#point-set-topology
1 messages · Page 55 of 1
is taking the cup product with a 0-cocycle for a path-connected space just the identity?
If U is relatively compact, find disjoint open sets respectively containing cl(U)-U and x
Not so shrimple as previously thought.
Unshrimplous indeed
ok nvm i figured it out
collapsing 0x{I} basically makes the collapsing points equivalent and we end up with circles who only intersect at 0 and some specific 1/n
whicn is just homeomorphic to a circle of radius 1/2n with center (1/2n,0)
so now this reduced suspension is homeomorphic to the hawaiian rings and therefore it has an uncountable π_1
Even on a path connected space, there’s more than one class of zero cycle. Can you name them all?
i mean every 0-chain is a 0-cycle
Have you solved it
Yeah, there are a lot of zero chains, which is why I asked you to name the classes, not the chains. I meant the homology classes
You aren’t supposed to take cup product with homology cycles, only cohomology. So you really should be naming zero all the H^0, not H_0. Luckily they’re dual, so it’s not harder. But it’s technically different. In particular, not every zero cochain is a closed
so by UCT we can identify H^0 with the dual of H_0
H_0 is generated by a single element
so the dual of that generator generates H^0
i think geometrically the intuition is that the generators of H^0 are the functions that map a particular PC of X to 1 and the others to 0
so for a pathconnected space, the generator of H^0 is simply the function that assigns a value of 1 to every point
and that function acts as the identity in the cup product
so my original claim is true?
^
(here we're assuming the ring over which we're taking cohomology is unital)
- ||π1(∑X) → ∏Z||
- ||π1(SX) → ⊕Z||
- ||Adding a cone is quotienting||
yeah the first 2 points are obvious
well the 2nd point
Here's a dumb question lol, can a noncontractible manifold have contractible one point compactjfication
One meme solution would be to use poincare.duality ig
Is that a question?
struggling with showing that PC^n is a 2n-manifold
i showed that P^n is an n-manifold by showing its homeomorphic to the n-sphere
struggling with using that here
but maybe i could?
you can pretty directly put a manifold structure on it with very concrete coordinates
you can write a point of CP^n as [a_0 : ... : a_n] with the a_i in C, yes? think about the subset given by a_0 =/= 0 (recall that this means you can scale the whole thing by 1/a_0, so that your point is [1 : a1 : ... : an])
can you get a homeomorphism from this to R^2n (or equivalently, C^n)?
P^n is not homeomorphic to the n-sphere for n>1
fuck
wait why not?
whats the issue with just sending a ray to the point in it with magnitude 1
there are 2 of them
oh fuck lmao
totally forgot about that
well either way each point should be locally homeomorphic to a neighborhood on the n-sphere which is in turn locally euclidian so it should still be fine?
or is that not the case
if P^n were homeomorphic for S^n we probably wouldn't really care about the former lol
Yes
if π : S^n -> RP^n is the canonical projection, you can find a cover of RP^n by opens U such that each π^-1(U) is taken homeomorphically onto U by π
indeed this is a covering map which is nice
no you don't
hatcher just states that the reduced suspension is the hawaiian earings
it's not actually a part of the question

how're you using SvK tho?
svk is for wedge sum
I don't see how the suspension of X is basically infinite wedge of circles
it's not an infinite wedge but it basically is
but "basically is" can mean a lot of things
start with a circle S^1 intersecting 0 and 1 and the converging points
join another S^1 intersecting 0 and 1/2 and have the paths from the converging points including 0 be in an equivalence relation with each other
then by svk its really easy to see that its Z*Z
we repeat this for all n
what's the difference between the unit disc and the unit circle in R2?
is the unit circle centered at the origin with diameter 1 or something
cuz like armstrong defines the unit disc as x^2 + y^2 \leq 1
but like he considers the unit circle differently
circle is just the boundary, x^2 + y^2 = 1.
Disk is the entire... disk, x^2 + y^2 <= 1
more generally this is the difference between a ball and a sphere
well for ball, depending on context, you need to be careful whether it's open or closed. But I think I've mostly (pretty much only?) seen disk for closed ball
nuuuuu
you can use induction

by this same logic the hawiian earings have a countable fundamental group
which we know is false
no it isnt the same
the construction we made has two points of where the circles overlap
I don't exactly understand your construction tbh
but SX isn't homeo wedge of infinite circles
not even homotopy equivalent
and you can't use induction
nope
i just imagined some point floating above the real line
and below
and then lines starting at the upper point, going through some 1/n and then the lower point
the problem is that open neighborhoods of 0 will touch infinitely many circles
what
whereas in the wedge sum you don't necessarily have that
damn
this just makes me sad
I really did just write whereas as where's as
dw it took me a bit as well
it's not super trivial
i thought that was basically the suspension
Why is the interval [0, 1/2) open in [0, 1)? how would we define the neighborhood about 0?
so [0, 1) is taken in the subspace topology, yes?
(-1/2, 1/2) is open in R, so [0, 1) cap (-1/2, 1/2) is open by the definition of the subspace topology, and the intersection is [0, 1/2)
yea sorry
oh ok thx
what exactly do they mean by "we can rotate any point of S^n into any other?"
i'm picturing n = 2 rn in my head
or right in front o fme
also what's the relationship between a homeomorphism and isomorphism? R^3 whose third coordinate is all 0 is clearly isomorphic to R^2 and here they're also homeomorphic if i'm interpreting it correctly
i feel like the relationship between the field and topological structures on R are tenuous
like |a - b| requires the field structure sure but it also requires an order to define absolute value
really just rotating
it doesn't matter which point we remove because we can rotate the sphere s.t. the removed point is at the top
and that transformation is a homeomorphism
A homeomorphism is an isomorphism in the category of topological spaces and continuous functions 🙃
obv they were talking about field isomorphisms (or maybe linear isomorphisms)
i had a feeling you'd type this
I mean this feels like the one context it's actually appropriate lmao
yeah but probably more confusing than just
keep isomorphism reserved for algebraic objects for now and just use homeo for topological spaces
Yes use homeo when you mean topology and iso for algebra
Because ofc not every homeomorphism is an isomorphism of, say, rings or vector spaces when we have both a topological structure and a VS structure
Hmm okay thanks
In this case yes ofc they are also isomorphic as vector spaces with the obvious inclusion map, but this generally does not hold
Hi, I had a question on this lemma in Bredon's "Topology and Geometry"
In the union which gives f^(-1)(-inf, α) wouldn't we also need the complement of U_1, if α > 1 ?
darq are you sure i cant do this
It's more of a basic question rather than a real topological one though I guess
What
That's not ∑X if that's what you mean
What do you mean by …
Induce weak topology on SX?
Cuz it's not that
Tho close
||That does induce the fundamental group||
i was trying to construct something homotopic to SX
Weak topology doesn't look homotopic
A point and a plane are homotopic
Any feedback?
||imma guess they're weakly homotopic||
I dunno what that means 
@novel ember do you need a hint?
meh I'll just leave it here coz I gtg
||try using SvK on the space itself, don't use induction or w/e, just vanilla SvK applied directly on SX||
Yes
Which doesn't allow us to conclude f^(-1)(-inf, α) is open for any α and hence deduce continuity of f, right?
This lemma is used to prove Uryshon's lemma right after
on god?
Which one?
You just need to do (-∞, α>1) separately
I see, thank you! Basically they forgot to mention the case 1<alpha was trivial
mathematicians going to an RV campsite, call that van Kampen
Nice
How to show these free product is isomorphic as groups?
$\mathbb{Z}\mathbb{Z}=\mathbb{Z}\mathbb{Z}*\mathbb{Z}$
kite
it is
Eh it probably is topology lol
this is something usually proven with topological machinery
It is not something usually proven
isn't Z * ... * Z n times just free group on n generators
The negation is proven using topological machinery though
(is this true?)
Must be right like you just chuck generators in with one another
Yeah potat don’t over think it
Lol phew
? am i tweaking
On topology free product = wedge
oh fr?
didn't know
Basically DarQ
also, don't go #groups-rings-fields anyways
A lot of statements about free groups can be proven using topology
oh brother i can't read
shits going down, there 
i instantly translated it into the subgroup statement
Essentially there is a nice correspondence between free groups and graphs
which can be proven using topology
The fundamental group of a graph is free, a covering space of a graph is a graph, and covering spaces correspond to subgroups of the fundamental group
are the kinda key ideas
that's next section in hatcher!
So like, if this statement were true, we could just find two spaces X and Y which are homotopy equivalent and have these as their fundamental groups (for example)
Unfortunately, it isn't true though
Lol

to clarify, the statement that the free group on 3 generators is iso to a subgroup of the free group on 2 generators is true
Yes
In fact then any two graphs with that fundamental group would be homotopy equivalent
Given some cw complex condition
and that's a classic
so my head made it into that, my bad
Actually maybe you don’t even need a cw complex condition
you see, :anyapat: would work so well here

well graphs are cw complexes right by definition
i'm not adding anime emotes
Okay you do because you could add in a long line or something
summary: graph theory is just a subset of algebraic topology
toki 
hello timo and shuri! 👋
is it true that given a partition A of a topological space X, if each set in the partition is endowed with the subspace topology, then the disjoint union across the partition is homeomorphic to X?
seems valid but im not sure
What is your definition of a partition?
a collection of nonempty disjoint sets whose union is X? not necessarily finite
right that would generate the discrete topology
Yes
well, shit
But here's a more general thing
A disjoint union of 2 of more spaces is disconnected
But there are connected spaces
So it'll never work for them
i havent done connectedness yet
Ah ok
Well basically think of it like
If I cut R into (-infty, 0] and (0,infty) and then disjoint union them
I have cut it into two pieces and broken many properties
right
For example, there is now a continuous function which is 0 for x <= 0 and 1 for x > 0
right
is the topology the whole (partition -> subspace -> disjoint union) process generates always finer than the original topology?
Yes, try to think why
actually a good exercise in playing with the various constructions
unless you want hints etc
say U is open in X. then U cap S is open in all subspaces S of X by definition. by the definition of the disjoint union topology U is open in the union.
that hold?
awesome yes
okay i was trying to use this sort of thing in a problem that i do now need a hint on
the problem is showing that if A is the set of all integers, collapsing A to a point (in R) creates a space homeomorphic to a countable wedge sum of circles
Ah sure nice
intuitively makes plenty of sense
the problem recommends expressing both spaces as a quotient of the disjoint union
the disjoint union of all [a, a+1) for integer a i assume
Not quite
So you basically want to cut R up and glue it back together properly lol
Note that if you do the [a,a+1) thing you can't really quotient out anything to get back R
At least, it's not clear how you'd do it nicely
Can I safely ignore anything covered about posets/partial orders/orders from my class in undergrad topology
Oh ok idk it didn’t seem motivated at all when I learned it but I’ll keep it in mind
Consider that any collection of subsets is partially ordered btw (so any power set, any topology, any ...), along with the real or rational numbers, the positive integers are partially ordered by divisibility even(!).... Once you get familiar with them you will recognize them all over the place
Do you need to study it on it's own? Probably not. But it's very useful to be familiar with at least the basics
so do you just take [a, a+1] for all integers a and then collapse all equal integers to a point
yes
if you collapse a collection of sets to points and then collapse all of those points to a point thats the same as collapsing the union across the collection to that point right
in which case collapsing some of the integers to a point to get to R and then collapsing all of the integers to a point is the same as just collapsing all of the integers to a point to start with
you can get countably many circles from the disjoint union of all [a, a + 1] for integer a by just collapsing the ends of each interval to a point
which are integers
so again collapsing all of the integers to a point is equivalent
so both spaces can be created by taking the disjoint union of all [a, a + 1] and collapsing all integers to a point, so they are homeomorphic
does this hold?
sounds real
i'm currently working through munkres topology but i feel like i don't REALLY understand what the definitons/theorems really really mean, is this a normal feeling or should i change something up in my learning style?
Munkres is pretty intuition and motivation bereft, so it depends on what you mean
What does "really really mean" mean
Do you know analysis / metric spaces
If so you can refer to this #point-set-topology message
An equivalent formulation of topology that may or may not be more intuitive https://en.wikipedia.org/wiki/Kuratowski_closure_axioms
by that i mean why things are defined the way they are, and that it would be impossible for me to come up with them on my own
I think that’s normal, topology comes at the end of hundreds of years of trying to formalize notions about space. The axioms we have now are very beautiful and flexible enough to encapsulate a huge range of examples
But as a trade off they’re probably pretty far from your (mostly manifolds/metric spaces based) intuition
i see, so then the confusion i have is normal? should i expect to get the intuition i'm looking for after a bit of work?
Yes you’ll get used to it
I also found the definitions of a topology very unintuitive at first, but appreciated them after studying for a few months
i see, thank you, this helped me a lot 
like what
is there something specific that bothers you
intuitively, a topological space is a set with a notion of closeness defined. Formally, what does the statement "x is closer to y than z is" mean then?
there's not really a way to make sense of it unless there exists a neighborhood of y which contains x but not z and not the other way around
however if you fix an ordered neighborhood basis of every point there is a way to make sense of closeness with respect to this choice
but intuitively it means "more neighborhoods of y contain x than contain z" although this is typically impossible to quantify
I'm following wikipedia on this intuition
yes
it's good intuition but there is not a way to quantify it
it's maybe better to say that a topology gives a quantification of which points are "very close" to a subset since it gives you a notion of closure of a subset
Can I say in a discrete topology, points are infinitely far away from each other. While in a trivial one, points are infinitely close to each other?
that's not how I think of it intuitively tbh
in my mind topology defines how points are "glued" to each other
You could say that if you want
nice, so the set is like vertices in a graph, and the topology is like the edges!
so like, for example, 0 is "glued" to the set {1/n for natural n} and the rational numbers are glued to all real numbers
and all the points in some interval are glued to each other but any two singletons aren't
I would say
I don't see how, if you take a point out of the real line you can find an open set that matches it any other point
it makes sense to talk about being "sufficiently close" to a specific point
does that mean all points are connected?
but it doesn't make sense to ask e.g. which of two points is closer to a specific point
or when two arbitrary points are sufficiently close to each other
when do you say x is sufficiently close to y?
some property P(x) holding for x sufficiently close to y would mean that there is a neighbourhood N of y such that P(x) is true for all x in N
and you should really note the asymmetry in this
ah, I see how this makes the definition for "isolated point" possible. Because a point is an isolated of a set when all sufficiently close points do not belong to the set
yeah (other than the point itself of course)
on top of this, you can think of all the various types of topological spaces like T_1, T_2, hausdorff, normal, regular, etc all add more requirements to the topology so that points being close (ie in neighborhoods of one another) becomes more and more meaningful. Like in general you can have topologies like the Zariski topology where all open sets are dense (aka very large) so being "close" in that topology isn't meaningful, but at the other end of the spectrum with metric topologies, being close actually means something (like within some e > 0 of one another).
proof no one remembers wtf the T_i axioms are:
I remember hausdorff, then I google the other ones if I need to lol
its always like "I can separate points" or "I can separate points from closed sets" or "I can separate closed sets"
lol just being able to draw bubbles around things more and more
until you can draw any kind of open set you want until you arrive back at metric spaces
which is like "hey that's where we started!" and you've recovered what "closeness" means
mmmm
lets see
what is completely regular (lol I have no clue but I remember the name)?
for 18.b shouldn't it be an isomorphism instead of just a homomorphism?
it just needs to be surjective (thats why they said onto)
at least to conclude uncountablility
my questions was about whether the two groups are isomorphic
||coz like, C is just the quotient of CSX which is just the quotient of I\times SX. And I think we can use the image of [0, 1)\times SX and (0, 1] \times SX to apply van kampen||
iirc it's when disjoint points and closed sets can be separated a continuous function --> [0,1]: x in X, F subset X with x not in F, then there exists phi: X --> [0,1] with phi(x)=0 and phi(F) = {1}

||[0, 1) \times SX is mapped to a subset that's contractible and (0, 1] \times SX is mapped to a subset that deformation retracts to \Sigma X||
oh wait, right
nice, lol better memory than me!
\pi_1(\Sigma X) isn't the infinite product of Z lel
I suppose I must be on the right track tho 
oh wait I'm so dumb lmao
DarQ
||and we already have a surjection from \pi_1(\Sigma X) to the direct product of infinitely man \bZ||

I was trying to construct the homomorphism directly 
@novel ember dont' click on the spoilers
you've been warned

how would i go about showing that the closure of a subgroup of a topological group is also a subgroup
well, you wanna show that multiplying two elements from the closure stays in the closure (inverse is similar). Note that if $f(x,y)=xy$ is the multiplication map (which is continuous), then letting $H$ be the subgroup and noting that $\bar H$ is closed, you have $$f(\bar H\times\bar H)=f(\overline{H\times H})\subseteq \overline{f(H\times H)}\subseteq \bar H$$
Nekory
yoiu can try the inverse similarly
(in fact taking f(x,y)=xy^{-1} serves the job quite simply)
could you make that last subseteq into an equality or does that not hold
since the image of HxH under f should be H
right?
what is H^*(A_n, Z) for n>=4?
|| (n!/2) copies of Z concentrated in degree 0 because it is a discrete space||
take x,y in the closure of H, and assume xy is outside of the closure of H. Choose a neighborhood U around xy which is disjoint from cl(H), and then choose neigborhoods U_x and U_y of x and y respectively such that U_xU_y is contained in U. But as x,y are in cl(H), those neighborhoods must intersect H, so use that to derive a contradiction
nets 
Unironically how I'd do it too lol
this is my default way of thinking about point set topology
I had an ultrafilters phase but doing some func analysis made me realise nets' power lol
no idea what nets or ultrafilter
they're like sequences but spooky
Limit point = limit of some sequence for 1st countable spaces
For general spaces you use nets
You’re such a troll
[0,1)U(1,2] would be a compact non-hausdorff connected topological space right?
uh it's the opposite of what you just said
lmao
uff
it's hausdorff, non-compact, and not connected
any subspace of a Hausdorff space is again Hausdorff. Necessarily, any compact subset of a Hausdorff space is closed. And in R, the connected sets are precisely the intervals (open, half open, or closed)
jeez
R is Hausdorff
how do i construct a compact non hausdorff connected space?
something like
my usual approach would be quotients
the easiest example i know is infinite but yeah
That is a good idea ^
A quotient of a compact (resp. connected) space is again compact (resp. connected)
2^-n : ncN
Such a space will always be Hausdorff
Because it is a subspace of a metric space
and metric spaces are Hausdorff
(In fact that is again the opposite of what you want oop)
||along with this, finite sets are always compact, so that might make it a bit easier||
are you familiar with the notion of gluing spaces?
oh wait
nah nvm
i mean i can imagine what it is
but tbh skipped topology and went straight into diff geo

its great and i get almost everything
but i gotta get some basic topology down
before contuinuing
algebraic topology has been the most difficult thing to get a grasp on for me personally
namely stuff im asking about rn
even harder than algebraic geometry
anyways for this, think of trivial examples
you want it non hausdorff
do you want it hausdorff or not
a hausdorff space has disjoint neighbourhoods for every 2 points
yes i know
so i gotta pluck something out
his question was what you are trying to construct
You asked about non-hausdorff examples
to be compact one does not have to take a closed interval in R
plucking something out of [0,1] can it even work?
no
or do i have to construct the entire set a diff way?
different way
If you just get rid of points in [0,1] and then give it the subspace topology that will never work
What about indiscrete on 2 points
Because subspaces of Hausdorff spaces are always Hausdorff. And you'd need to remove enough to make it a closed subset anyways
Done
how do i define that?
only open sets are empty and the entire space
it's kinda a "trivial" topology since it's the smallest you can have on a set
so i gotta construct it out of empty open sets?
I will just say to go back to the fundamentals
What are you learning topology from
You might wanna go to basics a bit if you’re unfamiliar with indiscrete, or suggesting using empty open sets
a diff geo book
it has it in the preliminaries and im trying to gain an intuition of all the topology classifications used
i dont think ill deal much with non hauserdoff spaces
it might be worth looking at a dedicated topology textbook for a few basics
but thought it would be good to see examples of non cause i had trouble imagining them
nothing needs to be fancy, but this would be like first section or two
are we at the point, where we can give straight answers
done already
i mean i get what u have to do intuitively
i havent checked anything but like a line with double origin should work
or the zariski topology obviously on something
maybe
is that disconnected? Seems like it should be intuitively but I can't think of what the disconnection would be
I believe you should look at a dedicated topology text for topology knowledge
also not compact
wait i though we needed a connected space
but I guess intersect [-1, 1] would work for compact
cant we just glue finite lines
Compact non-Hausdorff connected
okay yeah i thought it would be connected
anyways yeah indiscrete topology is the trivial example for this, on any set with at least two points
(since ya know, 1 point is trivially hausdorff)
thats really clever, didn't know u could define spaces like that, thanks
oh you can make all sorts of cursed spaces
yeah , for the record im learning diff geo for compsci purposes so not sure how this example will be of use but defo got a good intuition ou of it
so just to get it straight,
X=(⋃a>0(−a,0)∪{p}∪(0,a))∪(⋃a>0(−a,0)∪{q}∪(0,a))
this is a valid way to define a set?
i can replace the 0 with an arbitrary variable?
or well symbol
can you please write that in latex
All maps S^1 --> X homotopic implies X is simply connected is not as easy as it looks, right? After showing path connected, we would need to use that [f] = [g] in the [S^1, X] iff [f] and [g] are conjugates in the fundamental group, right?
ah, I see. So this is to make the "line with two origins" with p and q your origins. Yeah, since intervals are sets (subsets of R), as are {p} and {q}, this always will be a set. Defining the topology on it this way might be a slight(?) hassle compared to just "it's a quotient" but oh well
This won't be compact tho. You'd need to stop at some finite a
(in short: since R isn't compact itself)
yeah i got this from a paper
like i noted below
a can be some number and you just close it with [ on the edges
that would be compact correct?
so same thing written there but [-1,0) and (0,1]
not if you take the union over all a > 0. For simplicity, maybe we could just write it as [-1, 0) U {p, q} U (0, 1]
this should be compact
yeah basically what im saying
except im not sure u can put P and Q together like that
yeah , the alternative another guy wrote here also makes sense but the notation {1,R} and {0,R} with an equivalence relation seems a bit hasslesom to me
altho it gets to the point more clearly
No I think it should be as simple as it looks
all maps homotopic includes the constant ones
gotcha, will keep that in mind
Yes but homotopic means different things in [S^1, X] and the fundamental group
As homotopies in [S^1, X] are allowed to move the base points in between
oh I ssee what you mean. Fixing {1}
Yeah so even though the homotopy takes a loop at x_0 to another loop at x_0, it might move the base point in between
still tho, once you've gotten parts a, b, and c that should not be an issue
How so?
If all maps are homotopic, then clearly a) holds. So c) holds as well
by equivalence
Hence (after showing path connected) we have X simply connected
lmao
how tf would you even go about rigorously proving that h(x) is clearly one-to-one and onto
i guess visually speaking when n = 2 it's kind of easy to see
iirc parts (a), (b), and (c) are equivalent for all (higher) homotopy groups?
for this, I mean
Euclidean geometry
Intersection of a point and plane
faye's answer is morally correct but you can also just write down what the map is and its inverse
as in it's the "right" way to think about it
Morally correct 😎
also wtf is a half line
ray
ok
in order to check that U is open I have to find the formula for h(x) huh
bro this is algebraic topology not algebraic geometry
😭

Formulas are important stop complaining
you went to high school right
yeah but surely they expect all college students to forget all high school math

i thought this was being used in an idiomatic sense but now i realize it's literal
would this be a connected discrete space?
(2^1/n : neN)
2^(1/n) *
my apologies its
Discrete spaces are never connected
ic
Unless it's just a single point
so what i wrote would be an example of a discrete disconnected space that is compact?
{(2^(1/n) ): neN}
or i guess i could just do { U(n) : neN}
Not compact
nono my apologies
Exercise: show that a discrete space is compact iff it is finite
i emant just discrete space not compact too
in my example
for a compact one id have to think about
seems to me like u cant really define such a set within R
again, it would probably be more helpful to just read the first section or two (or three or ...) of a topology textbook for these. I'm not sure you fully understand these concepts
@ebon galleon Can you verify my solution for the problem? I don't think I've confused homotopies and basepoint preserving homotopies, but I thought I'd have someone else have a look.
Yeah those look good to me 
danke
i think my mistaken intuition is thinking of manifolds as embedded in an R^n space instead of looking at them as topological spaces on their own
taking that out, everything in the book seems to make a lot more sense now
Well the whitney embedding theorems exist
Every smooth manifold can be embedded into some R^n
yeah but this chapter is talking about topological groups
oh yeah those are generally not embedded into R lol
yeah there was my confusion
i was thinking of topological groups as manifolds by default cause everything before was about lie groups and manifolds
A simple counter example is to give a group the indiscrete or discrete topology lol
A nontrivial counterexample is the p-adic integers/numbers
it does touch on discrete topology once when defining G/H of and open subgroup H of G
but yeah like i said everything makes a bit more sense now
also ryx if i could ask you, are you familliar with topological data analysis?
not really, but I have heard that it's quite a trendy topic nowadays
lots and lots of research done there
planning my work to be centered around that + geometric deep learning
for malware analysis
so i guess i gotta get a topology book after this
to be fair this diff geo book is the most comprehensive thing ive read in a while
650 pages of beauty and full of interesting results and associations
Yeah I would imagine generally the topologies seen in TDA aren't as nice
What book are you reading
In fact, I know several people are even looking at abstractions of "topology" for it
But in particular, it should be on finite sets since finite data points
Grothendieck topologies?
nonono those are categorical lmao
I mean like cech closure spaces, pseudotopological spaces, ...
Ah lmao
tho you could then consider sheaves on them 
Jean Galliers Diff Geo
hes both a mathematician and computer scientist
dont think its very popular tho
Always the correct answer
Recently my enlightenment has been, working with locally ringed spaces uniformizes how to think about a lot of stuff
Suddenly differential geometry is just like algebraic geometry
This is the enlightenment I am looking forward to having this year. Make everything sheaves
You taking algebraic geometry next semester?
two semesters from now. But i'll be reading a bit from Mac Lane's "Sheaves in Geometry and Logic" and also reading Hartshorne over the next year
The most important stuff in hartshorne is in the exercises btw
A gentler introduction to algebraic geometry would be wedhorn imo
I'll keep that in mind, thanks. I also have Vakil if I don't like Hartshorne
I just need to finish up AM first lol
Whats AM?
Ah right
what’s the relation?
basically models that do the unpacking code for you
it's very cool
just how you would think about next-gen soc or any next-gen firewall etc
so cool how once u put math in something it instantly becomes cool
all malware are programs that can be represented as data in different ways
and since they are not random there is an underlying structure
capturing these structures in an elegant mathematical way through different techniques is the goal
in order to improve detection
also would love to do deep model interpretation and feature extraction work at some point in the future
so knowing higher maths is basically essential if u actually wanna make research advancements in that direction
i feel like more and more AI work is moving into higher maths territory
research wise at least
What’s the topological space there
i mean, ur asking me to tell you what im gonna find from my research before i even do it xd
atm im just learning as math as possible cause my grad studies start in nov
Oh I forgot TDA was a thing
I met someone who is a big guy in TDA and he told me it's 80% done and the problem is that it's not interpretable well. Sometimes I feel like people who do tda don't actually engage with the problem at hand because if they did they would be doing analysis. Signals are functions, it feels meaningless to talk about topological properties of a signal.
While reading this, I realize I don't understand what a topological subspace is. For R with the euclidean topology, if we consider the unit interval (0,1) as a subspace, what is the topology there?
I found the wiki for subspace topology, nvm
let S be a subspace of T, U contained in S, is cl_S (U) just cl_T (U) intersection S ?
Yes
Hi! Does anyone know how to prove that the 2-fold branched cover of CP² ramified along a degree 6 curve is Spin? I know this is used together with Furuta's 10/8 to obtain bounds on the genus, but Idk about this very specific fact. That is; how to show that its second Stiefel--Whitney class vanishes?
Nvm, found what I was looking for!
can I get a hint for calculating the fundamental group of the real line with two origins?
wait is it trivial?
It's the integers
Iirc there's a weak homotopy equivalence between it and the circle
i have a slight suspicion the fundamental group of the real line with double origin is abelian
aka it should be enough to calculate the first homology group
or maybe just use van kampen
i see no reason its fundamental group should be any different from that of
-o-
circle with 2 ends.
ahhh wait no no no
it is different
There are loops which "go around" an infinite number of times
@broken nacelle ^
it definitely isnt trivial. The loop starting at 1, goes to -1 (passing through origin 1), goes back to 1 (passing through origin 2) isn't homotopic to 0
wait what do u think the group is
Z
hmm
oh wait are there
ahhh maybe not
i feel like u can, but id have to write explicitly to be sure
yh im fairly sure u can
ok, but
see my proof
this is gonna blow your mind
start with a circle with two origins
and take your basepoint outside the origin
this space got \bZ * \bZ as its fundamental group
just lemme explain
u mean two 1's?
yea
I dont believe you, but continue
it doesn't matter what the origin is
ok so
we're gonna use SvK
choose the two open sets to be the whole space minus one of the origins
which is just a circle
and the intersection of those is simply connected
so you're not quotiening by anything
hence \bZ * \bZ
again, u have an "infinite loop" i believe but let me see...
can u paste me a formal statement of svk
yea
i want to check all conditions hold
here are the inclusion maps defined
but you don't actually need this since the quotient part is irrelevant
since pi_1(A_1 \cap A_2) is trivial so gamma is just gonna be homotopic to the constant loop
I need some time to think about this supposed infinite loop.
I don't think there is an infinite loop
coz like
wait hmm
yea
so like, suppose the basepoint is outside the origin
maybe its not possible because of some compactness thing?
everytime you traverse your loop you're gonna traverse some line segment infinitely many tiems
actually yeah should generally be impossible
I don't think you can do that
i dont follow
If a set A has no open subsets then what is Int(A)?
empty set
i just remembered, it shouldnt matter how big a hole you make in a space be
like, if your baseopint is at -1
then each time you go around the double origin you're gonna traverse [-1, 0)
why
-1, 1, -0.5, 0.5, -0.25, 0.25,
But this idea doesnt work cus compactness
I'm smokin
thingy
i think

im now not convinced it doesnt exist apart from R^2 without the origin is homeomorphic to R^2 without some closed disc
since this is the case it definitely isnt possible to infinite loop
im just like thinking - what if u force a topologists sine curve kinda thing
but ig its not possible for that to be a loop
loop needs to be homeomorphic to circle
I still think my proof is valid 
well yes it is
oh nvm lol only in uh spirit
its continuous map from circle 
so anywho
the kicker is
the real stupid part
to calculate the fundamental group of the line with two origins
instead of taking the two removed points from the origin
take one from the origin and one somewhere else
the open set that results from taking a point from somewhere else is homotopy equivalent to the line with two origins
and the intersection again is simply connected
so, if you call the fundamental group of the line with two origns G, you get
[\bZ * G \cong \bZ * \bZ]
DarQ
and then just quotient by \bZ 
I don't know how to state it precisely, but I feel like I want to prove its impossible to have infinite loops
like this feels like an important fact 
like ig disc without origin homeomorphic to annulus is enough but hmmm
It feels like my analytic intuition is lacking for me not to immediately see this must be the case 
maybe start by trying to state it precisely
claim: a loop can only visit each point a finite number of times?
false
r u thinking if it stays somewhere for a second before moving away or
if it was true then "looping infinitely many times" would never happen
what loop visits some point an infinite number of times
coz that'd require to go through the basepoint infinitely many times
but the hawaiian earings exist
im now confused, but actually ive suddenly remembered an example
hawaiian yeah
u able to paste that example, i wanna check it again...
for example you can traverse [0, 1] in 0.5 times
nvm ive got it, one sec
[0, .5] in .25 time
what's happening
why can't you loop through the line with two origins infinitely many times
wym loop though

we re tryna figure the fundamental group of line with 2 origins
pretty sure its Z
but then i couldnt see why
u rnt allowed loops
that loop around an infinite number of times
I couldnt construct one explicitly, sure.
this was the initial idea, but it doesnt work
but uhm. im not geometrically/analytically convinced its impossible yet
these defo exist
then surely its fundamental group is not Z
probably
ehh??? im now ultra confused
I don't think they do
Take the plane without the origin
at least they can't stop at the origin
this is homeomorphic to the plane without a closed disc
what basepoint
These 2 must have the same fundamental group
the point where your loop starts
oh w8 they have to stop somewhere
:cereal2:
I has thinking like R
if the basepoint is one of the origins it stops at the same origin
im now thinking back to hawaiian earing and think none of these loops are infinite here then????
w8 actually

why
wdym
can u write me an explicit loop that does so
uh ok
it's t1 isn't it
it constructs every loop in \prod \bZ actually
if it was at 0 at infinitely many isolated points then one of them would be a limit point
oh with the earings, you have a point at infinity kinda, allowing the infinite loops
the issue is that
earings is not compact is it
suppose your loop is infinite as you imagine it would be, right?
as you approach 1
right
ofc your loop starts at the origin
you can find a sequence that approaches 1 but is mapped through your loop to the other origin
wait wait i think this might make it clearer still:
the fundamental groups of:
R^2 - open disc
R^2 - closed disc
I reckon theyre not the same?
what
you can deformation retract both to a disk that's slightly bigger
and wtf do you mean so much better?
R is so stupid, who made this stupid set
just how bad is armstrong???
analysts
I dunno shuri, it was kinda game over for infinite loops once we calculated its first homotopy group
it's just better
also
yea
can we make a more minimalistic example where there are infinite loops
that embeds into R^2
sure
simpler than earrings
R^2\{1/n}
hmmm
wait no
this looks legit?
the one I proposed deformation retracts to earrings lel
well yes
but ill take it
so this is like weird
u can get infinite loops by poking an infinite number of holes
hmmm
like whyyy 
well anyway it's a fucked up space
but not in the important ways here
like it will have a universal cover
hi
not true
so try finding it
can
o
cant, with finite. yes?
no
is what i mean
I dunno what that is
it is covering space ting and useful for studying fundamental group and stuffs tbf
potet 
hi
potet, why can our loops go around a point infinitely only if we poke infinite holes into R^2
wdym
R^2 - finite set
uhhh
R^2 - {1/n}
Slightly vague saying "go around a point infinitely many times"
not that
you can't go round a specific point infinitely many times ever
So I guess really all you need to do is consider removing one point (and hence the circle ig lol)
And then it is gonna be compactness ig
why is $f^{-1}(Y) = X$? in topology, do we always assume that f: X --> Y has an inverse?
okeyokay
it's because functions are defined on the whole space

@unreal stratus it's munkres
he prolly skipped the set theory nonsense
good.
not tryna be russell or zamelo-frenkel
now close munkres

do not skip the set theory, the only sane part of it all
what the hell is lee
R is nonsense!
i thought everybody loved munkres
bruh
intro to topological manifolds
No i meant like
I don't
oh okay
I have a slight suspicion that darq doesn't either
is theorem 18 like near the start of the topology in the book lol
i thoguht munkres was the go to algebraic topology book
it's poitnset
lol
a lil bit of algebraic topology at the end
It's not algtop
Tell what lol
there's a lot of alg top by the end lmao
it's mostly used as a reference for pointset tho afaik
anywho, this is just insane

anyways, every x in X satisfies f(x) in Y
the go to is indisputably hatcher
So f^{-1}(Y) = X
who the hell is hatcher
ah
And it's useful to know that inverse image preserves complements, unions, and intersections
whereas forward image only preserves unions from those
yea but what if f is not surjective
okay
it's irrelevant
wait
f(x) is still in Y
oh








