but what does
#point-set-topology
rough cedar
"let p_0, ..., p_n, q in P^n such that no n + 1 points lie within a hyper plane" mean
languid patrol
It is the wrong channel
languid patrol
"let p_0, ..., p_n, q in P^n such that no n + 1 points lie within a hyper plane"...
It means let p_0, ..., p_n, q in P^n are points such that no n + 1 points lie within a hyper plane
rough cedar
Idk alg geo or sth
I'm scared of being yelled at in that channel 
asked in #groups-rings-fields
umbral panther
Why do you think that?
I’m not saying you need to do the opposite, but why not?
fickle hamlet
X is topological space, A is subset of it. closA = A U A', where A' is the set of all limit points. Can anyone give me an example where the closA!=A'? I know the proof for the earlier statement relies on considering elements not in A that are in closA but idk how to come up with a example for the A' only being a subset of closA.
stark fog
X is topological space, A is subset of it. closA = A U A', where A' is the set o...
perfect sets, for example Cantor's tertiary set is one of these
it is a closed set in which all of its points are limit points
(and a lot of another unintuitive properties)
fickle hamlet
perfect sets, for example Cantor's tertiary set is one of these
it is a closed ...
but isnt this closA=A'?
Im asking for closA!=A'
gritty widget
[0, 1] \cup {2} is an example. 2 is not a limit point, despite being in the closure
stark fog
Im asking for closA!=A'
oh, I didn't notice the !, never saw this notation
fickle hamlet
[0, 1] \cup {2} is an example. 2 is not a limit point, despite being in the clos...
Ah ok thanks, so kind of exploiting the fact that neighbouhoods must contain points other than the limit point
are there any other ways to exploit things to come up with such example?
but ig that is the fundamental difference
cuz im lookin at another thing and it says if every neighbourhood contining x intersects A iff x in closA
so if that detail isn't there its exactly the same
mint rose
So no ideas?
long remnant
Hey guys, new here so hope this isn't the wrong channel... Have a question...
Are there hilbert space filling surfaces (2D) / volumes (3D) / (N-D)?
If I understand correctly Hilbert space filling curves only map N dimensions to 1.
However, I'm looking for a similar concept, that maps N dimensions to X dimensions. (My choice of how many).
And ideally has the same nice properties as Hilbert space filling curves (good locality preservation).
Is there some generalization of Hilbert space filling curves?
This would give greater granularity over the traversal of the state space / degrees of freedom / dimensions.
It would allow compression of 100 dimensions down to 50, or 25, etc.
Why am I interested in this?
I want to create a tool to allow me to traverse the state space of all possible "sounds" continuously, but with the chosen level of granularity. I want to choose the level of compression.
So 44,100 dimensions -> 20,000 dimensions -> 5,000 dimensions -> 1,000 dimensions -> 500 dimensions -> 100 dimensions -> 50 dimensions (this is where I might start to add some knobs to control the output) -> 25 dimensions -> And so on.
Why no data?
I don't want to use an autoencoder, nor "bias" the state space. I'm not looking to traverse the state space of all sounds in my dataset. I'm looking to traverse the entire state space of all possible sounds.
willow viper
Hello I'm trying to prove that every subspace S of a second countable space X is also second countable. So the basis for X is just $\beta = {B_n : n\in\mathbb{N}}$. Is it sufficient to create a basis $\beta_S = {S\cap B_n : B_n\in\beta}$ because $B_n$ is already countable by definition?
gentle ospreyBOT
b3s4d
quasi steppe
Well, knowing there exists a bijection/ surjection from N to our original basis, can you construct a bijection/surjection to beta_s ?@willow viper
willow viper
o_o
quasi steppe
Well just surjective
Because what if your basis is finite
Unless by countable you also mean infinite
quasi steppe
You mean infinite or finite countable right?
But whatever you mean, surjection will cover it
willow viper
Yea I believe that second countable means finite countable
quasi steppe
it can be finite or infinite
willow viper
Oh
quasi steppe
Your definition of beta allows for infinite.
quasi steppe
Yes
willow viper
how
quasi steppe
A fun lemma is to show surjection from N to a set X implies a bijection from a subset of N to X
Have you proven surjection before?
willow viper
A fun lemma is to show surjection from N to a set X implies a bijection from a s...
well
willow viper
wait I'm thinking xD
quasi steppe
yea I think I get it why
If you’ve proven a function is surjective before you have all of the tools you need to prove your original problem
quasi steppe
but what about 1,2 -> a, 3 -> b, 4 -> c for example?
Just drop the 2
Define a new function with a smaller domain
You have all the power
You can do what you want as long as it makes sense
willow viper
Oh right
that's why S -> N injective also implies bijective because I can just drop the last numbers
willow viper
Ok wait
I might get it now
nope
Like I get that S \cap B_n is a subset of {B_n}_n\inN right
quasi steppe
Construct your function from N to beta_s
willow viper
I just said that N -> \beta_s, n \to S \cap B_n
quasi steppe
Like I get that S \cap B_n is a subset of {B_n}_n\inN right
That is true
willow viper
And I tried to rewrite it but I don't know how
quasi steppe
Yeah so you have your function f
Show it is surjective
Use your original surjection in defining your new one
If we have g:N -> beta, then we make f:N -> beta_s with f(n) = g(n) cap S
So show f is a surjection
quasi steppe
Well there exists B in beta such that B cap S = A
quasi steppe
Np
willow viper
Np
So in reality it's not a surjection to N but to \beta which is a subset of N so this works right?
or not subset that's stupid but has a lesser cardinality
quasi steppe
We have a surjection f:N to beta_s
quasi steppe
So in reality it's not a surjection to N but to \beta which is a subset of N so ...
I don’t quite understand
If you are saying a surjection from beta to beta_s
Then that is another way of proving it
willow viper
Ok
quasi steppe
The answer I gave before wasn’t the end of the proof
quasi steppe
Well there exists B in beta such that B cap S = A
This one
We know that we can find an n such that g(n) = B
By surjectivity of g
But then f(n) = g(n) cap S = B cap S = A
Thus for any A in beta_s we can find an n such that f(n) = A
quasi steppe
So in reality it's not a surjection to N but to \beta which is a subset of N so ...
You know what, you could do it that way too
A surjection from beta to beta_s
That’s less complicated actually
willow viper
right
so for all A in beta_s exists B in beta with B cap S = A. So f(beta) = \beta??
chrome ridge
Why is it defined as the sum instead of just picking one lift?
rapid patrol
Why is it defined as the sum instead of just picking one lift?
hi! could you screenshot the entire page if possible?
chrome ridge
hi! could you screenshot the entire page if possible?
It's from hatcher; appendix chapter 3.
chrome ridge
Just wondering why did they define it this way; is it not well-defined if we just pick one lift ?
rapid patrol
okay so
i mean im going off definition, but like isn't it supposed measure the extent to which elements of a subgroup H of a group G can be conjugated by elements of G?
also from stackexchange, it's defined as a sum of certain conjugates of element H
chrome ridge
I guess you're talking about the general case of transfer homomorphisms. I am just talking about this case of maps on induced on homology. Is there a specific reason why we take the sum of the lifts instead of pick a single lift? If we pick a single lift, does that induce a map on homology groups or not?
rapid patrol
I guess you're talking about the general case of transfer homomorphisms. I am ju...
I'm so sorry, that's as far as my knowledge goes 🥲
chrome ridge
gritty widget
mb
wintry stratus
I guess you're talking about the general case of transfer homomorphisms. I am ju...
I'm not great at topology, but I think what is happening here is that a single lift of a our map may not induce enought of a well-defined map. This is because there could be multiple lifts of the same map that differ by just a continuous homotopy, and these different lifts could induce different maps.
I don't really have a mental picture of what is described however, so I don't know how useful this information is.
chrome ridge
I'm not great at topology, but I think what is happening here is that a single l...
I think by unique lifting property, once you fix a point in the fiber, the lift is unique
wintry stratus
Right, ok, I see that now rereading how the problem defines the homomorphisms that take place. I suppose, then, that I am wondering all the same questions as you.
My apologies for not being able to help you get through this problem. Very best of luck on it!
chrome ridge
yeah it's fine. Thanks
bleak path
Hi, not sure if it belongs better here or in #diff-geo-diff-top, but would anyone happen to have questions on sheaf theory? I'm taking a course on it this year, and it's the first year running, and the lecturer doesn't provide questions on it, so I'm pretty much a week away from the exam without practice questions. Would appreciate any help.
gritty widget
honestly the best place for that is probably #algebraic-geometry
bleak path
Alright, I'll post there too, thanks 🙂
grim knot
hey guys, regarding R with the cofinite topology, which sets are open? Is it only the infinite sets, the empty set and R?
willow viper
hello. Can someone explain why the last element in H^n = {(x1,...,xn) \in R^n : x_n >= 0} must be greater zero?
coarse night
that's definition
coarse night
that's the prototype for manifolds with boundary, you could've taken any coordinate to he ≥0 or ≤0 and the end result will be the same as they're homeomorphic (diffeomorphic)
willow viper
Ah ok, but it must be ≥ 0 or ≤ 0 so that...?
coarse night
= c for your favourite real number
willow viper
And c is then kind of the boundary?
where all elements above c are interior points?
coarse night
Yes
willow viper
But why is H^0 = R^0 = {0}?
coarse night
Expand the definition
coarse night
hello. Can someone explain why the last element in H^n = {(x1,...,xn) \in R^n : ...
^
willow viper
Oh so I don't put n = 0 into that definition but make a new one
willow viper
I don't really understand this picture
languid patrol
I don't really understand this picture
The yellow part is the interior, the green part is the boundary
The pink part is a coordinate, and the whole space is defined by that coordinate being \geq 0
trail charm
try drawing it out for R1, R2, and R3
in R2, instead of (x,y), write (x_1, x_2)
in R3, instead of (x,y,z), write (x_1, x_2, x_3)
as topos said, you'll notice that the last coordinate being \geq 0 defines the half space
willow viper
Ooh ok so x^n >= 0 is just an axis in the coordinate system
trail charm
if you think of x_n as an axis, then x_n \geq 0 is an inequality
ie, the space “above” the x_n axis
willow viper
Ok I understand
Hello I'm trying to prove that "Suppose M is an n-dimensional manifold with boundary. Show that ∂M is an (n−1)-manifold (without boundary) when endowed with the subspace topology. You may use without proof the fact that Int M and ∂M are disjoint." Do I need to prove that φ: U ⊆ ∂M^(n-1) -> V ⊆ H^(n-1) is a homeomorphism so if φ bijective, continuous?
willow viper
Wait does V ⊆ H^(n-2) then??
nimble portal
It can’t be
You also can’t say your chart goes from boundaryM^(n-1) since you’re trying to prove that the dimension is n-1 in the first place
Did you already show that it’s Hausdorff & second-countable?
Try those first then save finding a chart for last
willow viper
well I just said that since ∂M is a subspace it has those 2 properties in the first place
languid patrol
Yes
nimble portal
Yes
willow viper
So I don't need to find a phi?
nimble portal
You do, that’s part of the condition for a top space to be a manifold
willow viper
Ok
willow viper
You also can’t say your chart goes from boundaryM^(n-1) since you’re trying to p...
So I can't assume that φ looks like φ: U ⊆ ∂M^(n-1) -> ∂H^(n-1)?
nimble portal
I misread the question lol don’t mind me, I think you can yes
languid patrol
So I can't assume that φ looks like φ: U ⊆ ∂M^(n-1) -> ∂H^(n-1)?
In general you have to give what your definition of a manifold with boundary is
For some definitions this is pretty trivial
willow viper
that the manifold with boundary is locally euclidean to a subspace of Rn
languid patrol
that the manifold with boundary is locally euclidean to a subspace of Rn
Okay but what do you mean by subspace?
Like a vector subspace?
willow viper
Like an open subset
willow viper
So an open ball or the whole Rn, ... I believe
languid patrol
So an open ball or the whole Rn, ... I believe
Yes but that isn't true at the boundary
what you just said is a definition of a manifold
willow viper
ah wait
languid patrol
But since your manifold has boundary you have to give a different sort of chart near the boundary
willow viper
you're right the whole time I've worked with the manifold property
So, every p has a nbhd. U homeomorphic to an open subset of R^n or to an open subset of H^n
So the only thing changing really is that it's not an open set but a neighbourhood right
languid patrol
So the only thing changing really is that it's not an open set but a neighbourho...
Yeah now it's a specific type of closed set
nimble portal
Doesn’t this amount to showing bd(H^n) is homeomorphic to R^(n-1)
languid patrol
Doesn’t this amount to showing bd(H^n) is homeomorphic to R^(n-1)
Yes basically!
You want to show that under every chart the boundary bd(M) maps to the R^{n-1} inside of H^n
so essentially that homeomorphisms take boundaries to boundaries
nimble portal
Hmm
Is it something like
Take a nbhd of bd(M)
Then by defn there’s a chart phi: U —> bd(H^n) s.t phi(U) cap bd(H^n) is not empty
Agh I gotta pick up my sister brb
nimble portal
Then by defn there’s a chart phi: U —> bd(H^n) s.t phi(U) cap bd(H^n) is not emp...
So take a nbhd in this intscn
Then by defn of bd(H^n), the x^nth component of any pt in this intersection is 0
hollow geyser
willow viper
willow viper
Yeah now it's a specific type of closed set
can u pls explain what u mean by that
willow viper
Doesn’t this amount to showing bd(H^n) is homeomorphic to R^(n-1)
does this mean that bd(H^n) = bd(M^n)?
nimble portal
So it’s got (n-1) nonzero components
Can projection maps take multiple components?
Because if so then can’t you just take the projection of the first (n-1) components into R^(n-1)
willow viper
does this mean that bd(H^n) = bd(M^n)?
ok definitely not xD
Ok I get it now. phi: U ⊆ bd(M^n) -> V ⊆ R^n-1 or bd(H^n). So then R^{n-1} is homeomorphism to bd(H^n)
nimble portal
Issue is you have to find a homeomorphism to call it homeomorphic LOL
nimble portal
Because if so then can’t you just take the projection of the first (n-1) compone...
But I think the projection works
So we have a boundary pt (x1, x2, …, xn) of M, by defn that means it’s in bd(H^n), which means it gets mapped to (x1, x2, …, 0), then we can just take the projection of the first (n-1) components onto R^(n-1)
it’s injective surjective cntns ggez no?
willow viper
I'm so lost
nimble portal
I don’t know if I’m right or not lol, but if you still want I can try to answer any questions you have
willow viper
I'm so lost I don't even know what to ask XD
nimble portal
willow viper
oh no a league player
nimble portal
Okay so the idea is we need to show that any nbhd in the bd of your manifold is homeomorphic to R^(n-1)
Well if we have a pt in our manifold bd
willow viper
yea I take a point in bd(M) and get a boundary chart (U, phi)
nimble portal
Then by defn there’s a chart on a neighborhood U containing that pt so that phi(U) intsct bd(H^n) is nonempty
willow viper
yea
nimble portal
So take any point in that intersection
Well by definition of bd(H^n)
The pt looks like (x1, x2, …, 0)
So it’s got (n-1) nonzero components and the nth component is 0
So I was thinking we can take the projection of the first (n-1) components, which goes to R^(n-1)
It’s injective since for two points in R^(n-1) to be the same ALL their components have to be the same, so if f(x) = f(y) then all the nonzero components have to be the same, which makes them the same point in bd(H^n) since the last component is 0
It’s surjective since given a pt in R^(n-1) we can always find a pt in bd(H^n) that maps to it as the same first (n-1) components and a 0 nth component
And I’m p sure the inverse is cntns in the subspace topology?
willow viper
Then by defn there’s a chart on a neighborhood U containing that pt so that phi(...
neighbourhoods are always open?
nimble portal
Yes a chart maps open neighborhoods
willow viper
Ok
urban zinc
Why is the letter F always used for closed sets?
gritty widget
the french word for "closed" is "ferme"
urban zinc
ahh ty
gritty widget
fermé
e.g. "F_\sigma" is "fermé somme" or "closed union"
that's where i learned it
urban zinc
TIL second-countable <=> separable in a metric space, this makes total sense idk why I never thought about it
Also Lindelöf's theorem is nice
cosmic socket
there are lots of different versions/definitions of spectra. Is it sufficient for most purposes to understand what a \Omega-spectrum is?
echo oyster
If X\A is contained in X\Int A, then Cl(X\A) is also contained in X\Int A ?
gritty widget
X \ int(A) is a closed set containing X \ A and Cl(X \ A) is the smallest closed set containing X \ A, so...
echo oyster
Yes I was thinking same, ty
urban zinc
ooooh alexander's subbase theorem is cool
cosmic socket
there are lots of different versions/definitions of spectra. Is it sufficient fo...
ok I realized the answer might be no
probably no
only fibrant spectra correspond to \Omega-spectra
and some important spectra like Thom Spectra aren't
willow viper
hello, let A subseteq B subseteq X. If A is dense in X then cl(A) = X. Why is cl(A) = B \cap cl(A)?
coarse night
closure of A in the subspace topology of B
echo oyster
The set { [a,b) : a,b belongs to Q } doesn't generate the lower limit topology right ? Since interval like [0,1) which have irrational end point not part of it
gritty widget
i feel like [0, 1) has rational end points
echo oyster
Really ? I thought if we are excluding the 1 then it can be done like that
How else should I try to show that it isn't generating Rl
gaunt linden
You could show that [pi,4) is not open in the topology generated by your set.
languid patrol
what are good examples to use van Kampen on
You can write S^1 as a union of S^1 and (0, 1/2) to show that \pi_1(S^1) = \pi_1(S^1)
odd flame
uhhh
odd flame
typo maybe?
languid patrol
That's pretty classic
languid patrol
typo maybe?
no it was a joke
odd flame
odd flame
scary word nope
languid patrol
scary word nope
But yeah a wedge of circles is good
Or like R with a circle attached at every integer
odd flame
Or like R with a circle attached at every integer
oh i remember this one
it's a covering space of something no
languid patrol
it's a covering space of something no
it's a cover of the wedge of two circles
odd flame
right right
solemn oar
what are good examples to use van Kampen on
Klein bottle
odd flame
identify circles with one S1 and the lines connecting them to the other S1
same question but for mayer vietoris
like what are good examples to practice on
languid patrol
the same examples work
Why not calculate the fundamental group and homology of every surface
hidden crag
odd flame
yeah figures
odd flame
Why not calculate the fundamental group and homology of every surface
i'll probably do this shortly
novel acorn
what are good examples to use van Kampen on
CW complexes!
Those are always neat
odd flame
actually on a related note to that
novel acorn
Why tf did i get danned
odd flame
can i get a hint on this pls
solemn oar
Just because a space is CW doesn't make it a good example of SvK.
odd flame
i barely get cellular homology
solemn oar
can i get a hint on this pls
Have you tried writing down the cullular chain complex for this space?
abstract saffron
i barely get cellular homology
Dw, you're not alone
odd flame
novel acorn
can i get a hint on this pls
Hint: what are the attaching maps for the 1 and 2-cells
Also writing out the cellular chain complex shouldn't be hard
Just count each cell
odd flame
ok i need to review my basics than bc idk some of these words tbh
like i do but not enough to do the problem
can i do this for a bit 
odd flame
Hint: what are the attaching maps for the 1 and 2-cells
what are these maps
i have another problem i'll do on my own but would appreciate guidance on this one
novel acorn
what are these maps
You know those aba^{-1}b^{-1} relations when talking about the attaching maps for the torus
That's what I mean here
quick bough
ive got a question to this proof of showing that if X is regular and A closed, then X/A is Hausdorff:
i dont rlly get the point of A having to be closed
for the first case: can't you separate x and y either way? you know that any subspace of a hausdorff space is hausdorff, so X \ A is Hausdorff, you also have that x and y are in X \ A, so separate them with two open disjoint neighborhoods
but i never used the fact that A is closed, no?
or what am i not understanding correctly
novel acorn
but i never used the fact that A is closed, no?
You did
Second paragraph
coarse night
How about [0,1]/(0,1]
quick bough
You did
Second paragraph
where exactly
novel acorn
End of second line
quick bough
sorry, could you say which part exactly
"separate them with two open disjoint neighborhoods"?
odd flame
You know those aba^{-1}b^{-1} relations when talking about the attaching maps fo...
no 0 cells?
maybe im overcomplicating this but how do you thinnk about cells
novel acorn
"separate them with two open disjoint neighborhoods"?
Ye
novel acorn
maybe im overcomplicating this but how do you thinnk about cells
Cells are lines and disks
coarse night
How about [0,1]/(0,1]
Is this Hausdroff?
odd flame
fair enough yeah
quick bough
Is this Hausdroff?
im not looking for a counterexample, ik, there's a problem with my argument, but idk where exactly
novel acorn
im not looking for a counterexample, ik, there's a problem with my argument, but...
Ok so
coarse night
Oh okay
novel acorn
Ok but tell me in English what that means
Regular means there are open neighborhoods that do what
novel acorn
Great
And by your definition also the open nbhds can separate disjoint points
quick bough
yeah
novel acorn
So you can take two disjoint open nbhds
Such that one contains x and the other y
And guarantee by T3 that they're also disjoint from A
(this guarantee comes from the fact that A is closed)
coarse night
but i never used the fact that A is closed, no?
A becomes a point in X/A, it’s the quotient topology not set minus. How would you separate [A] now a point in X/A with another point of X?
novel acorn
still dont get the last part
Let's assume that A is open
And take x to be a limit point of A not in A
Here there doesn't exist an open nbhd of x that is disjoint from A
quick bough
A becomes a point in X/A, it’s the quotient topology not set minus. How would yo...
are we not considering A to be a subset of X in this case?
coarse night
Where? X/A?
quick bough
okay, so my problem is
they are choosing open neighboorhoods in X
not in the quotient
coarse night
What’s the connection between open sets of X and X/A
quick bough
im guessing open sets in X/A correspond to open sets in X that contain A
coarse night
No
quick bough
oh
or
yikes
for the quotient map q : X -> X/A
U in X/A is open iff q^{-1}(U) is open
coarse night
Hm
odd flame
Cells are lines and disks
ok so there are three groups sin the chain
last of which is trivial
one disc (the filled in interior) means one generator
how do we get homology from there
novel acorn
ok so there are three groups sin the chain
yeah so you have Z \to Z^4 \to Z^2 \to 0
coarse night
It means yes
odd flame
yeah so you have Z \to Z^4 \to Z^2 \to 0
Z for the disc in the interior but why Z^4 and Z^2
oh is Z/4 for the four total edges identified
novel acorn
Z for the disc in the interior but why Z^4 and Z^2
those aren't quotients...
novel acorn
lol yeah ok
you take one Z for each cell
so you have 4 1-cells
so that term is Z^4
odd flame
but yeah Z^4 for the four lines being identified
and the two lines with no identification is Z^2?
novel acorn
no
OK lol
Let's learn the basics of cellular homology
Take some CW complex
finite
the idea is that
you count how many cells you have
in each dimension
in your case
you have one 2-cell
odd flame
repost to make it easier
novel acorn
so that means one generator of H_2(X_2, X_1)
odd flame
oh you find homology doing the normal ker/im stuff
odd flame
so that means one generator of H_2(X_2, X_1)
but are there four 1-cells or six
odd flame
4
odd flame
then what's Z^2
novel acorn
how many 0-cells are there
novel acorn
0-cells are the ends of the lines
you can see those little dots in the corners of your shape in the image you sent
odd flame
oh those are dots
novel acorn
yea
odd flame
i thought we got that from the identifications too
novel acorn
we do
But you still need to see how the identification identifies them
start labeling
:3
I already told you the answer (2) but I want you to learn how to count :P
(sry if I'm being too jokey I'll stop if it's annoying)
novel acorn
lol yeah np
since I'm not in uni till October I need stuff to do lol
odd flame
i invite you to take my final tm
novel acorn
lmaooo
I already took a final from this stuff lol I don't need a round 2
odd flame
going back to it i dont see the two points
novel acorn
OK
coarse night
Same uni?
novel acorn
don't try to do it in ur head
It's harder than it seems
Rather draw it on paper
coarse night
Cool
odd flame
is this like a cut n paste thing then tho
novel acorn
OK OK
where were we at I went to the toilet
Oh yeah learning how to count
on the LHS there are three points
call the one on the bottom like p or something
now
Which other points are also p
odd flame
ok the beginning of each line b is obviously the same point
same with the line a
the unidentified edges are what idk
novel acorn
the unidentified edges are what idk
they don't get identified
novel acorn
ok the beginning of each line b is obviously the same point
ok ok cool
What abt the end point
novel acorn
yeessss
odd flame
the three vertices involved in b are the same
but then that's also identified with the one starting at a
i can count 
novel acorn
so now do you see how we have the complex 0 -> Z -> Z^4 -> Z^2 -> 0
odd flame
like the number of cells gives us a skeleton
novel acorn
like the number of cells gives us a skeleton
ye
But idk if it's really helpful
You can certainly try to draw it lol
ok so I usually start from the back
since im = 0 there
so we can get the homology semi easily
odd flame
ok brb i'll ping in a sec 
thank you sm for the help
oh you meant like
0 -> Z
whoops
novel acorn
not sure what ker is
yeah so remember that $H_n(X) = \frac{\mathrm{ker}(d_n)}{\mathrm{im}(d_{n+1})}$
odd flame
actually hold up
gentle ospreyBOT
Irony Incarnate
odd flame
ok the map Z -> Z^4 takes the generator of Z to (1,1,1,1) right
novel acorn
here we label our maps as $0 \xrightarrow{} \mathbb{Z} \xrightarrow{d_2} \mathbb{Z}^4 \xrightarrow{d_1} \mathbb{Z}^2 \xrightarrow{} 0$
gentle ospreyBOT
Irony Incarnate
novel acorn
ok the map Z -> Z^4 takes the generator of Z to (1,1,1,1) right
nope
Cellular homology is more complicated
formally this is how the map is defined
in practice
you learn how to use it lol
odd flame
yeah that's
novel acorn
Joking
the important part is that attaching maps bit
the quotient map and degree stuff comes into play if we have like 3-cells and stuff then you have to worry about orientation
but here
your attaching map is like
acadb^2
odd flame
oh so it depends on the diagram whoops
novel acorn
abelianizing
we get
2a+2b+c+d
so the map is
1 -> (2,2,1,1)
and you can see that this is clearly injective
so ker = 0
and so H_2 = 0
novel acorn
it's ok this is usually the hard part
odd flame
ok homology is affected because the boundary chain complex are given by different attaching maps
odd flame
your attaching map is like
acadb^2
why do you include c if it has no identification
novel acorn
hm
odd flame
oh i guess because it's still a cell
novel acorn
I mean it's still attached to that cell
odd flame
yeah
so the map Z -> Z4 corresponds to the attachment of the 1 cells to the 2 cell right
odd flame
why do you include c if it has no identification
also i meant no orientation
novel acorn
so the map Z -> Z4 corresponds to the attachment of the 1 cells to the 2 cell ri...
yes
novel acorn
also i meant no orientation
uhhh see
I guessed the orientation lmao
Tbh it says filled in?
Otherwise idk lmao
odd flame
wdym
novel acorn
tbh
I'm unsure what's going on with those two lines
because they don't give an orientation
but orientation is very important
so tbh this problem is weirdly stated
odd flame
odd flame
novel acorn
<:bleak:877093280905592842>
Uhhh yeah so idk what to say lol
Let's just continue the way we started idk
odd flame
#diff-geo-diff-top maybee
abstract saffron
That thing came before GR, wdym modifed for a specific theory, lmao
And yes, #diff-geo-diff-top
odd flame
Uhhh yeah so idk what to say lol
Let's just continue the way we started idk
ook well i think the map you proposed earlier was right irony
coarse night
@odd flame what was your problem again?
odd flame
novel acorn
ook well i think the map you proposed earlier was right irony
:3
odd flame
trying to understand the boundary maps of the cellular chain complex
the chain being Z -> Z^4 -> Z^2
why monke
novel acorn
trying to understand the boundary maps of the cellular chain complex
yeah
Personally I view it mainly as the attaching map which is not totally correct but it works for this case
novel acorn
when does it not work
uhhh ok so
coarse night
feel like it should be
1 → (2, 2, 1, 1) →
novel acorn
the case for higher cells is more complicated
coarse night
aah
odd flame
feel like it should be
1 → (2, 2, 1, 1) →
how come
coarse night
first map is easy
novel acorn
the second map is
(1,0,0,0) -> (1,-1)
(0,1,0,0) -> (0,0)
(0,0,1,0) -> (-1,1)
(0,0,0,1) -> (-1,1)
odd flame
could you do this the other way too
like start with the 0 cells and see what it's like to attach the 1 cells to that
dinner break i'll be back in a bit
viscid nebula
Can someone please explain this to me simply i am losing my mind
abstract saffron
@nimble portal your turn to shine
solemn oar
Why are you double counting the image of a?
nimble portal
Uh
coarse night
<@124978310479347712> your turn to shine
Lol
nimble portal
Can someone please explain this to me simply i am losing my mind
Do you know what a cover is?
coarse night
Why are you double counting the image of a?
2 a
solemn oar
There is just 1 a
gritty widget
Can someone please explain this to me simply i am losing my mind
"K is said to be compact if every open cover of K has a finite subcover"
it's literally shorter and clearer in words. who tf wrote this
coarse night
cellular homology is fun theory not in practice 
odd flame
abstract saffron
cellular homology is fun theory not in practice <:monkey:651816857153699840>
This is why ppl only use simplicial in TDA 
viscid nebula
Do you know what a cover is?
If X is a top. Space the cover is the subsets of X whose union is the entirety of X?
gritty widget
the definition of an open cover is given to you in the screenshot, by the way
nimble portal
A* cover
nimble portal
Sniped
viscid nebula
Oh lol
odd flame
how are you going about calculating this though
nimble portal
I visualize it with balls
odd flame
i dont see how it follows from just "attaching maps"
odd flame
I visualize it with balls
visualize deez fucking nuts bozo
coarse night
first thing came to mind
novel acorn
why do you sum up two a's?
nimble portal
And find a finite subcover (so a finite cover inside the larger open cover)
novel acorn
they're the same 1-cell
odd flame
he corrected it
nimble portal
Then our space is compact
coarse night
how are you going about calculating this though
orientation, for example a goes from x to y so map is y-x
novel acorn
ahh sry
coarse night
*boundary
nimble portal
The idea is that we can use finitely many sets to cover our space, so our space is “finite” in a sense
novel acorn
we're still talking about Z -> Z^4?
we're talking abt Z^4 -> Z^2
fuck ok there are too many ppl again
nimble portal
Which is why we call it compact
abstract saffron
I visualize it with balls
Here, from Hausdorff measure, you're welcome
nimble portal
A good example is intervals in R
novel acorn
Calculating homology
nimble portal
The interval [2, 3] is “finite” compared to (-infty, infty) right?
viscid nebula
Yes
nimble portal
Right
And (not) coincidentally, [2, 3] is compact whereas the other interval isn’t
Does that help a bit?
nimble portal
You can also think about it as generalizing the idea of a space being closed & bounded
coarse night
((0,1) not compact tho)
coarse night
wrong channel
abstract saffron
And (not) coincidentally, [2, 3] is compact whereas the other interval isn’t
Heine and Borel: 
nimble portal
abstract saffron
Thank you for the help!
@nimble portal congrats! welcome to the teachers' gang
nimble portal
I think it’s a lot easier to understand compactness when you see it used
Or rather seeing how it’s used
And what would change if something wasn’t compact (compared to how much nicer it is when things are compact)
abstract saffron
I always maintain several definitions of closed sets in my mind
With sequences, with delta-epsilon, with open sets, etc.
It's easy to see how point-set topo makes sense once I have that picture in my head
A very good example to make use of this generalised definition of compactness is proof of Dini's theorem
Like, everything works almost magically because of this definition
nimble portal
Yeah compactness is goated
coarse night
Compact Hausdroff is
abstract saffron
Can someone please explain this to me simply i am losing my mind
Although, a French will put you on guillotine and call this quasi-compactness
coarse night
or locally compact Hausdroffs are
nimble portal
Although, a French will put you on guillotine and call this quasi-compactness
The French are invalid 👍
Man I’m so burnt out from school I don’t want to study anything
Tomorrow’s my last day at least
coarse night
me irl
abstract saffron
It's good to keep in mind the need for Hausdorffness. It makes limits of sequence unique, and makes the sequential characterisation of closed sets (and compact sets? I'm not sure) that we know and love
gritty widget
The French are invalid 👍
as a non-french canadian, i must say...
nimble portal
LMFAO
gritty widget
abstract saffron
Man I’m so burnt out from school I don’t want to study anything
You tell me, I have exam on numerical ODEs this Thursday
nimble portal
abstract saffron
Tomorrow’s my last day at least
God bless you
grizzled ibex
It's good to keep in mind the need for Hausdorffness. It makes limits of sequenc...
nets>
nimble portal
You tell me, I have exam on numerical ODEs this Thursday
Numerical ODEs? Is that just like Runge-Katta and similar?
grizzled ibex
im starting to read Allen Hatcher
abstract saffron
Yeah, Adam-Bashforth, Moulton, Crank-Nielson, and all these mfs
abstract saffron
im starting to read Allen Hatcher
My condolences
nimble portal
grizzled ibex
Do you think it's a bad start?
nimble portal
Doesn’t everyone love Hatcher’s book for alg top
grizzled ibex
it's my first book for alg top
nimble portal
Except for the “visual proofs”
abstract saffron
I hate it with passion
abstract saffron
He made no clear distinction between singular and cellular, and I was sooooo confused
coarse night
I hate it with passion
VBKT one is good ngl
grizzled ibex
im already struck by the creativity of the "house with two rooms"
abstract saffron
Plus, who starts with cellular? Bruh
nimble portal
coarse night
im already struck by the creativity of the *"house with two rooms"*
skip
coarse night
not necessary
abstract saffron
VBKT one is good ngl
VBKT?
coarse night
vector bundles and K theory
grizzled ibex
basic logic lol
nimble portal
im already struck by the creativity of the *"house with two rooms"*
I think you’d love to learn about Hilbert and his hotel ;3
coarse night
so is spectral sequences, but you never read the proof
abstract saffron
Tried reading it multiple time, remember all these defs, but never been able to use them
grizzled ibex
I think you’d love to learn about Hilbert and his hotel ;3
nah
house with two rooms >>
coarse night
Tried reading it multiple time, remember all these defs, but never been able to ...
may I suggest you some references
abstract saffron
house with two rooms >>
Try the Chinese Room
abstract saffron
Clearly not a CS folk
coarse night
the book way is indeed cumbersome
nimble portal
CS
abstract saffron
what the hell
abstract saffron
may I suggest you some references
Oh yes, please
I'm desperate and hopeless with homology and diff geo
grizzled ibex
should i learn at least the basics of category before alg top?
abstract saffron
should i learn at least the basics of category before alg top?
You don't need it, you'll pick it up along the way anyway
Although, do know what a product is
And an example to distinguish product and co-product, you'll thank me later
coarse night
Oh yes, please
First try reading the spectral sequence part from Ravi vakil's book on algebraic geometry, The rising sea. it'll make you familiar with basic working, i e for double complexes which is enough most of the time. You'll be able to prove universal coeff easily with that, even balancing tor and ext stuff
to get a pictorial understanding of what's happening, read 3blue1brown vlog on puzzling through spectral sequences, also given Ravi Vakil
First is enough for mostly all the cases because most of them arise from double complexes for example equivalence of De Rahm and Cech is just spectral sequence at play
There are some stuffs on exact couples which you can read later
abstract saffron
One thing I struggle with is I don't know where to go after simplicial homology
the next step is trivially cellular, but no books discuss it well
Hatcher is just horrible
grizzled ibex
What should i take into my heart from Hatcher?
i mean
you guys have some experience with this book
abstract saffron
The only thing I took from Hatcher was a simple explanation for Lefschetz's fixed-point theorem
novel acorn
the next step is trivially cellular, but no books discuss it well
Dieck maybe?
abstract saffron
But apparently i'm an exception, almost everyone likes Hatcher for some reasons
novel acorn
Idk yeah I started liking Hatcher after a while
coarse night
But apparently i'm an exception, almost everyone likes Hatcher for some reasons
not really
novel acorn
it's useful as a guide
He doesn't explain stuff too well
But if you do problems you'll figure it out
cosmic socket
Hatcher discusses cellular homology
grizzled ibex
except for hatcher who i should read then?
coarse night
I'm also not a big fan
grizzled ibex
😭
cosmic socket
I like hatcher
abstract saffron
except for hatcher who i should read then?
What do you know already?
grizzled ibex
general topology
novel acorn
how much algebra do you know?
coarse night
hatcher is the best choice prolly
grizzled ibex
i studied till algebras over a field
tidal cedar
I hate it with passion
same
cosmic socket
Best to see the geometric picture before you deep dive into the categorical perspectives
In my opinion
Though both are important
novel acorn
tru
Hatcher likes his pictures
tidal cedar
very category theory heavy
which is useful for going further
novel acorn
Do note that some of Hatcher's explanations aren't too good
So it's good to also use something else in conjunction
cosmic socket
Spanier is also good I think, but very dense
grizzled ibex
how much algebra do you know?
i saw free groups too
abstract saffron
general topology
I always recommend Nakahara for simplicial homology. He derives everything slowly and rigorously, breaks things down so that even a physics student can understand. He even starts with graph theory and Euler's characteristic for planar graphs first before going to Euler-Poincare theoren.
tidal cedar
Spanier and May are both breakneck speed but good references
abstract saffron
Like, I literally gave a 2-hour lecture on simplicial homology based on that book
tidal cedar
yeah Nakahara simplicial section is nice
coarse night
spaniee is too slow for me
abstract saffron
Spanier and May are both breakneck speed but good references
May is God
coarse night
also notations 
grizzled ibex
I always recommend Nakahara for simplicial homology. He derives everything slowl...
even physics student
abstract saffron
You read both books, and you can become a prof in AT
coarse night
Bredon is good tho
tidal cedar
Bredon is what my actual AT class used
abstract saffron
You'll have a stroke
Confirm this, first hand exp
coarse night
homological alg I'm aware of
abstract saffron
also notations <:kongouDerp:483988392099446794>
Try Hilton & Wylie
Took me a while to figure out the notations. 1960s folks are weird
grizzled ibex
*A First Course in Abstract Algebra (2000), *
coarse night
too many books, I'm fine with bredon honestly
grizzled ibex
An Introduction to the Theory of Groups (1995),
i think i read the last one
too many things to keep in my brain
coarse night
Took me a while to figure out the notations. 1960s folks are weird
oh lol you meant sarcastically
abstract saffron
oh lol you meant sarcastically
Oh, I meant it literally
grizzled ibex
can i borrow another brain from mother nature plssss
abstract saffron
I think some ppl got mad and they changed the norm for notations somewhere between 1960s and now
abstract saffron
Lang
grizzled ibex
rip
abstract saffron
Who reads Bourbaki anyway, lmao
You know when you know something but you don't know where to refer exactly
coarse night
a friend of mine leart it the hard way
abstract saffron
So you find a random book that mentions the topic, give it a quick read to make sure it's correct, then add it to the reference
abstract saffron
a friend of mine leart it the hard way
Wow, someone actually learned from Bourbaki?
quick bough
how does one show (c)?
coarse night
no
abstract saffron
how does one show (c)?
Am I tripping, or is a) Hann-Banach?
Literally extension of linear operator
quick bough
is it?
quick bough
^
abstract saffron
My mind works very peculiarly
quick bough
anyway, no clue how to do (c) lmao
take determinant as continuous map
but no clue how to show that it is equal on a dense subset
abstract saffron
Hint: invertible matrices are dense
quick bough
oh
abstract saffron
Not sure if it helps but that's the only thing making sense to me in this case
quick bough
yeah
i completely forgot about dense subsets of R^n, i was thinking too much of R^n analytically
so i would have taken Q^n
coarse night
ah it's a nice way to prove NGL 
abstract saffron
All roads lead to Rome
quick bough
do i fix B and consider the map to be in A
abstract saffron
do i fix B and consider the map to be in A
I guess just WLOG A and B to be both invertible, then play around with the formula
Looks like it's simple enough famous last word
quick bough
hahah
does this work for other fields
other than R
prob not right
F_2 surely is a counterexample
abstract saffron
Euhm...
There's something to do with the characteristic of the field iirc
Fields of char 2 are particularly nasty when it comes to vector spaces and matrices
But I'm not sure
coarse night
in GL(Finite field), any HD topology makes it discrete and any dense subset it the whole space itself so it's true
novel acorn
tf is a HD topology
Is there a 4k topology then 
quick bough
hahah
coarse night
High definition, meaning you can separate out each points individually
abstract saffron
Why did I always think Grothendieck came up with cohomology?
Strange, TIL it's not true
Ah, he came up with Sheaf cohomology, predicted by Artin for Weil's conjectures, OK
dry jolt
Sheaf cohomology was introduced by Leray
abstract saffron
What? 😄 Ok, something is very wrong with my memory
ocean narwhal
Ah, he came up with Sheaf cohomology, predicted by Artin for Weil's conjectures,...
etale cohomology
odd flame
what's the fundamental group of the connected sum of projective planes
i got it (a, ... a_n | a^2 ... a_n^2)
now i need to prove it kek
novel acorn
what's the fundamental group of the connected sum of projective planes
<a1,...,a_n | a_1^2... a_n^2>
odd flame
actually i'll do T # T
novel acorn
You can do it by drawing the fundamental polygon for these shapes
odd flame
nnaw got it do it the real way with SVKT
then MVT for homology
gotta practice
on a different note
is there any intuition for that retraction or should i just like
remember it
also on a different different note, what's the intuition behind reduced homology
round dust
Anyone happen to know the relationship between the homology groups of a space and one of its covering spaces?
Namely trying to show that if H_i is trivial for a covering space (not necessarily connected, I think) then H_i for the base space is annihilated by the number of sheets
warm quiver
Sounds like the transfer sequence would be useful. See pg 175 in Hatcher and section 3.G.
warm quiver
also on a different different note, what's the intuition behind reduced homology
The reduced homology of a space is the same thing as relative homology with respect to a point. So the reduced homology can be interpreted as the homology that the point can "see." This explains why only 0 dim homology can be affected since points must "see" loops and such but can't "see" themselves which corresponds to killing off a factor of Z in the regular homology of the space.
echo oyster
L = (x,-x) x is real
Then subspace topology on L from Rl2 is the discrete topology
What will be the topology on L from Ru2
bitter smelt
L = (x,-x) x is real
Then subspace topology on L from Rl2 is the discrete topolo...
could you ask this in a more coherent way
for example: what is Rl2? What is Ru2?
bitter smelt
is L just one point? Or is it all ordered pairs of the form (x,-x)? (i.e. the line y= -x)
echo oyster
It’s the line yes
bitter smelt
You know the definition of subspace topology?
echo oyster
A topology induced from the given topology to its subset
echo oyster
Intersection ?
bitter smelt
Can you write it down
echo oyster
I tried to do it like this ....bases in Rl2 is two sided closed, two sided open rectangle & intersect it with y=-x
echo oyster
I drew like this
bitter smelt
Great, so now do the same with the euclidean topology
What you are calling the usual topology
bitter smelt
Sure, this is equivalent to the usual euclidean topology
it should be noted that the usual base for the standard euclidean topology is given by open balls
echo oyster
Yes
bitter smelt
How does your picture look now
bitter smelt
Yes
echo oyster
Ah I see, the question was show that both of them generate same topology, so i was confused
Maybe it's a mistake
bitter smelt
Could you post the question?
bitter smelt
Are you sure $\bR^2_\bullet$ is the euclidean topology?
gentle ospreyBOT
Migilonomy
echo oyster
No, i just assumed
I'm not familiar with ' . ' looked kind of like u , so tried with the usual topology
bitter smelt
If the question is "correct" then we have demonstrated that it is not the usual topology
As your line (indeed, any line) with subspace topology from the standard euclidean topology on R^2 is equivalent to the standard euclidean topology on R
Which is certainly not the "same" as the discrete topology
bitter smelt
maybe its the discrete topology?
echo oyster
If it's discrete topology, then basis are singleton sets, & intersection with the line is also singleton sets
Then gives discrete topology right ?
echo oyster
Maybe its that
bitter smelt
🤷
surreal hornet
I am trying to show that H_0(X) = Z for a path-connected X, but I'm stuck with one part of the proof. We know that H_0(X) = ker(∂_0)/im(∂_1) = C_0(X)/im(∂_1) and now if we define e : C_0(X) -> Z mapping the sum a_iσ_i to the sum of a_i's, then apparently ker(e) = im(∂_1). Could someone elaborate on why this equality should hold?
odd flame
The reduced homology of a space is the same thing as relative homology with resp...
i think i understand this definition but im not sure when to put it into practice
omg ryu 
coarse night
I am trying to show that H\_0(X) = Z for a path-connected X, but I'm stuck with ...
sebb might be able to help ^
gentle ospreyBOT
The current time for stμ₂dying is 04:52 AM (EDT) on Wed, 10/05/2023.
coarse night
i think i understand this definition but im not sure when to put it into practic...
reduce homology is much easier to compute because you don't have that annoying Z factor ar the 0'th place
MV sequence gets easier to deal
odd flame
is there a good example to practice on
i was gonna redo pi1 and H groups of like
T # T
S1 v S1 stuff like that
coarse night
can you show H_1(∨Sⁿ)= ⊕_n Z
coarse night
It gets easier if you use reduced
odd flame
actually super quick question
two things actually
what is that map v(x) it's phrased super weird
coarse night
is it spanier?
odd flame
massey
coarse night
god who uses E^n for closed disx
odd flame
yeah idk 
coarse night
that map is given in hatcher nicely
odd flame
of course it is
coarse night
you join x to f(x) via a line and see where it intersects the boundary
odd flame
im so glad we used massey and not hatcher 
coarse night
map x to the point to the intersection point on the boundary
odd flame
so it's like a projection onto the whole of S^n-1?
oh yeah it iis
no fixed point assumption so every point will create a line to a different point on the boundary
ok my second question: isn't the proposition that there is no retraction from D^n -> S^n-1 actually part of BFT
coarse night
sorry?
odd flame
hm silly question perhaps
odd flame
ok that's it's own conclusion then got it 
coarse night
at least one of the proofs
odd flame
yeah i was practicing proof in dimensions \geq 2
coarse night
you get no retraction from H(boundary) → H(D) → H(boundary) being indentity but middle being 0
coarse night
you can maybe try this also, M be matrix with all positive entries, then it has an eigenvalue
say in dim 3 for simplicity
it's just BFT
coarse night
I am trying to show that H\_0(X) = Z for a path-connected X, but I'm stuck with ...
if that sum is zero means you can always make a collection of path whose boundary is 0 chain you choose
think about why that should be
odd flame
just realized i forgor the hureiwicz thm
how is the hairy ball theorem related to degree
ik it's kinda relate to brouwer too right
unreal stratus
The hairy ball theorem can be proven by showing any nonvanishing vector field on any sphere leads to a homotopy between the antipodal map and the identity
Then degree theory allows you to show that is possible only on odd dimensional spheres
odd flame
what about a connection to brouwer tho? is brouwer a corollary
odd flame
for this problem is the right chain Z -> Z^3 -> Z^2?
im pretty sure it's a mobius strip
but im getting H2 = 0, H1 = 0, H0 = Z
coarse night
doesn't look like Mobius
odd flame
you can do a cut and paste
coarse night
what do you get?
odd flame
wdym
coarse night
after cut paste
odd flame
damn i had it on a chalkboard but i dont recall atm
im 99% confident in that it's a mobius strip though
visually at least
coarse night
okay
odd flame
but im getting H2 = 0, H1 = 0, H0 = Z
but that doesnt agree with thiss does it?
novel acorn
for this problem is the right chain Z -> Z^3 -> Z^2?
Ye
odd flame
novel acorn
So the first map is 1 -> (2,0,1)
odd flame
yeah i got that
novel acorn
Which is injective so H2=0
novel acorn
0,0,0 -> 0,0 no?
Also here you mean (0,0,1)
novel acorn
:3
odd flame
is the process the same if there's a 3-cell in the skeleton?
novel acorn
Uh similar but there are a few caveats