Well you first have to know what set you want to define a topology on.
In your case, the set of 2x2 matrices (over a fixed field k, which I assume, in your case, are the reals) can be identified with k^4, and you can define the zariski topology on it.
#point-set-topology
sturdy notch
saying that something is a subset of something else does not make it a closed set
to show that the set of invertible matrices is open in the zariski topology, you have to show that it's complement, the set of non-invertible matrices is closed
by definition of the zariski topology, that just means you have to show that the set of non-invertible matrices is the zero locus of a set of polynomials
wanton terrace
Ok, I have in my notes that "The Zariski topology on an affine variety X is the topology where the closed sets are subvarieties. The open sets are compliments of closed sets"
which makes me have 2 questions:
-
In this example it says the topology is defined over the affine variety (in the problem I have I understand the affine variety to be the set of 2x2 matrices). But you seem to imply that the topology is instead defined over the field that defines the affine variety.
-
I had read online that a subset of an affine variety is itself an affine variety. So is a subvariety something different? If not, why is it not sufficient to just show that the variety of non invertible matrices is a subset?
sturdy notch
- this is the correct definition yes, when you say "affine variety" you typically mean an algebraic variety which can be embedded in affine space.
By affine space one usually talks about the set k^n with the zariski topology
2)A subset of an affine variety is not always a variety, for instance the zero locus of y-sin(x) is not an affine variety in R^2 as it cannot be cutout by a set of polynomials.
(though R^2, with the zariski topology clearly is an affine variety)
wanton terrace
Ah 2) clears up a lot for me, thanks! But you didnt totally answer 1). Is a topology defined on an affine space, or on a field?
sturdy notch
No offense, but I feel like you're missing quite a bit of elementary point-set topology.
Here's how you typically define the zariski topology on k^n (note that here n may equal 1, in which case the underlying set is just the field)
First, you define the zariski topology on k^n by declaring that the closed sets are exactly those which are cut out by polynomial equations
Then you define the zariski topology on affine varieties (subsets of k^n cut out by polynomials) by taking the subset topology from k^n
Later on you might talk about projective varietes or schemes, but we're quite far off of that right now
If you don't know what the subset topology is (or what a topology is for that fact) you should definitively read up on that before reading about the zariski topology
Btw saying this is the typical definition is not quite accurate, but it's close enough for our purposes
wanton terrace
Yeah, ok. I will read up on that a bit more. Is the question I'm asking nonsensical somehow? Because it seems like a pretty straight forwards question to me. Is a topology defined on a field or an affine space? Because I have written in my notes something that implies it can be defined on an affine space, but I want to know if that actually implies its defined on the field that defines the affine space.
sturdy notch
yeah so you first have to learn about point-set topology before doing classical ag or related stuff.
A topology can be defined on any set, the definition has nothing to do with a field or affine space.
You should at least have a grasp of the subjects in the first half of munkres's topology before starting with classical ag
wanton terrace
ok, any set, thanks
vague kite
theres two topology channels i just guessed knots werent a differential geometry thing
unreal stratus
I think using R^2/Z^2 is nice, as any (closed) path in R^2/Z^2 lifts (uniquely, after picking a basepoint) to some path in R^2, which you can use to define your map
Though they are likely other methods too
long grail
What sets are clopen in the usual topology on Q?
gritty widget
What sets are clopen in the usual topology on Q?
Intervals with irrational endpoints
For example
unreal stratus
Any open set O in Q can be written as U \cap Q for some U open in R. We can write U as a (countable) union of disjoint open intervals. However, we can also do the same with the complement of O in Q, so I'd wager that the clopen sets are exactly the countable disjoint unions of open intervals with irrational endpoints (?)
Hm
gritty widget
Huh. They form a basis so that can't be true
Otherwise every open set would be clopen
gritty widget
I don't think so no
Pretty sure (-infinity, 1) can be written as a disjoint union of such intervals by considering increasing sequence of irrationals which converges to 1
And it's not clopen
gritty widget
@unreal stratus I think bounded clopen sets are a finite union of these though?
No, this is false too
So yeah. Not much to say other than, every interval with irrational endpoints is clopen and every clopen set is a countable disjoint union of them, of positive distant apart each, but not necessarily conversely
So the characterization of clopen subsets of Q ends here it seems
gritty widget
So yeah. Not much to say other than, every interval with irrational endpoints is...
there are both clopen and not clopen examples which satisfy this i.e. being infinitely countable sum of disjoint such intervals of positive distance apart (each can be made to be bounded)
long grail
Ah I understand
So since there are uncountably many irrational numbers, there are uncountably many clopen subsets of Q with the standard topology? Or am I way off
Or, wait
Each interval contains a rational, so it'd be countably many
gritty widget
So since there are uncountably many irrational numbers, there are uncountably ma...
Yes, there is continuum many of them
you were right the first time
long grail
you were right the first time
Thanks!
unreal stratus
(Sorry for the large bit of text but) what's the easiest way to see that this homotopy of fφ induces a homotopy of f? It seems Hatcher is skipping over some details but perhaps I'm being silly
gentle ospreyBOT
potato
echo dove
We're assuming $f$ already takes $X^{k-1}$ to $B$. The homotopy we're constructing will be defined on $X^k$, and it will be rel $X^{k-1}$ (i.e. it will fix that subspace pointwise) so all we need to do is define a homotopy on `the rest' of $X^k$, which amounts to defining it on each $k$-cell rel its boundary.
gentle ospreyBOT
daveamayombo
echo dove
But yes, I think it's fair to say this uses $\left(X^{k-1} \coprod D^k \right) \times I \cong X^{k-1}\times I \coprod D^k \times I$, together with the fact that $X^k$ is a quotient of that space, and the map we're defining passes to the quotient.
gentle ospreyBOT
daveamayombo
queen shale
But yes, I think it's fair to say this uses $\left(X^{k-1} \coprod D^k \right) \...
does $\coprod$ in this context mean disjoint subsets of D^k?
gentle ospreyBOT
Daifeng
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
queen shale
if so, use $\sqcup$
gentle ospreyBOT
Daifeng
queen shale
or is it really the coproduct?
echo dove
Fwiw, something like $X^{k-1} \amalg \coprod_\alpha D^k_\alpha$ would have been more accurate, where $\alpha$ indexes the $k$-cells, but yeah, I meant coproduct.
gentle ospreyBOT
daveamayombo
echo dove
I'm on my phone, so latex-fu is limited. 🙂
queen shale
it's just i've seen people use the coproduct symbol to denote a disjoint union
echo dove
What's the difference in this case? We're talking about topological spaces.
queen shale
What's the difference in this case? We're talking about topological spaces.
understood. I just sort of got confused with the notation
echo dove
Ok! Got worried for a sec I was missing something.
queen shale
Ok! Got worried for a sec I was missing something.
though are you implying that coproducts and disjoint unions are interchangeable? 😲
echo dove
I mean that a coproduct of a family of spaces (in Top... no basepoints, no homotopy, nothing fancy) is the disjoint union of those spaces.
queen shale
I mean that a coproduct of a family of spaces (in Top... no basepoints, no homo...
ahhhh just a definition
that's neat
or is it a lemma?
echo dove
Sure, you can use that to define what disjoint union of spaces means, I suppose. But I think it's common to describe / define the disjoint union and its topology directly, and one can then show immediately that it satisfies the universal property of a coproduct.
Given a family $X_\alpha$ of spaces, a point in the disjoint union is a pair $(\alpha, x)$ where $x \in X_\alpha$. And the topology is... hopefully what you'd expect.
gentle ospreyBOT
daveamayombo
queen shale
makes sense...
unreal stratus
But yes, I think it's fair to say this uses $\left(X^{k-1} \coprod D^k \right) \...
Sure, thanks
Whitehead's book has another proof of this lemma which I may look at too tbh
sturdy notch
a morphism from Z^2 to Z^2 is uniquely determined by where you send (1,0) and (0,1)
suppose (1,0) gets sent to (2,0) and (0,1) to ( 0,3 ) (for simplicity's sake, you can generalize this afterwards), how would you go about defining a map R^2 -> R^2 which induces this map once you take quotients and fundamental groups? (Think abt matrices)
unreal stratus
@plain raven Silly question, but you mentioned with proofs like this that it can help to draw a picture - how would you go about it when the diagrams like these are 3D? lol
Especially since we're working with pairs
I suppose this is the sort of thing where purely syntactic reasoning would be more useful lol
wise ruin
Just a quick sanity check: if phi is bijective and phi and phi^-1 both map basis elements elements (of the topology) to basis elements, then phi is a homeomorphism. Correct?
gritty widget
Just a quick sanity check: if phi is bijective and phi and phi^-1 both map basis...
Yeah
That phi does that means phi^-1 is continuous etc
sturdy notch
Yes thats the basic idea
You should try to think about linear transformations of R^2 which are well defined on equivalence classes mod Z^2
Lets call q: R2->R2/Z2 the quotient map
So fir my example, you can look at the linear map R^2->R^2 given by the matrix( (2 0)(0 3)), call that one phi.
You can compose these to get a map qophi : R2->R2/Z2
Using the universal property of quotient maps, this induces a map R2/Z2-> R2/Z2 (you should check this)
You should be able to check relatively easily that this map induces the desired map on the fundamental group, you should also try to visualise what this map is doing
plain raven
<@340405629111894016> Silly question, but you mentioned with proofs like this th...
if you want i'll try and draw a picture later
idk i would have to sit down and read through it
long grail
Compact metric spaces have a dense countable subset, is that true?
unreal stratus
Yes. Use the fact that compact metric spaces are totally bounded
unreal stratus
If we have some pushout square and I replace each space with its product with I (as in X x I, A x I etc) and use the "obvious" maps X x I -> A x I etc, does it remain a pushout?
It seems so: if I have some other space Z and the necessary diagram commutes, then for each t in I I'll get some map Y x {t} -> Z; then the only thing we need to show is that the overall map we get from stitching these together is actually continuous, which is easy if I've not done something silly
I'm asking just because tom Dieck seems to appeal to something like it here
(pls ping if you respond aha)
bronze ruin
random question: how does one go about proving that two different metrics induce the same topology on a space?
my problem is specifically asking about three metrics on R^n but i'm simply not sure how to do this in general
little hemlock
random question: how does one go about proving that two different metrics induce...
in general, you have to show that everything that is open in one metric space is open in the other and vice versa (a set inclusion of topologies).
In a metric space, this is equivalent to showing that for all epsilon > 0, there is delta > 0 such that B_d(delta, x) \subset B_d'(epsilon, x) and vice versa switching the roles of d and d'
(and the above is exactly the statement that the identity map is a homeomorphism between (X,d) and (X,d'))
tender halo
in general, you have to show that everything that is open in one metric space is...
note that by doing this in one direction you will essentially prove that one topology is finer than another
so proving it both ways will tell you that both of them are finer than each other and therefore equal
being finer is also equivalent to the identity map being continuous
graceful sequoia
Hello. Is my understanding of the standard topology on R correct? Let T be the standard topology on R. Then T = {U (a,b)}, U is meant to be union, be it infinite or finite.
vocal wharf
the open intervals are a base of the standard topology of R
gritty widget
Hello. Is my understanding of the standard topology on R correct? Let T be the s...
Yeah. The topology consists of unions of open intervals (the union might be empty)
lament needle
If U is a manifold properply embeded in M and M is a manifold embeded in N then is U embeded in N
More generally is embedding a transivitve property
gritty widget
just compose lol
long grail
If f: [0,1] to X, where X is an ARBITRARY metric space and f is continuous map, then f([0,1]) is closed. This is because [0,1] is equal to its closure and f is continuous hence image of a closure is closure of the image.
tender halo
if f is continuous, then the image of a closure is contained in the closure of the image, not necessarily equal
long grail
When is image of compact also compact? I know if X was R it would be
tender halo
image of a compact is always compact
long grail
Ok
tender halo
if X is Hausdorff, then any compact subset is closed
tender halo
yeah
long grail
Thanks
Image of compact is always compact when f is continuous because any open covering of the image would have an open covering in the pre-image that admits finite subcovering. So would any open covering of the image have a finite subcovering that is the image of the finite subcovering of the pre-image?
lament needle
continous functions in general do not have to take open sets to open sets. For example f: (0,1) \raw {0} is continous
long grail
Yeah
tender halo
but it will produce a not-necessarily-open finite refinement of the original cover
from which you can easily get a finite subcover
tender halo
yoinking an example from the internet
no, take the cofinite topogy on a countable set
its T1, but every subset is compact
long grail
Will someone critique this proof?
gritty widget
Will someone critique this proof?
Quickly without looking at the proof, consider $C = C\cap \overline{A} \cup C\cap \overline{A^c}$. If the sets $C\cap \overline{A}$ and $C\cap \overline{A^c}$ were disjoint, then this would contradict that $C$ is connected, since they are non-empty by assumption. But $\overline{A}\cap \overline{A^c} = \partial A$ by definition, so $\partial A\cap C \neq \emptyset$.
gentle ospreyBOT
Blitz
unreal stratus
Will someone critique this proof?
Where did you get U from?
gritty widget
Will someone critique this proof?
About the proof, I don't understand it. What's the purpose of this set $U$? How does what you wrote prove that $U$ intersects $\partial A$? $U$ doesn't have to be open in $X$
long grail
Well I was thinking that if a is in the boundary of a then every neighborhood of a intersects A and A complement
gritty widget
That's true. But there's no obvious choice for "a"
long grail
All right thanks for the feedback. It didn't feel right
gritty widget
Hey, I don't think I really understand definition of cw complexes. I'm trying to show that the interval (0,1) is a CW complex, but how can it be constructed by cells? I realpy don't see it. For [0,1], we can starz with the end points and then attach the line in between. But what about (0,1)???
Also, what are some simple examples of spaces that are not cw complexes?
How to see which spaces are and which aren't?
unreal stratus
I think it's easier to see that R is a CW complex. Think what the 0 skeleton could be
(Which is enough, as R is homeo to (0,1))
gritty widget
Well I guess we could take infinitely many points and connect them
Perhaps at integers
unreal stratus
Yeah exactly
unreal stratus
Yes
gritty widget
At all real numbers?
unreal stratus
Just do at Z
gritty widget
With such reasoning any space would be cw complex
unreal stratus
Since you need to have the right topology
You can't just say take every point as a 0 cell
gritty widget
Because the weak topology would be wrong?
unreal stratus
Yeah
gritty widget
Ok thank you
coarse night
Also, what are some simple examples of spaces that are not cw complexes?
do you know that all finite cw complexes are compact?
gritty widget
I did not know that, but it seems reasonable from what I was thinking
coarse night
you can try to prove it. In general X,Y is compact Hausdroff and A is a closed subset of X, then X ∪_A Y is compact (Hausdroff)
use induction to show finite cw complexes are compact
gritty widget
Ok thanks, i'll try it in the morning
coarse night
(tho it only tells you you it's not finite cw complex)
gritty widget
I think it's a nice result
coarse night
tho i think you can give (0,1) cw structure
gritty widget
Would it be ok to choose points 1/2, 3/4, 1/4, ...
And then connect them?
@wooden falcon can you explain to me what's wrong with taking all the points in any sets to be 0-cells?
As compared to taking all integers
And comnecting them with 2 cells
Which is what we do to show that R is cw complex
Well, we can say that it is a 0-dimensional cell complex then?
Ok fair
gritty widget
Do you have any tips on how to learn algebraic topology on my own
Im only going through hatcher now
I need to learn mostly about homotopy theory
For my thesis
bitter smelt
thats a fine starting point
coarse night
are you liking hatcher?
if you don't there are other books/refs
not everyone likes hatcher
gritty widget
I like parts of jt
But some parts I find pretty confusing
The main topics I need are homotopy equivalences and weak homotopy equivalences
coarse night
you can try rotman, read chap 0 1 3 and jump to ch 11 ( homotopy groups)
gritty widget
Ok I'll take a look at that
coarse night
probably not a good advice
grave maple
Are there well-know examples of inductive limits in topology?
plain raven
are you talking about filtered limits specifically
a CW complex is the inductive limit of its n-dimensional skeleta
grave maple
are you talking about filtered limits specifically
Nope. Any examples from general topology?
gentle ospreyBOT
Nobody
plain raven
That's a good example. Similarly with the sphere S^\infty, which is the inductive limit of n-dimensional spheres S^n.
plain raven
Nope. Any examples from general topology?
If you have a space $X$ whose underlying subset is equipped with an equivalence relation $\sim$, then the projection $X\to X/\sim $ can be expressed as a coequalizer
gentle ospreyBOT
diligentClerk
grave maple
That's a good one.
plain raven
A very simple example is that if you have two spaces $X,Y$ then their disjoint union $X\coprod Y$ is usually equipped with a "coproduct" topology, where $U\subset X\cup Y$ is open iff its intersection to $X$ and $Y$ is open. this is an inductive limit
gentle ospreyBOT
diligentClerk
plain raven
Any n-dimensinal manifold can be expressed as the inductive limit of a diagram of open subsets of R^n.
grave maple
The coproduct being an inductive limit follows from the fact that inductive limits are colimits and coproducts are colimits, so yeah.
Where did this concept of inductive limit originally come from? Was it in topology?
I wonder if there are any good examples from functional analysis.
plain raven
Inductive limits are found in topology and algebra. There are some examples in analysis like sometimes you can represent a frechet space as an inverse limit of banach spaces
I mean an inverse limit is dual to an inductive limit so maybe that's not a great example
grave maple
OK, I found an analysis book with some examples. It says that that coproduct of a family {Xi : i in I} of convergence vector spaces is the strict inductive limit of all the finite products of the Xi.
Oh, there is a more complicated example involving distributions and test functions.
I suppose the reason for using directed sets is because you need to be able to find a bigger space for any two spaces you pick in the system.
If you have a procedure to glue spaces, you can always make a bigger space by glueing the smaller spaces together.
plain raven
yeah. the directed sets are not necessary to for the concept to make sense but they do confer technical benefits
odd flame
is that {a,b} or [a,b]???
ornate berry
The latter
gritty widget
Cool exercise
odd flame
ok curly brackets would've been really weird anyways
gritty widget
||This is just about (a, b) being non-empty and not having maximum and minimum||
tiny obsidian
Well e.g. the closure of (0,2) in N is not [0,2]
gritty widget
Honestly, this is just an application of the following characterization of closure: $$x\in \overline{A}\iff \text{for all neighbourhoods }U\text{ of }x, U\cap A\neq \emptyset$$
gentle ospreyBOT
Blitz
gritty widget
This, applied to points a and b.
tender halo
||This is just about (a, b) being non-empty and not having maximum and minimum||
as edward pointed out, it can have maximums and minimums, a natural non-discrete example being ordinals
gritty widget
as edward pointed out, it can have maximums and minimums, a natural non-discrete...
?
For that example, the closure of (0, 2) isn't [0, 2]
sleek thicket
What is a "concrete" description of a stable infinity category whose homotopy category is the derived category of a ring R? Like how spectra are the stabilization of Top
lunar yoke
wouldn't it be the infinity category of chain complexes, presented by the usual model structure on chain complexes?
im pretty sure the htpy category of that is the usual derived category
lunar yoke
take this with a grain of salt tho, im still learning infty categories ^^
sleek thicket
But yeah just take the usual model structure on unbounded chain complexes
lunar yoke
well if its bounded complexes, then every one is qiso to a bounded complex of projectives iirc
so what effect does restricting to perfect complexes have in the unbounded case?
i dont think i've ever worked with unbounded complexes tbh
oh i see
agreed
also fits nicely with dold kan then
if by simplicial abelian groups we mean functors Delta^op -> Ab, then we get an equivalence of categories between simplicial abelian groups and bounded below chain complexes, induced by the normalized moore-complex functor. This functor is also homotopical and descends to an equivalence of homotopy categories
So I'm used to simplicial objects not having negative degrees
we covered this in a course recently
took us 3 lectures, and the only reason we did this was to get that every simplical / topological abelian group is ismorphic to a product of EM-spaces in the homotopy category
so a literal ablian group structure is "too boring"
and then we proceded to prove Segal's theorem that very special Gamma spaces are infinite loop spaces for the rest of the whole lecture course lmao
"Homotopy Coherent Algebraic Structures"
wild course
gentle ospreyBOT
Nobody
lunar yoke
wait is that just notation or is this actually related in some way?
I know the suspension spectrum as S^n smash X in degree n
does that have anything to do with abelianization?
ohhh
gentle ospreyBOT
Nobody
lunar yoke
is moore spectrum the same es EM-spectrum, just a different name?
I've only heard the latter
gentle ospreyBOT
Nobody
lunar yoke
that is indeed very neat
coarse night
are CW complexes are metrizable in general?
lunar yoke
well cw complexes are normal
but i dont think they have to 2nd countable or the like
oh hm there is this metrization theorem saying that a space is metrizable iff it is paracompact hausdorff and locally metrizable
but yeah i guess in general cw complexes are not necessarily metrizable
hidden crag
a connected CW complex is metrizable iff. it's locally finite apparently
lunar yoke
but thats not really a problem
i guess it depends on what you do
but cw complexes are really nice spaces
lunar yoke
the problem is that you take lots of quotients in building them
coarse night
since any point in cw, there is a open set that lies in exactly one open cell e_n
they are locally metrizable?
coarse night
i don't think either
lunar yoke
if you pick the point as the 0-cell
coarse night
yes
i was thinking abt S², same conclusion
x lies in unique open cell, not necessarily every nbd
lunar yoke
yep
gentle ospreyBOT
coarse night
gentle ospreyBOT
Phil
coarse night
$\bigvee_{n \in\mbb{N}} (-\frac 1 n, \frac 1 n)$
gentle ospreyBOT
coarse night
it's a nbd of the special point *, does it contradict something somehow
maybe uncountable products??
ok I think I have a proof why it's not metrizable. It's kinda same proof of why R inf with box top is not metrizable. Take ∨ [0, ½] in ∨S¹. Then Π [0, ½] is compact and quotient is again compact so ∨ [0, ½] compact, but not sequentially compact.
@lunar yoke
lunar yoke
ok
ill just believe you
a year ago I still knew pset topology
I've only done topology since
but only the category-perspective kind really
every pset topological thing is swept under the rug nowadays
we just work in a nice cartesian closed category and everybody is happy 🙂
coarse night
pse is more of a tool than a field now
ppl have a strong metrization theorem and they're happy with it
lunar yoke
to me its more set theory than topology tbh
lunar yoke
"Homotopy Coherent Algebraic Structures"
but yeah i gotta study for this exam i have on tuesday now
odd flame
is that {a,b} or [a,b]???
can i just say that if x is in the closure of (a,b) it is either between a and b or equal to a and b?
gritty widget
Just say that [a, b] is a closed set
odd flame
being rigorous about the equality condition is weird tho
i can tell when it's not equal - is it only equal when (a,b) is precisely an open interval in X
ehhhh nvm, if a=b in N, then (a,b) = empty set but cl(a,b) = a = b
i think im confused by what (a,b) could be
like could it actually be clopen? or does that fact that it's written like that mean that we're only considering the closure of open intervals a to b
odd flame
ohhhhh so the bar over it changes how i should read it
gritty widget
Huh?
odd flame
like the parentheses in this case arent saying "it's an open interval"
gritty widget
(a, b) = {x: a < x < b}
gritty widget
Not sure how useful the limit points perspective is
odd flame
it's equivalent to the set of all points with open balls intersecting X tho right
gritty widget
We don't have open balls here
It's not a metric space
Just open rays and open intervals
Try writing what it means for a to be in cl((a, b))
odd flame
i mean i was thinking about it as open balls but i hadnt thought abt metric space, prof hasnt introduced that yet technically
is it more helpful to think of closure as the intersection of closed sets containing (a,b)
also since you said you need a metric to consider open balls, does that eliminate the idea of neighborhoods as well?
im assuming those are equivalent ideas right
gritty widget
also since you said you need a metric to consider open balls, does that eliminat...
No. But instead of neighbourhoods try thinking of "basic neighbourhoods"
Like how you were trying to reason with balls, we can reason using open intervals and open rays
odd flame
i mean now that i think about it, prof defined neighborhood of a point as an open set containing the point
so i can see how that might be different from an open ball
gritty widget
It's good practice to think about neighbourhoods as arbitrary sets which contain an open set which contains your point
Both definitions are common though
If you take open interval (c, d) with c < a < d, then (c, d) intersecting (a, b) means the same as open ray (<~, d) intersecting (a, b)
So a being in cl((a, b)) means that for every c > a there exists a < d < c such that a < d < b
I've assumed this is a total order
So that means those points sit "densly" to the right of a
For all c > a there is d with a < d < c
odd flame
If you take open interval (c, d) with c < a < d, then (c, d) intersecting (a, b)...
where did b come from here
gritty widget
Or, equivalently, open ray (a, ~>) has no minimal element
b was always here though
odd flame
oh it's still just the interval (a,b)?
gritty widget
Yes
odd flame
ok i think i see what you mean before but how does that get to equality
gritty widget
I've written out what it means for a to be in cl((a, b))
Similiar statement holds for b
cl((a, b)) = [a, b] means precisely that a and b are in cl((a, b))
odd flame
oh im definitely overcomplicating this then 
long grail
It's good practice to think about neighbourhoods as arbitrary sets which contain...
What's the advantage of this line of thinking instead of just saying that a "neighborhood" is an "open set containing the point"? In Munkres he talks about both conventions but chooses the latter without explanation
gritty widget
What's the advantage of this line of thinking instead of just saying that a "nei...
Look up the point-less topology
It gets used a lot
long grail
Ok I will
gritty widget
In which topology
odd flame
order topology
long grail
odd flame
ok that's what i thought
gritty widget
What's the advantage of this line of thinking instead of just saying that a "nei...
I was thinking of neighbourhoods in functional analysis when we often use that they can be taken to be closed, convex, balanced
For the first definition such statement doesn't make much sense
long grail
Ah ok. I've yet to take FA but plan to
gritty widget
Can (a,b] be clopen in itself
Every set is clopen in itself
gritty widget
what is the closure of (a,b]
Depends on the order
odd flame
sorry for belaboring an otherwise simple question but i dont see then when cl(a,b) is not equal to [a,b]
gritty widget
It can be [a, b] or (a, b]
long grail
So there is an order topology for which the closure of (a,b] is (a,b]
odd flame
it's gonna be when a and/or b are not in cl(a,b) ofc
but in what ordering is that the case
ooh wait
(0,1] in N
gritty widget
yes, for example
odd flame
closure is just 1 right...?
gritty widget
Is there an immediate predecessor of 1 in N with your ordering
odd flame
i mean just the natural ordering on N
gritty widget
closure is just 1 right...?
Yep
long grail
Finite sets are closed inT1
gritty widget
In what spaces?
long grail
Yeah
I'm just thinking about order topology on N
As per @odd flame question
You guys might be surprised but I did great on my first topology test
gritty widget
good job
long grail
But most of the questions were about topology on metric spaces which make a lot more sense to me
gritty widget
I'm just thinking about order topology on N
If you liked it then I have homework for you
When does subspace topology coincide with order topology on a subspace?
long grail
I will answer your question
Well
A basis for an order topology is all open intervals (a,b) in X, all intervals [x,b,) where x is the smallest in X, and all intervals (a,y] where y is the largest element in X
According to my book
Sooooooooo
Open rays are the open sets in X
An order topology is generated by open rays
Or
The topology generated by open rays contains the order topology
tender halo
metric topologies are a scam
odd flame
could i do c) by induction using b) as a base case?
long grail
So intersect the subspace with the open rays and get order topology on a subspace
tender halo
imagine living only in first countable spaces
long grail
What are they
I don't have to imagine living in a metric space. Because I do. Or do I?
long grail
could i do c) by induction using b) as a base case?
I think c might mean uncountable union
gritty widget
could i do c) by induction using b) as a base case?
induction (the usual kind involving the naturals, at least) only works if the indexing set is finite. since (c) is true for any indexing set, you shouldn't assume this
long grail
Also
gritty widget
HOWEVER! if you have a finite indexing set, you actually have equality
tender halo
induction (the usual kind involving the naturals, at least) only works if the in...
transfinite induction is pretty common in point-set topology, so i wouldnt confuse them
but yeah, here you wont get anywhere with induction in this one
gritty widget
edited for clarity
there's always gotta be someone who brings up transfinite induction 
odd flame
tender halo
im sorry, #foundations is leaking
gritty widget
you're fine
i may be taking a set theory course this semester so i should probably get acquainted with such things
tender halo
if i dont bring up transfinite induction every 5 minutes my blood sugar goes down actually
odd flame
long grail
Tell me more about transfinite induction
odd flame
it's finite (but not really)
long grail
(b) is true?
odd flame
yeah
gritty widget
HOWEVER! if you have a finite indexing set, you actually have equality
(this is what follows from part b and an easy induction over naturals)
odd flame
so do i just have to do c) by let x be in this, it must also be in that
tender halo
Tell me more about transfinite induction
its somewhat technical, but let me tell you a topological example. you have your open and closed sets, say, in R, and you want to know what subsets of R you can actually get from those sets
for example, intersections of closed and open sets are in there
and unions of families of closed sets (which are not necessarily closed if the union is infinite)
then you might take those sets and consider their unions and intersections
and so on and so on
long grail
Oh ok
tender halo
for reasons, we restrict ourselves to just countable unions and intersections
so X_1 = set of all closed and open sets, X_2 = set of all countable unions and intersections of sets from X_1, X_3 = ...
and so on and so on
long grail
I GET IT
tender halo
the point it is that its not enough to take X_n for natural n
because what if you take one set from X_1, one from X_2 and so on, then their union and intersection wont be in any of X_n's
long grail
Ok...
tender halo
so you set X_N = union of X_i, X_{N + 1} = set of unions and intersections from X_N
tender halo
and so on until X_2N, and then X_iN for every i
and then you will get X_{N * N}
which looks like this
in the end if you repeat this enough you will get a set that is closed with respect to unions and intersections
and those are called borel sets
tender halo
and the formal way of speaking of those "and so on and so on" is called transfinite recursion
proving things about them is called transfinite induction
icy schooner
A contractible space is homotopy equivalent to a point
Let’s say X is contractible and V consists of a point v_0
The only map f: X to V is the constant map
g : V to X is defined by g(v_0)
long grail
I see
icy schooner
fg = id_V no problem
gf(x) = g(v_0)
so need to compose a homotopy h
h_0(gf)=gf
h_1(gf)= id_X
but then gf has a left inverse
so it must be injective
and gf maps everything to a point
so X is a point?
There are definitely contractible spaces that are not a point so something is wrong here
odd flame
hushed marlin
Hi there! Can I get a double check on a proof? We're only on the topological properties on the real line right now
gritty widget
did you mean in the second line to write that the segment is contained in A_m?
you also wrote that on the final line
gritty widget
"contained in" as in "is a subset of", not "is an element of"
gritty widget
the "and for all a in A_m, a in r" is wrong
and you need r to contain x
which also means you can delete the entire following sentence ("then by definition, ...")
hushed marlin
oh you're right I looked at the definition wrong
gritty widget
you might also want to clarify that r should be an open segment
whatever that means
hushed marlin
ah I see
okay I'll keep working on it then. Thank you for the pointers!
Okay I think I have something closer
is that more in the right direction?
tawny smelt
is this not just part of the definition of a topology? any arbitrary union of open sets is itself open?
gritty widget
you are correct
odd flame
gritty widget
they're being asked to prove that this is the case for open sets in R
defined using open intervals and the like
(they're doing countable unions for some reason, but the proof goes through verbatim for arbitrary indexing sets)
hushed marlin
yeah correct. We're only on the real line right now taking baby steps. We haven't actually defined what a topology is yet (I believe we will this week though)
gritty widget
Okay I think I have something closer
this is good, but needs a bit of cleaning up
instead of the third sentence just write "Since A_m is open, there exists a(n open) segment r contained in A_m containing x"
restating the definition of an open set is unnecessary
just apply it with A_m and its element x
hushed marlin
great thank you! 🙂
I'm being intentionally explicit to drill the definition in lol. But I understand! I should clean it up before turning in
gritty widget
could i do c) by induction using b) as a base case?
This doesn't require any sort of induction, just use the property a)
gritty widget
So intersect the subspace with the open rays and get order topology on a subspac...
That's wrong. The subspace topology can have more open sets
gentle ospreyBOT
Blitz
gritty widget
For a subspace remove 0 from it
dry jolt
If X and Y are simply connected, is pi_n(X v Y) isomorphic to the direct sum of pi_n(X) and pi_n(Y)?
gentle ospreyBOT
Nobody
dry jolt
Oh nice, thank you. Couldn't figure out an explicit counterexample on my own
gentle ospreyBOT
Nobody
hard wind
In this long exact sequence, i* is induced from an inclusion i, and j* is induced from a quotient map j (on chain complexes). Does this mean that i* is injective and j* is surjective or is this sometimes not the case?
unreal stratus
No, in fact you can tell when this is the case by examining the long exact sequence
unreal stratus
For example: R^2 is contractible, but you can have subspaces with interesting homology, so the map induced by inclusion certainly isn't always injective
wise ruin
I think I'm having a blindspot with this problem. Could someone possibly point me in the right direction?
wise ruin
I think I’m having trouble understanding what T_1 looks like
wise ruin
Indeed, so for a sanity check, the disjoint union of any open ball in V1 with any open interval in V2 (not contained in the ball) will be open in that disjoint union topology but not in the topology of V?
shadow charm
Not quite, for example the interval in V2 could also be contained in V1, and so taking the preimage in V1 U V2 wouldn’t give an open set
Ngl though I’m also kinda stumped
wise ruin
Nobody, were you able to dream up an example or was your hint based on an intuited direction to go with this problem?
wise ruin
Thank god, I think my previous possible example was a crazed attempt to make the problem work
shadow charm
Alright thank fuck I’ve been thinking about this nonstop too and couldn’t figure out why I was being dumb
wise ruin
My prof’s hw assignments. Though he just pointed out the solution - The open unit ball around (-1,0) union with the open unit interval at 0
(He’s saying that it’s open in the quotient topology but not open in V) I’m pretty sure he’s correct, but I swear I’ve considered this example before but figured it was open in V
tiny obsidian
Oh of course it's not
I thought of it as well and also discounted it for some reason
It's because of the point at the origin
You can't get an open set containing it but not any of the rest of the boundary of the unit ball
wise ruin
It’s funny cuz this was the first example I thought of when doing the assignment a few days ago
icy schooner
If \Phi([f])=\Phi([g])
Then f and g are homotopic disregarding the base points
WLOG f(s_0)=x_0
g(s_0) is some different point in X
there is a path h between x_0 and g(s_0)
so that hg\bar{h} is homotopic to g freely and homotopic to f relative to the base point x_0
however I don’t see some element in pi_1(X, x_0) conjugating them
h is just a path it is not a loop
small hemlock
Perhaps a low question, but say we're looking at elements in the K-topology. Would, for example, (-3, 0) U (1, 4) be in that topology? (-3, 4) obviously is, but since we can also take away K from any of the intervals, is this how it would be perceived?
(-3, 4) and (-3, 0) U (1, 4) are different elements* obviously
Oo no that's definitely wrong, oops. Scratch that
Because K = {1/n, n \in N}, so (a,b)\K is just removing the specific points of K, right, not (0, 1)?
pastel linden
homotopy doesn't require it to be non-intersecting
just contract the circle to a point
flat plinth
Hi! Can someone help me with this?
I am unable to show that the closed unit ball and its surface are no homeomorphic
both are compact and connected
so what approach should i go for here?
plain raven
useful trick
if X is homeomorphic to Y, say by f : X -> Y
and A is a subspace of X
then f is a homomorphism from A onto f(A) and X\A onto Y\f(A)
so if you want to show that X is not homeomorphic to Y
you can show that for some carefully chosen subspace A, X\A cannot be homeomorphic to Y\f(A) for any conceivable choice of homeomorphism Y
Let A = {-1, 1} be a subspace of S^1 containing two distinct points. Then || X-A has two connected components. But deleting any two points from the closed ball \overline{B}(0,1) gives a space which is still connected. ||
@flat plinth
flat plinth
I am confused on one thing
I can only show that I can only show the closed B(0,1) is still connected after removing two points only by showing that it is path connected
Oh wait path connected always implies connectedness
Then it's fine
swift fjord
You got it
unreal stratus
Yes - it may be helpful to keep in mind that it's often easier to show stuff is path connected since you can sometimes just write down a path
or draw a picture
flat plinth
useful trick
I applied it to the next case for \overline{B}(0,1) and \R\times {0}
they can't be homeomorphic and my reasoning is as follows
This can't happen since if this were true then the restriction of the homeormphism to $B(0,1)$ would be a homeomorphism to its image. So, by transitivity $R^{2}$ is isomorphic to a necessarily connected subset of $R \times {0}$ i.e. an interval or a half line but this can't happen for as above deleting a point from $R^{2}$ keeps it connected but this is not true for an interval.
gentle ospreyBOT
QNovus
flat plinth
Is this correct?
arctic relic
I dont understand why (in cellular homology) H_m(S^k) = H_m-k(S^0) ie how the k fold suspension of a top space affects the Hom functor
Is it as simple as diagram chasing for cellular homology (like the ones in Hatcher) or is there something deeper?
Also is there a reference on equivariant chain complexes that are motivated by lens space? I dont want to work this out myself
Ok never mind on the the first 2 questions; still in need of the reference
I will probably work it out but in a reluctant manner
cursive vigil
What is top?
gritty widget
category of topological spaces
cursive vigil
Oh
woven sundial
both are compact and connected
cant you just use that the image of a compact set under a continuous function is continuous
oh nevermind i misread you.
your reasoning for R x {0} works, yeah
I don't think you need to mention R^2 though. the open unit ball also remains connected if you remove a point
fallow shale
Noob question: If V is a subspace of X, are the closed subsets of V closure the same as the closed subsets of V (subspace topology?)
tiny obsidian
No
If V isn't closed, V closure itself is a counterexample because it's not even a subset of V
fallow shale
you are very right
tiny obsidian
If V is closed, then V closure is V so then trivially yes
And all sets closed in V are also closed in V closure because a subspace of a subspace has the same (subspace) topology regardless of which set you consider it a subset of
fallow shale
right, but proper closed subsets in V closure need not be so in V
tiny obsidian
Yes
Although any counterexamples I can quickly think of aren't in V at all
I think it's any counterexamples actually
fallow shale
they are closed subsets of X intersected with V closure right, so only the ones without points in the boundary should be in both
tiny obsidian
Yes
fallow shale
I suspect the answer is no but, are they at least in 1-1 correspondence?
tiny obsidian
Closed subsets of V closure without points on the boundary are entirely in V
gritty widget
A Seifert surface of a knot is a surface whose boundary is the knot. The genus of a knot is the minimal genus among all the Seifert surfaces of the knot. Is there an algorithm to find the genus of a knot?
little hemlock
is there some way to approach this using long exact sequences?
devout sorrel
how do you show that for an open set X, cl(X) \cap cl(X^c) = cl(X) \ int(X)?
little hemlock
how do you show that for an open set X, cl(X) \cap cl(X^c) = cl(X) \ int(X)?
X is open so X = int(X) and cl(X^c) = X^c
little hemlock
ye
devout sorrel
got it ty
another q, (X is open again) why is int(X') always empty? (X' is the set of all limit points of X)
gaunt linden
That doesn't sound true. If X is R, then X' is also R, which definitely doesn't have empty interior.
devout sorrel
hmm, youre right
woven sundial
do you mean the boundary of X?
devout sorrel
int(cl(X) \ X) = Ø
im trying to understand this line
where its given that X is open
devout sorrel
do you mean the boundary of X?
wouldnt the boundary here be X' itself?
woven sundial
the boundary is exactly cl(X) \ X. this is not the same as the set of limit points of X
wait, no it isn't.
it's cl(X) \ int(X), which is the same for X open
devout sorrel
is cl(X) = X \cup X' not true?
woven sundial
it is, but X and X' are not disjoint in general
devout sorrel
yeah, got it. my mistake
woven sundial
equivalently, the boundary is the set of limit points of X which are not interior points of X. suppose there is an open set contained within the boundary - what does this imply?
devout sorrel
points of that set will be interior points?
gritty widget
how do you show that for an open set X, cl(X) \cap cl(X^c) = cl(X) \ int(X)?
that's true for any set
those are both equivalent definitions of a boundary
gritty widget
equivalently, the boundary is the set of limit points of X which are not interio...
The first part is only true for open sets
gritty widget
points of that set will be interior points?
If there is a non-empty open $U\subseteq \partial X$, then $U\cap \overline{X}\neq \emptyset$ so that $U\cap X\neq\emptyset$. But this is impossible for open $X$, since then $\partial X = \overline{X}\setminus X$
gentle ospreyBOT
Blitz
gritty widget
in general, int(bd(X)) can be non-empty
odd flame
why is it useful to know that a topology is contained in another
gritty widget
why is it useful to know that a topology is contained in another
because it gives us various implications about how compactness, closed/open sets, continuity, convergence of sequences, etc. relate to each other with respect to both topologies
not much else to say
flat plinth
HI can someone please help me with this?
i get one side
if V has the property that any bounded sequence in V has a convergent subsequence
then since any sequence in S is bounded
it follows that has a convergent subsequence
oh wait
but it doesn't necessarily need to be in S
so i can't prove sequential contintuiy this way
ig
gritty widget
Why is everyone posting functional analysis questions here
#advanced-analysis is the place for that
gritty widget
The theorem follows from Riesz lemma
gritty widget
We seem to be talking about different things
gritty widget
Riesz lemma says that if $Y\subset X$ is a closed proper subspace, and $t\in (0, 1)$, then there exists $x\in S_X$ such that $d(x, Y) > t$.
gentle ospreyBOT
Blitz
flat plinth
Riesz lemma says that if $Y\subset X$ is a closed proper subspace, and $t\in (0,...
yeah got it!
thanks i'll look into it
gritty widget
Using it we can construct a sequence $s_n$ in $S_X$ such that $|s_n-s_m|\geq \frac{1}{2}$ if $X$ is infinite-dimensional.
gentle ospreyBOT
Blitz
gritty widget
And from this the exercise will follow
gritty widget
Oh sorry. I thought this exercise were saying that S_X is compact iff X is finite-dimensional
flat plinth
nah
flat plinth
Using it we can construct a sequence $s_n$ in $S_X$ such that $\|s_n-s_m\|\geq \...
i got scared when you mentioned an infinite dimensional
result just to get this
😦
that's a relief
gritty widget
Yeah this one is easy.
flat plinth
any hint tho?
gritty widget
Taking a subsequence you might assume that the norms converge
If they converge to 0 you are done
opaque cloud
i got scared when you mentioned an infinite dimensional
yea infinite dim spaces are scary 
gritty widget
If not, divide by the norm
This is how you prove it assuming S_X is compact
Alternatively you can just say f:[0, 1] x S_X to B_X given by f(t, s) = ts is continuous, so B_X is compact
Now x_n is contained in rB_X for large enough r
flat plinth
oh ok
gritty widget
First part of this exercise is basically, B_V is compact iff S_V is compact
B_V being the unit ball and S_V the unit sphere
And one part of it follows from this map f being continuous and onto.
Second follows from S_X being a closed subset of B_X
flat plinth
First part of this exercise is basically, B_V is compact iff S_V is compact
i get this cuz any bounded sequence can be scaled to be in $B_V$ so B_V is (sequentially) compact iff S_V is compact
gritty widget
One can prove it more explicitly but using f is the clever solution
gentle ospreyBOT
QNovus
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
sleek thicket
Let $B$ be an extremely disconnected compact hausdorff space (sorry) and suppose we have a sequence of closed subsets $S_1,S_2,\ldots$. If $B = \liminf_{n\to\infty} S_n = \bigcup_{N=1}^\infty \bigcap_{n=N}^\infty S_n$ is there some $N$ such that $B = \bigcap_{n=N}^\infty S_n$?
gentle ospreyBOT
shamrock
sleek thicket
Recall that an extremally disconnected space is one where the closure of every open set is open
By the baire category theorem there is some N where the intersection has nonempty interior
But idk where to go from there
sleek thicket
I thought I could construct a counterexample using free ultrafilter stuff but it's not clear to me hmm
So here's something
It would suffice to show that for every x in B there is a nbhd of x contained within some \cap_n>=N S_n
Then pass to a finite subcover and etc
A galaxy brained thing to try here
Would be to look at the c* algbera of functions on this space
And try to apply uniform boundedness somehow
nah, I don't see it
Well hmm
Maybe I can try to do partition of unity things
So we can find functions φ1, φ2,... such that supp φi < B \ Si and Σi φi = 1
And then uh
Oh yeah so if x in Si then φi(x) = 0 necessarily
So for every x there is some N such that for k > N, φk(x) = 0
So can I show from this that there is some N such that for k > N we have φk = 0 identically?
sleek thicket
It would suffice to show that for every x in B there is a nbhd of x contained wi...
Actually maybe I should take this and suppose not
So take an x
And suppose that for every neighborhood $U$ of $x$ there is a sequence ${S_{n_k}}{k=1}^\infty$ and $y_k\in U \setminus S{n_k}$
gentle ospreyBOT
shamrock
sleek thicket
hmm
pick N large enough that x in S_n for n > N
and wlog n1 > N
hmm
U \ Snk is always open
so we have all these open sets
none of them contain x
none of them are nonempty
gritty widget
**shamrock**
Consider unique continuous surjection $f:\beta\mathbb{N}\to {0}\cup {1/n} = \alpha\mathbb{N}$ defined using any bijection between $\mathbb{N}$ and $\alpha\mathbb{N}$. Let $A_n = {0}\cup {1, 1/2, ..., 1/n}$, then $f^{-1}(A_n)$ is a strictly increasing sequence of closed sets whose union is $\beta\mathbb{N}$
gentle ospreyBOT
Blitz
sleek thicket
Ugh, thank you for the example
This makes things very frustrating
This means two colimits of condensed sets that I wanted to be the same are (probably) not the same
gritty widget
it was an interesting question
for a moment I even started believing it's true myself
sleek thicket
Haha yeah it seemed pretty plausible!
Ah hm this might actually not give me a counterexample
Here's the actual thing I was wondering about
sleek thicket
**shamrock**
Say B is extremally disconnected and you have two sequences of functions {fn}, {gn} from B to a compact hausdorff space X. If you set Sn = { x in B | fn(x) = gn(x) } and this holds, does the conclusion hold?
So I had thought every closed set in a compact hausdorff space was the zero set of some function, ie compact hausdorff spaces are perfectly normal, but it seems like that isn't true?
Hm it's possible I'm being silly though?
If you take a compact hausdorff space X and collapse a closed subset C to a point you still get a compact hausdorff space, since the subset (C × C) union Δ will be closed in X×X
So now
Take sets Sn that form a counterexample like you did
Let $Y = \coprod_{n=1}^\infty B/S_n$ and let $\hat{Y}$ be a compact hausdorff space containing $Y$
gentle ospreyBOT
shamrock
sleek thicket
Then we can define $f_n = B \to B/S_n \to Y \to \hat{Y}$
gentle ospreyBOT
shamrock
sleek thicket
And g_n is just constantly the point fn(Sn)
So fn(x) = gn(x) iff x in Sn
Cool (not cool)
dawn sage
is $\pi_0(G)$ of a topological group $G$ abelian?
gentle ospreyBOT
Potitov06
dry jolt
is $\pi_0(G)$ of a topological group $G$ abelian?
This sounds false, the simplest counterexample that comes to mind is if G is discrete, then pi_0(G) is just G
candid hedge
show that for any matrix A,B det(AB-lambda id)= det(BA-lambda Id)
i posted it here because it uses topological properties and not eigen values or something liek that
unreal stratus
Is there a nice description of this map q? I find it very hard to visualise (and not sure if Hatcher is handwaving a bit since another text does a diagram chase to define q)
eager vigil
Hey, could anyone check my solution, please?
west spindle
is the 1st homology group of RP^2 equal to Z/2Z
outer crescent
is the 1st homology group of RP^2 equal to Z/2Z
how do you want to show it?
west spindle
i made RP^2 into a delta-complex and computed its first homology group "by hand"
unreal stratus
But yes that is correct
I guess one standard way of doing it would be to use cellular homology, which gives you all the groups
west spindle
my work (messy and probably missing details)
echo dove
Not so messy -- looks all correct!
echo dove
Maybe the thing with d isn't really needed tho? It's homologous to c.
echo dove
Well, that d - c is the boundary of the lower-right 2-simplex, so c and d represent the same homology class.
echo dove
wait so c=d modulo im del_2? or something? is that what you're saying
Yeah, exactly. 🙂
echo dove
And that being the case, you probably don't need to introduce d in the first place.
west spindle
hmm...
but i'm computing the quotient while working inside the group of 1-chains
so i feel uneasy just hastily identifying everything mod im(del_2) for fear of accidentally missing something
echo dove
Well, d is defined to be -b-a, right? So it's sort of a convenience for referring to that element. That's all I mean -- you could keep calling it -b-a or something.
I'm splitting hairs -- I still think it was correct.
echo dove
Yep! In any case, H_1 is generated by c, or by d, or b-a.
west spindle
good to know i am able to compute homology groups in such simple cases
i was also able to calculate the second homology group of S^2 constructed as two triangles sewn together
and got Z
echo dove
One thing that might simplify the argument: <c, a-b> = <c, c+a-b>.
echo dove
i was also able to calculate the second homology group of S^2 constructed as two...
Yeah, nice!
unreal stratus
*exactly
unreal stratus
Tbh I think I'm fine rn but just checking what I wrote above is fine lol
Just realising a lot of stuff on, say, Hopf invariant one doesn't define the maps like precisely but that's probs okay
unreal stratus
Yeah
unreal stratus
Yeah and then u gotta attach the like last D^k x D^ell bit
sleek thicket
basepoint × right union left × basepoint
sleek thicket
yee
unreal stratus
I guess this is just how like product of cw complexes works though so this is more basic than i imagined anyway
Just a bit sus as when people say like
sleek thicket
Well let's think about what the product structure should be
unreal stratus
"the" standard cw complex structure on the n-sphere they can mean a couple of things
cause like you can also put the one where you take S^n and then attach two hemispheres to give S^{n+1}
sleek thicket
I'd expect that you get n * m cells
You take all pairs
and attaching maps will be like
It'll be sort of exactly this map won't it?
unreal stratus
Yeah exactly, so I guess here you just get 4 cells but that's fine because you just have a zero cell, then the two you mentioned above, then this thing as the last step
unreal stratus
yh
sleek thicket
But what you want is something from a higher dim sphere
unreal stratus
Yeah but I think that just comes from the fact that like
the final cell is of dimension k+l innit
sleek thicket
Yes
unreal stratus
And then take k=l and we have a map S^{2n-1} \to S^n \times S^n$ which lets us have the wh product, sexy
Yeah
sleek thicket
And then the general case of cw complexes followed
unreal stratus
ye
sleek thicket
Okay okay so
unreal stratus
Wait like I mean I don't think I have any more qs, just wanted a check aha
sleek thicket
Ah sorry
unreal stratus
Because yeah it's just product of characteristic maps innit
sleek thicket
It seems reasonable to me
Well sort of
You do that
But you precompose with S^(n+k-1) -> S^(n-1) × S^(k-1)
right?
And we were trying to figure out what this map is
Idk maybe I'm confusing myself
sleek thicket
Because yeah it's just product of characteristic maps innit
But this won't have the right domain
sleek thicket
**potato**
Ah I think the way you were writing it here makes sense(?)
You have two bits of the boundary
you just ignore the other factor on each
And run the attaching map
At least I think so
unreal stratus
Yh I guess like what I'm saying is like
U take product of char maps to get the new char maps
And then restriction to boundary is the attaching map
sleek thicket
Right
unreal stratus
(Where im talking in Hatcher terminology)
But ye come to think of it this is just a theorem in Hatcher anyway I guess lol
Nice
Ah and yes the attaching maps for the sphere are just the quotient maps D^n to D^n/boundary
So it all fits together anyway, sexy
sleek thicket
noice!
unreal stratus
Anyway thanks lol I think I just skimmed over stuff on pointset top of cw cpxs at the beginning and am having to pay for it aha
like hadn't dealt w products lol
Thank
graceful sequoia
Hello. I've been given a theorem by my teacher which I have to prove, just wondering if it's really true since I'm having trouble proving it.
Definition of continuity we are using:
f: (X,T_1) --> (Y,T_2) is continuous if U is an open set in Y, then f^-1 (U) is open in X.
My idea: For all B in B_2, we have that B is in T_2 and hence open in (Y,T_2) which implies that f^-1 (B) is open in (X,T_1) and so f^-1 (B) is in T_1. How do I conclude f^-1 (B) is in B_1?
vocal wharf
(i misread originally, i think you identified the issue)
i think you can play around with some examples
more specifically consider X = Y but choose different bases and look at the identity function
rancid umbra
graceful sequoia
So lets say I take f(identity function): (X_T(with basis B_1)) --> (X_T(with basis B_2)), where B_1 and B_2 are entirely different. Assuming f is continuous, take b in B_2. Then f^-1 (b) = b is not in B_1, hence a contradiction?
grizzled hound
Hello. I've been given a theorem by my teacher which I have to prove, just wonde...
i may be sleepy, but that theorem does not seem true to me. if (X,T_1) and (X,T_2) are both the real numbers with the standard metric topology, but B_1 is the set of open intervals and B_2 = T_2, then clearly the identity is continuous (since the topologies are identical), but not every open set is an open interval
in fact, i think that counterexample works for any two distinct bases of the same topology on the same space
grizzled hound
So lets say I take f(identity function): (X_T(with basis B_1)) --> (X_T(with bas...
yeah your counterexample is correct here, there's no way to conclude f^-1(b) being in B_1.
graceful sequoia
i may be sleepy, but that theorem does not seem true to me. if (X,T_1) and (X,T_...
Thank you.
jagged sage
To prove any subset of discrete topological space is closed, is it as simple as noting the complement of any subset is open by definition of discrete topology?
gritty widget
how do you define discrete topological space?
You could define it as every subset is open. Or every singleton is open
For example
jagged sage
Subset open
jagged sage
Alright thanks. Seemed too simple
pseudo coral
question
how do I show the map from open unit interval onto the reals given by the following is onto
gentle ospreyBOT
MyMathYourMath
pseudo coral
cause if r is in R then this has two solutions
if we set $r = \frac{2x-1}{1-\vert 2x - 1 \vert}$ then solving for $x$ we get two solutions
gentle ospreyBOT
MyMathYourMath
lunar yoke
but that literally means you hit every r
twice at that
assuming you've solved the thing correctly, I didn't check
gentle ospreyBOT
MyMathYourMath
pseudo coral
$x=-\frac{1}{2r-2}$
gritty widget
The calculation should be wrong, this is a bijection
gentle ospreyBOT
MyMathYourMath
pseudo coral
thats where im stuck @gritty widget
gritty widget
anyway. This is not exactly topology
pseudo coral
sorry it was given to me in a topology course to show open unit interval and R are homeo
is it conitnuous since the denominator is never zero and its just a rational function thus continuous
lunar yoke
denominator is zero for x=1 though
lunar yoke
oh right forgot you restrict to there
lunar yoke
well you did it the right way
lunar yoke
onto does not mean 1-1
pseudo coral
i know but the map is certainly 1-1 , is it not?
lunar yoke
by unit interval do you mean (0,1) or (-1,1)
pseudo coral
(0,1)
lament needle
i feel like surjective and injective are more clear terms LOL
gritty widget
You are given that 0 < x < 1
lunar yoke
then its bijective
gritty widget
so it's not just that you can say "yes those two solutions work"
lunar yoke
well yeah for positive r one of them is negative for example
ok not exactly but you get what i mean
man
this pointset & analysis stuff is really not my thing anymore
pseudo coral
lol
lament needle
where are the diagrams
all math is triangles change my mind
lunar yoke
no i agree
lament needle
where are the diagrams
"this diagram commutes bc just look at it"
pseudo coral
@lament needle can you help me out lol
lament needle
what is the question, the surjection onto R?
pseudo coral
yes for this map
pseudo coral
**MyMathYourMath**
this
lunar yoke
well you basically need to figure out for which r your solutions of this equation lie in your domain
like do you always have at least one of them in (0,1)
then its surjective
pseudo coral
i see, so it has to do with restrictions on r?
lunar yoke
you want to show surjecitvity
given r in R
you need x that hits it under the function
lament needle
in general to show a map $f: X \rightarrow Y$ is surjective you show forall $y\in Y$ there is some $x \in X$ such that $f(x) = y$
gentle ospreyBOT
matthew
lunar yoke
yes
and X = (0,1), Y = R for your example
i mean you gotta do the actual computations yourself im too lazy for that lol
pseudo coral
lol
lament needle
Any actual computations are beyond me
lunar yoke
so true
lament needle
does this make sense what we are saying
lunar yoke
ppl think its bad that in infinity category theory you basically cannot ever construct anything, but its actually a blessing in disguise
lament needle
its best to understand what it means for a map to surject in whatever context its asked then hopefully its clear how this occurs
pseudo coral
i know what it means to be surjective
lament needle
so I need restrictions on r to show only one of my x's hit it
this is for a bijection?