So maybe it's an excision argument on the inclusion of the two endpoints?
#point-set-topology
plain raven
digital wraith
that feels like it
plain raven
or one or the other. because if you delete one or the other endpoint the result is contractible
digital wraith
huh I think I was just using the wrong subspaces
like I was looking at H(SX,X) but I think you might actually want to do H(SX,CX)
plain raven
The inclusion of the two points into SX should certainly be a "good pair" btw
or like
yeah
(SX, CX) is a good pair
etc
digital wraith
then excise the endpoint
oh wait is this just trivial induction?
like let C^n(X) be the union of n cones CX with bases identified
take H(C^nX,CX) and excise the endpoint and then that deformation retracts onto H(C^{n-1}X)
and then H(CX) has all trivial homology groups since it is retractable
although what does that really get you
plain raven
i think you might have to put it into the long exact sequence to get more info
digital wraith
well I'm doing that and I'm not sure what info I get
since you get a LES where every third group is just 0
wait but this feels wrong then
since this also proves S^n is retractable 
or more likely I am just dumb
$$...\to H_n(CX) \to H_n(C^kX) \to H_n(C^kX,CX) \to H_{n-1}(CX) \to ...$$
gentle ospreyBOT
Namington + mniip 2024
digital wraith
excise and you get
$$...\to H_n(CX) \to H_n(C^kX) \to H_n(C^{k-1}X) \to H{n-1}(CX) \to ...$$
gentle ospreyBOT
Namington + mniip 2024
digital wraith
except that can't be right
stone cipher
I'm stuck in int(F)=empty set. Out of ideas.
quasi forum
Okay, so i can use a bit of help. I need to show that if h is homotopic to h' and k is homotopic to k', then h o k is homotopic to h' o k'.
My thought is to try to show that h is homotopic to h o k (although I cannot guarantee this is true off the bat). If I can do that, then the rest is pretty simple
If I can't do that, then I am not entirely sure what I should do instead.
plain raven
this doesn't sound like it's formally possible
it's like
this doesn't type check right
have you drawn a picture?
i'ts not meaningful to say "h is homotopic to hk"
quasi forum
Yea, I had a hunch :(
plain raven
Do you know that homotopy is transitive?
quasi forum
Yes, that's what I was trying to get to (if I could use it)
plain raven
like if F1 is homotopic to F2 and F2 is homotopic to F3 then F1 is homotopic to F3
yeah.
ok so
prove hk is homotopic to hk'
and then prove hk' is homotopic to h'k'
quasi forum
Ah, I see what you are getting at. So how might I do the first one? Homotopies are still a bit new to me
plain raven
try drawing a picture!
if your spaces are $X, Y, Z$, with $h, h' : X\to Y$ and $k : Y\to Z$
gentle ospreyBOT
diligentClerk
plain raven
pick some point in $X$
gentle ospreyBOT
diligentClerk
plain raven
the idea is to give a path $kh(x) \to kh'(x)$ in $Z$, which is assigned in a continuous way as a function of $x$
gentle ospreyBOT
diligentClerk
plain raven
and what you have is a path $h(x)\to h'(x)$ in $Y$ (from the homotopy $h\to h'$) and a continuous map $k : Y\to Z$
gentle ospreyBOT
diligentClerk
plain raven
how can you combine them?
digital wraith
Hatcher 2.1 problem 20
quasi forum
Okay, let's go back a bit to the picture. I have a good idea as to how to visualize path homotopies, but not homotopies themselves
plain raven
yeah so just fix a single point in $x$ and think of it as a path homotopy from $khx$ to $kh'x$ and then think about the bigger picture after that
gentle ospreyBOT
diligentClerk
plain raven
if you solve the problem for a single point $x\in X$ it gives a path ${ x } \times I \to Z$
gentle ospreyBOT
diligentClerk
plain raven
then if you solve it for all points in $X$ it gives a map $X\times I\to Z$ as $X = \bigcup_x {x}$
gentle ospreyBOT
diligentClerk
quasi forum
digital wraith
bruh does TeXit not work in threads
quasi forum
This is a lot to process
plain raven
yes math is hard
stone cipher
I'm stuck in int(F)=empty set. Out of ideas.
<@&286206848099549185>
quasi forum
I appreciate the help, but I am so lost right now
stone cipher
ok, I solved my problem.... 😐
plain raven
great
let that be a lesson to the rest of you
when somebody yells at you for asking questions
just solve the question
rancid umbra
ok, I solved my problem.... 😐
what was the soltution?
plain raven
and then it won't be an issue anymore
stone cipher
what was the soltution?
The set $F=\left{a_n::n\in\mathbb{N}\right}$ inherits the properties of local compactness and Hausdorff from $X$, because it's closed. So if none of $\left{a_n\right}$ is open, then $U_n=F\setminus\left{a_n\right}$ is open dense in $F$. Thus, by Baire theorem $\bigcap_n U_n$ is dense in $F$. But the intersection happens to be empty set.
gentle ospreyBOT
RaD0N
rancid umbra
gonna have to re-look up what baires theorem is
stone cipher
gonna have to re-look up what baires theorem is
stone cipher
oh. this is a nice solution then
indeed. I was stuck because I didn't make use of this property. I was trying to work with the bigger one, the X.
plain raven
baires theorem is op
quasi forum
Okay, back to the homotopy problem (I took a break).
So we definitely know that $h \simeq h'$, so then is the homotopy, $F$, for $k$ and $k'$ sufficient to show $k\circ h \simeq k \circ h'$?
gentle ospreyBOT
dackid
quasi forum
Wait, actually, the map we are looking for is a function from XxI into Z, so my thought process clearly won't work :/
Actually, hold on a sec! $F: X\times I \to Y \times I$ given by $F(x,t)=(f(x),t)$ is continuous. And there is a function $G: Y\times I \to Z$ that is continuous. So $G\circ F$ is continuous.
plain raven
bravo
gentle ospreyBOT
dackid
quasi forum
Oh and this is the exact map we need to show k(h) is homotopic to k'(h)
So how about the other part. That k'(h) is homotopic to k'(h'), I don't think the same idea works for this.
gentle ospreyBOT
dackid
quasi forum
Then transitivity gets the job done.
obtuse meteor
ye
quasi forum
Heck yea 🥳
Okay, next question: what is a simple way to describe a star shaped region in R^2?
plain raven
wydm
quasi forum
Well, I need to find a star convex space in R^2 that is not convex, and the thing that jumps out at me is a star shaped region. Not only is it in the name, but it is not a convex region
plain raven
this counterexample works.
are you writing this up
if you're not writing it up just find a simpler example.
HINT
idk there's lots of examples
here's one easy answer|| The union of the x and y axes. ||
quasi forum
Oh good heavens, that is super easy xp
Okay, so the next thing is to show is that star connected spaces are simply connected. So i have an idea of what to do.
Since any loop for x_0 can be represented as two separate paths: a path from x_0 to x_1, and a path from x_1, to x_0.
Though I am not sure how this helps us get to creare a path homotopy with the trivial loop
plain raven
ok this is a good exercise and worth doing
so i'm going to bounce and let you think about it
quasi forum
Oh wait, any path from a star convex point to any other point is path homotopic to a straight line between them
Then you can shrink that line to a single point.
limpid leaf
i dont think there is a fact about homeomorphisms fixing 0 that im forgetting so I guess I'm asking if removing 0 is actually important
sry let me type this so its readable lol
gentle ospreyBOT
jesse
limpid leaf
Oh okay so I could just compose it and be fine
yes this makes sense
right yeah thanks very silly
coral pawn
Thanks. I figured out what the answer should be and then just stated that adding the three things up gives us that. Hopefully my professor doesn't look too closely at the computation
quasi forum
Okay. I am not too sure how to use the assumption of a space being star convex to show a loop is in the trivial class...
I don't see a workable connection
Actually, I think this is just a straight-line homotopy
All points can be mapped to the star convex point by a straight line
Let a_0 be the star convex point, and f be a loop centered at a_0.
Then, F(s,t)=(1-s)f(t)+a_0 s is the straight line homotopy between f and the trivial loop.
quasi forum
Alright, having a bit of trouble with this one.
In case that notation is not standard, here is the definition for a hat.
So if I understand the problem correctly, the goal is to show that if $\pi_1(X,x_0)$ is abelian, then $\hat{\alpha} \simeq_p \hat{\beta}$.
gentle ospreyBOT
dackid
quasi forum
I believe that is what the equality represents here.
obtuse meteor
consider the loop alpha * beta^{-1}
quasi forum
That concatenated with beta* alpha^-1 gives you the identity loop
But that doesn't tell us much :/
quasi forum
I'm catching on to that, just not sure how to approach it differently.
Truth be told, this needs to be path homotopic to the identity loop to get what we are looking for.
gritty widget
sorry to pop in but
obtuse meteor
Then I dont know where to go from here
so you should think about it like
[alpha bar] * [f] * [alpha] = [beta bar] * [f] * [beta] if and only if...
quasi forum
Yea, that's what we are trying to get to. I understand that
Btw, when you say beta^-1, do you mean beta bar?
quasi forum
[alpha bar] * [f] * [alpha] = [beta bar] * [f] * [beta] if and only if...
Ohhh, we should star f with alpha*beta bar
That gives you this
Well, that doesn't quite seem like what we are looking for 🤦♂️
quasi forum
So this approach seems like a dead end
gentle ospreyBOT
Namington + mniip 2024
Namington + mniip 2024
Namington + mniip 2024
Namington + mniip 2024
digital wraith
oh whoops typo with the subscript there
and hopefully whatever technique you use would generalize nicely to spaces where you identify more cones
quartz edge
fading vale
https://media.discordapp.net/attachments/496785752936546304/910645809324896267/u...
This idea is pretty close but the best way to do it is to identify SX with CX/X = (CX, X)
then just use the LES for relative homology
with X -> CX -> (CX, X)
cone is contractible so it dies and you just get isos
fading vale
What do u mean identifying more than 2 cones
digital wraith
look at the problem
digital wraith
I mean CX/X only gives SX, not 3 cones with the bases identified
fading vale
the two cones are given by X x [0, 1/2] /X x {0} and X x [1/2, 1]/ X x {1}
digital wraith
unless it's a trick and all the spaces are homotopic to SX
digital wraith
the two cones are given by X x [0, 1/2] /X x {0} and X x [1/2, 1]/ X x {1}
wait what exactly do you mean
fading vale
The question is asking you to define an iso on homologyt
you can use the fact that H_n(CX, X) cong H_n(CX/X) cong H_n(Sigma(X))
fading vale
wait what exactly do you mean
What is your definition of the suspension
digital wraith
X times I with the ends identified
fading vale
Right
fading vale
X x I/(X x {0} U X x {1})
CX = X x I/(X x {1})
so
identify X in CX with X x {0}
and kill
digital wraith
what I'm asking is for
compute the reduced homology groups of the union of any finite number of cones CX with their bases identified
fading vale
Oh i see
So you mean like disjoint union of cones with X x {0} in each quotiented together?
digital wraith
yeah the other part is clear with the "identify CX/X with SX" idea
digital wraith
So you mean like disjoint union of cones with X x {0} in each quotiented togethe...
yeah
or like
digital wraith
https://media.discordapp.net/attachments/496785752936546304/910645809324896267/u...
this
just read the last bit of the problem
fading vale
I see
Same principle id think
let Y be our final space
its given by disjoint union of CX quotiented by disjoint union of the X x {0}
so apply LES
am i reading this correctly
digital wraith
it's the end where you're not quotienting
fading vale
Ok this is pretty hurbed, its not disjoint union of CX quotiented by the X x {0}
Uhhh hmmm id probably use CW homology or something
theres not a nice representation of this as a quotient by a subspace
afaict
digital wraith
it feels like there should be some kind of induction you could do
but I'm not immediately seeing it and am also probably going to bed now since it is late
fading vale
Ok solution: let Y be the disjoint union of the cones with the bases identified. Then Y/X x {0} cong wedge sum of suspensions
right
digital wraith
what is X x {0} here
fading vale
The base of each cone
digital wraith
is that the end where you have the "point" or not
fading vale
No, the base
digital wraith
ok right
fading vale
Sorry if its a bit ambiguous : p
fading vale
Ok solution: let Y be the disjoint union of the cones with the bases identified....
so this identification makes sense?
digital wraith
yeah that's a wedge sum of k suspensions if you had k cones
fading vale
Right
So we apply LES
we get
$\ldots \to H_n(X \times {0}) \cong H_n(X) \to H_n(Y) \to H_n(\bigvee SX) \to \ldots$
gentle ospreyBOT
when the yass pills hit
digital wraith
well H_n(X\times {0}) isn't iso to H_n(X) though right?
it's iso to k copies of H_n(X)
fading vale
Uh remember that we're working in Y, so the X x {0} are all already identified
this is a quotient of Y by X x {0}
not of the disjoint union
fading vale
Ya so we have that the $H_n(\bigvee SX) \cong \bigoplus H_{n-1}(X)$
gentle ospreyBOT
when the yass pills hit
digital wraith
since $H_n(X) \to H_n(Y)$ is the zero map
gentle ospreyBOT
Namington + mniip 2024
fading vale
Yea similar, just for a general # of cones
fading vale
Uh we factor through the wedge sum to get direct sum
Yeah
And then apply the first part
digital wraith
yeah right I see that
gentle ospreyBOT
when the yass pills hit
fading vale
Ur correct that H_n(X) -> H_n(Y) will be the 0 map
digital wraith
right
fading vale
hence H_n(Y) -> bigoplus is injective
(kernel = image of last map = 0)
so basically you need to compute the kernel of the map bigoplus_k Hn(X) -> X
that will be the image of H_n(Y) -> bigoplus_k Hn(X)
hence isomorphic to H_n(Y)
digital wraith
and what is that kernel exactly?
fading vale
well like, we have all these cones sharing a base X x {0}
digital wraith
yeah
fading vale
Uh im not going to prove this exactly but I think its not unreasonable to see that if f: Delta^n -> X generates the nth homology of X then the generator for each cone is going to be like Cf: C Delta^n = Delta^n+1 -> CX
which kinda makes sense intuitively i think
digital wraith
I think I might be getting too tired to follow
fading vale
Yea thats fair
before u sleep the intuition is basically that the generator for each component of our bigoplus_k
they all share a boundary, the generator for X
so that all but one of them dies
digital wraith
ok but like
fading vale
Hence the kernel is going to be bigoplus_k-1 H_n(X)
digital wraith
right so that's the image of Y?
obtuse meteor
Yea, that's what we are trying to get to. I understand that
If you star by beta on the left and beta bar on the right I think you get something nice
digital wraith
(only took me banging my head against a wall for 2 days and then having the solution be the one thing I didn't think to try)
digital wraith
anyway i hope that helped gamma
yeah that was very helpful thanks
digital wraith
this is my first real problem set for homology
fading vale
simplicial homology is an evil demon from hell
digital wraith
isn't this singular homology?
fading vale
Singular too
digital wraith
simplicial homology is just like combinatorial
fading vale
Yea
digital wraith
you don't have these uncountably infinite chain groups
oh wait I have one more question
digital wraith
that's what I thought. It was just funny to get that problem in like 30 seconds after spending hours on the previous one
fading vale
Lol yeah thats usually how it is with homology
the geometry is the hard part
the algebra is very very nice
digital wraith
well it seems like here it's just a question of what subspaces to quotient by
fading vale
Usually anyway...
digital wraith
and I just got bamboozled since it turns out the trick was to not actually look at the subspace whose homology you were trying to compute
fading vale
Day one of algebraic topology: haha loops are so cute
day 500: my E_14 page hasnt degenerated yet
digital wraith
and just incidentally form it as a quotient so it was in the right place in the LES
digital wraith
Day one of algebraic topology: haha loops are so cute
day 500: my E_14 page hasn...
E_14?
fading vale
Spectral sequence meme
digital wraith
that's like a homology thing right?
fading vale
Uh its certainly related
digital wraith
well like a tool for computing homology
fading vale
Yes it is
digital wraith
how far into alg top are you Moth?
fading vale
Loaded question 
Idk
I think on the threshold of transitioning into more advanced stuff
fading vale
Probably not all of the appendices but i think all of the main content yeah
digital wraith
idk yeah I realize this is a weird question
since progression isn't really linear
fading vale
Once you finish hatcher i highly highly highly recommend finding some source that treats homotopy theory in a less... schizophrenic for lack of a better word?? way
a lot of very intro AT stuff is illuminated by some basic categorical relations
In a way that will make the transition so much less painful
digital wraith
my instructor is giving me some of that
digital wraith
oh I don't know that stuff
fading vale
They are very very nice
digital wraith
what is the duality between loop and suspension
fading vale
basically Maps(X, Omega Y) cong Maps(Sigma X, Y)
digital wraith
oh wait I think I have done this
fading vale
This happens because Omega Y = Maps(S1, Y) and Sigma X = X smash S1 and smash products are tensors in the category of pointed spaces 
fading vale
Theres just nice things like
digital wraith
my advisor for my reading course is so funny
I'll like ask him a question and he'll go on a 30 minute long tangent using math I barely know
fading vale
Theres a certain kind of space called an H space (you have probably seen them) where like Maps(X, Y) has a natural group structure when Y is an H space
and a dual notion called coH space where X coH -> Maps(X, Y) has a natural group structure
And basically suspensions are always H spaces and loops are always coH spaces and a lot of intro pi_n homotopy theory falls out of this really fast
digital wraith
yeah this seems like exactly the kind of thing he'd go on about
and also the kind of thing I would be very interested in hearing more about if I wasn't so tired
I want to say something more about my reading course instructor but it might be a bit of a self dox idk
fading vale
oh a fun consequence of this: pi_2(X) = Maps(S^2, X) = Maps(Sigma Sigma S^0, X) = Maps(Sigma S^0, Omega X). you get two group structures on pi_2(X) then, one from Sigma S^0 being an H space and another from Omega X being a coH space. there are certain theorems (eckmann hilton) that say that when 2 group structures exist under certian conditions theyre the same and always commutative
digital wraith
not really since it mostly just reveals what school I go to
fading vale
so this is a cute way to show commutativity of higher htpy groups
digital wraith
which I think I've said in the past
digital wraith
but it was like a while back
fading vale
anyway you should sleep! the math will be waiting in the morning
digital wraith
indeed. My Goldwater scholarship application will also be waiting in the morning 
*applcation
fading vale
digital wraith
lmao typo that made it seem like I actually won it
when I very much did not
and almost certainly will not
since I am starting the application late and have no idea how to write a research essay
fading vale
well good luck regardless
i am also going to go to sleep now since i have school in the morning
digital wraith
oh yeah you're still in highschool 
fading vale
Pain 
digital wraith
what are you even going to do for college Moth
like you know a good deal more than is taught in the average math major already
fading vale
Vibe
Idk take a bunch of cool courses and stuff hopefully
Maybe i will jsut drop out and then live in a hole in the ground like a hobbit
r/nonbinarylivingspace
ok i am going to sleep for real, good night
marsh ibex
Let $E$ and $E'$ two normed Vector Spaces. $\ f:A\subset E \to E' \ \lim_{x\to a}f(x) = b \iff \forall V$ neighborhood of $b$ in $E'$, $\exists U$ neighborhood of $a$ in $E$, such that: $f(U \cap A) \subset V$. i didn't get it honestly.
gentle ospreyBOT
Salah
quasi forum
I could use a little help with 3).
So if I understand this correctly, the goal is to show B has k homeomorphic copies in E.
Although I am not too sure how to show that.
gritty widget
since B is connected, maybe you can do a "clopen set proof" with {b \in B: p^{-1}(b) has k elements}
quasi forum
Well, since B is connected, we can take an open neighborhood of b_0 that contains some b. Then the pre image of this has k homeomorphic copies, and since p^-(b) is contained in this nbhd, we have k elements
Actually no, not sure that works. But since B is connected, maybe we can show that B is evenly covered by p
empty grove
teppa is suggesting take the set of all elements with fibre having k elements and proving that this set is clopen
quasi forum
Why would that be effective?
gritty widget
what's a clopen set in a connected space
quasi forum
Ah, the space itself or the empty set
gritty widget
and you know this one's non-empty
quasi forum
But we already know it is not empty
gritty widget
bingo
quasi forum
0kay, interesting. So how can we use the connectedness of B to show that our set is clopen?
gritty widget
🤨
quasi forum
Sorry, let me rephrase
gritty widget
for showing it's clopen i'd probably play around with the definition of a covering map
this i have not done
quasi forum
Well this is the definition I have to work with.
gritty widget
quasi forum
Okay, I guess the best way to extend this is to take an evenly covered nbhd for each b in our set and take the pre image of the union
quasi forum
Not sure honestly. It just seems like in order to use our definition, we need a nbhd that is evenly covered to make much use of it.
gritty widget
surely
empty grove
pick a point with m points in its fibre, show that is has a neighbourhood in which every point has m points in the fibre (hint: ||evenly covered neighbourhood||)
apply this with m = k and m \neq k and you get both open and closed
gritty widget
dammit sniped
empty grove
quasi forum
apply this with m = k and m \neq k and you get both open and closed
Oh good lord...that's easy 🤦♂️
Thanks guys, that makes a lot of sense. Fair warning: I am gonna need a bit of help today. Algebraic topology is a bit difficult for me
pearl holly
I haven’t done any topology whatsoever in like a week now
quasi forum
Sometimes I wish I could say the same. It is very interesting, but also quite difficult
pearl holly
Ye it do be hard 
quasi forum
Okay, so now I need to show that Example 3 is a covering map. Then I need to generalize and show that p(z)=z^n is a covering map of S^1
One thing I am noticing is that this is an n-fold covering of B, since there are exactly n values of z s.t. p(z)=z^n
empty grove
hint: small enough basic open sets are evenly covered
actually remove the "small enough"
quasi forum
Yea, it doesn't actually matter. The only relevant thing is the nbhds are disjoint.
pearl holly
oh is this the second part of Munkres?
quasi forum
Sure is
pearl holly
oof I see, I haven't read the second part tho
quasi forum
Okay, so I think the main issue is to show that p(z) is continuous
Then the rest is rather straightforward.
empty grove
quasi forum
Doing it with epsilon delta sounds like hell tbh
empty grove
ye don't
quasi forum
Is there a nicer way to do it?
empty grove
yes
quasi forum
Show me da wae
empty grove
multiplication C x C to C is continuous
if you have to prove that too
all the best 
you are on your own
quasi forum
Okay easy. Multiplication CxC to C is continuous, and since p(z) is just restricted multiplication, it is also continuous
empty grove
😌
it is composition of the diagonal map S^1 → S^1 x S^1 and the multiplication map S^1 x S^1 → S^1
quasi forum
Wait, ya lost me on the diagonal map part
empty grove
diagonal maps are always continuous, in fact homeomorphisms onto their images
diagonal map is x maps to (x,x)
quasi forum
Oh, we are still talking z^2 here.
empty grove
ye
quasi forum
But I do see how this generalizes to n rather easily
empty grove
😌
quasi forum
Okay, lastly, since we want disjoint neighborhoods, we take an open nbhd with a difference in argument of pi/n
Then the preimages will be disjoint.
empty grove
They should be disjoint even without that
as long as the neighbourhood is not the whole circle
quasi forum
I wanna play it safe, because I am not so sure
Okay, I need help on 6b.
Actually hold up
empty grove
oh boi I remember spending a lot of time on this one
quasi forum
I almost have it. My only issue is to ensure that the preimage of the image of a finite subcover is an evenly covered neighborhood
If I can do that, then I am set.
empty grove
That doesn't sound semantically true, because preimage of anything is in E and the evenly covered neighbourhoods are subsets of B
quasi forum
Because if you take a cover of evenly covered neighborhoods of B, you can take a finite subcover, say U_1,..., U_n. Then the preimage of this cover definitely gives finitely many open sets that cover E.
But of course, we need to start with an arbitrary open cover of E.
And that is where my issue is.
empty grove
yep
You get a cover of B from an arbitrary cover of E by taking all the evenly covered neighbourhoods of B whose preimages are contained in a finite union of elements of the cover
With this, the only thing left to show is that this is an open cover of B
empty grove
You get a cover of B from an arbitrary cover of E by taking all the evenly cover...
edited the claim
quasi forum
Now 1) how would you do that, and 2) why is that the last step?
Ah, I understand 2) now
empty grove
Nice
Here's how you can prove 1. Take an element b of B. Let F be its fiber. Because C is a cover and F is finite, there are finitely many open sets in C such that their union contains F. Try to prove that this union contains the preimage of an even covered neighbourhood.
quasi forum
I haven't the foggiest :/
And here I thought this one was not gonna be too bad
Looks like someone has been waiting to use that emoji for a while
empty grove
lol this one took both Saketh and me hours each when we were taking the topology course
and we came up with different solutions
I could also send you his after this
So have you seen the tube lemma yet?
quasi forum
Yes?
empty grove
Nice, you should notice that the statement here is very similar
You have a cover of a set F, and you need to show that this actually covers a neighbourhood of F of "uniform width"
Does that make sense?
quasi forum
Somewhat. I do understand that description w/ the tube lemma
empty grove
Nice. So this is actually a consequence of tube lemma itself
But if you are not able to see that immediately, you can try and mimic the proof of tube lemma for this
quasi forum
Oh, look at ExB
empty grove
I don't think that would work
The compact space here would be a finite discrete set, with cardinality that of F
So here is one useful way of viewing covering spaces
A map p: E → B is a covering map iff for each b in B, there is a neighbourhood U of b, such that p restricted to preimage of U looks like projection from U x F to U where F is a discrete set
The idea is that disjoint union of a space a bunch of times is the same as taking product of that space with a discrete space
quasi forum
You're starting to lose me :/
empty grove
oof ye
Let me try something with less definitions
So here's what we want to prove
Given an open subset V of E containing F (the preimage of a point b, finite), there is an evenly covered neighbourhood of b, U, such that preimage of U is also contained in V
quasi forum
Here's how you can prove 1. Take an element b of B. Let F be its fiber. Because ...
Wait... the definition of our cover is that the pre image of the evenly covered nbhd is contained in finitely many sets
quasi forum
You get a cover of B from an arbitrary cover of E by taking all the evenly cover...
That is what we are looking at here.
empty grove
Yes, but the union of finitely many sets is also an open set
quasi forum
No questions there
empty grove
And finite union of finite unions of open sets is again a finite union
so we may replace the cover C with this cover of finite unions if we want
ie C' = { finite unions of elements of C }
so we may assume that C is closed under finite unions without loss of generality
quasi forum
Are we calling C the cover for B?
empty grove
C was the cover of E, D the cover of B
quasi forum
Oh goodness, okay sure. I have U as my cover for E and V the potential cover for B.
quasi forum
So how does any of this show that V, or D, is an open cover of B?
empty grove
For any point b, we are able to find a neighbourhood of it whose preimage is contained in a finite union of elements of U
So there is a neighbourhood of b contained in V
quasi forum
But I'm not sure we've shown there is a nbhd for each point in b.
empty grove
ye that is what we need to prove
I would suggest going through the rest of the proof first and seeing if everything seems fine
quasi forum
Wait. E is a covering map. So we know there is an evenly covered nbhd around each b, and the pre image is contained in finitely many elements of our cover, since the fiber is contained in finitely many elements of the cover.
empty grove
The last bit of reasoning does not seem fine to me
The fiber of b is smaller
The preimage of the neighbourhood may very well end up too large to fit in finitely many elements of the cover
quasi forum
Ugh. When you're right, you're right
empty grove
empty grove
I would suggest going through the rest of the proof first and seeing if everythi...
ye so the reason I am suggesting this is that this last part can be a bit annoying and it will be easy to get lost in the words
it would be better if you can clear up any doubts regarding the rest of the proof first
quasi forum
This is what we have so far: I have no issues with it
quasi forum
so we may replace the cover C with this cover of finite unions if we want
I dont think replacing the cover is helpful, since we arent showing the original cover has a finite subcover
empty grove
The claim is that C' has a finite subcover iff C does
This is fairly easy to see from finite union of finite unions being finite unions
quasi forum
Oh I see, it is just to condense things
empty grove
ye
It is just to make it easier to say what the cover of B is
instead of finite union we may just say a single element
quasi forum
Gotcha. I am following
empty grove
Cool
So there are a couple ways of proving the claim: either we realise that this is an instance of the tube lemma and apply that, or just do it in a hands on way
quasi forum
Which ever way is more efficient
empty grove
I would definitely recommend the former because it will be a more useful view in the long term
and yes more efficient too
So first notice that a disjoint union of X with itself |F| many times is homeomorphic to X x F where F is given the discrete topology
quasi forum
What is |F|?
quasi forum
Ummm, sure
empty grove
The reason that this is useful is that the preimage of an evenly covered neighbourhood U of B is homeomorphic to the product U x F then
where F is the fiber of any point in U (we know all fibers of points in U have the same cardinality)
And p restricted to the preimage of U really looks like the projection U x F → U
And this is the other definition of covering maps that can be useful if you have theorems about product spaces but not for disjoint unions
empty grove
And p restricted to the preimage of U really looks like the projection U x F → U
Applying tube lemma to this instantly gives us what we need in our case since F is finite (therefore compact) and we get a tube around b whose preimage is contained in a finite union of elements of the cover of E
quasi forum
How is this contained in finitely many elements of E?
empty grove
Let me know if this is overwhelming, I can tell a hands on solution too
empty grove
How is this contained in finitely many elements of E?
U x F technically isn't, but we are identifying it with a subset of E via a homeomorphism
quasi forum
Not sure what you mean
I understand everything except how this relates to the cover E
empty grove
The preimage of U is a disjoint union of open subsets of E, each of which are homeomorphic to U via p
And this disjoint union is then homeomorphic to U x F
There is a subtlety here because we have to check the commutativity of a certain diagram
quasi forum
Oh goodness.... E isnt the cover, it's the space 🤦♂️
quasi forum
But okay, I still do not understand how we get a finitely many elements of the cover containing our nbhd
empty grove
The union of finitely many elements of the cover contains the preimage of b, therefore by the tube lemma contains a tube around the preimage of b
That's it
quasi forum
I think we are still running into the issue that we dont know this is true for every b
empty grove
b was taken to be arbitrary in the beginning
quasi forum
Oh okay, sure. So how does the tube lemma help us here?
empty grove
Because we have a projection U x F → U, and we are taking a point b in the image and saying that the preimage of b (which is a vertical line) is contained in an open set and therefore a tube around this vertical line is contained in this open set
The last part coming from the tube lemma
quasi forum
And how does this translate over to the cover?
empty grove
This seems like there's a lot going on here, but really all of the work is in recognising this alternate definition of a covering map
empty grove
So here's what we want to prove
> Given an open subset V of E containing F (the ...
this is what we have proven
quasi forum
Okay, I have one last question. I really think I just need to move on to that one for now. We can come back to this after. This is a lot
Okay, I need to do 4). I have never really worked with functoriality before, so I am not sure how to translate the continuous map to the homomorphism r_star is surjective.
empty grove
if f: A → B and g: B → A are maps of sets, and fg is identity, then f is surjective and g is injective
r being a retraction really means that ri = identity where i is the inclusion of A into X (which is omitted in the above statement)
quasi forum
I assume you mean the identity of A
quasi forum
Hold on. J is injective, sure. But does that mean j_star is injective?
quasi forum
$r\circ j =id_A \Rightarrow r_{\star}\circ j_{\star}=id_{A_{\star}}$
gentle ospreyBOT
dackid
pearl holly
⭐
quasi forum
Should just be pi_1(A,a_0)
empty grove
It is supposed to be a map
quasi forum
Well, it is the identity map of that
empty grove
yes
empty grove
if f: A → B and g: B → A are maps of sets, and fg is identity, then f is surject...
Now apply this
quasi forum
Surjectiveness I get, but I may need a bit of convincing as to why j_star is injective
empty grove
if a and a' map to the same image under j_star, they map to the same image under r_star j_star
so they map to the same image under identity
quasi forum
Ah and since it is the identity map, a=a'
I'm gonna stop there. The hw is due soon, but I need a lot more time to process 6b. So I'll take the L and I'll go over it with you more later.
Thank you so much for the help Moldilocks
You nearly melted my brain today
quasi forum
I'd prefer to talk about it after class. Otherwise, I'd pretty much be copy pasting the problem, and I don't feel right about that
Plus, I'm not ready to think about it right now. Brain needs to simmer down xp
empty grove
sure, feel free to ping (though I may fall asleep at any point now)
quasi forum
It might not be today, but I'll try to remember to ask about it over the weekend.
gritty widget
Is neighbourhood of every point on an unit circle homeomorphic to R?
I had to explain why this is true on exam
But I'm not even sure if it's true
lol
gentle lark
How do you define neighborhood?
If it's just an open set that contains the point, then the answer is no
There is an epsilon for which every ball of radius smaller than epsilon is homeomorphic to R in the circle, though, if that's what you're looking for
Metaquestion for the regulars here: how is point-set topology advanced mathematics?
(not that the division of topics is necessarily wrong, but I'm curious cause my guess is this is probably country-dependent)
ivory dragon
Metaquestion for the regulars here: how is point-set topology advanced mathemati...
its a course thats mainly taken by mathematics students (and some physicists)
thats where we put the cutoff
roughly speaking
gentle lark
Hmm
I'd argue that relates more to applicability outside mathematics than actual complexity of a subject
ivory dragon
i mean, are you in a country where point-set topology is done in first year?
gentle lark
No, but that's just Calculus+Linear Algebra+Discrete Math
It is a third semester course, though
ivory dragon
right, and #calculus , #linear-algebra , #discrete-math are all in the early uni section
gentle lark
It *is* a third semester course, though
Just as probability, for which you need Calculus II!
ivory dragon
the lines are vaguely defined, but the idea is that the Advanced channels are the ones where you can expect students to be at least a bit mathematically mature
gentle lark
ODEs are fourth semester, though, and you could also just never take a course on PDEs here and get a PhD
ivory dragon
like, responding to a long question with "take equivalence classes under [relation]" or "induct on k" would often be sufficient here
expecting the reader to be mature enough to fill in the gaps
which you cant expect in the early uni channels
gentle lark
Hmm I guess point-set topology might be a weird in between then, at least in my mind
I feel like that was a transitional course for me in that regard
ivory dragon
point set topology is typically taken after a first analysis course in the US
not always but typically
gentle lark
Analysis course as in some calculus or as in measure theory?
ivory dragon
proof-based calculus, so like baby rudin
gentle lark
Two analysis courses are a pre-req for topology here
ivory dragon
to be clear, this server isnt using "advanced" to mean "graduate level"
99% of this servers audience is well, well below that
"advanced" is a relative classifier
gentle lark
Oh, yeah, I do have that in mind
digital wraith
Two analysis courses are a pre-req for topology here
wait topology is a 3rd semester course and has 2 semesters of analysis as a prerequisite?
gentle lark
Uni degrees are far more specialized here than in the US, I think
ivory dragon
the goal of these channels is basically to isolate the "serious math talk" from the chaffe
digital wraith
99% of this servers audience is well, well below that
are they?
like the regulars
gentle lark
You sign up for a BSc in Mathematics from the get-go and mostly just do math
ivory dragon
you know, the kind of stuff that a math grad student might actively want to discuss
ivory dragon
like the regulars
yes.
digital wraith
what do you define as "graduate level"
ivory dragon
vague
digital wraith
like does Hatcher count?
gentle lark
Some of the stuff that gets asked here is graduate level imo
digital wraith
because it seems like a lot of people here know Hatcher
ivory dragon
because it seems like a lot of people here know Hatcher
your exposure is very very biased
digital wraith
I mean the regulars
digital wraith
not the people who just come to ask for help
gentle lark
How many graduate students are there in this server? (sorry if I'm derailing the channel btw)
digital wraith
I think quite a number, but still very much a minority
digital wraith
Nami, nG, Moth
ivory dragon
thats probably a high estimate
ivory dragon
grad student
digital wraith
how many profs do we have
gritty widget
There *is* an epsilon for which every ball of radius smaller than epsilon is hom...
How
ivory dragon
mniip is an undergrad
ivory dragon
icy, buncho, ashura have phds
plain raven
mniip is an undergrad? wow
ivory dragon
probably someone im forgetting
digital wraith
mniip is an undergrad
you can't fool me Nami
plain raven
knows a lot of proof assistant stuff for a UG
ivory dragon
he is
digital wraith
or wait am I dumb
ivory dragon
hes a physics undergrad
digital wraith
and confusing mniip and gomez
gritty widget
How does that prove its homeomorphic to R :/
gentle lark
Ah
You want the homeomorphism between R^n and a ball in R^n?
ngl I think it's best if I leave that as an exercise
gritty widget
Take unit circle in R2, take some point on it, prove it has neighbourhood homeomorphic to R
gentle lark
You can just write an explicit map
ivory dragon
construct homeomorphisms ℝ → (0, 1) → part of a circle
gentle lark
^
btw while we're on the topic, are there any graduate students here who're going down the road of specializing in differential (geometry/topology)/relativity/geometric analysis?
digital wraith
probably better to ask in #discussion
stone cipher
is any (topological) n-manifold compactly generated?
orchid forge
every locally compact Hausdorff space is
stone cipher
every locally compact Hausdorff space is
since n-manifolds are locally compact Hausdorff separable, they are compactly generated. thx
remote beacon
mniip is an undergrad? wow
There's a few high schoolers in this server that know a LOT too
TLDR there's a lot of math nerds on this server
digital peak
but he also has a full time job
working half time right now
quasi forum
TLDR there's a lot of math nerds on this server
Shocking
I was totally expecting this to be a server filled with athletes.
bleak path
mathletes
shy moss
how can i prove that $\pi_n(RP^k)=0$ when $n<k$?
gentle ospreyBOT
Or x1
flint sonnet
Use that $\mathbb{R} P^k$ is a quotient of $S^k$ by a finite group, and that $\pi_n(S^k) = 0$
gentle ospreyBOT
polikuj2
shy moss
Use that $\mathbb{R} P^k$ is a quotient of $S^k$ by a finite group, and that $\p...
thanks
obtuse meteor
specifically use that S^k is the universal cover of RP^k
dusky vault
Can someone help with this?
I was recommended to ask here when I got no responces
hopefully that's cool
ivory dragon
Specifically I don't understand why: x in int K
definition of neighbourhood. K is a neighbourhood of x, so a subset of K is an open set containing x, hence x must be in K's interior
why the two sets at the end are disjoint.
set theory
U and V are disjoint
long coyote
Prove that if $\phi : B^n \to X$ is a characteristic map for an n-cell $c^n$ in a
Hausdorff space $X$, then $\bar{c}^n = \phi(B^n)$ is the closure of the open cell $c^n = \phi(int(B^n))$
in $X$.
ivory dragon
intersecting U with the interior of K means that its elements must be in both U and int K (hence in K)
gentle ospreyBOT
亜城木 夢叶
ivory dragon
unioning V with (X \ K) means its elements are either in V (hence not in U since U, V are disjoint) or in X but not in K (hence not in int K)
so elements of both U and intK cant be in V or in (X \ K), and vice versa
so these sets are disjoint
Does that make sense? @dusky vault
long coyote
pick a point $x\in\bar{c}^n$, and let $U$ be its neighborhood, how do i show that $U\cap c^n\neq\emptyset$
gentle ospreyBOT
亜城木 夢叶
dusky vault
Hello Just got back to this chat
Hey your arguments make sense
The definition I have for neighborhood of x is that it is an open set containing x
I don't they are using that here
ivory dragon
thats fine, rephrase the argument to use "Let K be a compact set containing a neighbourhood of X"
x is still in intK and K is still closed
which is all you need
dusky vault
Ah ok thank you, nice reference btw
digital wraith
are integral homology groups always Z-modules?
this feels like a really obvious question
plain raven
for me, a Z-module is equivalent to an Abelian group
and integral homology groups are defined as a kernel/im in a diagram of abelian groups
you can prove that taking kernels and quotients in the category of abelian groups always gives abelian groups again
it's closed under all these notions
there's not too many operations i can think of on abelian groups that gives a non-abelian group other than like, the automorphism group
vocal anchor
there's not too many operations i can think of on abelian groups that gives a no...
Forget to Set and take free group 
plain raven
yes well if the definition of the operation involves "leave the category of abelian groups" then indeed you will leave the category of abelian groups
coarse night
ok so say I send the outer part of S2 to R^2\{0,0} trivially, where should I send the inside part of S^3?
rancid umbra
is this a T/F question?
rancid umbra
you cannot possibly have a homeomorphism, sorry thats what i was thinking
coarse night
using rough terminology here
coarse night
you cannot possibly have a homeomorphism, sorry thats what i was thinking
yes but we are asked to find a onto mapping not bijective so it's fine
homeomorphism isn't possible that I can understand
rancid umbra
nvm that wont work
coarse night
yeah
rancid umbra
can we map R^2 - S^1 to R - {0} in the same fashion?
coarse night
|r| will work here
rancid umbra
|r|?
gritty widget
rancid umbra
what value of r in R^2 - S^1 is there for which |r| < 0?
coarse night
oh yeah lol
coarse night
can we map R^2 - S^1 to R - {0} in the same fashion?
oh it's R^2-S1 now, u changed it, now my answer is outdated and stupid
rancid umbra
i think for R^2 - S^1, you wanna send the inner part to the negative half of the real line and the outer part to the positive half of the real line, both in a continuous manner
rancid umbra
oh it's R^2-S1 now, u changed it, now my answer is outdated and stupid
my bad, i typed the wrong thing the first time
rancid umbra
i think for R^2 - S^1, you wanna send the inner part to the negative half of the...
we should be able to generalize this idea
coarse night
like we can try to map |r|>1 to R+ and |r|<1 to R-
the continuity is still bugging me
rancid umbra
bro... what is |r| ?
coarse night
x^2+y^2
rancid umbra
like we can try to map |r|>1 to R+ and |r|<1 to R-
??
coarse night
the outside of S1 to +ve reals and inside of S1 to -ve reals
since it have to continuous, there are limited choice
rancid umbra
lol okay. i thought you were using a function i wasnt familiar with. you keep writing |r| < 0
think you meant |r| <1
coarse night
lol x^2+y^2 is too long to type
coarse night
think you meant |r| <1
yeah already edited it
rancid umbra
the problem kinda reduces to finding a continuous surjection from R^2 to R
coarse night
the problem kinda reduces to finding a continuous surjection from R^2 to R
does it?
rancid umbra
pretty sure. both inner and outer parts are homeomorphic to R^2
coarse night
yeah
but you can map inner or outer to R+ or R-, because they are connected
so there isn't any cont mapping from inner/outer to whole R (surjective)
rancid umbra
um. okay
coarse night
so like inner has to go to R+ or R- and to be surjective outer must go to R- or R+
rancid umbra
project R^2 - S^1 onto R
just take the first component function
this is a continuous surjection
oh dude i got it
take R^3 - S^2 to R^2
then use R^2 to R^2 - {(0,0)} via polar coordinates
then compose the two maps
coarse night
polar: (0, 0) \maps to (0, 0) 
rancid umbra
nah, dont include (0,0)
rancid umbra
no this should be really close i feel like
yea, you can just map R^2 to (0,infty) x R and then use polar
so, p : R^3 - S^2 --> R^2 given by p(x,y,z) = (x,y) is a continuous surjection
then take f : R^2 --> (0,infty) x R by f(x,y) = (e^x, y). continuous bijection
then use g : (0,infty) x R --> R^2 - {(0,0)} by g(r,t) = (r sin(t), r cos(t)). continuous surjection
then you do g o f o p : R^3 - S^2 --> R^2 - {(0,0)}
gritty widget
gofop
rancid umbra
gofop is fs the way to go here lmao
digital peak
I guess I'm trying to lift a path intro a covering space
more specifically, given $p : E \to B$ a cover, given a path from $x$ to $y \in B$, I'm trying to define a "canonical" map $p^{-1}(x) \to p^{-1}(y)$
gentle ospreyBOT
mniip
digital peak
I can kinda intuit what's going on, but I'm struggling with the technical details
the definition of a covering space gives us for each $x \in B$ a neighborhood $U_x$ such that $p^{-1}(U_x)$ is composed of some disjoint opens that are homeomorphic to $U_x$
gentle ospreyBOT
mniip
digital peak
so if we have a path between $y, z \in U_x$, given a choice of $p^{-1}(y)$ we can lift it to a path in $E$ (in the respective part of $p^{-1}(U)$)
gentle ospreyBOT
mniip
digital peak
If the path doesn't "fit" in one U_x, I figure I gotta use the compactness of the interval somehow? Find a finite cover of the path by sets of the form U_x
However there's a problem: how do I know that the lifting is coherent on the overlaps?
if I have path in $U_x \cap U_w$, how do I know that the "$x$ lift" coincides with the "$w$ lift"
gentle ospreyBOT
mniip
digital peak
I guess that's just contravariance of p^-1?
Really what the covering space is saying is that for every $x$ there exists $U_x \ni x$ open, a cardinal $\kappa_x$, a family ${V_{x,\alpha}}{\alpha \in \kappa_x}$ of opens, such that $\bigcup{\alpha \in \kappa_x} V_{x,\alpha} = p^{-1}(U_x)$, and such that $\left.p\right|{V{x,\alpha}}$ is a homeo with $U_x$. And the opens are disjoint, meaning there is a map $h_x : \left(\bigcup_{\alpha \in \kappa_x} V_{x,\alpha}\right) \to \kappa$ such that $e \in V_{x,\alpha} \iff \alpha = h_x(e)$
gentle ospreyBOT
mniip
coarse night
then you do g o f o p : R^3 - S^2 --> R^2 - {(0,0)}
digital peak
actually does this help
rancid umbra
just do gofop and it works @coarse night
digital peak
connectedness of the interval gotta play a role too
digital peak
ok yeah
If I have a connected space $Y \subset U_x \cap U_w$ with $y \in Y$ and a choice of $e \in p^{-1}(y)$, that fixes $h_x(e)$ and $h_w(e)$. I claim that $p^{-1}(Y) \cap V_{x,h_x(e)}$ (a homeomorphic image of $Y$ in $E$) lies within $V_{w,h_w(e)}$ and vice versa
gentle ospreyBOT
mniip
digital peak
even though it's not necessary that $V_{x,h_x(e)} = V_{w,h_w(e)}$
gentle ospreyBOT
mniip
digital peak
wait this suddenly got very sheafy
obtuse meteor
I mean
etale space
it shouldn't be too surprising that covering space theory is sheafy
digital peak
ok so how the fuck do I glue things
I can kinda see how to lift a path
but how do you lift a square?
plain raven
mniip i think hatcher's argument is pretty readable
for the homotopy lifting property of covering space maps
plain raven
jesus i thought you were talking to me
i was like "Damn that was a heavy dose of sarcasm"
solar raptor
what are the pre-requisites to learn algebraic topology? i don't know if pre requisite questions are allowed here sorry if it is the case
plain raven
you need to know point set topology well, you could read munkres or take a course in real analysis
beyond the topology that's in a good real analysis book, you need to know about identification spaces and the quotient topology
and connectedness
you need to know basic algebra, group theory, you need to know what a group is and how to do computations in a group
for homology you need some basic facts about abelian groups like the fundamental structure theorem for abelian groups, the fact that a subgroup of a free Abelian group is free, and the fact that finitely generated free groups have a well defined "rank"
you'll need to know more algebra later on, but that's enough to get you started.
eventually you'll need to know about rings and modules, the tensor product, graded rings and modules, exterior algebra... etc but i guess you can learn that when you get to it
orchid forge
you need to know point set topology well, you could read munkres or take a cours...
the first few chapters of munkres, to be clear :D
solar raptor
so what course should i do? i am doing abstract algebra,but i dont know if real analysis is necessary
orchid forge
besides algebra, the best course would be a course in point set topology
which will most likely include the fundamental group at some point
you could get by with only reading (selected portions of) the first few chapters of munkres and diving right into homology
solar raptor
does algebraic topology have other applications in computer science ,instead of only distributed computing that i have found
orchid forge
certainly
one example i happen to be aware of is using sheaf theory as a model for data fusion problems
sheaves are a topological object that describe "local data" and "gluing local data to see a bigger picture," so you can use them to describe e.g. a network of sensors, each one seeing a local picture, and gluing that data together into one big picture
then you can interpret results from sheaf theory in that practical context
plain raven
there's something called topological data analysis as well
solar raptor
i thinking about doing my undergraduate final assignment about applications of abstract algebra or other math topic i want to learn in cs ,because i want to learn more math topics and also get some applications from it
solar raptor
there's something called topological data analysis as well
i am gonna check this
plain raven
No the discrete topology is not very interesting from the point of view of topology.
orchid forge
you can sometimes assume that your discrete data is sampled from a continuous distribution
solar raptor
i deleted it wasnt a good question
obtuse meteor
one example i happen to be aware of is using sheaf theory as a model for data fu...
do you have a good source on this? It sounds pretty interesting
orchid forge
digital peak
Theres more to it than it seems
Why is this lift independent of the finite subcover chosen?
Not just the subcover, you're choosing some points too
How do you show that homotopic paths lift to homotopic paths? You meed to lift a square. Lifting a square is even more complicated wrt the choice of an open cover and the points at which you choose to glue
When lifting a path all you care about is staying connected. When lifting a square you care about simply connectedness
plain raven
did you look in hatcher
digital peak
Nope sorry
digital peak
hmmmm
empty grove
✓
digital wraith
Yeah Hatcher does have a proof of this exact fact
Also mniip how do you know category theory but not the homotopy lifting property
plain raven
I like to call a proof a "just do it" proof to intimidate people.
Like, to me it's so simple i can't even understand how to describe breaking it into smaller parts
you just do it
you prove it
you proceed to understanding in one single, atomic, indivisible step
Blakers-massey is a good example of a "just do it" style proof
digital peak
Also mniip how do you know category theory but not the homotopy lifting property
I started doing cat theory when I was 16 or so
digital wraith
mniip was the high school category theorist all along
empty grove
What do those reactions mean
digital peak
\begin{tikzcd}
~ \arrow[d] & ~ \arrow[l] \
~ \arrow[ru] &
\end{tikzcd}
gentle ospreyBOT
mniip
empty grove
ivory dragon
ah yes, the supercommutative diagram
empty grove
but why is toki caling you bald
pearl holly
Oohh shit lmao
I didn’t see that the emote was a bald person sorry lmao
I just wanted to emote with a “general person”
coral pivot
Mniip is bald from all the categories
digital peak
that's the problem, I understand the statement intuitively, but I am not satisfied with the technicalities
plain raven
I ❤️ Spanier
however it is encyclopedic to a fault lmao
i got so bogged down in the homology chapters
i really spent like a year and a half of my life on the chapters on homology and cohomology
and i didn't even get that far into the chapter on duality in manifolds
just the first few sections
i honestly had like
moments of existential crisis
where i was just staring out the window going
is this how i spend my 20s
reading Spanier
bleak path
Hi, my friend and I disagree on this statement, can someone help clarify for us?
For any continuous map f: X -> Y, and for any compact subspace K \subseteq Y, the inverse image f^{-1}(K) is compact
Is this statement true or false?
I think it is false because I believe the statement is only true for closed maps, am I right in my reasoning?
orchid forge
this is the definition of a proper map
empty grove
Take both spaces discrete, X infinite, Y finite, f any map X → Y, K = Y
orchid forge
or i guess, this says that every continuous map is proper
which is not true in general
bleak path
I have not heard the term proper map before, is there a specific definition you would recommend?
orchid forge
a map is proper if the preimage of a compact set is compact
bleak path
Take both spaces discrete, X infinite, Y finite, f any map X → Y, K = Y
I see, and this would make it not-compact
Is there a term for not-compact btw, or is it just called not-compact
bleak path
Ahahhaha
orchid forge
mpact
bleak path
Reminds me of back in linear algebra
Where we had a whole definition for linear independence as "a span of vectors that was not linearly dependent"
empty grove
do you mean set instead of span
rancid umbra
Hi, my friend and I disagree on this statement, can someone help clarify for us?...
another nice/visual example imo. let pi_1 : R^2 —> R be the first component projection. this is continuous. look at the pre-image of [-1,1]
bleak path
another nice/visual example imo. let pi_1 : R^2 —> R be the first component proj...
Need a bit of explanation for this one, sorry. Subspaces of R^n are compact iff they are closed and bounded, yes? So what would the inverse image of this be? I imagine that it could be open along the second component projection and so not compact? Does that make sense?
ivory dragon
its unbounded
your preimage is {(x, y) | x in [-1, 1], y in R}
this is obviously unbounded, its an infinite vertical rectangle
so the continuous preimage of compact set [-1,1] is not bounded, ie not compact
fading vale
Maybe im missing something obvious but I'm not really sure where to go from here with this exercise
like hurewicz will get you that X is n-1 simply connected sure and that pi_n(X) cong H_n(X) cong Z
So i see why "take the spheroid f: S^n -> X representing the generator" is possible
but i dont see why it should induce isos on pi_n+k for k > 0
I guess to add on i know that X is homotopy equivalent to a CW complex with one vertex and no cells of dim 1 to n-1, and that pi_n(X) cong Z has a system of generators given by n cells and of relations given by n+1 cells
But i dont think thats particularly useful here bc ofc there are lots of different CW structures that will get you pi_n(X) cong Z lol
somehow the fact that the higher H_i are 0 has to be useful ehre
Hmmm maybe i can use cellular homology here somehow
quasi forum
Okay, so on the first part, I had to try and sketch a graph of this path.
I just wanna know if this path is okay (it took some thinking to sketch out).
Note: dotted path means out of view and normal path is in view
I also have no idea how to go about finding a lift for this, so I can certainly use help with that
It looks funky, but really the path is just wrapping around like a candy cane stripe
digital peak
taurus
quasi forum
Whoops
digital peak
do you know what p and/or pxp looks like in coordinates?
quasi forum
-_-, I flipped the order, didn't I?
digital peak
?
quasi forum
I treated the first coordinate that represents the angle about the larger circle, and the second coordinate the angle w.r.t. the circle made by the first coordinate.
But I think it is flipped
digital peak
that's hardly important
quasi forum
Okay, then I will just say no as I am not sure
digital peak
from the fig it looks like the first coordinate is the small circle
quasi forum
Yea, that's how it should be, but I did the opposite
digital peak
also that's not where f(1) is
quasi forum
Oh sorry, that was f(1/2)
digital peak
but like ok, do you know p in coordinates?
quasi forum
When you say p, do you mean polar coordinates? If not, then no
digital peak
you have a map p in your exercise
digital peak
right
now what is p^-1(a, b)
supposing (a, b) lies on the circle
supposing it has the form (cos 2pi t, sin 2pi t)
quasi forum
(2pi t, 2pi t)x(4pi t,4pi t)?
quasi forum
Wait, that does not make sense
digital peak
the question is, for which x is it true that p(x) = (cos 2pi t, sin 2pi t)
quasi forum
Sorry, I have to eat. This is gonna have to be put on pause. I will be thinking about the answer to your question
To answer your question, it should be an element of the quotient group R/Z
long coyote
Prove that if $(X,C)$ is a CW complex, then the zero-skeleton $X^0$ is discrete.
gentle ospreyBOT
亜城木 夢叶
long coyote
starts from weak topology $X_w$, pick $A\in X^0$, we have $A\cap F$ is closed where $F$ is an element of $X_w$. i don't see how this gives me $X^0$ is discrete
gentle ospreyBOT
亜城木 夢叶
cedar pebble
Okay, so on the first part, I had to try and sketch a graph of this path.
I jus...
nails
quasi forum
So I think I understand the lift portion. If I am not mistaken, it is just a straight line
Of course, the endpoints are a bit more particular than that, but the path itself is just a straight line segment. I know what they are, it is just hairy to explain
coarse night
is it true that any uncountable subset of R must have a rational limit point? the original question - "is there a perfect set of R¹ that contains no rational number"?
rancid umbra
is it true that any uncountable subset of R must have a rational limit point? th...
can u just shift the usual middle thirds cantor set over to the interval [e,pi] or something?
coarse night
how can I be sure that there are no rational numbers? cantor contains a rational number,
rancid umbra
yea but everything in the shifted set is like (e+pi) times a rational number with denominator a power of 3
rancid umbra
oh do we not know that e + pi is irrational or not?
orchid forge
nope
rancid umbra
uhhh