point-set-topology

[sterile fossil]

and $B(f(a), \epsilon) \subset V$

[gentle osprey]

Leu

Leu

[sterile fossil]

thus $f(y) \in V$

[gentle osprey]

Leu

[sterile fossil]

by definition of $\subset$

[gentle osprey]

Leu

[sterile fossil]

that one first

[sterile fossil]

then this one

[gentle osprey]

Leu

[limpid sandal]

I agree with those steps besides 7. How do you know B(a,delta) is a subset of f^-1(V) is that’s what we’re trying to show

[sterile fossil]

sorry

7 is actually

7 - By (1) there is $\delta > 0$, such that $x \in B(a, \delta) \subset f^{-1}(R^m) \implies f(x) \in B(f(a), \epsilon)$

you agree with that

?

[gentle osprey]

Leu

[sterile fossil]

inverse-image :v

that's just U btw, I do a lot of typos mb

1 - f is continuous in U

2 - U is open

3 - V is open

4 - $a \in f^{-1}(V)$

5 - By (4) $f(a) \in V$

6 - By (3) there exists $\epsilon > 0$, such that $B(f(a), \epsilon) \subset V$

7 - By (1) there is $\delta > 0$, such that $x \in B(a, \delta) \subset U \implies |f(x) - a| < \epsilon$ this follow from the definition of continuous function

ALL good now

[limpid sandal]

I think we need $x \in B(a, \delta) \intersection f^{-1}(V) \subset f^{-1}(R^m) for the implication right?

[gentle osprey]

Leu

[sterile fossil]

Nope

7 is the definition of continuity

Sometimes instead of $B(a, \delta)$ being subset of U you can say that $x \in B(a, \delta)$ and $x \in U$

[gentle osprey]

Leu

[sterile fossil]

But once U is open there is r>0 such that $x \in B(a,r) \subset U$, then you can take $\delta' = \frac{r+ \delta}{2}$

[gentle osprey]

Leu

[sterile fossil]

Then you have 7 with $\delta'$ instead of $\delta$

[gentle osprey]

Leu

[limpid sandal]

Wait but if x is not in f-1(V) then f(x) isn’t in our open ball B(f(a),Epsilon) since f(x) would be outside V but our open ball is inside V

[sterile fossil]

That's the definition of continuity dude

For every Epsilon

7 is tru

True

If f is continuous

[sterile fossil]

You mean that $B(a, \delta')$ would not be inside the open ball in V?

[gentle osprey]

Leu

[limpid sandal]

Nah B(f(a),Epsilon). I agree that’s the definition of continuity, I’m just questioning about the requirement for B(f(a),epsilon) to be inside V

[sterile fossil]

I seee

Because the ball was chosen inside V

And such a ball exists because V is open

And $f(a) \in V$ because $a \in f^{-1}(V)$

[gentle osprey]

Leu

[sterile fossil]

f: U -> R^m is continuous iff for every a in U and for every epsilon > 0 there is delta > 0, such that x in B(a, delta) and x in U => f(x) in B(f(a), epsilon)

Then

[limpid sandal]

Okay I see that now

[sterile fossil]

This

It makes sense?

[limpid sandal]

Yeah but now im at the conclusion that x in the intersection of B(a,delta) and U implies x is in f^-1(V) or nah

[sterile fossil]

Yup

[limpid sandal]

But that needs to reduce to x in just B(a,delta) implies x is in f^1(V) to show it’s a subset and conclude it’s open

[sterile fossil]

This

[limpid sandal]

Ah I see it. Doing that makes an open ball contained in U so then the intersection just becomes the whole new open ball?

[sterile fossil]

Yup

[limpid sandal]

Thanks for the help. The confusion was me using f^-1(V) instead of U in the definition of of f being continuous on U f me lol

[steel glen]

they are arbitrary closed compact sets in some topological space X

[unreal stratus]

If you want you can just start with K_1, as it doesn't matter whether you view them as subspces of a bigger space

but yeah I don't think this is particularly well-written

[winged viper]

I think I’m going crazy, can someone confirm if this is an equivalence notion of properness for first countable spaces?

f : X -> Y is proper if and only if for any sequence (x_n) such that f(x_n) converges to some y in Y, there exists a subsequence of x_n converging to some x in X such that f(x) = y

[unreal stratus]

I don't think this quite works. Your definition is equivalent (for first countable spaces) to f having compact fibres (preimages of points) and I don't think such maps need to be proper

[winged viper]

Not quite, if I take the natural numbers N and map that to a convergent sequence, that has compact fibers but doesn’t satisfy the property I gave

[unreal stratus]

but if you add in f closed then this works i believe

[umbral panther]

It is stronger than compact fibers. For example, it excludes open inclusions

[winged viper]

Funny how our two counterexamples were so different lol

[unreal stratus]

Yeah closedness should be part of the definition of properness anyway so you'll want to include smth like that

[umbral panther]

Nah, they’re the same. Inclusion of N in its one point compactification

This is a standard definition of proper in terms of nets. If you can replace nets with sequences, it should be fine

[unreal stratus]

I wondered if they end up being the same for the first countable case though

oh wait sorry that's what you mean, yes

[winged viper]

Right yeah first countable

Okay I think it’s the same too, I’ll write up a careful proof tomorrow when my brain is more functional

[winged viper]

Oh right, that’s smart

What even are spaces that people care about which are not first countable?

All I can think of is the weak topology on Banach spaces

[unreal stratus]

For me, natural examples come up in algebraic topology as soon as you start gluing lots of stuff together. For example, if you attach infinitely many circles to a single point (infinite bouquet of circles) or more generally certain infinite graphs

Or as you say, functional analysis type spaces

[unreal stratus]

This may sound silly, but they are very important spaces for some things

[winged viper]

Ah yeah, of course, loop spaces and whatnot are definitely not first countable

[umbral panther]

Aren’t they?

[rich trail]

is it right to write $\pi_1(S^1)={…,[\gamma_{-2}],[\gamma_{-1}],[\gamma_0],[\gamma_1],[\gamma_2],…}$?

[gentle osprey]

kevinhardy2

[rich trail]

where $\gamma_n$ is a loop that goes around $n$ times

[gentle osprey]

kevinhardy2

[noble prism]

yeah this is correct! and in this way its clear its the integers

[uneven bronze]

\textbf{Theorem} Let $f:X\to Y$ be a continuous map between two topological spaces $X,Y$ and let $X$ be connected. Then $f(X)$ is also connected. \

\textbf{Proof} Suppose that $f(X)$ is disconnected. Then there exist nonempty, disjoint open sets $U, V$ in $Y$ such that $U \cup V \supset f(X)$. Since $f$ is continuous, $f^{-1}(U), f^{-1}(V)$ are open, nonempty, disjoint sets such that $$ X=f^{-1}(U) \cup f^{-1}(V),$$ so $X$ is disconnected. It follows that $f(X)$ is connected if $X$ is connected. \

Is it correct to write $U \cup V \supset f(X)$? Why?

[gentle osprey]

Philip

[knotty vine]

No, that should be an equality: what if U was disjoint from f(X)?

[uneven bronze]

note, however, U,V are open sets in Y. I don't know if that changes anything...

if you write U cup V=f(X), then you're assuming f(X) to be open in Y, right?

[knotty vine]

You're right, U and V should be open in the subspace topology f(X)

Certainly f(X) is open in this topology on f(X)

[uneven bronze]

so should the correct proof read "...there exist nonempty, disjoint open sets U,V in f(X) such that U cup V=f(X)"?

I don't know, maybe that's the same as what is stated in my original post...

[knotty vine]

Yep, that's the definition of f(X) being disconnected. It's different from what you had in the original proof

[sterile fossil]

Actually you should say there is disjoint nonempty sets U, V which are open in the subspace topology of f(X) such that U cup V = f(X)

which is equivalent to say that there is U, V open in Y such that $f(X) \cap U, f(X) \cap V \neq \emptyset$ and $f(X) \subset U \cup V $

[gentle osprey]

Leu

[uneven bronze]

ok 👍 are f(X) cap U and f(X) cap V open sets in the subspace topology of f(X)?

[sterile fossil]

yup, because S is open in the subspace topology of f(X) iff S = U cap f(X), where U is open in Y

[unreal stratus]

I'd like to mention too, like

Being connected is an intrinsic property of the space. So you can just say WLOG f is surjective to make your life easier if you want

Rather than having to keep saying stuff is in the subspace topology everywhere

[empty grove]

Love me some vector bundies

[gentle osprey]

kolmotchima

[granite slate]

Can someone explain how they get the second row from the def of Tor?

[cedar pebble]

which part of this isn't obvious?

[granite slate]

Well when calculating Tor(A,B) you take resolution in A, but this seems you get the same resolution when doing Tor(B,A)

[cedar pebble]

take a projective resolution P of A and a projective resolution Q of B. Then one has Tor_i(A,B)=H_i(P\otimes Q)

but P\otimes Q is isomorphic to Q\otimes P

[granite slate]

Whats the point of this diagram then?

[iron kite]

(You mean the total sum of the double complex P \otimes Q, right?)

[cedar pebble]

I don't know, it's not really saying much

[cedar pebble]

yeah, specifically what is going on is you have quasi-isomorphisms like

P\otimes B <- Tot(P\otimes Q) -> A\otimes Q

and these induce isomorphisms

H_i(P\otimes B)=H_i(Tot(P\otimes Q))=H_i(A\otimes Q)

anyways yeah all that is going on here is that the usual tensor product commutes (up to isomorphism) and this will induce isomorphisms on Tor since these are left derived functors of the tensor product

I guess sometimes you want to actually keep track of these isomorphisms, since in general the braiding isomorphism for a tensor product can be a nontrivial interesting thing

[iron kite]

I think the point of this diagram is to give an ad hoc proof of symmetry of Tor without introducing double complexes. How is Tor defined in your book, exactly?

[iron kite]

Ok, so in the setting of your question F is a resolution of B and the second line is exactly the definition of Tor.

For the first row, the point is that any exact sequence 0 -> U -> V -> W -> 0 induces a natural long exact sequence Tor(Z,U) -> Tor(Z,V) -> Tor(Z,W) -> Z \otimes U -> Z \otimes V -> Z \otimes W -> 0.

Now take Z=A, U=F_1, V=F_0, W=B. Since F_0 is projective (I guess) you can check directly that Tor(X,F_0) is always zero, and thus you get to identify Tor(A,B) with the kernel of A \otimes F_1 -> A \otimes F_0.

[worthy olive]

What’s an inclusion map?

[unreal stratus]

Depends on context, but for topological spaces, if A is a subspace of X then the inclusion map is the function A -> X sending a | -> a

like "includes A in X"

[gritty widget]

What are the applications of numerical analysis to topology?

[sterile fossil]

Don't you mean the opposite?

[gritty widget]

No.

[sterile fossil]

I see

[hidden crag]

Id be surprised if there are any direct ones

But I guess you’d use numerical methods to solve linear equations that arise with coefficients in a field

Not sure how well these work over a non field

[winged viper]

Maybe something like computer programs that calculate examples?

Like computer programs that find knots for instance

Idk if that counts as “numerical methods” though

[zinc siren]

Applied topology! Topological data analysis stuff

[warped rover]

Even supposing TDA is a subfield of topology (which I find to be somewhat of a stretch), are there really applications of numerical analysis to TDA theory?

[opaque scroll]

I don't know where the bounds of numerical analysis lies. But certainly effective methods of computing various decompositions of matrices are important in TDA. And I guess that would fall under numerical analysis.

But I guess that's more using the tools developed by numerical analysis rather than using numerical analysis itself.

[warped rover]

yeah, I guess my issue is that I don't consider an application of a subject to be contained within that subject

So like I don't really consider TDA topology, I'd be willing to say ok the theory of TDA is topology but the development of TDA-based tools and libraries is not. (Not to disparage it, I just think it's categorically different.) Similarly I don't think the usage of numerical tools constitutes numerical analysis.

errr I phrased that last sentence wierd

[opaque scroll]

Well if applications of a subject don't count as applications of the subject, then I guess they don't have applications 😛

[warped rover]

I don't think the usage of numerical tools to develop TDA tools constitutes numerical analysis having applications to TDA, let alone topology

[opaque scroll]

Yeah I agree. But we're really reaching for these numerical analysis applications

[warped rover]

no clue if this is what carla was looking for, but ideally I'd like to see a numerical analysis theorem/method of proof used to prove something about topology

which maybe exists?

like idk it wouldn't shock me if there's some silly quadrature proof that exists somewhere

[iron kite]

~~Does Hensel’s lemma count? (Yes, it’s not algebraic topology, but it seemed to fit theme-wise…) ~~

[quartz edge]

bifurcation continuation is kind of nice

[prime elbow]

They define that , in Topology (X,T) a subset of X is said to be saturated when S can be written as an intersection of open sets of T.

In a finite set if every set is saturated then it implies that topology is discrete topology, right?

[paper wedge]

probably

just keep going deep ig and the process should enumerate literally all open sets of your topology

or idk

yeah nvm

[opaque scroll]

If there are only finitely many open sets then any intersection of open sets is a finite intersection

[paper wedge]

the singletons are closed

so

[opaque scroll]

clopen even

[paper wedge]

yeah

yeah yeah

pretty cool ig

[prime elbow]

Yes

I don't understand how I can show that if T has T_1 property then every subset is saturated

[paper wedge]

a hitn is

a finite union of closed sets is closed

[opaque scroll]

Take your set S. Then for every element x in S and y not in S, what does the T1 property give you?

[prime elbow]

X-{x} is an open set which contain y

[opaque scroll]

That's right. Though maybe you want the other way, if I said x was in S.

Anyway, then you take the intersection of those...

[prime elbow]

Don't get it but do this hint work for an infinite set?

[paper wedge]

if so then in like

like take [1/n,1]

in R

and take the infinite union indexed by n

[paper wedge]

doesn't matter then since u now know how to prove it

[prime elbow]

I don't get what meant by x was in S?

[prime elbow]

No I don't know how to prove it🥲

[paper wedge]

suppose you have some subset , call it Y

[prime elbow]

Yes

[paper wedge]

then Y = X-Y^c which is X-(U{y}) which is intersec(X-{y})

{y} is closed <--> X-{y} is open

X is your "universe" or whatevers

[prime elbow]

Y^c ?

[paper wedge]

complemen

complement

[prime elbow]

If I want to prove that S = {0,1,1/2,....,1/n,.....} is closed in Euclidean topology on R.

So if I say that union of (-∞,0) , (1/n, 1/n+1) (n runs over Natural number) and (1,∞) is open so S is closed in Euclidean topology On R, is this correct?

[tender halo]

yeah

[prime elbow]

okay, thank you

and discrete topology on uncountable set does not hold axiom of countability, because every basis of discrete topology has {x} x runs over X, so it has uncountable element in basis, right ?

[formal tide]

I've just realized I have a horrible understanding of the geometric intuition of homology, some questions:

• Two loops in the circle with different basepoints are identified in H1(S1), why?
• A loop in the circle that goes at constant speed gets identified with one that goes at variable speed, why?

I can't find the 2-simplex that identifies these loops

[stuck sleet]

Certainly this is correct: "if $A$ is an open subset of $\mathbb{R}^n$ and $\mathcal{O}$ is a dense subset of $C(closure(A))$ then $\left.\mathcal{O}\right|{A}$ is a dense subset of $C(A)$, where $\mathcal{O}|{A}$ are the functions of $\mathcal{O}$ restricted to $A$."

[gentle osprey]

deadpan2297

[stuck sleet]

haha sorry i realized i wrote that terribly the first time

[iron kite]

I think the important idea is that squares can work as simplices, in a pinch. In particular, a homotopy of loops H:[0,1] x [0,1] -> [0,1] induces two triangles (by really decomposing the square) whose sum of boundaries is exactly the boundary of the square?

[iron kite]

What’s the topology on C(A) here?

[stuck sleet]

@iron kite oh thats why I was getting confused. Ty for pointing this out

[iron kite]

This only explains the second point though. I need to think a bit more about the first point.

[formal tide]

oh nice point! it's a particular case that homology respects homotopical maps

ty

[iron kite]

In fact, this explains the first point because the boundary of the square (since it’s a homotopy of loops) is exactly the difference of the loops.

[formal tide]

right, same thing again, ty!

One last question, take X = S1 and A = a closed half hemisphese. How can I see that a loop is the same as a constant path in H1(S1, A)?

[iron kite]

You mean half-circle for A?

[formal tide]

yeah

[iron kite]

Are you sure it’s true?

[formal tide]

nope :p sorry if its not

[iron kite]

I don’t think it is, write the long exact sequence…

[prime elbow]

if i want to prove that every open set of R(Euclidean topology on R) can be written as union of countable closed sets.
so if i let first open set as of form (a,b) which can be written as (a +1/n, b-1/n) (n runs over N)
but if i want to (-infinte, a), then i take its complement {a, infinte) which can be written as intersection of countable open sets as (a-1/n, infinte) n runs over N, is it correct for (-infinte, a) ?

[unreal stratus]

I assume you mean countably many closed sets?

[unreal stratus]

Hm well (-oo,a) can be done more quickly

Well, just reverse your argument, it's the union (-oo,a-1/n]

One nice way to do this would be to use the fact that every open subset of R is a countable union of open intervals

[prime elbow]

yes

[unreal stratus]

(or you can say countable union)

But yeah seems you've got the right idea thus far.

One small hint to make this quick would be ||can you just find a countable base for the topology consisting of open intervals? Then you can easily write every open as a countable union of open intervals, and then you'll be done from the work you've just done||

[prime elbow]

i need to prove this one

[prime elbow]

i think (a,b) where a and b are rational number work, right?

let U be open set in R, then i know that for all x belongs to U, there exist an open interval containing ( a_x, b_x) is subset of U so i can write U as union of ( a_x, b_x ) x runs over U , and this a_x belongs to Q so {a_x |x belongs to U} is counatble, right ?

so then we know that open interval can written as countable union of closed sets so U can be written as countable union of closed sets, right?

[unreal stratus]

Perfect, yup.

[unreal stratus]

Yes

[unreal stratus]

Yes exactly

Well done.

[prime elbow]

Got it, thank you

[gritty widget]

true?

[unreal stratus]

As you've written it, A is not a subset of R

You keep adding unnecessary brckets around everything

{ (-1,2] } is the set with the one element (-1,2]

[gritty widget]

but A subset R

[unreal stratus]

No it isn't

(-1,2] is not an element of R

[gritty widget]

hmm question is wrong

[sullen bear]

i think what theyre trying to get at is that you have some extra curly braces

[unreal stratus]

.

[sullen bear]

also i think the interior of A would be [0, 2)?

[gritty widget]

no, ı see a book

[unreal stratus]

Okay it wasn't clear that T is the topology on A

[gritty widget]

and is not written in the question

[unreal stratus]

Well [0,2] isn't in the topology so it can't be the interior

[gritty widget]

yes, right

[sullen bear]

however negative numbers are not contained in any open sets, so i dont think they can be included in the boundary

[gritty widget]

but if A not subset R , question not want find interior etc.

i think question is wrong

[sullen bear]

A is a subset of R as written in this question. When you put the extra curly braces around the interval in your writing, it technically is not a subset

[gritty widget]

ohhhhh

thanks

[sullen bear]

np

[gritty widget]

ı understand

[sullen bear]

i love your I's lol

In this post https://math.stackexchange.com/questions/3043509/homology-of-quotient-of-two-copies-of-mathbb-s2
what do they mean by "As Z is free this means that it splits"?

Using the Mayer-Vietoris sequence, I can reach the point where H_2(X)/G = Z for some G isomorphic to Z x Z, but I don't see how this necessarily implies that H_2(X) = Z x Z x Z

[unreal stratus]

If you have an exact sequence $0 \to A \xrightarrow{i} B \xrightarrow{\pi} \mathbb Z \to 0$, then you can find a section /right inverse $s$ to $\pi$ (pick any $b$ with $\pi(b) = 1$ and consider the map $\mathbb Z \to B$ sending $1 \to B$) so $\pi s = \mathrm{id}$. Such a section makes it split - see the `splitting lemma'

[gentle osprey]

Süßkartoffel

[sullen bear]

thanks!

[unreal stratus]

Np. The same argument works for any free abelian group i.e. Z^I for some I

But not necessarily otherwise e.g. since there is an exact sequence 0 -> Z/2 -> Z/4 -> Z/2 -> 0

[worthy olive]

What are some applications of homology and algebraic topology in general?

[gentle girder]

what kind of application do you want? you get things like the jordan curve theorem from homology (to start) so that is pretty cool in its own right

[zinc siren]

Applied topology and topological data analysis is a thing

[eager vigil]

Hi, I would really like to read up on this paper on the homotopy type of the cobordism category as I will need it for a project in the future. I am studying algebraic topology and smooth manifolds currently from Lee for manifolds and Bredon and Rotman for algebraic topology.
Could someone guide me on resources to learn about the topics necessary to read this paper? I see a book by G. Segal mentioned in the abstract. Would this be a good resource to go with?

[eager vigil]

here's the paper in question, btw

[visual rock]

yes

[grim knot]

how do you find the triangulation of the torus? Is it only the square with vertices identified? What is meant with triangulization at all?

[knotty vine]

What do you mean precisely by a triangulization?

This is a simplicial set whose realization is homeomorphic to the torus, yes. But it's not a simplicial complex

[grim knot]

I don't have a definition, that's my problem: Find a triangulation of the torus and compute its homology groups using reduction methods

[knotty vine]

Oh if you just need homology then this is fine yeah

[grim knot]

do you know how to apply reduction methods? Can I for example just remove a?

[knotty vine]

Idk what is meant precisely by "reduction method". Isn't it just matrix reduction?

You certainly can't reduce this simplicial set any further

[grim knot]

nope it's this thingy

[knotty vine]

What you drew already has only 1 maximal simplex.

But they're also talking about an "abstract simplicial complex" which it might be important to note: what you drew isn't one!

[grim knot]

wait what's the difference

I'm confused

[knotty vine]

I wrote a little bit about the different notions of simplicial space before: #point-set-topology message

[grim knot]

Kinda funny that it was the answer to my question as well

[quiet thorn]

I'm gonna save that link

[knotty vine]

Notably: in a simplicial complex every simplex is uniquely defined by its set of vertices and each n-simplex has exactly n+1 vertices. In the picture you drew, both 1-simplices a and b have the set {v_0} as set of faces: this defies both properties I just mentioned!

[grim knot]

does it have to be this?

[unreal stratus]

i think that is the most efficient yes

[knotty vine]

I think that works, yes! But it might not be minimal

[unreal stratus]

well actually may not be most efficient yes

[knotty vine]

Here's a more difficult exercise: this thing is in fact an abstract simplicial complex, but it's not the torus!

[unreal stratus]

but it is rather common

[knotty vine]

It's "efficient" in the sense that it's efficient to reason about

[grim knot]

where can I find the definizion of triangulation? At least to be rigorous

[knotty vine]

wikipedia has one

Maybe this thing is the most efficient one? I dont recommend using it https://en.wikipedia.org/wiki/Császár_polyhedron

[jaunty summit]

so i know that the fundamental group of a topological group is abelian, but does this property extend to the other homotopy groups

[kindred cairn]

homotopy groups are all abelian except maybe for n=1

[plain raven]

In my vocabulary a triangulation is a homeomorphism between a topological space and (the geometric realization of) a "simplicial complex".
where the definition of simplicial complex is given from context.
Common definitions of simplicial complex are

• an abstract simplicial complex (see Lee, Topological Manifolds)
• a simplicial set (Gred Friedman, An Elementary Illustrated Approach to Simplicial sets)
• Delta sets (Hatcher)

So for a given space X and an abstract simplicial complex K, a "triangulation" of X by K is a homeomorphism X \cong |K|.

[grim knot]

I am so confused on how to apply this theorem to the above tiangulation

like what is the bestsimplex to remove? Does it depend on the choice at all?

[grim knot]

but here we have the two triangles bfe and bec, but they don't share an edge right?

[umbral panther]

There is no simplex in this complex to which that theorem applies

[grim knot]

That makes sense then, the following applies though right?

I could just remove the triangle in the middle right? v6 v7 v8

[knotty vine]

You could remove any triangle ye

[grim knot]

then I should get a punctured torus

Okay this makes much more sense now, sorry for spamming

[zinc siren]

...why?

[knotty vine]

Now that I look at it, it's a bit dishonest to call this an abstract simplicial complex: the edge from a to d in the top row is equal to the edge from d to a in the bottom row! (the same holds for the other vertical edges)

[zinc siren]

this i agree with

[west brook]

So what is it, if not an abstract simplicial complex?

[knotty vine]

It's certainly either a simplicial set, a Delta complex (a la Hatcher), or in the worst case a CW complex

It could be an abstract simplicial complex, but I just feel it's a bit dishonest two draw the same edge twice without labelling it as being the same edge

Technically there's nothing wrong with it

[zinc siren]

yes, Delta complex best complex

[west brook]

Hmm, I think it’s also an abstract polyhedron

[worthy olive]

In Mayer-Vietoris, how do you find a map $h_{*}$ from the direct sum of the homologies of two generating spaces to the Homology of the union of the spaces?

[gentle osprey]

CJ_:)

[umbral panther]

Functor plus addition
a oplus b |-> i_*(a)+j_*(b)

[umbral panther]

Or, depending on your conventions, maybe subtraction

[stuck geyser]

Assume $T$ is a Hausdorff-Normal topological space, and consider the ring of continuous functions $C(T,\mathbb{R})$, then an element $f$ is a zero divisor if and only if there is an open set that is mapped to $0$ under $f$. Proof: \ \

\textbf{Necessary}: Assume we have an $f$ such there is a non-empty open set $\Omega$ where $f(\Omega) = {0}$. For each point $t_0$ in $\Omega$, the closed sets $\Omega^C$ and ${t_0}$ are disjoint, therefore by $\textbf{Urysohn's Lemma}$, we have a function $g \in C(T, [0,1]) \subset C(T,\mathbb{R})$ such that $g^{-1}({t_0}) = {1}$ and $g^{-1}(\Omega^C) = {0}$. Then the function $fg$ is $0$ everywhere, and $f$ is a zero divisor. \ \

\textbf{Sufficient}: Assume $f$ is a zero divisor, then we have a zero divisor $g$ such that for all $t \in T$ : $f(t)g(t) = 0$. This implies for each $t \in T$, either $f(t)$ or $g(t)$ is $0$. However, since $g$ is continuous and isn't $0$ everywhere, the preimage $\Omega = g^{-1}({0}^C)$ is nonempty and open. Since $f(t)g(t) = 0$ everywhere on $\Omega$ and $g(t) \neq 0$ everywhere on $\Omega$, $f(t) = 0$ everywhere on $\Omega$

Is this a valid proof?

[merry geode]

What is Omega’ ?

[stuck geyser]

Complement of omega

[gentle osprey]

Mizalign #1 simp

[merry geode]

Ahh, yea that’s enough for this

[stuck geyser]

sick!

the literal only application of Urysohn's I've ever done aside from Partitions of Unity on Normal paracompacts

[prime elbow]

if X is a finite closed topology , then the collection B of the sets X-U, where U is finite is a basis for finite closed topology, right?

i want to prove that X has at least two element, the collection {X{x}| x belongs to X} is subbasis of finite closed topology.
so i take finite intersection of X{x_i}, which is equal to X-{x_1, x_2,........,x_n}.

is it correct?

[alpine nest]

yes, but also that's a description of the entire topology

well, you'd need to throw in the empty set

[prime elbow]

yes, but for basis empty set in basis is not necessary, right?

[alpine nest]

Yeah, I just mean your characterization describes all open sets apart from the empty one

[prime elbow]

okay, thank you

if X is an infinite set and i need to find subbasis for discrete topology such that there is no singleton set in subbasis.
so if i take collection C of sets {x_1,x_2}, x_1,x_2 are elements of X then does it work ?
i think its work but i think there is better answer

[tender halo]

it works yeah

[alpine nest]

And I don't think there's anything significantly better

[tender halo]

i can think of a subbasis where all the subsets are infinite

but not really necessary

in fact just take the set of all inifinite subsets of X

[alpine nest]

I don't even remember exactly what the definition of subbasis was

Ah yes, finite intersections form a basis

So yeah

[prime elbow]

means?

[tender halo]

it will be a subbasis for the discrete topology

[alpine nest]

So long as X itself is infinite

[prime elbow]

so for all x there is A and B belong to Subbasis such that intersection of A and B is {x}

[tender halo]

yeah

[prime elbow]

okay thank you

[tender halo]

also to prove that you need choice, funnily enough

[alpine nest]

To have the guarantee that an infinite has two (almost) disjoint infinite subsets?

[tender halo]

yeah

it is consistent with ZF that there is a set for which the intersection of any two infinite subsets is infinite

[alpine nest]

I've never really thought about it, but it sounds plausible

I've always just worked in the ZFC and not worried particularly

[round oyster]

Can anyone explain to me def of the compact set ?

[red yoke]

A topological space is compact if every open cover of the space has a finite subcover

[grave solstice]

doesn't in ZF the naturals inject into any infinite set?

[red yoke]

Compactness is useful because we can reduce problems about open covers to problems about finite covers, and finite is always nicer

[alpine nest]

And in metric spaces an alternative characterization is that every sequence has a convergent subsequence

[gritty widget]

No

See infinite Dedekind-finite sets

[red yoke]

For more context, open covers appear whenever you need to turn local "qualities" / "constructions" into global ones

[grave solstice]

why can't you do the obvious thing? Take a set A, and inductively form a sequence A_n of subsets of A by A_(n+1)=A_n cup {a} where a in A setminus A_n, then take B= bigcup A_n

[round oyster]

im not that good in topology , its my first year in uni , i can't understand these btw

[grave solstice]

well maybe you can't do that idk

[red yoke]

I'll try some examples

[gritty widget]

Need choice to create the full sequence

[sterile fossil]

In R^n a compact is set which is closed and bounded

[round oyster]

i know that definition , but what does it have to do with functions ? how can a function be continoius on closed interval be necessairly bounded , do we consider the imF as aset ?

and we apply heine theorem?

image(f) is a set right?

if its continious

[red yoke]

A closed interval is bounded, and it turns out images of compact sets in R are bounded

The short explanation is that the image of a compact set under a continuous function is compact, therefore bounded

[grave solstice]

I wonder, then in ZF if you look at the partial order A<=B iff A injects into B, how many minimal elements are there?

looking only at the infinite sets I mean

[red yoke]

Do you know what an open cover is?

[round oyster]

oh okay , how can we characterize it by sequences ? how do you think about sequences?

[round oyster]

no i don't

[red yoke]

On a metric space (such as R^n), a set is compact iff every sequence in the set has a convergent subsequence

[round oyster]

it feels like Bolzano weirstrass theorem , if a sequence is bounded then it admits atleast one convergent subsequence

[red yoke]

Yes, since the closed ball that bounds the sequence is compact

[gentle osprey]

Outsider

Outsider

[red yoke]

[red yoke]

Do you know what open and closed sets are

[round oyster]

yes i know

i'm taking real analysis 1

[red yoke]

An open cover is just a collection of open sets whose union is the entire space

"Covering" it

[round oyster]

like a set than can be covered by a union of open sets?

[alpine nest]

Yes, in fact every set can.

[round oyster]

Outsider , i barely understood your proof ha

i didint understand the presence of bn

[alpine nest]

It should have been y_n, sorry

[round oyster]

oh alright thanks

[alpine nest]

We take a sequence of elements of B of increasing magnitude.

And show that it has a bounded subsequence, so something went wrong.

[round oyster]

that is literally the exact same proof my teacher wrote

[red yoke]

In the most general case of a topological space (where open and closed sets are more fundamental concepts than sequences and boundedness), a compact set is a set that such that any open cover A has a finite subcollection that is also an open cover

[alpine nest]

It's a fairly standard argument.

[red yoke]

The cover refers to the collection (set) of open sets

[round oyster]

so a compact set if every open cover A of it has finite number of open sets?

[red yoke]

If every open cover has a finite subcollection that still covers it

[round oyster]

if every collection of open sets that covers A has a finite collection that covers every open set?

my brain isnt braining

[red yoke]

A topological space X is compect if every open cover A that covers X has a finite subcollection B that also covers X

[round oyster]

how can we define a toplogical space ?

[red yoke]

It's a set where open and closed subsets are defined in a "sensible way"

R^n (and its subsets) are topological spaces for instance

And all metric spaces are also topological spaces

E.g. If X = [0, 1], A = {..., (-2, 0), (-1, 1), (0, 2), (1, 3), ...}, then we can choose the finite subset B = {(-1, 1), (0, 2)} of A, which still covers X

[round oyster]

all of these are toplogical spaces?

so its a family of sets that their union contains X

a family of open sets

[red yoke]

yea

[round oyster]

we call that an open cover of X

[red yoke]

yea

[alpine nest]

https://en.wikipedia.org/wiki/Topological_space#Definition_via_open_sets

[round oyster]

ill check and read definitions and i'll be back so i can keep up

[alpine nest]

There are many ways to define a topological space, but the most common is using neighborhoods or open sets

Or metric

[red yoke]

The bare definition of a topological space might not feel very motivated, other than the fact it generalizes metric spaces

[gritty widget]

iirc aleph_0 is the only one up to isomorphism

[red yoke]

But all you need to know is it allows us to focus on "flexible" things like openness and compactness instead of "rigid" things like distances

[tender halo]

seems right

[grave solstice]

oh, I guess you needn't have minimal elements. I think a better question is then to ask how many "branches" does the partial order have, as in if you have two sets A, B that are not comparable then they lie in different branches

[tender halo]

no branches because foundation

well they might be there, but you cant see them inside the model

[gritty widget]

id expect there could be arbitrarily many

[grave solstice]

didn't we say there are infinite sets that are not comparable to each other? these are distinct branches

[gritty widget]

there can be

Its not provable in ZF

[tender halo]

well by branches do you mean upwards or downwards

[grave solstice]

right, then by my question was about possible models. Sry if I was sloppy

[tender halo]

i meant downwards ones

[grave solstice]

I think we are talking about different things. Foundation says that the partial order given by $\in$ is well founded, in that you cannot have an infinite string $\cdots \in x_3\in x_2\in x_1$?

[gentle osprey]

croqueta3385

[grave solstice]

or am I clowning? I don't really know

[tender halo]

it does

[grave solstice]

ok. But I was speaking about the ordering $A\leq B\iff$ there exists an injection $A\to B$, which I don't think is exactly the same? Anyway, my downward branches are allowed to end, they needn't be infinite. I think that was your observation

[gentle osprey]

croqueta3385

[grave solstice]

like in ZFC there is only one branch, because everything ends in N (and any two elements are comparable, I think)

but carla said there are no minimal elements (other than N), hence if they are different branches they needn't end, idk

[red yoke]

But ∈ is not ⊂

[tender halo]

well the point is that you may have a branch that can always be extended downwards but that isnt a subset of any infinite branch

[gritty widget]

i was interpreting croquetas question as antichains

[grave solstice]

any element belongs to a branch by definition? this is a tree. Like I think you could define a branch by a maximal totally ordered subset or something. Different branches may share elements tho

[gritty widget]

If M is an infinite dedekind-finite set, it neither embeds into N neither N embeds into it

[grave solstice]

but I'm just yapping I don't really know

[grave solstice]

well not a subset it would be a class or whatever, I'm waving hands

[red yoke]

What does ∈ have to do with ≤

[gritty widget]

It gives a partial order on the class of all sets

Maybe just preorder without foundation

[red yoke]

I thought we were talking about the preorder given by existence of injections?

[gritty widget]

oh idk how \in is relevant then

Maybe this discussion should be moved to #proofs-and-logic

[gritty widget]

Set theory is just discrete topology so its on topic here

[gritty widget]

lol true

[red yoke]

#proofs-and-logic message

[balmy field]

You responded to yourself.

[gritty widget]

Yes.

[worthy olive]

Why?

[worthy olive]

How do you find an induced map in Mayer vietoris homology sequence?

[paper wedge]

wdym

the connecting homeomoprhims?

[worthy olive]

Like how do I continue in this example

.rotate

[gentle girder]

,rotateccw

[paper wedge]

i dont see much honestly

what i can say is that most of the time u should use first iso theorem

😄

[dusk slate]

Is there a space which has an infinitely generated fundamental group?
I'm trying to find a space which has fundamental group as \mathbb{Q} (as an additive group or maybe Q* as multiplicative)...

[hidden crag]

There’s a general construction for this kind of problem

[visual rock]

yes there are spaces which have infinitely generated fundamental groups. For instance the wedge of a countably infinite number of circles (or the hawaiian earring, though these have different fundamental groups). Another fun example is R^2 \ Q^2. I don't know of an explicit example of something with Q as a fundamental group though such a space does exist

[hidden crag]

R^2 \ Q^2

[visual rock]

should have said 'fun'

[ebon hinge]

is anybody here good at vector

just some very basic vector stuff

[hidden crag]

#linear-algebra

[dusk slate]

Waitaminute what does R^2/Q^2 look like, is there a way to compute its fundamental group???

[visual rock]

R^2 \ Q^2 (not R^2 mod Q^2) is the plane but with a lot of punctures. I don't know if there is a nice description of the fundamental group, though you can say a lot about it, such as it being uncountable. Its abelianization is nicer (still ugly though).

[kindred cairn]

Has a lot of holes but not enough holes to make it not path connected !

Exercise: show that R^2 \ Q^2 is path connected

[dusk slate]

You can do it by taking component wise paths and adding them?

[proper fiber]

||we can find a line through the origin so that it dosent hit any Q^2 sth like y=√2 x||

[kindred cairn]

||A line through the origin will contain 0 so its meh||

But that's the idea

[proper fiber]

Actually we could claim that any set of countable points removal will still make it stay path connected?

[kindred cairn]

Well, you cannot join two points with a straight line. But you can do some trickery. Does this space even contain any straight lines ?

[kindred cairn]

This is probably true

[proper fiber]

Actually ||take those lines from the points you'd want to find a path for|| ||it's just the intersection of those two lines through the respective points||

[dusk slate]

The line y=b when b is irrational will be a straight line which has no intersection with Q^2

[kindred cairn]

It contains 0

So you need an irrational affine component in your line

[proper fiber]

y=b dosent contain Q^2

I mean no intersection

Both components need to be rational

[kindred cairn]

Ah, you meant constant line

[tender halo]

yeah you can just go in straight lines along irrational coordinates

[proper fiber]

Yeah

[kindred cairn]

So: any equation of the form ax+by+c=0 with coefficients linearly independent over Q should work

[proper fiber]

Do we explicitly need to state such lines

[kindred cairn]

No

[proper fiber]

Won't they just exsist because the points ins countable

[kindred cairn]

Your argument about countability works

[proper fiber]

Yeah

[dusk slate]

So for a path from (a,b) to (c,d) 1st assume b and c are irrational the the path (r1,b)+(c, r2 ) where r1 is a path from a to c and r2 path from b to d would work...

[kindred cairn]

For any two points, you know there exists a line going through each of them that contains no rational points. Actually, there is more than one. In particular the two lines are not parallel so they intersect somewhere giving you your path

[proper fiber]

Actually I had a question to ask but couldn't help myself with the ongoing discussion

---

---

What would be the example for such a sphere which is disjoint with the closure of the open ball with the same centre and radius

By sphere I think the author means every point is of the same distance from the centre no more no less

[tender halo]

well, trivial example, take the discrete metric, the closure of the 1-ball will be disjoint from the 1-sphere

[proper fiber]

1 ball means radius 1?

Yeah I guess you mean that

As it's closed there and also open

Thanks

[tender halo]

less trivial example uhh take the Baire space with the standard metric

both of those are kinda cheating since they are both totally disconnected

https://math.stackexchange.com/questions/108010/when-is-the-closure-of-an-open-ball-equal-to-the-closed-ball

you might find this interesting

[proper fiber]

Also I wanted to ask and this is out of context but how does one find stuff related to their question on stack exchange

Like it shows many results at once it's hard to determine which ones are relevant

There must be something with the right keywords to search for ?

[tender halo]

i just googled "closure of an open ball is a closed ball"

[proper fiber]

👍👌

[proper fiber]

https://media.discordapp.net/attachments/867389710132707378/1222210124698091630/855919932151562310.png?ex=661562e1&is=6602ede1&hm=2d896a2d571ed924be7c4255de6a119396ffa102e07401cc2b78d7b494447988&

How to do 6.43

I am expecting sth like zorn's lemma but I can't find the set up

[tender halo]

interior of the closure

[proper fiber]

Okay

I couldn't think of this

Mind sharing how you came up with this

[tender halo]

well i know the fact that the closure of an open set is the same as closure of (open set intersectected with D, where D is a dense set), its very commonly used

so its probably something related to that

[winged viper]

Let $E \to M^n$ be an oriented rank $k$ vector bundle over a closed oriented manifold. The Thom isomorphism for homology says that the map $H_{n+k}(E, E - M) \to H_n(M), \alpha \mapsto \pi_*(\alpha \cap u)$ is an isomorphism, where $u$ is the Thom class of $E$. In particular, it sends the fundamental class for $(E, E - M)$ to the fundamental class for $M$, up to a sign. I want to show that the sign is $+1$, assuming $E$ is oriented via the direct sum $T_{(p, 0)} E = T_pM \oplus E_p$. Does anyone know how I might approach computing the sign here?

[gentle osprey]

Frank

[winged viper]

Hmm maybe nvm, I found a note by Michael Hutchings where he says he doesn’t know a satisfactory way of nailing down the sign

Lol

[umbral panther]

If the dimensions are even, this should be trivial. If both dimensions are odd, you get a sign when you switch them, so you have to be careful not to do that

[winged viper]

Can you explain why it’s trivial

In even dimensions

[umbral panther]

Why are you defining it in that direction with cap products rather than in the other direction with cup product?

[winged viper]

Because I specifically need this version of the Thom isomorphism

It’s for a Poincaré duality thing

Which is why there’s a cap

[umbral panther]

Reduce to the case M=R^n
This requires generalizing to non compact manifold

[winged viper]

Like use a chart?

[umbral panther]

Once you have a version that is functorial, the number has to be the same for every n-manifold, so you just have to compute it once, in R^n

[winged viper]

Hmm I see

[umbral panther]

Maybe instead of non compact manifold, use compact manifold with boundary, so you can talk about relative homology, since that’s the language of Thom

[winged viper]

I’ll think about it thanks

Once I have a “functional” version, the general result should follow by a Mayer-vietoris argument? Or were you thinking of something simpler

[umbral panther]

MV induction seems pretty simple to me

[winged viper]

Great lol

[uneven bronze]

I have a basic question. In a topological group, why does one require that the group operations be continuous? Aren't topological spaces that are groups but that do not fulfil the requirement of continuous group operations interesting?

[umbral panther]

Do you have an example?

[uneven bronze]

well, I'm very new to this and haven't been able to find any motivation for the continuity requirement of the group operations, i.e. what it brings to the table. The group Z/3Z={0,1,2} under addition with the topology consisting of {empty set, {0,1,2},{1}} is an example where the continuity requirements are not fulfilled.

[umbral panther]

No, such examples are not interesting

[uneven bronze]

ok, that's what I thought

[alpine nest]

Just to expand on that: we don't just introduce a topology to a space because we feel like it (unless we specifically want to construct weird counterexamples); if there's some actual reason for your group to have a topology, it's very unlikely for that topology to be "incompatible" with the group structure.

So we require the group operations to be continuous because on the one hand we wouldn't be able to do much if they weren't, and on the other hand they are continuous in the vast majority of actual cases when you have both a group structure and a topological structure on some space.

[alpine nest]

Just another side note: observe that in R the topology is given by the metric d(x,y) = |x-y|, so it's explicitly defined using the group operation

Or at least it can be thus defined, of course you can also arrive at it from the order perspective (but the order is also defined to be compatible with the group operation)

Basically in R the analytical structure, the topological structure, the algebraic structure, they're all very closely intertwined and in many ways they're the same thing.

And then analysis, measure theory, algebra, topology, differential geometry, etc., generalize it into various directions

But when those directions can meet again, they should play nicely with each other in similar ways to the ones that happen on R

All of those branches came out from studying real numbers and focusing on some specific behaviors

[uneven bronze]

yeah, cool stuff

[alpine nest]

It is!

[fair idol]

Hey can someone give me a tip on how to finish showing that H^0(X,G) is the set of maps maps f:X->G such that f is constant on the path components?

I know if you take the singular cochains you have a sequence 0->Hom(C_0(X),G)->Hom(C_1(X),G)->... And so I know H^1(X,G) = ker f1:Hom(C_0(X),G)->Hom(C_1(X),G) but I can't tell what to do from here.

I'm also a bit confused because ker f is a subset or Hom(C_0(X),G) so I can't tell why these are maps from C. It looks like they are maps on chains C_0

[iron kite]

Can you write explicitly what C_0(X), C_1(X) and the differential are? What does it mean for a morphism C_0(X) -> G to vanish when pre-composed by the differential?

[fair idol]

I know C_1(X) is the free abelian group on the singular chains which are maps D^1->X. I think I'm supposed to just think of a singular chain D^1->X as some disk with potential singularities and potential self intersection inside X. Similarly for C_0(X)

Then the codifferential S^0:Hom(C_0,G)->Hom(C_1,G) given by precomposition S^0(f)=f d.

I guess if a cochains vanishes under this differential which precomposes another way to say that is it vanishes on boundaries In C_0.

I think I'm supposed to think of boundaries as literal boundaries of 1 dimensional submanifolds...

[iron kite]

What is D^1? What is D^0?

Can you describe which elements of the boundaries in C_0 are boundaries?

[fair idol]

Oh D^n is the standard n simplex so I think D^1 is a homeomorphic to a line segment and D^0 is a point.

Then should I think of singular chains f:D^1->X in C1 as paths in X?

I guess if p and q are in different components then no element in C0 that has nonzero coefficients for p and q is NOT going to be a boundary? So are the coboundaries B^0(X) in C^1(X) of the form c1p1+...cnpn in C1 where all pi are in the same component?

[uneven bronze]

I've come across two different definitions of local compactness. One states that every point in the set contains a compact neighborhood. The other one states that every point is contained in an open set whose closure is compact.

From what I understand, they are not equivalent (if you use the definition of a neighborhood of a point that is a set containing an open set that contains the point). I'm having a hard time accepting that these two definitions are not equivalent; is it OK to have two inequivalent definitions? As far as I understand, they are equivalent under the extra assumption that the space is Hausdorff.

[umbral panther]

Compact already has two definitions

[uneven bronze]

yes, but I meant local compactness

[tender halo]

whats the other definition?

[uneven bronze]

One states that every point in the set contains a compact neighborhood. The other one states that every point is contained in an open set whose closure is compact.

[tender halo]

nonequivalent definitions exist to make you read the opening paragraphs of a paper or a textbook to see which one they mean

[tender halo]

i mean the other definition of compactness

not equivalent to the usual one

[tender halo]

it happens all the time, just a nuisance really

one author has posets be strict, other has them to be reflexive, dont even get me started on monotone, increasing, nondecreasing and strictly increasing

[uneven bronze]

true 🙂

[tender halo]

or countable/at most countable/denumerable

[red yoke]

When authors decide to define "compact" and "first countable" for clarity and then proceeds to not define "locally compact"

[iron kite]

I don’t understand your notation (how exactly is a point an element of C^1?) but you’re on the right track. Note though that the set of “potential coboundaries” that you’ve given is not a group and generates all formal linear combinations of elements of X…

[rich trail]

Why is a loop that goes around the circle twice not homotopic to a loop that goes around once?

[kindred cairn]

their images are homotopic, because both are a circle. but the parametrisation sees more than the image

it sees how many times you wind around the circle

so just like a loop that winds around the circle once is not homotopic to a constant loop

[winged viper]

If X and Y are G-spaces, why is X x_G Y = (X x Y) / G denoted like a fiber product

[paper wedge]

lift them up

[unreal stratus]

Honestly I think it is just unfortunate notation lol

[empty grove]

It's analogous to the tensor product of modules over a ring

They both have a common generalization to the tensor product of a contravariant functor with a covariant one (same domain and codomain)

Whenever you can "move scalars across a comma", that thing is written with this notation

You define it by taking the "naive product/tensor product", and then quotienting by the relation which allows moving scalars past a comma

[true furnace]

Given a topological space X and a field k, by considering induced maps on homology we obtain a representation of Aut(X) on H_n(X, k). Is there theory about the interplay between the representation theory of Aut(X) on H_n(X, k) and the topology of X? If so, where can I read about this?

[winged viper]

Ahh that makes sense kinda thanks

[novel ember]

what the fuck is the square lifting property

[lime sable]

https://en.wikipedia.org/wiki/Homotopy_lifting_property probably this

[novel ember]

I see.

[merry geode]

What if only doing this just to avoid confusion with "quasicompact"

[worthy olive]

This might be off topic, but how do you pronounce simplical homology? Is it:

Simpli kal
or
Simpli shal

[umbral panther]

It is spelled simplicial with an extra i, so the first is right out. The second is the most common pronunciation. I think some people say -si al

[slender glen]

question: is there a classification for simply connected topological n-manifolds? Like, in 2d a simply connected topological manifold is either R2 or S2. In 3d we have poincare conjecture which says that there's only one compact simply connected 3-manifold. What if it isn't compact? What about in higher dimensions?

[umbral panther]

Do you have a guess about listing all simply connected 3-manifolds?

Surgery theory answers questions like this, but only in dimensions at least 5. It works for both smooth and topological manifolds. Casson and Freedman showed that surgery actually works for simply connected topological manifolds in dimension 4. It is unclear if works for general fundamental groups in dimension 4.

For every unimodular quadratic form over the integers there are 1 or 2 topological 4-manifolds. There are 2 if the form is odd and a unique one if the form is even

[merry geode]

I heard cohomology class induces a homomorphism on homology groups

How does this work? E.g. in singular cohomology, isn’t cohomology just a map C_n -> Z ?

[umbral panther]

Things are complicated if you work over Z, so let’s work over a field. Then H^n is the dual of H_n. It is exactly the homomorphisms from H_n to the field. If the groups are finitely dimensional, then the opposite is true, too

Did you know that cohomology forms a ring? Homology forms a module over this ring. This is what you’re asking about. But you can define this using the ring structure and the above identifications. Take a cohomology class x in H^i. We want to define a map from H_n to H_n-i. We can think of a homology class in H_n as a function f from H^n to the field. Define a new function in H^n-i called x.f. We define it as (x.f)(y) = f(x.y)

If you use de Rham cohomology, the ring structure is just wedging forms together. Duality is integration. The action of forms on homology classes probably can’t be made better than the formula above: think of homology cycles as domains of integration. To integrate a small form, wedge with the given form, then integrate in the big cycle

But if you use singular homology, you get the cup product and cap product which are not as nice and cannot be as nice because they are not quite commutative

[merry geode]

Ahh, so the ring makes the homology a module, and then the usual R -> End(M) follows?

And then you also have some grading.

[merry geode]

Ah. Not as nice.. (I was taking a lesson in AG where this came up, so it will be even less nicer)

[umbral panther]

I wouldn’t say being a ring makes it a module. I don’t know what you mean by R -> End(M). That sounds like it’s just rewriting the fact of being a module. But you have to produce the module structure at some point
I’m just saying that homology is the dual of cohomology, so is almost the same and the module carries over. It’s even true over Z, but it’s not so formal. If you delve into derived categories it should become formal

[merry geode]

I see.

Yeah, I see that module structure is not for free here

[umbral panther]

And I used finite dimension. It’s true without that hypothesis

[merry geode]

How do you define for infinite dimension, where H_n would not be a dual?

[umbral panther]

You could pursue this approach and say you get a map H^n -> k and have to prove that it comes from H_n

I think you could argue by approximating a big space by one with finitely generated homology to show this

[merry geode]

Hmmm.. sounds a bit involved

[umbral panther]

According to constructive math, all vector spaces are reflexive. Since I wrote down an element in the double dual of homology, it is an actual element of homology

[merry geode]

Wow

[limpid dragon]

for 4.2 here I'm missing something obvious. I want to say the arbitrarily large lengths of the attached rays break something, but I'm not actually sure what's wrong

[paper wedge]

is D contractible

[limpid dragon]

it should be possible to parameterise each line (nt, 1-t) and then draw back along it right?

[paper wedge]

yeah

probably has to do with path-connectdness

like the spaces between each line

in D

[limpid dragon]

i thought so, but you can do the same thing you can in the infinite wedge interval case, just make a small enough ball and you always get local path connectedness

distances between lines get arbitrarily small but we can just do the same to neighbourhoods right

[paper wedge]

idk what happens if you break the first line or just remove a point between (0,1) and (1,0)

[limpid dragon]

not much, i think connectedness is actually the same between them

[paper wedge]

probably has to do with countability

or nvm

[paper wedge]

u sure? even wiht like

removing points and shit?

yeah ur right nvm

[limpid dragon]

i hate topology counterexamples

[paper wedge]

yeah it has to do with the topology itself

as in

the open sets in D are intersections of Z with intervals

while with the cone u have Z x I

so you probably have osmething that is open in Z x I but not in D

[last gale]

Need a sanity check, this is wrong right? The image of [1,0,1]^T has dimension 1, so the 0th homology group should be isomorphic to F^2 instead of F?

[limpid dragon]

yeah i thought we could use the fact it's subspace somehow

[paper wedge]

i think this is #algebraic-geometry

[paper wedge]

an open set would be an open interval intersected with the integers no

so first lets assume they are homeomorphic

[last gale]

does algebraic topology fall under algebraic geometry?

[paper wedge]

idk but ig its more useful for you to ask in the alg-geo channel

[knotty vine]

Nah this is fine here

[paper wedge]

yeah

yeah any interval having non-integer distances

[limpid dragon]

wait a fucking second

[paper wedge]

intersected with Z is not open in D

[limpid dragon]

you cant include 1 in an open set in the cone

[paper wedge]

yeah

[limpid dragon]

fml

it's always something obvious in retrospect

[paper wedge]

so consider any interval intersected with an open subset that contains (0,1)

with non-integer distances

the image must be open no?

[limpid dragon]

yeah

ok thats it ty

[paper wedge]

yeah

np

didnt do much haha 😄

[limpid dragon]

(that is also what I'm getting, yes)

[last gale]

Thank you

[knotty vine]

Yeah I think this should be F^2

[last gale]

Appreciate it, I'm not very comfortable with homology group so wasn't sure if it was just me

[gritty widget]

anyone knows a proof of the fact that countable complete metric spaces have an isolated point, without using Baire's theorem?

[alpine nest]

There's a proof in Rudin (in chapter 2), which is for R^n but can be adapted to an arbitrary metric space

[gritty widget]

thanks

[alpine nest]

Theorem 2.36 is that the intersection of a descending sequence of compact sets in a complete space must be nonempty.

Oh wait, this proof requires the Heine-Borel property

So it won't generalize to arbitrary metric spaces, or at least not without adaptation

But maybe it will give you an idea

[gritty widget]

alright thanks

[tender halo]

its true in Polish spaces

oh well actually you need Polish-ness to prove that P has the cardinality of the continuum

[alpine nest]

Yeah, I know the theorem is true, but I was talking about the specific arguments used in the proof I included

[tender halo]

tbh really its just a restatement of Baire's theorem to the case when the nowhere dense sets are just one point

so its Baire in disguise

no one can escape

[alpine nest]

True, most proofs that don't explicitly invoke Baire do essentially the same thing anyway

Just embrace the Baire ||🐻|| and move on

[uneven bronze]

I have a silly question. I'm reading in a text that "...$\mathbb{R}/\mathbb{Z}$, $[0, 1)$, and $S^1$ are all homeomorphic (with the appropriate topologies; in particular, the topology on $[0, 1)$ is not the subspace topology)." Which topology is meant for $[0,1)$ if not the relative topology in this case?

[gentle osprey]

Philip

[uneven bronze]

the choice of the topology of [0,1) probably depends on the choice of topology on R/Z and S^1

[alpine nest]

No, on S^1 and R/Z the topology is standard

On [0,1) you want a topology that identifies 0 and 1

Since intuitively, the circle is an interval with the ends glued together

So the neighborhoods of points other than 0 can be as normal, but the neighborhoods of 0 will be different

But it's been a while since I did anything with quotient topologies so I can't make it rigorous offhand

[uneven bronze]

ok, makes sense

[sterile fossil]

You take the closed set S = [0,1] with its subspace topology, then you take the quotient topology of S/{0,1}

If B_1 is the set of open intervals in [0,1), and B_2 = {[0, a)U(b,1): a>0, b>a, b<1}, then B = B_1UB_2 is a basis for the topology of [1,0)

[nova fjord]

Dumb question. I'm trying to show the complex quadric defined by z0^2 + z1^2 + z2^2 + z3^2 = 0 (in homogeneous coordinates) is homeo to S^2 x S^2.

[gentle osprey]

Eternal Way

[cedar pebble]

if you have the quadric

x^2+y^2+z^2+w^2=0

consider the change of coordinates

a=x+iy
b=x-iy
c=-z+iw
d=z+iw

so this becomes the quadric

ab-cd=0

But this is precisely the image of the Segre embedding P^1 x P^1 -> P^3, so your conic is a P^1 x P^1 in P^3 (after homogeneous change of coordinates), which over C is homeomorphic to S^2 x S^2

[nova fjord]

Thanks!!

[rich trail]

Is the isomorphism from $\pi_1(S^1)$ to $\mathbb Z$ just $f([\gamma_n])=n$?

[gentle osprey]

kevinhardy2

[rich trail]

Where gamma n is a loop that goes around n times

[limpid fern]

pretty much

but like

the meat of the proof is showing why its a bijection

[chilly lynx]

Guys, how can I prove that int and fr of A are disjunct in R?

tried reductio ad absurdum but did not reach a contradiction

[limpid fern]

whats fr

[chilly lynx]

border

[limpid fern]

boundary?

[chilly lynx]

yes

[limpid fern]

uh how does your course define boundary of a set

this should literally be immediate with some definitions

and if not still quite short

[chilly lynx]

I thought that, but my teacher wants the greatest possible rigor

[limpid fern]

show your work please

okay its this definition

[chilly lynx]

starts from the 3rd line

Let's suppose ....

[limpid fern]

I see

okay so suppose x is both boundary and interior

lets draw a picture of the real number line

so theres some neighborhood of x, with radius eps such that this neighborhood is a subset of A

now if it is a boundary point we are saying for every delta we have (x-delta, x+delta) intersected with A has a point of A AND a point not in A

again draw a picture

focus on the latter part of the definition

[chilly lynx]

so is not necessary that eps = delta

and the proof is trivial by definition?

[limpid fern]

delta is arbitrary but eps is fixed

keep that in mind

eps is given

[chilly lynx]

:0

ty man

I have been delving into these definitions for analysis and topology. But it has cost me a lot, do you have any advice?

[limpid fern]

honestly just practice a bunch

it was quite tedious for me the first time around as well

[chilly lynx]

okay thank you very much

[stuck geyser]

This is a really stupid question, but is it relatively simple to show a compact metric space is separable

Like I can just, extract a finite subcovering for each n from the covering by ALL open balls of radius 2^-n, and just use their centers

Actually this should work for Lindelof’s too

More reasonable question

Let M be a metric space, is it possible to have a proper subset S of M where there exists a point such that every ball around M fits in S

I assume that is false since there must be some point y in M\S, and d(s,y) = r must be bounded so any larger ball can’t fit solely in S

[fair idol]

Say $D^n$ is the disk then in hatcher it says that $\partial D^n$ is not a retract of $D^n$.

Is it possible that $\partial D^n$ is a deformation retraction of $D^n$?

[gentle osprey]

HausdorffT1

[red yoke]

No

Deformation retracts are retracts

[unreal stratus]

In fact, showing it's not a deformation retract is easier because you can just say that D^n and its boundary have different fundamental groups

[round oyster]

,,\text{ By def a compact set A is compact if and only if all sequenes } a_{n} \subseteq A \text{ have an accumulation value ( converge ) }

[gentle osprey]

Kapa

[round oyster]

have an accumulation value

but

whats the essence of it. how could i understand it abstactly? , like whoever put the definition of it. why was it like this?

[gritty widget]

this is false

[paper wedge]

this is true only over metric spcaes

spaces*

[gritty widget]

or first-countable spaces

in general you need to replace sequences by nets

also having an accumulation value doesn't mean it converges

it means theres a subsequence (or subnet if you are using nets) that converges

[round oyster]

yes i forgot to add all sequences have a subsequnce that converge

a sequence having an accumulation value doesn't mean it converges

right?

[round oyster]

yes

[gritty widget]

right

[round oyster]

but why was it named Compact

and why is't defined this way ?

and extra question , if we apply a function that is countinous on a compact set A , would f(A) be compact

[gritty widget]

intuitively i'd say, if your space is contained in a bigger one

then as you walk inside your space, you can't get close to a point that is outside of it

[round oyster]

yes

hmm

so its like pressed within it self

that nothing inside it can leave it without having to get close to some element of it

[alpine nest]

Because it's not too spread out, basically

[alpine nest]

Yes, it's a very important result, and any textbook should have a proof (and it's even a fairly doable exercise)

[round oyster]

in R , can a set be spread out ?

[round oyster]

i'll try and do it as an exercsise thanks

[alpine nest]

Oh yes, even sets you'd think of as quite small, such as (0,1), since the concept is related to something broader than just diameter.

In R the only compact sets are ones that are both closed and bounded

[round oyster]

for example (0,1) is spread out , apart from definition , if we consider [a,b] the only diffirence is those boarder points that changes everything righ?

[stuck geyser]

Heine Borel implies closed and bounded sets are compact

Bolzano-Weierstrass implies closed and bounded sets are sequentially compact

Both are equivalent assuming choice for metric spaces lol

Kinda funny imo

[alpine nest]

Yes, R with the standard topology is a metric space, and I'm not some kind of sicko who would quibble with AoC

[stuck geyser]

Good luck working with sequences without at least dependent choice

[alpine nest]

Indeed, the inclusion (or ommission) of these two points changes a surprising number of things.

Topology can be weird like that

[stuck geyser]

Mmmm Hausdorff

[alpine nest]

Lovely boy

I mean, (0,1) is also Hausdorff

It's a perfectly normal space (literally)

[stuck geyser]

No I mean R is

[alpine nest]

Just not compact

[stuck geyser]

Precompact

Which is the statement usually made for analogous theorems like Arzela Ascoli

Which is kinda just like Bolzano-weierstrass’s functional cousin

[round oyster]

thats really weird , i guess i'll learn about it deeply when i take toplogy next year

thanks for ur hlp bro

[stuck geyser]

So chat, I haven’t done much topology actually I haven’t even finished real analysis and I’m bored so I’m massively going to jump ahead

I’m trying to think of a way to construct a function into R+ who’s fiber of 0 is a given closed set on a metric space

I have an idea

[unreal stratus]

Well think in terms of the metric specifically

[stuck geyser]

Let’s say we have a closed set S, for each x, we can define r(x) =sup{r : B(x ; r) \subset S}

If S isn’t all of the metric space

Then there is a point y outside of S, then d(x,y) bounds that r(x)

So it’s defined for each x

[unreal stratus]

Hm this r may vanish for every x though

Consider the Cantor set in R

[stuck geyser]

Ah.

[unreal stratus]

You want to do smth less intrinsic to S

[stuck geyser]

The problem is that I am concerned about the countable covering “overflowing” outside of S

so if I define a function off of it, the 0-fiber might “overflow” outside of S

[unreal stratus]

This is funny actually as I just said the cantor set but remembered my advisor giving a stack exchange answer about a function vanishing on only the cantor set lol

[unreal stratus]

Yeah

[stuck geyser]

That’s my main concern

Or if it’s even possible

@unreal stratus my ultimate goal was to try to prove smooth Urysohn’s

[unreal stratus]

Well I can give a hint lol, what would you do if the set is a point

[stuck geyser]

… what do you mean

[unreal stratus]

Sorry got distracted lol

If you know about Urysohn for metric spaces then the answer to this is very similar

[stuck geyser]

I haven’t seen the proof for it and I want to try to prove it myself first

[unreal stratus]

Sure

Then dw

[unreal stratus]

But ye think about this

[stuck geyser]

I mean I can send d(x, x_0) into a function smooth and zero only at 0, like e^-1/x^2

[unreal stratus]

Wait you didn't say your function had to be smooth lol

How would that make sense in general for a metric space

[stuck geyser]

Nah if it wasn’t smooth I could immediately just use normal Urysohn lmao

[stuck geyser]

*separable

[unreal stratus]

Well still

[stuck geyser]

Fair, I was about to invoke a linearity argument but it’s not a topological vec space always

[alpine nest]

Here's a hint: take a closed set A in a metric space, and take a point x that's not in A

[unreal stratus]

I thought you just said you didn't know the proof of urysohn

[alpine nest]

What can you say about the distance from x to A?

[unreal stratus]

Lol

[stuck geyser]

smooth Urysohn

for like R^n

[tender halo]

thats just distance to that set

[unreal stratus]

But I didn't say smooth

Ok lol

[stuck geyser]

Nah that’s on me

[alpine nest]

Right, I gave a hint, and baire cow gave the full answer

So you should be all set

[unreal stratus]

Can you just state the actual question then instead of smth else which misleads lol

[stuck geyser]

I know the regular proof of Urysohn, or at least a variant of it (I call it a “bullseye” route)

[unreal stratus]

For future

[stuck geyser]

Let me restate in a specific form:

For a closed set S in R^n, I am trying to prove there is a smooth function from R^n to R+ where the fiber of 0 is S

[unreal stratus]

Sure that makes sense and what I assumed from this

[stuck geyser]

I tried to generalize way to hard, and completely dropped the vector space structure which the notion of a derivative depends on

[alpine nest]

XY problem strikes again

[stuck geyser]

Anyway, the issue is that if I try to use a ball-covering argument, the “support” might overflow

For individual points like s_0 there are options, like e^(1/d(x,s_0)^2)

I was originally going to make a series out of the Lindelof property

Namely just series over scaled bump functions over the countable subcover such that it’s smooth but the support will then overflow out of S’s complement

I think now this won’t work

[tender halo]

whats the problem here

you take the countable cover of the complement of S

[alpine nest]

Yeah, that's the way to go

[stuck geyser]

Oh shit, it shouldn’t overflow then

I was thinking of the wrong direction lmao

[tender halo]

place a bump function that is nonzero in the ball and zero outside, take the sum, ur done

[alpine nest]

Well, possibly include some kind of convergence term 😛

[tender halo]

if they refuse to converge thats a they problem

[stuck geyser]

I need to chose my coefficients in a way that keeps shit smooth but I’ll toy away and try to base it off the geometric series and go from there

[alpine nest]

I like your thinking!

[stuck geyser]

Problem I need to deal with is the smoothness and the radii

but I think I can choose my coefficients in a cheeky way

I can keep my bump functions bounded above by 1, and just make it an almost geometric sequence for basic continuity of the series

[round oyster]

if we have a sequence Vn that is bounded and we define a sequence Un = supVp for all p >=n would it always be decreasing?

[tender halo]

yes

[round oyster]

what if Vn is increasing , then Un would be constant no?

[tender halo]

yea

it will be decreasing non strictly

or non increasing

whatever your class calls it

[round oyster]

ohh everything makes sense now bro , its from french book

[stuck geyser]

That’s actually how Bolzano Weierstrass is proved usually in one case lol

[round oyster]

its proved by peaks rght?

[stuck geyser]

“Trailing” peaks but yes

Points greater than all succeeding points

[round oyster]

we define a peak as a term where all terms after it are smaller than it

[round oyster]

yes but

the proof was all oral

it felt like a scam

[stuck geyser]

Let u_n be a bounded sequence, then v_n = sup{v_m, n > m} is a monotonic sequence (technically can be embedded as a subsequence if you “collapse” equal terms

[round oyster]

yes then we apply that theorem where every bounded sequence and monotonic is convergent

[stuck geyser]

yep

[round oyster]

thus there always exist a subsequence that converge

[stuck geyser]

There’s like three statements of BW that have increasing specificity

[round oyster]

there is another statment ive seen where there is a finite number of peaks

then it must be increasing(non decreasing) again after the last peak

i suppose it must converge to some value l where l < last peak

[stuck geyser]

Like for example

Monotone bounded sequence => Cauchy holds in any Archimedean ordered field

Which can be proved from Archimedean property rather intuitively

Assume x_n is monotone increasing, and bounded above by M. assume x_n isn’t Cauchy, then there exists an e such that for all N, there is an n > m > N where x_n - x_m >= e

If we keep adding up these “e-large gap terms”, eventually from x_n, they will exceed M due to the Archimedean property, a contradiction

[round oyster]

so whenever we have a sequence that is monotonic and bounded it must be cauchy in any archimedian field ?

[stuck geyser]

Succinctly stated, if we have a bounded increasing sequence that isn’t Cauchy, then there is a gap length such that no matter how far out in the sequence you go, there are terms after it with gaps greater than your gap length. However starting at x_0 if we keep adding up these gap terms, by Archimedean property, it’ll eventually exceed our bound, which is a contradiction

[round oyster]

also a question , when we add these terms they will eventually exceed M yes , but what does archimedian property have to do with it?

[stuck geyser]

For any x and y, there is an N such that Nx > y

[round oyster]

yes

[stuck geyser]

So if we add up the gap terms N times, it’ll be greater than x_0 + Ne

But by Archimedean property, there exists some N such that Ne > M - x_0

[round oyster]

ohh ok got it

[stuck geyser]

The sum of the gap terms are all greater than x_N - x_0 in essence

So it implies that x_N exceeds M

Which can’t happen

It’s a bit annoying to do formally

[round oyster]

i got it , if a sequence isn't cauchy , there is a small gap of lenght that the diffirence of terms can't be smaller than that gap right

[stuck geyser]

There is a gap length such that for each N where there are term pairs both after N who’s indices differ greater than our gap length

$\exists \varepsilon : \forall N [ \exists n, m : |x_n - x_m| \geq \varepsilon]$

[gentle osprey]

Request a new nickname

[stuck geyser]

Which holds for any Archimedean ordered field

But it can also be proved since every Archimedean ordered field embeds into R and it holds for R so

[round oyster]

$\exists \varepsilon : \forall N [ \exists n, m =< ? N : |x_n - x_m| \geq \varepsilon]$

[gentle osprey]

Kapa

[round oyster]

what do u mean by it embeds into R , i know that the archimediean property hold for R , do you mean you could construct some sort of homomorphism between the two fields?

[stuck geyser]

This is more analysis than topology

[stuck geyser]

Monomorphism from ordered Archimedean field F into R

I can elaborate more in the analysis channel in about 20 minutes

I need to drive to class

[round oyster]

oh got it so they share similar structure

[round oyster]

ok sure bro

good lck

[stuck geyser]

My proof of that did the following rq

I recreated the construction of R from Q in the Archimedean field F, and embedded that field F into its completion. You can show that completion is isomorphic to R because Q lies in F. Then you can prove the embedding of F into R by that isomorphism

I didn’t go off of a textbook for this shit, I tried to prove it on my own so it could be wrong but it seems fine to me

Q is a prime field and every ordered field is char 0 so Q is in F by default

[round oyster]

i wish i could understand most of the proof , but it feels correct , good job , i'm still not that advanced , ill look at ur proof in the future ig

[fair idol]

Hey I'm confused by what exactly are abstract cohomology theories

Is an abstract cohomology theory a collection of functors h^n:Good pairs-> Ab groups with a bunch of axioms like excision and dimension (maybe)?

Also is the example of the thick A and the thin A deformation retraction from hatchers algebraic topology text an example of a good pair?

[cedar pebble]

right, you're axiomatizing the usual properties that singular cohomology satisfies

if you include this dimension axiom, then essentially any collection of functors satisfying these conditions is the same as singular cohomology

if you drop the dimension axiom, you can get other more interesting "generalized" cohomology theories

sometimes like complex K-theory (which is defined in terms of complex vector bundles on your space) satisfies all these axioms except the dimension axiom for instance

[gritty widget]

What is the point of non-generalized cohomology theories if they behave the same as singular cohomology?

[cedar pebble]

well there are other cohomology theories that turn out to be the same as singular cohomology but in a way that isn't obvious from their definition

a good example is something like de Rham cohomology (for smooth manifolds)

it satisfies all the same axioms of ordinary cohomology theories, and it turns out to be the same as singular cohomology, but this is an actual theorem you have to prove

having some characterization of ordinary cohomology theories like this helps you prove theorems like this

it's also just kind of clarifying that the dimension axiom is exactly what distinguishes ordinary cohomology from these more general cohomology theories

just because every ordinary cohomology theory is "the same" as singular cohomology doesn't necessarily mean you want to just treat everything as singular cohomology

like there are certain extra structures that exist on de Rham cohomology that aren't so obvious for singular cohomology, and some extra structures on singular cohomology that aren't so obvious on de Rham cohomology for example

for example de Rham cohomology for smooth manifolds ends up being isomorphic to singular cohomology with coefficients in R. But for singular cohomology you could always take a smaller field/ring of coefficients, like coefficients in Q or even coefficients in Z (modulo torsion issues)

this means that singular cohomology carries a natural Q-structure or even a natural Z-structure which you can transport to de Rham cohomology

whereas if you start with de Rham cohomology alone it's not clear at all how this could carry a natural Q-structure

conversely if you have something like a complex manifold then de Rham cohomology carries some extra structure like a Hodge decomposition coming from how you can decompose differential forms into dz's versus d\bar{z}'s

and this structure is kind of invisible to singular cohomology alone

[round oyster]

mind if i ask but what subject is this ?

[cedar pebble]

algebraic topology

[round oyster]

oh okay thanks

[fair idol]

Thank you! Are the ones without the dimension axiom also sometimes called extraordinary cohomology theories but I'm not exactly sure if thats a standard term

[cedar pebble]

yeah historically these were called extraordinary cohomology theories but maybe these days they are not so extraordinary anymore...

the modern technology for studying these kinds of (extraordinary) cohomology theories is the notion of a spectrum in stable homotopy theory

any spectrum represents a cohomology theory in this sense

[coral pawn]

Thinking of G_m as motivic spectra, is there a way to extend the multiplication map on G_m to make G_m into a ring spectra?

[fair idol]

Hey I wanna show the collection of coefficients $h^n:\hbox{good pairs}\to \hbox{Ab}$ defined by $h^n(X,A):=\hbox{Hom}(H_n(X,A),Z)$ is not an abstract homology theory.

I think this fails additivity I guess cohomology theories are supposed to turn disjoint unions into products and so since homology turns disjoint unions into direct sums we have $h^n(X):=h^n(X,\emptyset )=\hbox{Hom}(H_n(X),Z)=\hbox{Hom}(\sum {\alpha} H_n(X{\alpha},Z)\not = \prod {\alpha} H_n(X{\alpha},Z)$ .Is this instead an abstract homology theory?

[gentle osprey]

HausdorffT1

[unreal stratus]

You're missing some brackets

It should actually turn disjoint unions into products

And idk what you mean by abstract homology theory

And I assume somewhere you meant to say cohomology

[slender glen]

What does "it follows that H(1)=0" mean here

What is 0?

[gritty widget]

the zero morphism?

[slender glen]

Ah right, that makes sense

[gritty widget]

Ah right, that makes sense

[slender glen]

Carla~

[chilly lynx]

I have a pretty stupid question, I come down to this definition of derived set

all points isolated

Is the delta fixed?

[gritty widget]

wym by fixed?

[chilly lynx]

Or is it literally any number greater than 0?

[gritty widget]

also this would be more fitting in #calculus than here

[chilly lynx]

ah oks

srry

[gritty widget]

i dont understand your question

lets look at an example

{0} u [1,2]

here 0 is isolated because

(0-1/69, 0+1/69) intersects that set only at 0