show it. I just assumed that the product of closed sets is closed...

[gritty widget]

if you take complements then you should obtain something like a union of open sets pretty easily

$x\notin \prod A_i \iff \exists_i x_i\notin A_i$

[gentle osprey]

Blitz

[gritty widget]

So if you got the projection maps p_i, this is a union of sets which are inverse images of open sets by p_i

I think this is the easiest approach

[mystic reef]

hmm ok. Still quite difficult to understand but Ill give it a go

[gritty widget]

x_i = p_i(x) so last formula is really just

[gentle osprey]

Blitz

[gritty widget]

I think there should be no problem in writing this in terms of inverse sets

[mystic reef]

ahh ok, yea that makes sense. Thank you

[gritty widget]

np

[gritty widget]

I'm having a brain rot with this. I'm embedding a "ring", so {x : 1 <= || x || <= 2} into R^n using function h. And I have a ball K in the bounded component of h({x : ||x|| = 2})^c which doesn't cross this ring. And I'm taking a ray from center of ball K. How do I prove it must cross h({x : ||x|| = 1}) ?

[empty grove]

Prove that K lies in the unit ball, and then on the ray you have the norm function which starts less than 1 and goes to infinity, apply intermediate value

[gritty widget]

What do you mean by unit ball

[empty grove]

{x : ||x|| < 1}

[gritty widget]

It doesn't have to lie in a unit ball

[empty grove]

Is h an arbitrary embedding?

[gritty widget]

Yes

Topological one

[empty grove]

Let H_r = h{x: ||x|| = r}. Suppose we have a point z in the bounded component of H_2 complement, and suppose this doesn't lie in the image of the annulus. Then it is also in the bounded component of H_1 complement. Any ray from this point thus starts in the bounded component and goes to the unbounded component, so must cross H_1

[gritty widget]

How can you prove it's in the bounded component of H_1?

[empty grove]

Had some handwaving in mind which doesn't work lol

let me think

[gritty widget]

Yeah. Going one direction from circle to the annulus is easy but the other direction doesn't work without using some further properties

[empty grove]

Either H_1 lies in the bounded component of H_2 or the other way around. In the second case, the bounded component of H_2 is contained in that of H_1 so we are done. In the first case, the bounded component of H_2 is the union of the bounded component of H_1 and the annulus, and the point doesn't lie on the annulus

This is the reasoning I had in mind, just trying to see if the second case works

Wherever I say component of H_r I mean of H_r complement ofc

[gritty widget]

I need to think about this. It's really annoying me

[empty grove]

[gritty widget]

This is the que

[gritty widget]

Soln

Can anyone explain why did they choose epsilon like that ?

[empty grove]

∫|f| = 1 - e for some e > 0,
Then for g ∈ B ͚(f, e), ∫|g| < 1

Because ∫|g| < ∫(|f| + e) = 1

Since |g-f| < e ⟹ |g| - |f| < e

Basically if you add a constant e to your function everywhere, the integral increases by at most ∫e

And anything in B ͚(f, e) differs from f by at most e

[gritty widget]

Oh okay got it

Thanks !

[gritty widget]

,tex |
\begin{equation*}
H^{k}{dR}=\begin{cases}
\bR & k=0,1 \
0 & \text{otherwise} \
\end{cases}
\end{equation*} \ or
\begin{equation*}
H^{k}{dR}\cong\begin{cases}
\bR & k=0,1 \
0 & \text{otherwise} \
\end{cases}
\end{equation*}

aiohgoaiwhgioawh

[gentle osprey]

simp

[gritty widget]

wait

im dumb

i mean of R

like

I guess technically isomorphic, but what does it matter

$H^{k}{dR} = H^{k}{dR}(\bR)$

[gentle osprey]

simp

[gritty widget]

i forgot to add the (R) my bad

[marsh forge]

The deRham cohomology of R should be 0 in degree 1

because R is contractible

Anyway, you can use = or \cong

doesn't matter

[empty grove]

I define the real numbers to be the 0th de Rham cohomology of R

[marsh forge]

then degree 0 is for sure wrong

wait

no its not

lmfao

thats so funny

[gritty widget]

im confused

is it wrong

or

[marsh forge]

ignore moldi

it is wrong

[gritty widget]

am i being trolled

[marsh forge]

but if you choose to use \bR to denote 0

oh wait no its wrong either way

lol

sorry im having a MOMENT

[empty grove]

[marsh forge]

TLDR: H^1(R) =0

not R

[gritty widget]

alr

what is H^0 (R)?

[marsh forge]

R

[gritty widget]

oh

[marsh forge]

H^0 is correct

H^1 isn't

[gritty widget]

but moldi said 0th

and u said moldi is wrong

[marsh forge]

ignore moldi

i didnt say moldi was wrong

i said ignore

[gritty widget]

alr

[marsh forge]

(not in general moldi was just trolling)

(also I misread moldi at first)

(I thought moldi was defining R to be the first cohomology of R)

[empty grove]

Same

[marsh forge]

im having a stroke

[hollow harbor]

why did you me moldilocks

[empty grove]

A rice popped out of nowhere and started crying

[hollow harbor]

I'm a #point-set-topology lurker

[gritty widget]

Angry birds

[empty grove]

I have decided after learning the meaning that this joke is funny

[hollow harbor]

I'm no angry bird.

[empty grove]

I am 4 years in and I am not laughing

[marsh forge]

im so worried that at some point in my talk im going to say something that is not true 1-categorically

[hollow harbor]

You're the butt of the joke.

[marsh forge]

i am just pretending to use 1-categories for this whole talk

the beginning of the talk has a all limits and colimits are homotopy limits and colimits disclaimer

[empty grove]

Is this for the blue book?

The last talk?

[marsh forge]

no hahahaha

i have to give two talks this month

[gritty widget]

@marsh forge do you mind pointing out whats wrong with my reasoning (for learning purposes)
$Z^1(\bR) = ker(d_1 : \Omega^1 \rightarrow \Omega^2) = \Omega^1 \$ $B^1(\bR) = im(d_0 : C^{\infty}(\bR) \to \Omega^1(\bR) = \Omega^1(\bR) \$
$\frac{\Omega^1}{\Omega^1} \cong 0$ wow i'm actually so stupid I just realised I got the quotient wrong... my bad lol

[gentle osprey]

simp

[marsh forge]

both on topics that are much better discussed with infinity categories

[gritty widget]

so its iso 0

whoopsie

I thought the quotient was omega1

not 0

[marsh forge]

im glad u figured it out bc tbh i cannot compute deRham cohomology

it just happens to be isomorphic to singular cohomology

[gritty widget]

hot 🥵

[marsh forge]

@empty grove i barely understand what a differential form is

[empty grove]

same but I can compute with it

[marsh forge]

lol

i see no reason to learn it

seems like a much worse version of singular tbh

[empty grove]

It is thing to be computed on manifolds 😌 all I understand

[marsh forge]

especially since everything is a vector space

you can just abuse LESes

[empty grove]

I like it it is very nice

It was my first exposure to homology

[marsh forge]

that is pedagogically unhinged

[empty grove]

had fun

I loved it

[gritty widget]

ong I just think of them as matrix's or smooth functions lol

brainiac

[marsh forge]

that could be good or bad

i have no idea

[gritty widget]

lol

like tensors not matrix's

[marsh forge]

my brain is pure of analytic thoughts

[gritty widget]

🙏 pure

[marsh forge]

I mean I think of it as just like

smooth functions tensor an exterior algebra

and then everything is just kinda formal

but i have no intuition for what the fuck a dx is

[gritty widget]

i read somewhere that a 1-form is an element of the cotangent bundle so I jsut go like hmmk so a 2-form is just a wedge b for a,b in cotangent bundle hmm yes I understand

[empty grove]

hmm yes

[gritty widget]

alternatively you can compute it using Grothendieck theorem for smooth affine varieties, I think?

genius

[empty grove]

but it should be a section of the cotangent bundle right

[gritty widget]

so the cotangent bundle is like an amazon deal right

[empty grove]

not an element

[gritty widget]

wikipedia scammed me

[empty grove]

oh ye are you talking about the algebraic de Rham cohomology

Lurie has a talk on that

It is great

[hollow harbor]

differentiability imposes extra structure that you can exploit. e.g. people study harmonic forms on riemannian manifolds (hodge theory) and this gives you extra topological control. idk if you can get that from singular

[gritty widget]

yes, I've watched Lurie's video, it was so cool, I love it

[empty grove]

It was screened in one of our weekly college seminars when the speaker had to cancel his talk at the last moment

Full watch party 😌

[gritty widget]

but I think the coefficient ring needs to be C for this to work, so we can't use Grothendieck theorem

[empty grove]

Ye

[marsh forge]

Oh that’s kind of interesting

That breaks homotopy invariance though

Presumably anyway

[hollow harbor]

yeah it depends on the metric, it's not strictly topological

what i should have said was "differentiability lets you exploit extra structure"

[empty grove]

Imagining MaxJ interrupting a surgery and going that's not a homotopy equivalence

[hollow harbor]

sippy

[marsh forge]

Everything i do must be model independent

spaces are a crutch

[plain raven]

You guys are killing me

hahahahaha

lmfao

[plain raven]

i've been trying to think about this myself.

not recently

i kinda put it on the back burner with all the coq shit

What i mean is that i'm trying to come up with an exposition for how the complex of sheaves of differentials is constructed and why it works that way

[marsh forge]

i've essentially given up on understanding analysis notation

[plain raven]

lmfao

[marsh forge]

i am truly not smart enough

Shamrock tried to explain what the fuck |dz| means in complex analysis

[plain raven]

i feel like if i came up with an exposition you'd understand it tho, like it would be in your kind of language

Hahaha

[marsh forge]

I stg everyone always tries to dunk on topologists for being unrigorous

but that shit is unhinged

[plain raven]

i have a nice understanding of d as something that like

you can construct in a monoidal category

with duals

i wrote this down somewhere

[marsh forge]

i like this

[plain raven]

but i haven't figured out an elegant way to get the whole complex of forms from this by some kind of bar construction

[marsh forge]

hm

[plain raven]

please don't take bar construction quite literally here

[marsh forge]

seems too big yeah?

Like a "bar" construction is usually huge

[plain raven]

yeah maybe

[marsh forge]

i guess so is the deRham complex

actually the deRham complex is like, uncountable right?

or wait not

[plain raven]

uh sure yeah it's made of real vector spaces which are infinite dimensional

they're function spaces

[marsh forge]

C^infinity functions are determined by a dense countable subset right

[plain raven]

hmm

[marsh forge]

so theres countably many?

[plain raven]

no because any constant real function counts

so there's at least |R| many

[marsh forge]

rite

yeah its just not bigger than |R| i think

[plain raven]

i think what that argument shows is it's not bigger than R, yeah.

It's not like |R|^|R|.

[marsh forge]

i remember discussing this qual problem with a friend and forgot the conclusion lol

[plain raven]

i will talk to you about d in a monoidal category another time but it's on my tablet in handwritten notes and i don't feel like looking it up rn

[marsh forge]

sure np

[plain raven]

what kind of derivations arise in the stuff you look at

or are those like, in a different corner of hom alg then you usually think about

[marsh forge]

Theres like

this square-zero extension business in homotopy coherent algebra

[plain raven]

hmm

[marsh forge]

I haven't spent any real time with it

but its important

I don't spend much time with it in the concrete hom alg sense

[gritty widget]

can someone help me with understanding komi-san?

what specific disorder does she have

[plain raven]

@ivory dragon

[marsh forge]

this is not topology

[ivory dragon]

sorry guys

[gritty widget]

exactly,.

[gritty widget]

false ban

😠

[gritty widget]

anyway if anyone knows help would be enjoyed 😃

[plain raven]

i think you misunderstood me

i didn't ping nami to get you banned

i pinged nami because i think he'd be most suitable for answering your question

[marsh forge]

fwiw I think Komi just has severe social anxiety, I don't think the diagnosis is anything more specific than that

[gritty widget]

I see

wait this is a stupid question

is {1,1} just {1}

nvm they are the same

unless multisets

[marsh forge]

This channel is for topology questions

[gritty widget]

and anime

[fading vale]

can you not post nonsense in this channel

and/or stuff that obviously doesnt belong here

[tough imp]

I have a big potatoe that I haven’t cut

[mossy ermine]

have you tried cutting it

[elder loom]

If you look at the channel description, it lists anime.

[plain raven]

Namington has already pointed this out in the screenshot above.

[gritty widget]

'anime'

checkmate

[plain raven]

We've now had this pointed out thrice

The explanation of the joke is starting to grow wearying

[coarse night]

[rugged rock]

Anyone know how i can show that $A = {(x,y,z) \in X | z = 0 }\subset X = {(x,y,z) \in \mathbb{R}^3|x^2 + y^2 + z^2 = 1}$ is not a retract of $X$? Hints are preferres over full answers 🙂

[gentle osprey]

Évariste Galois

[rugged rock]

Is the no-retraction theorem applicable here? I feel like $X \cong S^2$ and $A \cong S^1$.

[gentle osprey]

Évariste Galois

[empty grove]

A retraction induces a surjection of homotopy/homology groups

[empty grove]

yes

No retraction can be applied

[rugged rock]

But is $A \cong B^2$?

[gentle osprey]

Évariste Galois

[empty grove]

What is B^2

[rugged rock]

the unit disk

[empty grove]

No

A is S^1

[rugged rock]

Ok, and so how can the no-retraction theorem be applied?

On my book it is specifically from B^2 to S^1

[empty grove]

X is S^2, look at one hemisphere of it

[rugged rock]

Ok ill give it a try

thanks

Ok, so let me just make sure my argument is correct. I just showed that there is a continuous bijection from the upper hemisphere of X=S^2 to B^2. So A is not a retract of the upper hemisphere of X, by the no-retraction theorem; and so A is not a retract of X either.

[gritty widget]

Homeomorphism in fact

[rugged rock]

yup

[gritty widget]

Yes, that's the idea

[rugged rock]

But do I have to argue why A not being a retract of B^2 implies that A is not a retract of S^2?

if so, how?

[gritty widget]

You restrict the retraction to upper hemisphere

[rugged rock]

Intuitively, I would say that since B^2 is essentially a symmetrical half of X, which cant be retracted onto A, then neither can two copies of it.

[gritty widget]

If it exists

[rugged rock]

Ah ok, so it isnt trivial

[shy blaze]

does the bockstein homomorphism commute with the suspension isomorphism?

[pearl holly]

the Z/2 one does at least

[high haven]

Is the predix "co" in cohomology used to indicate that the functor is contravariant instead of covariant?

[gritty widget]

maybe

[high haven]

I have always wondered about the naming convention and the common denominator that I have found is the one above

Nice, I really like that naming convention

[pearl holly]

convention

[stable kite]

This is humoring me way more than it should, haha

[unreal stratus]

Is my argument okay? Suppose we can write $\mathbb R$ as above.
Then $\mathbb R$ is connected, so X and Y are.
$\mathbb R$ minus a point is disconnected, whilst X x Y minus a point (say $Z = X \times Y - {(a,b)} $) isn't: fix a basepoint $(x_0,y_0) \in Z$ and and suppose $f: Z \to Disc({0,1}) $is continuous. Then for all $(x,y) \in Z$ we have $f(x,y) = f(x,y_0) = f(x_0, y_0)$ because the restrictions of $f$ to ${x} \times Y \cong Y$ and $X \times {y_0} \cong X$ are constant (by connectedness), and so $f$ is constant; as $f$ was arbitrary $Z$ is connected

[gentle osprey]

potato

[unreal stratus]

(which leads to a contradiction lol)

[golden gust]

I'm not sure but I think in your chain of equalities f(x,y) = f(x,y_0) = f(x_0,y_0), you have to be careful that you don't pass over the point (a,b)

[empty grove]

Wait till you hear about coconuts

[unreal stratus]

Well (x,y) and (x,y0) are both in X x {y0} which is contained in X - {(a,b)}

since y0 isn't b

similarly for the other

[golden gust]

ah yeah

[unreal stratus]

And these aren't paths anyway so not rly 'passing over'

but ye

[golden gust]

pretty sure that works then

[unreal stratus]

noice

I suppose you can probably use path connectedness too but that'd wind up being almost the same proof

Just take a path from (x,y) to (x0,y0) avoiding (a,b)

(in fact the same path in mind with my argument)

Ok cheers

I wonder if there's any other way to argue this but connectedness was the intended one contextually lol

[empty grove]

You could say that X and Y in your notation are connected subspaces of ℝ, so are intervals, so ℝ is homeomorphic to the product of 2 intervals. And then argue that this isn't possible using connectedness

[unreal stratus]

Sure yeah noice

I think a slightly over the top way might be like

[empty grove]

Literally the same argument as yours except I just added that X and Y are intervals lmao

[unreal stratus]

Lol

No

[empty grove]

[marsh forge]

is there a proof by the universal property of a product

[unreal stratus]

I had smth else in mind

[marsh forge]

there should be

but i cant think of one

there should be some issue with the projection maps not being able to exist

@empty grove an exercise for u bc i need to do work hahaha

[empty grove]

I feel like that shouldn't be possible lol

Or at least would be very complicated

[golden gust]

fix $(a,b) \in X \times Y= \bR$. we must have $(X \times {b}) \cap ({a} \times Y) = (a,b)$. but $X \times {b}$ and ${a} \times Y$ are both closed intervals in $\bR$, so as $\bR$ is noncompact the only way they can intersect at a point is if one of them is a point, and one of them is $\bR$.

does this work?

[gentle osprey]

xdres

[golden gust]

forgive the bad notation where I identify X x Y with R, so that (a,b) is a point and not an interval

(they are intervals because they are connected subsets of R, and they are closed because X and Y must also be hausdorff)

[empty grove]

I didn't get why they must be closed

If they are compact then you're already in trouble because the product would be compact

I feel like some invariance of domain argument should prove that they are open

[golden gust]

if X x Y is hausdorff, does this imply X and Y are both hausdorff?

[empty grove]

Oh are they closed because they are the preimage of a singleton under projection

[empty grove]

Yes

[golden gust]

ok then yeah

[golden gust]

the only way they can intersect at a point is if one of them is a point
turns out this is not true, e.g. for (-\infty,0] intersect [0,1]

[gaunt linden]

It doesn't seem to make sense to say that "X×{b} and {a}×Y are both closed intervals in R" -- they're not even subsets of R.
Oh, I missed the context.

[gritty widget]

If X and Y are non-empty then yes, because they can identified as subsets of X x Y.

If not, then you can put X = empty space and Y = any non-Hausdorff space

[sour geode]

Is there anything where you can attach algebraic structure to a topology and talk about seperation axioms through it?

Similar to how you can talk about connected components with homology as an example

[marsh forge]

There might be, but I would wager that it would have to be really complicated

[marsh forge]

Suppose I have functors G:C--->C' and F:C--->D

they have all colimits and whatever

Does knowing that G is essentially surjective make computing the left Kan extension of F (along G) easier?

I honestly am not sure how to compute this regardless

[visual fulcrum]

given an embedding of compact manifold f:X->Y there is a wrong way map f! in K-theory taking K(TX) to K(TY). I'm trying to see how to get f!g!=(fg)!, in Atiyah's first paper on the index theorem it says it follows from transitivity of the Thom class, but when i write out the definitions it doesnt seem that simple... anyone know a good approach?

[mint sage]

how to do that? S3 is symmetric group

[plain raven]

Hatcher ch1 should be helpful here, I think

the section on van kampen's theorem and the subsection on applications to cell complexes

[sturdy notch]

Try writing down a presentation of the group S3 and recall how attaching 1 and 2-cells changes the fundamental group using van kampen

[zenith bay]

Is hatcher the best introductory text for algtop?

I know it's de facto the go-to

[gritty widget]

many people don't like it

it's more about geometry and intuition, I think
just not a standard way of teaching math

should be good for beginners

[pearl holly]

A lot of people here say that a good way to learn at is to do some Hatcher first and then some May or something

[zenith bay]

Mm fair enough

I did an algtop class about 5 years ago but I only really recall the vague intuitions for certain definitions

Guess I'll check them out

[marsh forge]

The standard takes on hatcher are not great imo

It is not "geometry and intuition" with maybe like 3 exceptions

It is entirely rigorous and a pretty good intro

[empty grove]

Hatcher first chapter is pretty good

Others depend on taste

[marsh forge]

I think May's concise is kind of outdated at this point

[empty grove]

I liked Rotman

Less content than Hatcher

[marsh forge]

It tried to be too modern before the modern setting was actually finalized

[empty grove]

Equally introductory

[pearl holly]

What other sources are there apart from may, Forman and hatcher?

[empty grove]

Rotman good

Rotman does acyclic models

Which is really cool

[pearl holly]

Maybe I should take a second round at AT in rotman then

[marsh forge]

there are like 100 books on intro AT

Massey has two of them for some reason

Spanier

Tammo

[pearl holly]

Oh right

[empty grove]

But then proceeds to not use it to prove eilenberg zilber

[marsh forge]

I think they are all kind of "eh"

[empty grove]

Is this Dieck

[pearl holly]

So what are acyclic models?

[marsh forge]

tom Dieck*

[empty grove]

I would not call that introductory

[marsh forge]

iirc it technically is self contained

[empty grove]

Hatcher lemma 3.1 but for functors from any category to the category of chain complexes instead of just chain complexes

[pearl holly]

Wait lemme look it up

[empty grove]

You can use it to prove construct isomorphisms of homology theories very quickly

[pearl holly]

oh okay I see

[empty grove]

Example is that on the category Top x Top, there are 2 functors, one that maps (X, Y) to S(X x Y) and one that maps it to S(X) tensor S(Y) (tensor at each index). Both of these functors are "free" and "acyclic" and they agree on the 0th homology so they are like free resolutions of the 0th homology in a way, so their homologies are isomorphic

S(X) is singular chain complex of X

[pearl holly]

right I see

[mystic reef]

So Im having a difficult time showing the following: Let $X=\mathbb{Z}$ be equipped with the topology $\mathcal{T}={U\subseteq\mathbb{Z}, |, \mathbb{Z}\backslash U$ finite or $U=\emptyset}$. Show that $(\mathbb{Z},\mathcal{T})$ is connected. Do I show its connected by showing that it is not disconnected, or how should I go about it?

[gentle osprey]

Ursus1234

[empty grove]

Suppose there is a disconnection, and show a contradiction

[gritty widget]

https://math.stackexchange.com/q/1914400
then what do you think about this question

[marsh forge]

"I don't agree that Hatcher is non-rigorous, but I will not argue with the rest (taste is taste). Δ
Δ
-complexes are indeed rarely used but are actually quite good pedagogical tools, since it allows one to think simplicially without having to actually find some (necessarily larger than you'd like) triangulation of your space."

i will quote user940583094

except i think "delta complexes are rarely used" is false dep. on what you mean

i think a lot of classes teach them

but they don't show up much in the literature

this isn't because they are not useful, but rather because all of the computations people do with them were already known by the time hatcher introduced them

The language on this MSE post is very annoying

[gritty widget]

But doesn't Hatcher have proofs that leave lots of details untouched?

[marsh forge]

People are conflating rigor (good) with spelling out all of the details for the reader, which is not the same thing

Hatcher does not give any proofs that are incorrect

he occaisonally expects the reader to fill in some details

this can always be accomplished

hence the proofs are by any reasonable standard rigorous

(In fact, there are essentially very few math books with unrigorous proofs at all. Some books have incorrect proofs, they usually have errata)

[gritty widget]

That'd annoy me a lot, there can be errors hidden under lack of details

[marsh forge]

You will not enjoy reading research papers then hahaha

[gritty widget]

I don't actually, at all

[empty grove]

Ye Hatcher is perfectly rigorous

[marsh forge]

Yeah I mean I think one has to like

get used to "sniff testing" proofs that omit certain details

[gritty widget]

I was presenting an article at Friday, it's only 3 pages but 2/6 proofs took me 2 hours already to explain in detail

[empty grove]

I don't like the details he decides to leave out tho

[marsh forge]

or be able to fill them in

[gritty widget]

I was sweating a lot

[marsh forge]

Well, I obviously don't know much about the presentation

but I would argue in general that talks should give even fewer details

unless the purpose of the talk is the details themselves, which does happen

[gritty widget]

There were some huge leaps in logic in the proofs, so I'd feel like I'm leaving everyone confused without those.
After all, they're not written in the article itself

[marsh forge]

it does seem like people have varying opinions on how annoying the specific details are. I only recall like one or two instances where I really felt pressed trying to finish the results

[marsh forge]

I think you might be conflating your experience with the material and your audiences

for sufficiently advanced material, following the details of a proof live is essentially impossible

Or at the very least requires a sort of unpleasant approach to the presentation

It's usually a lot more productive to cover broad strokes ideas, maybe do some examples, discuss broader applications, and overall motivate the material and convince the reader to look into it themselves

[empty grove]

My entire batch hated the book (it was the recommended one in our course) though I cannot say how much I influenced them with my opinions

Everyone used Rotman instead and failed to solve the exam problems that were almost straight out of Hatcher

[marsh forge]

lmao

hatcher has good exercises imo

[empty grove]

Ye that I agree on

[gritty widget]

Yeah, I guess I might have been too detail oriented for my own good

[marsh forge]

hatcher chapters n, n>0 have good exercises

[empty grove]

n < 5

ch5 has no problems

[marsh forge]

and hence they are all good

[empty grove]

Go back to #foundations

[plain raven]

👀

[marsh forge]

its been so long since ive read a book with exercises

i miss it tbh

[empty grove]

Never too late to read Rotman

[marsh forge]

Higher Algebra Exercises when

lol

i dont think id get much out of rotman

[empty grove]

I know

But it is never too late regardless

[marsh forge]

that is true i guess

I'm gonna write an intro AT book

that does not mention spaces at all

only kan complexes

[empty grove]

Yes pls

Goerss Jardine

[marsh forge]

no

[empty grove]

So unreadable

[marsh forge]

Goerss Jardine works with model categories

my book will be purely based on higher category theory

[empty grove]

Intro AT book doing simplicial homotopy theory in infinity categories?

[marsh forge]

Pretty sure this is like

impossible lol

[plain raven]

chapter 1

[marsh forge]

theres no way you can do all the relevant computations without models

[plain raven]

coends

[marsh forge]

chapter 1: homotopy colimits

We define a colimit to be a colimit in the infintiy categorical sense. For the classical definition, we will use the term "bad colimit".

[plain raven]

Hahahaha

[empty grove]

I have decided that it is finally time to google what a homotopy colimit is

[plain raven]

I understand how to construct them but idk what they are

[empty grove]

Nice homotopy of diagrams

I shall one day read Hovey past page 15

That day shall not come before August

[gritty widget]

Hi, if $x_0, x_1 \in X$ (a path connected space), $ \pi_{1}(X, x_0) $ is abelian, and $ \alpha $ a path from $ x_0 $ to $ x_1 $ does that mean that $ \hat{\alpha}([f]) = [\overline{\alpha}] * [f] * [\alpha] = [\overline{\alpha}] * [\alpha] * [f] = [f]$ (sounds too good to be true)

(f is a loop of x_0)

[gentle osprey]

kaz

[gritty widget]

[\alpha], [\bar{\alpha}] aren't elements of any fundamental group, so i don't know about that. also, the left-hand side is in \pi_1(X, x_1) and the right-hand side is in \pi_1(X, x_0), so this can't work

the proper thing to show is: if $X$ is path connected, $x_0,x_1\in X$, and $\pi_1(X, x_0)$ is abelian, then $\hat{\alpha} = \hat{\beta}$ for any two paths $\alpha,\beta$ from $x_0$ to $x_1$

[gentle osprey]

TTerra

[gritty widget]

Yeah, that's what I was trying to prove and this came on my mind, but you're right about [\alpha], [\bar{\alpha}] thanks!

[zenith bay]

Ok i'm having a dumb here, what exactly is an induced homomorphism

As in the statement of Van Kampen's theorem

Wait hang on

[gritty widget]

the maps on fundamental groups induced by the inclusions?

van kampen has a lot of maps going on lol

[zenith bay]

Yeah yeah

haha

The inclusion maps are homomorphisms right?

[gritty widget]

the inclusion maps are continuous maps that induce homomorphisms

[zenith bay]

Oooooh I see where I'm getting confused

The inclusion maps between spaces induce homomorphisms between the fundamental groups of said spaces?

[gritty widget]

right

this is a more general case of: if $f\colon X \to Y$ is a continuous map of topological spaces and $f(x) = y$, then you get a homomorphism $f_\colon \pi_1(X, x) \to \pi_1(Y, y)$ given by $f_([\gamma]) = [f \circ \gamma]$

[gentle osprey]

TTerra

[gritty widget]

something something \pi_1 is a functor on the category of based spaces

[zenith bay]

Aaaaaaaah

Perfect

Yeah I guess I was just confused what the induced homomorphism looked like

[γ] is some equivalence class of loops, correct?

as opposed to just paths

[gritty widget]

equivalence class of loops at x

yeah

[zenith bay]

been too long since I've done algebra. I think I just forgot paths are functions and therefore composable with other functions

lmao

[gritty widget]

lol

you might want to prove that this f_* is well-defined

[zenith bay]

well defined in the sense that [a] = [b] implies f_([a]) = f_([b]) i assuume?

[gritty widget]

exactly

[zenith bay]

alright cool

[gritty widget]

(just if you haven't seen it before)

[zenith bay]

If I have it will have been years ago, so a good exercise

Hmm I mean this seems trivial but perhaps i'm being naive

a = b implies f∘a = f∘b right?

or have I truly forgotten my whole undergraduate lmao

[gritty widget]

that's certainly true

but it's not entirely trivial

remember what the equivalence relation is

[\gamma] = [\gamma'] if \gamma and \gamma' satisfy...

[zenith bay]

yikes

I must've skipped over that paragraph or something

However I'd assume the composition of gamma with the reversal of gamma' should be the trivial loop

Seems like that should be necessary

Sufficient though? I'm not sure

[gritty widget]

the equivalence relation is path homotopy (with the endpoints fixed)

[zenith bay]

Ah okay

[gritty widget]

$[\gamma] = [\gamma']$ if and only if there is a continuous map $H\colon [0, 1]^2 \to X$ with $H(s, 0) = \gamma(s)$, $H(s, 1) = \gamma'(s)$, $H(0, t) = x = H(1, t)$

[gentle osprey]

TTerra

[gritty widget]

the statements meaning "for all s" or "for all t" respectively

[zenith bay]

Yep that makes sense

So well definedness requires proving the existence of such a map H between f comp a and f comp b

Where a = b

[gritty widget]

not where a = b, where such a map exists for a and b

[zenith bay]

OHHH okay

Right of course

I was trying to prove the much weaker a = b implies f*([a]) = f*([b]) lmao

[gritty widget]

https://cdn.discordapp.com/emojis/889939602507460719.gif?size=96&quality=lossless

[zenith bay]

as I said, it's been a long while

Does it not suffice to compose $H$ with $f \circ \gamma$?

[gentle osprey]

josh the 🐀

[gritty widget]

how are you going to compose these two?

you're close

[zenith bay]

Mm

No that's backward

$f \circ H$ gives a map $[0, 1]^2 \rightarrow Y$

[gentle osprey]

josh the 🐀

[gritty widget]

👀

[zenith bay]

Does it not?

hmm

[gritty widget]

it does

does that give you what you want?

(it does i just want you to check it)

[zenith bay]

Well, denoting H' = f ∘ H

H'(s,0) = f(gamma(s)) = f ∘ gamma(s)

Similarly for H'(s,1)

H'(1,t) = f(x) = y

H'(0,t) = f(x) = y also

[gritty widget]

ratjam

[zenith bay]

And we get continuity from the continuity of f and H

[gritty widget]

perfect

[zenith bay]

based