point-set-topology

[novel acorn]

So singleton sets of prime ideals aren't always open

Okay well

In general for rings

How did you define the Jacobson topology?

[heady skiff]

in terms of kuratowski's closure operation

like bruh what is the problem even saying

We show that any open set of Prime($\mathbb{Z}$) can be expressed as an arbitrary union of elements of Prime($\mathbb{Z}$). Let $O$ be open, so that $O = \text{Prime}(\mathbb{Z}) - C$ for some closed set $C$. We claim that $O = \bigcup_{p \in \mathcal{P} - \mathcal{C}} \hspace{.05cm} p\mathbb{Z}$ where $\mathcal{P}$ denotes the set of all primes and $\mathcal{C}$ denotes the set of all primes in $C$. Indeed, if $K \in O$, then [\bigcap_{I \in C} \hspace{.05cm} I \not\subset K] so $I \not\subset K$ for all $I \in C$ (otherwise, for some $J \subseteq K$ we would have $\bigcap_{I \in C} \hspace{.05cm} I \subseteq J \subseteq K$).

[gentle osprey]

okeyokay

[heady skiff]

to make sure I got things right, this is saying that the final topology is the intersection of all collections which contain subsets of Y such that f_alpha is continuous for each subset right

there's gotta be some better way to write this right

nvm

[rancid umbra]

more succinctly, its the smallest topology making all of the functions f_a continuous

[tiny obsidian]

It's the collection of subsets whose preimage under any f_a is open, which is equivalent to intersecting each collection of subsets for a fixed f_a

(which as already said is the coarsest topology where each f_a is continuous)

[heady skiff]

got it thanks

how is $\varphi_{b - a, a}$ a restriction of $\varphi_{a, b}$? for instance they don't agree on 1

[gentle osprey]

okeyokay

[tiny obsidian]

you're right that is isn't a restriction

I have no idea what they meant here

[sonic crane]

Wtf textbook is this

It looks like if topology was taught by a highschool textbook

[junior granite]

Not sure if this is the right channel, but below is the start of a proof I am writing: \ \

Let $P$ and $Q$ be posets with partial orders $\leq_{P}$ and $\leq_{Q}$, respectively, such that every subset of $P$ has a least upper bound in $\leq_{P}$ and every subset of $Q$ has a least upper bound in $\leq_{Q}$. Let $\leq_{prod}$ be the product order on $P \times Q$.
Let $S \subseteq P \times Q$ such that the elements of $S$ are ordered pairs $(p,q)$ where $p \in P$ and $q \in Q$. By hypothesis, $P$ has a least upper bound and $Q$ has a least upper bound. \ \

How do I show that $P \times Q$ has a least upper bound? It feels so obvious and yet not obvious.

[gentle osprey]

Vic

[junior granite]

This is for topology course I just don't know if this is the right category to ask this question.

[rancid umbra]

P has a least upper bound p* w/r/t <_P, Q has a least upper bound *q w/r/t <_Q. what can you say about (p*,q*) in P x Q?

[junior granite]

Can I conclude that it's an (maybe not the least) upper bound in P×Q? Or am I jumping the gun?

[rancid umbra]

why do you have any reservations?

the product order is compared component-wise

[junior granite]

Because I wasn't sure of this. I was second guessing myself.

[junior granite]

So if that is what I can conclude, how can I explain that any subset of P×Q has a least upper bound? Do I just say "by hypothesis, any subset of P and Q has a least upper bound, so P×Q has a least upper bound, therefore any of its subsets must have a least upper bound."

[gentle osprey]

Vic

[junior granite]

I have a new question:

"Let $X$ be a set and view the power set $\mathcal{P}(X)$ as a poset via the partial order $\subseteq$. Prove that every subset $\mathcal{A} \text{ of } \mathcal{P}(X)$ has a greatest lower bound." \

My proof so far:
Let $X$ be a set with the power set $\mathcal{P}(X)$ as a poset via the partial order $\subseteq$. We want to show that every (non-empty) subset $\mathcal{A}$ of $\mathcal{P}(X)$ has a greatest lower bound. Let $B \subseteq \mathcal{P}(X)$ be the greatest lower bound of $X$. Now we need to show that this lower bound is a subset of $\mathcal{A}$. Well, let $A$ be the set of lower bounds in $\mathcal{A}$ such that $B \subseteq A$. Then, $\bigcap_{A \in \mathcal{A}}=B$. \

Am I going about this correctly?

rewritten because i made a mistake

[gentle osprey]

Vic

[uncut tartan]

think you forgot to add the A at the end of the intersectoin but i believe so. Since if $B\subseteq A$ for all $A\in \mathcal{A}$. Now to make sure that its the glb is probably the only thing thats not in this proof

[gentle osprey]

!Pymamba™

[uncut tartan]

which shouldnt be too hard i think if any subset $C\subseteq A$ for all $A\in \mathcal{A}$ then by definition it all $x\in C$ are also in $A$ for all $A\in \mathcal{A}$ which by definition of the intersection means $x\in B$ so $C\subseteq B$

[gentle osprey]

!Pymamba™

[uncut tartan]

you can probably say that in a more concise and less akward way

[tribal palm]

so i'm looking over some of my notes from last year, and i'm skeptical of my past self's "proof" of the closure under intersections here

[uncut tartan]

but the point is that you can do a standard subset proof to show C is a subset of B

[tribal palm]

in particular how can i justify going from |script A| < infty to |script E| < infty ?

[tribal palm]

oops and that plain E should of course be a script E

[junior granite]

I wound up rewriting it because I was told my assumption that $B \subseteq \mathcal{P}(X)$ is the greatest lower bound of $X$ is not a valid assumption to make.

[gentle osprey]

Vic

[uncut tartan]

i mean yeah you shouldnt assume it. but you can prove. like mentioned above

[junior granite]

I wrote this:

Let $X$ be a set with the power set $\mathcal{P}(X)$ as a poset via the partial order $\subseteq$. We want to show that every (non-empty) subset $\mathcal{A}$ of $\mathcal{P}(X)$ has a greatest lower bound. Well, let $\bigcap_{A \in \mathcal{A}}=B$ for subsets $A \subseteq \mathcal{A}$. By definition, $B \subseteq \mathcal{A}$, meaning that $B$ is a lower bound of $\mathcal{A}$. Suppose we have a $B'$ such that $B' \subseteq \mathcal{A}$ is another lower bound. Then $B' \in B$, making $B$ the greatest lower bound. Thus any subset of the power set has a greatest lower bound in the poset defined above.

[gentle osprey]

Vic

[uncut tartan]

$B \subseteq \mathcal{A}$?

shouldnt it be B is a subset of A, for all A in \mathcal{A}

[junior granite]

oh

oops

yes

wait no?

[uncut tartan]

the starting is correct here but it goes wrong when you say B'\in B

[junior granite]

wait you're right i misread my own work lol

[uncut tartan]

i figured lol its alr

[junior granite]

I meant to say B' \subseteq B. would that fix the issue?

[uncut tartan]

well you need to show that

but it is correct

that is p much the only error besides the typos

like the intersection not saying \bigcap_{A \in \mathcal{A}} A=B and instead saying \bigcap_{A \in \mathcal{A}}=B

i think thats the only mistake i didnt mention before hand

[junior granite]

pain i'm so groggy this morning how did i miss that lol

wait so is B' a lower bound because it lies in the intersection? and B is the greatest lower bound because it contains B'?

no wait

gahhh

How would I go about showing that B' \subseteq B?

[uncut tartan]

what is the definition of a lower bound

for this set

for this poset*

[junior granite]

a set that is a lower bound for every set in \mathcal{A}?

feels cyclical but that's what i got

[uncut tartan]

which means what?

[junior granite]

it bounds the set from below...?

all other sets in \mathcal{A} are greater than or equal to that set

rather the glb is contained in all those sets

[plush folio]

If $\mathcal{A}$ is finite then $\bigcap \mathcal{A} = \bigcap_{i = 1}^n A \cap U_i = A \cap \bigcap_{i=1}^n U_i$ where $U_i \in \mathcal{T}$ unless I'm mistaken. Sets of sets is pretty confusing, so I think it's easier to explicitly write out the contents of $\mathcal{A}$

[gentle osprey]

sheddow

[uncut tartan]

what?

[uncut tartan]

what is the definition of that

[junior granite]

i'm not sure i know how to explain that. i guess that is where my studying shall begin this evening. thank you for the help!

[uncut tartan]

the definition of a lower bound of a set is an element that is less than or equal to every element of the set. in this case B' should be a subset of all sets A in \mathcal{A} since thats how we defined this poset.

so by definition if x \in B then for all A in \mathcal{A}, x is in A

[junior granite]

i misused set instead of element.

sigh i tricked myself into thinking i didn't know what a lower bound was.

i had to turn in the assignment but i will use this for corrections/exam prep. thank you again.

[robust drum]

if f: X -> Y is continuous, X is locally compact, and Y is Hausdorff, is f a closed map?

this is definitely false right

like [0, 1) is locally compact isn't it

ok yes this is obviously false every point has a small enough open interval around it so its closure is compact so [0, 1) is locally compact

and then you can get continuous injections from it to the circle

[uncut tartan]

is there a name for metrics on rings or fields that satisfy:

1. d(x,y)=d(x+a,y+a)
2. d(a * x, a * y)=|a|* d(x,y)
3. d(x * a,y * a)=d(x,y) * |a| (if the ring isnt commutative)

i know norms as far as i can tell satisfy this

but are there others

im guessing compete fields or rings might have metrics similar to the l^p metrics

[rancid umbra]

are you sure you don’t mean d(ax,ay) = |a| d(x,y)

otherwise the distance could be negative

[uncut tartan]

uh yeah thats probably a good thing to change it into

i dont really have a concrete idea of what these spaces are like

im just comparing them to what the standard metric does on R or R^2

[rancid umbra]

the second (and third) condition(s) turns it into a normed vector space in the case that it’s a field

[robust drum]

I mean to do this you need to have a norm in the first place

Yeah

[rancid umbra]

d(x,0)

i’m sure there’s a concept for a normed ring/module

[robust drum]

Yeah then you just recover that this is some scalar times the metric induced from the norm bc d(x, 0) = ||x||d(1, 0)

Wait

I’m being dumb

Lol

[rancid umbra]

it happens lol

[rancid umbra]

li think we were both being dumb

u need translation invariance to turn it into a norm

[rancid umbra]

i think in the case of a field, the metric d would induce a norm if and only if it is translation invariant and homogenous of degree 1

well

absolutely homogenous ig

[rancid umbra]

also, look up normed ring. having a normed ring means that you can reasonably and meaningfully study normed modules

so you can weaken the statement to:
the metric induces a norm on a commutative R-Module if and only if it is translation invariant and absolutely homogenous of degree 1

[low flame]

whats this zarski topology?

BTW i understand your hint now 😛

@alpine nest whats with the reaction?

[alpine nest]

That's my default reaction to the mention of non-Hausdorff topologies

[low flame]

ah its non hausdorff, so no disjoint unions

[alpine nest]

I know they're necessary to some people, and I do not envy those people.

[low flame]

i dont even know what it is, im trying to google now

im using viro, so its just problems haha

[alpine nest]

I don't either, TBH, I just see the name mentioned often, usually as the example of a non-Hausdorff topology that is genuinely useful.

[low flame]

so, i can just prob move onto the next problem safely?

if thats the case 😛

[alpine nest]

Apparently the Zariski topology on R is just the cofinite topology

So it's not very complicated

I.e. a set is open in the Zariski topology if and only if its complement is finite.

[low flame]

ah okay

[alpine nest]

Well, in addition the empty set is open, of course

[low flame]

wouldnt this just be empty then in the interior?

[alpine nest]

Yep, since any nonempty subset of (0,1) has infinite complement

[low flame]

haha yes, i thought so 😛

[low flame]

thank you brother

[prime elbow]

Intersection of two open dense set is open dense set.

Since the intersection of two open sets is open and let U and V are given open dense set.
And U intersection V is non-empty.
Take any open set S≠\nonempty, then S intersect U is non-empty set and they V intersection U intersection V is non-empty so U intersection V is dense set.

Is it correct?

To show d:X×X -> R is continuous we can use product metric on X×X then take any sequence (x_n,y_n) convergent to (x,y) then x_n convergent to x and y_n convergent to y.

So d(x_n,y_n) ≤ d(x_n,x) + d(x,y) + d(y_n,y) so we get | d(x_n,y_n) -d(x,y) | < e.

Is it correct?

[rancid umbra]

you should def make this as clear as you can.
but it looks like the idea is there

U,V open --> U cap V open

U,V dense --> for any open S, S cap U and S cap V is non-empty, so S cap U cap V is non-empty

[rancid umbra]

you need the reverse triangle inequality as well to bound d(xn,yn) below

so that
d(xn,yn) >= d(xn, x) - d(x,y^n) (reverse triangle)
>= d(xn,x) - (d(x,y) + d(y,y^n)) (triangle)
= d(xn,x) - d(x,y) - d(y,y^n)

[prime elbow]

Yes

[prime elbow]

Or we can interchange d(x_n,y_n) with d(x,y)

[prime elbow]

But S cap U and S cap V is non-empty not necessarily S cap U cap V is non-empty, we need U cap V be open, but yes there is

If metrics are equivalent then they give same open sets, right? But Converse is not true.

Let X = R then take the usual metric on R and another one is d_1(x,y) = | x-y| (1 + |x-y| ).

One is bounded another one is not so they are not equivalent but they give the same open sets.

Correct?

[rancid umbra]

no. these metrics are equivalent. you should prove it

if they give the same open sets, then they are equivalent. that’s exactly the condition you want for metrics to be equivalent: that they generate the same topology on your space

[prime elbow]

But they define two metric space are equivalent when there exists M>0 such that 1/M d_1(x,y) ≤ d(x,y) ≤ Md_1(x,y)

[rancid umbra]

yes. you can show that there is such an M

[prime elbow]

But one is bounded another one is not bounded so d_1 is not equivalent to d

[rancid umbra]

oh shoot yes, ur right,
these are not strongly equivalent

[heady skiff]

for c), could i show that any open ball of any radius in the topology detrmined by this metric is not an open subset of Prime(Z) udner the jacobson topology?

we haven't gotten to it in class yet but something tells me Urshoyn's metrization theorem sounds applicable

[tiny obsidian]

urysohn's wouldn't be really helpful I think since it only gives sufficient conditions to come from a metric, not necessary ones

I would instead think of what topological properties all metric spaces have, and see if the jacobson topology satisfies them

[robust drum]

I think there is a theorem which gives an iff

I’m not sure it’s urysohn

[tiny obsidian]

bing nagata smirnov

[robust drum]

But it does exist

Yeah that’s it

That’s probably overkill for this problem tho lol

[tiny obsidian]

yeah

especially since to show it doesn't apply you still show a certain topological property isn't satisfied

[heady skiff]

yeah i was able to figure it out

does normal space imply hausdorff?

since for any point we can just take singleton set

[alpine nest]

Normal and T1 implies Hausdorff (because T1 gives you closedness of singletons)

https://math.stackexchange.com/questions/3547027/example-of-a-normal-t-0-space-which-is-not-hausdorff

Unless your definition of normal space includes being T1, which I think is sometimes done; then it does indeed imply Hausdorff

[heady skiff]

i see thanks

do we require that V =/= V' in this definition? for example, could we consider the emtpy set and the empty set

also isn't example 3.6 wrong, X and the empty set are disjoint closed sets

also am i missing something or is there a reason they included the converse in a remark

does that mean Urysohn's is an iff?

[alpine nest]

Then the empty set and the empty set wil be a pair of disjoint open sets containing them

[heady skiff]

ye

[alpine nest]

And X and the empty set are a pair of disjoint open sets containing them

[heady skiff]

oh so they just didn't mention that case

i got confused for sm reason then i guess

[alpine nest]

Yes: https://en.wikipedia.org/wiki/Urysohn's_lemma#Formal_statement

[heady skiff]

ye i did some googling too after lol

thanks

[alpine nest]

It's just that one direction is almost trivial

[heady skiff]

got it got it

[tiny obsidian]

they did, they have or when saying V1 or V2 are empty

I did have to reread it twice to catch that though

[sonic crane]

Any tips on showing that hawaiian earring is closed in R^2?

[opaque zodiac]

How are you defining it?

literal union of circles?

[sonic crane]

Yeah

Union of those circles

[tiny obsidian]

I would personally show it's an intersection of closed sets

[sonic crane]

Ok i will try to think about how to do that

[tiny obsidian]

Actually showing the complement is a union of open sets (the ||moon shaped areas between the circles||) might be even easier

[sonic crane]

Ok true thanks i think i can see how to do that. I will go work on it

[opaque zodiac]

honestly, I think it might be slightly easier to just show it's closed under convergent sequences

a convergent sequence either ends up hanging out in some circle or it's limiting to the one special point

[sonic crane]

Cantor set has empty interior 😢

Man how do i even think about the cantor set?

How do you guys think about cantor set?

[tiny obsidian]

with immense difficulty, intuitively as just the construction after only a few steps

that's kind of the point though, that it has weird behaviour

another useful characterisation is as the set of numbers in [0,1] that have a base 3 expansion without the digit 1

note I do mean a base 3 expansion, e.g. 1/3 is 0.1 in base 3 but also 0.022222... and hence in the set

[sonic crane]

Is showing cantor set have no interior we can just use that decimal argument

Any epsilon radius around some point in cantor set would have some decimal that has a digit 1 in it

[tiny obsidian]

you'd need to prove that characterisation if you don't already have it

an alternative is to show that the cantor set has 'length' 0 and thus can't contain an open interval (which is necessarily of positive length)

[bronze wadi]

Im not sure why we can assume the last line

I thought about taking the intersections from k=n to inf but you could have Bn’s be the empty set every other set

Oh wait

Right this isn’t an ordered tuple

Sets already filter out and then it’s fine at some point

I mean requiring the Bn to contain x is not necessary right

They would just not be useful

Yeah I got that

This seems wrong no? Maybe the inclusions are supposed to be the other way

In the context of a metric space you’d need the distance to go to 0

And that does not have a minimum

Oh wait no I just like

Thought about it the wrong way by myself oops

Ok yeah just take the intersection up to n right

That’s why u wanted me to assume the sets contained p

Otherwise that doesn’t make sense

Alright thx

[low flame]

let A be a subset of a topological space X. if I let F_i = X - A_i, can I just go \cap_i F_i is the smallest closed set containing all of A. i mean is that how you would prove this

i can use demorgans here right?

(union_i A_i)^c = intersection_i (A_i)^c = intersection_i F_i

[tiny obsidian]

what are the A_i?

also how was closure defined

[low flame]

closure is defined as the smallest closed set that contains A

i was just thinking of Ai as the indivual open sets in A

[tiny obsidian]

the intersection of the X - A_i doesn't contain A though

the problem directly tells you what the intersection you want is

It is the intersection of all closed sets containing A

[low flame]

i thought closed sets were the compliments of open ones im sorry

so my idea was to take all of these closed sets that are compliment of A

and then shrink it down, but yeah i see your point

[tiny obsidian]

they are, you're just taking the wrong open sets if you want to write this as an intersection of complements of open sets (which you don't need to do)

[low flame]

alright, well i dont wanna get bogged down on one question, how should i approach this then?

[tiny obsidian]

let F_i indexed by i in I be all the closed sets containing A, and take their intersection

show that this always exists (i.e. at least one such F_i exists) and is equal to the closure

[low flame]

alright ill get on it, i appreciate it

Can i say C is the intersection of all these closed sets

and then lets call A* the closure of A

obviously A* subset C

and since C is a closed set containing A then it obviously contains A*

thus C = A*

[tiny obsidian]

you've said $A^*\subseteq C$ twice, not both inclusions

[gentle osprey]

Edward II

[low flame]

okay C subset A*

i mean A* is the closure of the set, therefore its the smallest closed set containing A. C cannot be smaller than A*

Every set in the intersection must contain A*, IE it cannot exclude any points of the closure of A.

Then doesnt that mean that C is a subset of the closure of A?

[tiny obsidian]

I would be concluding A* subset C from what you've just said, not the other way around

'C is not smaller than A*' means C is not a proper subset of A*, but that doesn't tell you that C is a subset. It does tell you that if C were a subset of A*, we must have equality

'Every set in the intersection must contain A*' implies that the intersection contains A*, i.e. A* subseteq C

[low flame]

yeah i understand, i just dont see how to get the reverse

[tiny obsidian]

An intersection is a subset of each set you intersected, so C is a subset of each F_i

That is, C is a subset of every closed set containing A

A* is such a set

[low flame]

ah wow okay

[low flame]

thank you

[heady skiff]

i don't really understand, why do we have to check this forms a topology if we already know that the induced topology is a topology?

i.e. the set of all preimages of open subsets is a subbase so there's nothing to check right, we know it generates a topology

oh are they saying that the topology generated by this subbase actually just looks like the subbase elements

[civic verge]

Are these your notes?

[heady skiff]

i wish lol

but no, somebody took these notes in the class that i'm taking before me

[tiny obsidian]

It's saying that we have a sub-base S, and not that it generates a topology (though ofc it does) but rather that it is one already

[tiny obsidian]

so yes

[heady skiff]

ye i figured it out

thanks

[uncut fulcrum]

If supposed to show that if $X,Y$ are sets that $f:X_{fc} \to Y_{fc}$ is continuous if and only if it is finite to one or continuous.

I am a bit confused as to why this is true however. If $X$ is infinite and $Y$ is finite, the $Y$ gets the discrete topology and $X$ does not. In particular, $f^{-1}(U) \subset \mathcal{T}_X$ for all $U$ in $2^Y.$ That means, if $U={y}$, then $X\smallsetminus f^{-1}({y})$ is finite, so that $f^{-1}({y})$ must be infinite. What am I not getting?

[gentle osprey]

Zander

[gritty widget]

What is X_{f_c}?

[uncut fulcrum]

Finite complement topology

[gritty widget]

Is the "or continuous" part a typo?

[uncut fulcrum]

yes. or constant i meant oops

[thorny agate]

studying for an exam yay

how does the "in particular" follow?

[gritty widget]

For each x not in K, choose such an U_x. Take the union of all the U_x's

[thorny agate]

ah

thanks

[gritty widget]

I wonder if it can be proved without axiom of choice though

It can

Just consider all the collection of all such possible U's

I wonder if there's some natural condition on topological spaces such that it + KC implies Hausdorff

[thorny agate]

KC?

[gritty widget]

A topological space is called a KC space if every compact subset is closed

[thorny agate]

Ok not sure how to do (2)

I've done this exercise so pick your favorite definition of LCH:

[tiny obsidian]

For every x in U = X\K, use condition (4) and pass to a basic subset of V

I think

[heady skiff]

why does this fail/what hypothesis is violated in the tietz extension theorem?

is it that X is not normal?

[tiny obsidian]

Are you asking what fails in 3.27? It's the codomain

[heady skiff]

oh yeah oops

[sonic crane]

For a countably infinite product of lets say R, given box topology , is the boundary of some subset equal to the product of boundary of each factor subset?

[alpine nest]

That's not the case even for a finite product, consider [0,1] x [0,1] (as a subset of R x R)

The boundary of the product is the perimeter of the square, and the product of the boundaries is just the 4 corners.

[sonic crane]

Dang

I get messed up with products

[sonic crane]

Ya this just definitely wouldnt make sense lolol

[sonic crane]

I also realized complement of a product is not product of complements

So basically. When there are products pls think carefully

But i am able to solve the question i had now

On homework

[sonic crane]

Hello, i am missing some detail.

If you have two different metrics on the same space, i know they may or may not generate the same topology. However, i am confused because given an epsilon ball from one metric, couldnt you just stuff a ball with the other metric using a radius that is supremum of distances in the other ball?

Basically i am like couldnt this make us able to always stuff balls in each other

Lol

“Stuff balls in each other”

[opaque scroll]

Maybe consider an example like the discrete metric versus the standard metric on R.

In the discrete metric {0} is an open ball.

How are you suggesting to stuff a ball in that one?

[sonic crane]

yea i see that cant be done, what is the underlying reason here?

sup of distances there is 0?

[opaque scroll]

Yeah, pretty much

[sonic crane]

hm

so besides that edge case it can work?

[tiny obsidian]

there's loads of other counterexamples e.g. there are different metrics you can endow R^N (real sequences) with that lead to different topologies

[opaque scroll]

Well calling it an edge case is perhaps a bit strong. That is pretty much the typical case

[sonic crane]

Btw, the question i am trying to solve rn is that metric space with d and d' generate same topology when d' is d/1+d

so i was trying to fit epsilon balls

[opaque scroll]

Like either a set contains an open ball or it doesn't. If it does then you're good, if it doesn't then the largest open ball it contains has radius 0

[sonic crane]

i feel like this question isnt that hard but now im just a bit confused i think

[opaque scroll]

Do you know what in particular is confusing you?

[sonic crane]

Not yet

[opaque scroll]

What do you have so far?

[sonic crane]

Oh ok wait i think whats happening is that im starting with trying to fit 1 ball inside the other, but then when i try i end up thinking its wrong because that choice doesnt make the other ball fit the other way, but i dont have to worry about this right now because im only considering one case at a time

[tiny obsidian]

what do you mean by "that choice doesnt make the other ball fit the other way"

[sonic crane]

OK so starting with a d' ball, a d ball fits inside using same radius that you started with from the d' ball

that is true right?

[tiny obsidian]

yes

[granite slate]

if X is a topological space and $X \times X$ a product space, can one regard $x \mapsto (x, 0) \text{ and } x \mapsto (0, x)$ to be two "different" inclusion maps $\iota: X \rightarrow X \times X$ ?

[gentle osprey]

dompa

[sonic crane]

i mean that from what i said about d ball fits in d' ball, its not true that also d' ball fits in d ball

like containment is only one way

[tiny obsidian]

ah

[sonic crane]

but i was thinking i was doing something wrong because in my head i wanted both to fit inside each other

[tiny obsidian]

correct

[sonic crane]

but we are considering these cases separately

yeah

[tiny obsidian]

I don't quite understand your question, because these aren't the same map and so yes are different? and yes both are inclusions

(slight note: you didn't say what '0' is though from context it's obvious it's an arbitrary fixed element of X )

[prime elbow]

If X is an infinite subset of metric space which has no limit point then X is discrete space?

[tiny obsidian]

I think so

[sonic crane]

Starting with a d-ball, how can I fit a d' ball inside? All im thinking of is take radius for d' to be sup of d distances

[opaque scroll]

Any topological space without limit points should be discrete

[sonic crane]

wait this doesnt even work at all ?

[opaque scroll]

Well the d' ball of a given radius is contained in the d-ball of the same radius right, so that direction is the easy one

[sonic crane]

I thought i already did that case

starting with d' ball, a d ball fits in with same radius

now im starting with d ball and want d' ball to fit

[opaque scroll]

maybe I swapped which one was d and which was d'

[sonic crane]

d' was the smaller bounded above metric

d' = d/1+d

[opaque scroll]

Okay, right. So then d = d'/(1-d')

So all you have to show is that by restricting d' to sufficiently small you can make d small as well

[sonic crane]

Ok right because I am trying to basically control the size of d by controlling d' now

control the size of d' by controlling d is immediate because d' is always less than d anyway

I am probably taking longer on this question due to my inexperience in analysis tbh

[prime elbow]

Yes

[sonic crane]

ok so you have a d-ball centered at x with radius e. If you take e' = e/(1+e) then anything in the ball d' < e' will make d < e so its in d ball

[granite copper]

As per https://en.wikipedia.org/wiki/Cantor_space#Examples,
if V={0,1}, we can map the (here binary) sequence x in V^N to the real
2·\sum_{k=0}^\infty a_k / 3^{k+1},
and this gives a homeomorphism onto the Cantor set.
If now V is a larger finite set, is there a similar sum, mapping to another Cantor-like subset of the reals?
Maybe bumping the denominator to something large enough beyond |V| does the trick and I take whatever the range of this thing is. But strangely I don't manage to google this seemingly natural generalization.

[alpine nest]

You're on the right track; note specifically what the mapping you've given does to sequences starting with 0 and 1 respectively: it maps them into two disjoint intervals which are the first step of the construction of the Cantor set. Therefore if V consists of n elements, you'll want to do something similar except with the Cantor set constructed using n intervals at every stage

Try to first do this with V = {0,1,2} specifically

Comedy option, all Cantor spaces are homeomorphic so you can always map V^N to the standard Cantor set, but the mapping will be less nice

[tender halo]

you are just choosing a specific topological base of the cantor space

it doesn't strike me as particularly interesting, to me cantor space is the unique stone space of a countable atomless boolean algebra, from that point of view it is most natural to use base 2 approach

[tender halo]

(because Cantor space is a free object in the category of boolean algebras, the coproduct of countable amount of 2s)

what is category of boolean algebas called anyway

[thorny agate]

Struggling to prove (2) => (1). Not sure how to prove basically any part of the definition of a filter or ultrafilter

(ignore deleted messages) I've only shown that the empty set cannot be in U

[tender halo]

obviously X is in U and empty set is not

[thorny agate]

yea

but say A, B in U. How do I show A \cap B in U?

or if A in U how do I show that A^c is not in U

seems basic but I have 0 ideas

[tender halo]

take the partition A, A^c

[thorny agate]

ahhh then by uniqueness can't have A^c is in U

wow

I was trying to partition into 3 nonempty sets >_>

oh wait ok then I see how to do intersection

[prime elbow]

Does R complete under every metric space?

Yes

[plush folio]

R under the metric d(x, y) = |arctan(x) - arctan(y)| is not complete

[tender halo]

i think a metrizable space is complete under every metric iff its finite

[junior granite]

hi folks, i am looking to prove that something is a topology but i'm not sure how to go about that in this instance.

[tender halo]

whats the problem?

[junior granite]

Is there no image attached to the message?

I thought I sent the question, my bad.

[tender halo]

there is

what i mean is what are you having difficulty with

[alpine nest]

Sure, for a category theorist it's probably completely unimportant, but on the other hand, in something like symbolic dynamics you're often interested in the specific basis/metric associated to your alphabet.

The fact that all Cantor spaces are homeomorphic to each other is certainly useful, but it can also be convenient to work with some specific version rather than another.

[junior granite]

Oh oh what's MY problem lol I am just confused how to show that T is closed under arbitrary unions and closed under intersections of finite open sets.

[tender halo]

well uhh my reaction would be saying "obvious", not sure you should write that in your assignment

[plush folio]

hint: T is a topology when X = {a, b}, but not if X = {a, b, c}

I assume they mean Ø and X are in T, but they're not necessarily the only sets in T

[junior granite]

right

[junior granite]

i'm just not sure how to show the other two properties of a topology. is it true that if X={a,b} then it is closed under arbitrary unions? because idk X is an arbitrary union or something

[plush folio]

X only contains 2 elements. You can just list all possibilities for T and check it's a topology in each case

[granite crane]

@prime elbow hi.
Are you still using Carothers? If i spell correctly

[sonic crane]

how can i show closed disk is closed

i wanna use it in my homework but im like uhh can i just say its closed

x^2 + y^2 <= 1

just that closed is complement of open set lol

its early in the class

[unreal stratus]

Well then consider the complement {(x,y) | x^2 + y^2 > 1} and use a similar method to showing open disks are open

It is hopefully clear from a picture heh

[sonic crane]

i just think open disks are open because they are the basis for the topology anyway

do i really need to pick a point in the complement and stuff a ball in there

like do i really need to get down and dirty with the coordinates and shit

[unreal stratus]

Depends what you are doing lol, maybe you would have to on your first homework

But it is also yeah obvious from a picture

[sonic crane]

lol yeah

[sonic crane]

i literally emailed prof if i have to show that lmao

To be fair he actually said if ur not sure if u have to show something ask him

lol

[fierce lily]

only one question here, why finitely many values of n instead of infinite?

[sonic crane]

For R^N and subset A, what do the open sets of R^N\A look like ??? (R^N has box topology)

[sonic crane]

U is a neighbourhood of x and xn converges to x, so necessarily at a certain point in the sequence, ALL xn's lie in U. So, there are only finitely many that arent in U

[tiny obsidian]

open sets of R^N (box) with A cut out if needed, or you can pass to basic subsets being boxes with A cut away

arguably the box topology is the more intuitive generalisation of the topology on a finite product in terms of which sets are taken to be basic open, so the open sets will look similar to what e.g. the open sets of R^2\A look like for A subset R^2: the open subsets of R^2 and cut away A, corresponding basic subsets are rectangles with A cut away

(box is not the nice generalisation in that a lot of the properties of the product aren't actually preserved so really you lose a lot of the intuition other than this)

[prime elbow]

Yes

[fierce lily]

I know all xn's lie in U, but I don't know why it should be finite instead of infinite? convergence implies at n>=N, xn converges to x, N is infinitely large here

[sonic crane]

No, all xn’s do not lie in U

[robust drum]

I mean point set topology at the start (actually in general) is a lot of these annoying verifications

You should probably actually check it

[sonic crane]

Poop

Yeah true

[robust drum]

Especially if you’ve never done it before

It’s just using triangle inequality in a clever way basically

(To show the complement is open that is)

[sonic crane]

The blurb was saying xn is not in U for finitely many

[robust drum]

I think |x| - 1 should work for the size of the ball you want

Around some x in the complement

[sonic crane]

Yeah to be fair i didnt actually give it a shot

[sonic crane]

Oh yeah i guess so lol

[fierce lily]

you mean xn is not in U until n>N?

[sonic crane]

Yeah

So then that means finitely many xn are not in U

[fierce lily]

for this part, why Ux subset f^-1(V) implies f^-1(v) can be written as union of open Ux? My understanding is: for each V contaning x, we have Ux subset of f^-1(V), so it means every f^-1(V) has a open Ux containing x, hence it satisfies that f^-1(V) is open. is that correct?

[sonic crane]

What part are u looking at?

4?

Oh nvm sorry

[sonic crane]

The way you are writing this is unclear

f^-1(V) is open because it can be written as a union of open sets

It can be written as a union of open sets by using property (4). x in f^-1(V) means f(x) is in V (so V is a neighbourhood of f(x)) so by property (4) this means that there is an open set Ux containing x so that f(Ux) is in V, so Ux is in f^-1(V)

[fierce lily]

ok, then why Ux in f^-1(V) implies that f^-1(V) can be written as union of Ux?

[granite crane]

i hope so it makes sense

[prime elbow]

Converse is correct I am not sure about the first one. Is it true that if U is open in (X,d) then there is an open ball such that U \subset B(x,r)

[granite crane]

are you talking about forward direction?

[prime elbow]

Yes

[granite crane]

B_r^d (x) is open for M with respect to d (since every open ball is open in corresponding metric space)

[prime elbow]

Yes

[granite crane]

But the hypothesis is that every open set of (M,d) is also open for (M,d')

[prime elbow]

Yes

[granite crane]

so B_r^d (x) is open for (M,d') as well and so we can find an open ball with center x contained in B_r ^d (x).

[prime elbow]

Are you sure?

[granite crane]

yes? isn't it the definition of open set? (in metric space)

[prime elbow]

Yes😭

[granite crane]

LOL

[granite crane]

How exactly the topology look like? $\mathcal{T}_Y={A\cap Y: A\in \mathcal{T}_X}$

[gentle osprey]

Afzal

[quick delta]

Yes

[granite crane]

ah thank you so much

[sharp panther]

What bounded subset of R^2 can't be divided into 6 equal parts with 3 straight lines such that the lines intersect at some point?

It is a challenge problem in my topology homework set. It may seem irrelevant to the channel. I have shown that a bounded subset of R^2 can be divided into 2 equal parts using intermediate value theorem. You can even divide it into 4 equal parts with two perpendicular cuts; however, it shouldn't be true in general that you can divide a bounded open subset of R^2 into 6 equal parts with 3 cuts.

[red yoke]

You mean what bounded measurable subset of R^2 can't be divided into parts of equal measure with 3 concurrent lines?

[sharp panther]

exactly

[red yoke]

I bet 7 unit disks placed along a large circle works

Wait it doesn't

[sharp panther]

Also, let us assume that the bounded open set is connected. Otherwise, the problem doesn't make much sense

[red yoke]

Let's assume that we have a continuous family of lines parametrized by slope θ that cut the shape into equal areas

Taking polar duality this is a loop on RP² - origin parametrized by polar coords in R/π (allowing radius in R+∞-0)

This loop should be a generator of fundamental group of RP² (?)

[sharp panther]

oh, I haven't really studied algebraic topology. But looks like a great idea to tackle the problem

Thank you

[red yoke]

I'm just writing out thoughts to help me visualize the problem

[red yoke]

I think this can be fixed

Place 7 unit disks along a sufficiently large regular heptagon

And add a sufficiently small disk sufficiently near the clockwise side of each unit disk

And if you want this to be connected you just add an arbitrarily thin circular wire

Any line cutting the shape in half must pass through two unit disks or a unit disk and a small disk

3 lines each passing through 2 unit disks cannot be concurrent

2 lines each passing through only 1 unit disk cannot bound 1/6 the total area

So there is 1 line passing through only 1 unit disk and 2 lines passing through 2 unit disks

And this forces only one possible configuration (up to rotation) which is not concurrent

[sharp panther]

why do we do this?

[red yoke]

Idk you said we assume the set be connected

[sharp panther]

So I take a heptagon and place 7 unit disks on the edges and to make them connected i add arbitrarily small wires right?

[red yoke]

And 7 very small disks

[sharp panther]

to make them connected but with area of epsilon for arbitrarily small epsilon

[red yoke]

Yea

[sharp panther]

I don't see how its true

[red yoke]

3 diagonals of a heptagon are not concurrent

[sharp panther]

oh wait

right

my bad

cuts should go through the "cake"

weirdest cake idea

[red yoke]

Oops yea I'm assuming all lines cut area in half

[sharp panther]

It's actually called cake cutting problem

[sonic crane]

Because we were able to take any element in f^-1(V) and create an open set around it

Since u can do this for any element in f^-1(V), then from that we can write the set as a union of all the open sets we picked

@unreal stratus My prof said thats a basic fact and I dont have to prove it hahahaha

the closed disk being closed

[unreal stratus]

Yeah lol i was also thinking like this is smth usually done in an analysis course

[heady skiff]

does anybody know what $B_0$ and $B_1$ are here? I'm assuming something like $B_0 = \tilde{g}^{-1}({-1})$ and $B_1 = \tilde{g}^{-1}({1})$

[gentle osprey]

okeyokay

[tiny obsidian]

I think so, though it doesn't really matter for the rest of the proof since you can just replace that union with B

(my guess is it, like [0,1] above, is from an earlier version of the proof and wasn't corrected upon revision)

[heady skiff]

i see, thanks

[heady skiff]

also how does $|\tilde{g}(x)| = 1$ force $|\tilde{g_0}(x)| = 1$

[gentle osprey]

okeyokay

[heady skiff]

oh are we just considering the values on A

[tiny obsidian]

well both delta and g0 tilde have norm at most 1, so both factors of g tilde need to have norm 1 if g tilde has norm 1

[fierce lily]

Does it mean the map Z can commute with projective map pi X: XY->X and PI Y: XY->Y?

I mean if someone can explain what does unique map Z commutes with projections mean here?

[tiny obsidian]

It means that doing the unique map then the projection pX (for example) is the same as just doing the map Z to X directly

I'm not sure "commutes with" is the usual terminology for this, although saying that "the diagram commutes" is perfectly standard

By 'not sure' I mean I've never seen it, although I haven't really been exposed to the field much

[fierce lily]

so it means Z->X is equivalent to X*Y->X?

[tiny obsidian]

Z to X x Y and then X x Y to X

if you start with an element in Z, no matter what sequence of arrows you take in the diagram, if they end in the same place you will have applied the same function

or in general any start point, though this diagram only has Z as one where that means anything

[plush folio]

Where is this from? Kinda weirdly written, can you say "one experiences mathematical statements via diagrams" in english?

[fierce lily]

it is from my lecturer's notes, I read his notes and always feel not easy to understand compared to jame's munkres textbook

[civic verge]

Oooh, you could give it to me too :3 if you don't have a problem.

[plush folio]

I don't think Munkres even defines the universal property of the product, so in that sense the notes are better than nothing. But they're a bit sloppily written it seems. You could supplement the notes with Topology: A Categorical Approach though

https://qcpages.qc.cuny.edu/~jterilla/TopologyBook/index.html

[heady skiff]

sure thing: https://dfoiler.github.io/notes/202A/notes.pdf

why do we care about the open intervals contained within [0, 1]? If we give [0, 1] the subspace topology (which I'm assuming we are) then don't we only care about the first intervals they talked about

[fierce lily]

can someone explain what is the sup norm and euclidean norm refer to and why it induces the same topology in fininte dimensional vector space?

[tiny obsidian]

the section you sent is convincing you that sets of the form [0,a) and (a,1] do actually form a subbase, by explicitly showing we can intersect to get the usual basis (which is those + strictly contained open intervals)

[opaque zodiac]

Suppose $X$ is a CGWH space. Fix a subset $A \subseteq X$ and a point $b$ such that $b \in \mathrm{cl}(A)$. Is it always true that there's a compact Hausdorff subspace $Y \subseteq \mathrm{cl}(A)$ such that $b \in \mathrm{cl}(A \cap Y)$?

I'm assuming the answer is no, but in the context of sequential spaces the answer is yes, so it's a little hard to try to reason out a counterexample.

[gentle osprey]

Exomnium

[opaque zodiac]

(CGWH of course means compactly generated weakly Hausdorff space)

[limpid fern]

they probably meant "expresses"

lee ITM has the universal property of product too for what it's worth

[prime elbow]

In topological space X, if A is subset of X and sequence of x_n in A converges to x then x in cl(A), because if we take any open set containing x it has an element x_n, A has non-empty intersection with every open set containing x

Correct?

[prime elbow]

If I want an example such that x in cl(A) but there is no sequence in A which converges to x.

Any hint?

Yes we don't need metric space

[opaque zodiac]

Examples of this are a little subtle when you are first learning topology

Do you just want to know an example or do you want to try to figure one out yourself?

[prime elbow]

I am trying

Is it a natural example or need to learn more about topology?

[opaque zodiac]

There are what I would consider to be fairly natural examples but they're can be pretty hard to come up with on your own

[prime elbow]

I saw some examples on MSE but it is not natural

[opaque scroll]

It's fairly natural, but the space needs to be not first countable. I.e. the point in question cannot have a countable neighborhood basis.

So it would need to involve some complicated uncountable things

[tender halo]

easiest example i can think of is omega_1 + 1

[fierce lily]

One quick question: if there is a map q from X to X/~. Can I argue that there is an open set U in X/~ implies q^-1(U) must be open in X directly? or need any justification( I only can argue that q must be a surjective map)?

[novel acorn]

What is the problem saying exactly?
I'm a bit confused by what you wanna prove
Do you wanna prove there exists a single open U for which q^-1(U) is open or do you wanna prove that q is continuous?

[fierce lily]

I want to know: 1. if it is natural to just argue that a map from X to its equivalence class is surjective. 2. why U susbet of equivalence class can imply q^-1(U) is open here?

[novel acorn]

The openness of U and its inverse image depend on the topology given to X/~
And in the proof they don't claim this
They define it
As in they just say we give X/~ the topology such that U is open if and only if q^-1(U) is open

[fierce lily]

so we don't need to prove why U open in equivalence class imply q inverse is open in X?

I mean: is it unnecessary to prove this?

[novel acorn]

We don't because this is the definition of U being open

It's an axiom of the construction

[fierce lily]

just because this is equivalence class( construction on its own), and define U open, so inverse naturally open right?

[fierce lily]

do you think the part(2) of the diagram give us the condition that f:X -> equivalence X is a bijective map?

[novel acorn]

I mean it's natural to define it this way

[novel acorn]

It doesn't
It just says that X/~ is unique up to isomorphism

[fierce lily]

won't isomorphism -> bijective map? or what does [x] unique up to isomorphism mean here?

[fierce lily]

I am also confused about why f:X->Z is same as f|q^-1(x) here, because I don't see that q:X->[x] cont->q^-1(x)=X here?

[tiny obsidian]

It's not saying the maps are the same, it's saying the condition in (2) "whenever x~x' then f(x) = f(x')" is equivalent to "f|q^-1(x) is constant for any x in X"

the condition says that whenever x ~ x' in X, i.e. whenever x and x' lie in the same equivalence class, that f(x) = f(x'), i.e. that f is constant on this equivalence class

equivalence classes are by definition precisely the preimages of single elements in X/~ under q

[prime elbow]

Any hint to show that in the first countable topology space if x in cl(A) then there is a sequence in A which converges to x?

Yes I know we need to use nested local basis but how I don't know

[rancid umbra]

limit point + nested local basis

[tender halo]

there is only one thing to do

just do it

[fierce lily]

first of all, preimage of elements in X/~ should be in X, do you mean elements in X is equivalence class(because I think elements in X/~ is equivalence class? ( I think i may be wrong here?)

[fierce lily]

second, why f|q^-1(x) is constant? I mean why it relates to constant here

[tiny obsidian]

An equivalence class is the same as a set q^{-1}(y) (which I call the preimage of y) for some y in X/~, directly from definition of q and X/~

I'm swapping to y for notation reasons

Now the condition is saying that if x~x', i.e. if both x and x' lie in some equivalence class y which is the same as both being in q^-1(y), then f(x)=f(x')

i.e. no matter what element of q^-1(y) you send through f you get the same result

which is precisely what it means for f to be constant on q^-1(y)

[tiny obsidian]

this isn't really saying much though it is a little confusing: if x~x' they are in the same equivalence class y = {x, x', ...}, where y is itself a single point in X/~, and q^-1(y) = y as a set, though now we're treating it as a subset of X

I am slightly abusing notation, technically I should be saying q^-1({y}) but I suspect few people bother

[fierce lily]

so the constant here mean numbers like 1,2,1.2 etc or just saying sending some elements to the same result is treated as constant ?

[tiny obsidian]

we say a function is constant if it takes only one value

because it doesn't change as you move around the domain

[fierce lily]

One more question: is it correct to say that f:X->Z is the same as map f:X/~->Z?

[tiny obsidian]

technically no

A lot of people I personally could abuse notation and use f to mean both maps

But they have different domains

And are absolutely not the same

[tiny obsidian]

the map f:X/~ to Z you refer to is precisely the f tilde the proof is showing is continuous btw

so it is very useful to distinguish them sometimes, as in this proof

[fierce lily]

for this f tilde, I am confused why its inverse image is f^-1(U) here. The reason I doubt is because I wonder if we have x ~ x', and x' might not really be in X but is in inverse image of U under f tilde.

[tiny obsidian]

what do you mean by "x' might not really be in X"? of course x' is in X since that's the domain of f

[fierce lily]

if we have q:X->X/~, we might get y={x.x'}, so do you mean x and x' must definitely be in X in this case?

[tiny obsidian]

x and x' are in X by definition, since ~ is an equivalence relation on X

[fierce lily]

in this case, X and X/~ both contain the same elements here, and the only difference is that when y={x, x'}, X and X/~ is not the same domain right?

[tiny obsidian]

X and X/~ do not contain the same elements; the elements of X/~ are subsets of X.

I don't what you mean by "when y={x,x'}, X and X/~ is not the same domain" since y has nothing to do with X and X/~

[fierce lily]

I mean can you list some examples to illustrate that X/~ has different elements from X?

I want to see how X/~ is proper subset of X

[tiny obsidian]

it's not a subset

For example let X be the interval [0,1]

and take ~ to be the equivalence relation where 0 ~ 1

(and also each x~x for reflexivity)

then X/~ has as an element the set {0,1}, and also {x} (which is not the same as x) for each x in (0,1)

to illustrate what I was saying earlier, in this case q^-1({0,1}) = {0,1} which is a subset (and not a point) of X

the condition on f requires that since 0~1 we have f(0) = f(1), which is precisely what it means for f restricted to {0,1} = q^-1({0,1}) to be constant

[fierce lily]

for part(3) proving homeomorphism part: is universal properties apply to all unique continuous map such that phi cont and phi inverse is cont at the same time?

And also, how to show that phi is bijective here?

[rancid umbra]

q’ is such that for any f : X —> Z is continuous and agrees on the fibers of q’, then there is a unique continuous function bar(f) :Y —> Z such that bar(f) o q’ = f

[heady skiff]

could we adjust this for arbitrary closed intervals [a, b] just by translating everything in the proof?

[unreal stratus]

Of course

Can just compose with homeomorphisms between [0,1] and [a,b] to generalise it

[heady skiff]

how would you do that, assuming the proposition is true

[unreal stratus]

Well given a function A -> [a,b] you compose to get a map A -> [0,1], extend that to X -> [0,1] and then compose with inverse to get X -> [a,b]

and clearly the last map extends the first map

||Well, technically there is a special case where a = b but that case is obvious||

[heady skiff]

how do we know that delta has norm at most one

[tiny obsidian]

because it takes values 0 and 1

[heady skiff]

oh yeah I forgot inverses exist lol

thanks

[heady skiff]

oh is X = A u B?

[tiny obsidian]

oh actually good point

I forgot the details since this is from yesterday

[heady skiff]

yeah nw

i'm kinda in a rush so normally i would think about it more

[tiny obsidian]

is 3.8 not the existence of a function to [0,1] with certain properties?

I thought it would be Urysohn's lemma judging by its use

[heady skiff]

yeah that's what 3.8 is

[tiny obsidian]

well in that case since delta is coming from there then the norm of delta(x) is also between 0 and 1

[fierce lily]

I think I know this part, but I need to prove phi homeomorphism, so I need to know inverse of phi is cont and why phi is bijective

[fierce lily]

Here is what it explains, so I ask if universal properties applies to all unique cont function( or even cont is adequate)

Can someone explain why ex 4.0.4 the resulting space is homeomorphic to S^1? and how it defines the equivalence class in the next part?

[prime kindle]

is there a topology study group? id like to get started on munkres soon and it would be nice to do it with other motivated peeps

[fierce lily]

it is a forum to discuss relevant topics about topology

[prime kindle]

ya im aware, but i recall (on an old account, some time ago) that we had study groups where people would meet every now and then to discuss their progress on a book/topic

[fringe thorn]

perhaps this picture is convincing

also, the quotient space defined in the next part is not related to the one you asked about

[fierce lily]

yes, i mean how it wants to define?

[fringe thorn]

here's a GIF for the quotient space homeo to the 2-torus

https://upload.wikimedia.org/wikipedia/commons/6/60/Torus_from_rectangle.gif

[fringe thorn]

I don't understand the question

[fierce lily]

seems like a plane that can be homeomorphic to 2 torus right?

[fierce lily]

probably you can just explain what does the equivalence class discussed next to 2 torus is talking about.

[fringe thorn]

the equivalence relation is saying to take the left boundary of I and identify it with the right boundary of I, and then take the bottom boundary of I and identify it with the top boundary of I

[fringe thorn]

as done in this GIF

the left and the right boundaries become the same thing under the identification, as do the top and the bottom

[fierce lily]

may I ask what does identification mean?

just shaping like the same circle?

I mean when left and right boundary stick together, it shapes a circle, and same stuff happens to top and bottom boundary

[fierce lily]

so it should be a typo here, we should have (a,b) related to (a', b') when (a,b)=(a', b') right?

for all other points

[fringe thorn]

informally, to identify two points in I means to treat them as the same point

the idea behind a quotient space X / ~ is to take a space X, identify a bunch of points together (we do this formally with equivalence classes), and then glue the points in each equivalence class together into one point (this is formally done by sending every element x in X to its equivalence class)

[fringe thorn]

yes, I believe so

[fringe thorn]

so in our example, we have X = [0, 1] x [0, 1], and we identify each (x, 0) with (x, 1) (i.e. every left boundary point to its opposite right boundary point), each (0, y) to (1, y) (i.e. every bottom boundary point to its opposite top boundary point), and every other point to itself. this splits up X into a set of equivalence classes, where every point on the boundary of the square is "the same" as its opposite point (meaning, they belong to the same equivalence class). now we just map every x to its equivalence class [x] to get our quotient space, and this "glues together" all the points in any particular equivalence class

[prime elbow]

okay

[fierce lily]

for part 3 about homeomorphism part: it writes universal property furnish mutually inverse functions which are both cont. But I am confused about if that universal property applies to all unique continous map? I know this phi is unique continuous here, but want to know how phi inverse is cont.

And also: why this phi is bijective?

[prime elbow]

for a prime p, S_p = {n in N | n is a multiple of p }
now take S = { S_p | p is a prime } union {{1}}, then S is a subbasis of N.

yes because union S = N.

but they ask for open sets, i know that open set looks like a set of all multiple of n, where n in N. i know there are more open sets so how can i write ?

[prime elbow]

i am wrong here

finite intersection of subbasis elements is set of multiple of squarefree integer

[granite crane]

Wait wao is it from Carothers, interesting it contains interesting problem

[limpid fern]

let us suppose that Y is another space satisfying the universal property

we have a map q' from X to Y, so by universal property of X/~ q' factors uniquely through X/~ as q' = tilde q' circ q

now swap their positions

then q factors uniquely through Y as q = tilde q circ q'

tilde q' is a map from X/~ to Y and tilde q is a map from Y to X/~

what happens if you compose them? you get a map from X/~ to X/~

now if you apply the property of X/~ to itself you see that the identity also factors as

id = id circ q

so it follows that the composition is the identity

this argument can be rephrased slightly to prove that any construction given by a universal property is unique up to isomorphism (eg with the product topology)

also I recognize those lecture notes but I will not mention who wrote them here to avoid doxxing

[prime elbow]

no, but it helps you to prove that the set of primes number in Z is infinite

[granite crane]

I see.

[prime elbow]

wait sorry it is not correct

i was thinking about Furstenberg topology

[prime elbow]

i dont see why countable intersection of open dense set is non-empty, then dense ?

of R

usual topology

[vocal wharf]

if you want to prove this yourself, my hint is ||you will have to use the fact that R is a complete metric space||

if you want to look it up, this is called ||the baire category theorem||

[granite crane]

Isn't it Baries category Theorem

[prime elbow]

yes

[prime elbow]

okay

to show that if X is metric space and it is separable then X is second countable.
let D be countable dense set, then say D as { d_1,d_2,....,d_n,....}
so we take the set { B(d_n,1/m) | n,m in N }, then it is a countable basis it will work right?

[granite crane]

I proved it some time ago but still now sure that was correct lol

[rancid umbra]

yes, that should work

[heady skiff]

is the order topology a generalization of the standard topology on R

[unreal stratus]

In one way sure

The topology on R can be viewed as coming from an order or from a metric and if you generalise the first definition then yes

[heady skiff]

got acool thx

[fierce lily]

for id circ q, it seems to map from X to X/~ to X/~, why this equality is true?

[fierce lily]

and which kind of construction must satisfy universal property? Like given cont map from X->Y, can I say all maps like X->X/~ or map from X/~->Y all satisfies that universal property?

[limpid fern]

my bad this should be q = id circ q

[limpid fern]

the universal property essentially says that maps from X into Z correspond bijectively with maps from X/~ into Z

[fierce lily]

so you mean since X->Y is construction given by universal property, so unique up to isomorphism implies its bijective and also automatically satisfies its inverse cont right?

and also, I am still confused why the argument can prove that any construction given by a universal property is unique up to isomorphism ?

probably one more question that i need to ask: why universal properties furnish mutually inverse functions which are both continuous?

[languid patrol]

Because of uniqueness

the universal property means that for any Z satisfying P there is a unique f: Z \to X_P satisfying some property

usually the identity from X_P \to X_P is one such map since X_P satisfies P, which means it's the only such map

[limpid fern]

Suppose X is a thing and Y is another thing with a map p: X to Y such that for any other thing Z and map f: X to Z, there is a unique map tilde f: Y to Z such that f = tilde f circ p.

now suppose Y' and p' are a thing and a map that also satisfy the above property

put Z to be Y' and put f to be p'.

do it analogously

[heady skiff]

how does one come up with a homeomorphism from (a, b) to R? I understand intuitively that we have to have some limiting behavior in the denominator, something like (b - x) in order to reach very large values, however the other parts i'm confused about/need a hint on

[languid patrol]

can you get one from (a, b) \to (a, \infty)

because then you can do it step by step.

[heady skiff]

ah yeah that makes more sense

thanks i'll try that

[prime elbow]

let x in cl(A), then for every open set of x A has non-empty intersection with it.

now there is a nested local basis at x, so B_1 \superset B_2 ..... so A has non-empty intersection with B_i, take x_n in A and B_n.

and for any open set U containing x, there is a B_m such that B_m \subset U and because they are nested so pick m, for all n>m x_n in U.

correct?

[tender halo]

yea

[prime elbow]

now i want to show that if X is first countable space then it has local nested basis at x

[fierce lily]

so because we can know that phi is a (unique) continuous map, and it is a construction satisfying universal property, then we can say its inverse is continuous as well?

[languid patrol]

the point is that let's say that X satisfies some universal property P in the category of topological spaces (morphisms are continuous maps)

then for any space Z satisfying P, there exists a unique continuous map f: Z \to X

[cedar totem]

what is a morphism?

a mapping?

[languid patrol]

then suppose X' is also universal for P, then there exists a unique continuous map g: X \to X' because X satisfies P, and a unique continuous f: X' \to X because X' satisfies P

but then f \circ g: X \to X is the unique continuous map X \to X induced by the fact that X satisfies P

and this should be the identity of X

so then f \circ g is the identity

but X, X' are totally symmetrical so by the same argument g \circ f is the identity on X'

so there is a unique isomorphism X \cong X' determined by the universal property

here is an example:

If A, B are two topological spaces, then there exists a unique space A \times B which satisfies the property that there exists continuous maps \pi_X: A \times B \to X for X = A or X = B and for every Z, a map f: Z \to A \times B is determined uniquely by specifying the maps \pi_X \circ f for X = A or B.

To make the universal property more formal: the set of maps {f: Z \to A \times B} is in bijection with the set of maps {f_1: Z \to A, f_2: Z \to B}

for any Z a topological space

the universal property shows that there is a unique map f: A \times B \to A \times B such that \pi_A \circ f = Id_A, \pi_B \circ f = Id_B, thus this map f has to be the identity map on the product.

now the argument above shows that the product of two spaces is unique up to unique isomorphism

[languid patrol]

technically this is a bad way to explain it, because usually P is not a property of the space, but instead an extra structure. But I don't know if belaboring that technical detail will be useful for you.

[quartz horizon]

i do also like the yoneda perspective on this with representability

sometimes it’s a little nicer than the initial/final perspective, imo

[fierce lily]

for this part, because f circ g identity map on X, g circ f is identity map on X' so it is unique isomorphism right?

[languid patrol]

f and g are unique because of the universal property, they are isomorphisms because we verified that they are each others' left and right inverses

[cedar totem]

what is an isomorphism and a diffeomorphism? what is the difference?

[fierce lily]

so it means f and g are isomorphisms or isomorphism between X and X'?

[fierce lily]

they are all bijection, but diffeomorphism will require the map and its inverse are smooth

[languid patrol]

g: X \to X', f: X' \to X so it seems to me that your two statements are the same, unless you wanted to ask something else.

[fierce lily]

I am actually confused about what does product of two space is unique up to unique isomorphism mean?

[languid patrol]

it means that if Z is any space with morphisms f_1: Z \to A, f_2: Z \to B such that for any space S the map of sets {g: S \to Z} \to {g_1: S \to A, g_2: S \to B} is a bijection (natural in S) then there is a unique isomorphism t: Z \to A \times B

such that \pi_A \circ t = f_1, \pi_B \circ t = f_2

[fierce lily]

so in order to unique isomorphism, for any Z satisfying universal property and Y->Z, we can find two maps f1: X->Z, f2:Y->X such that f=f2 circ f1, it will show that map f is unique isomorphism?

[quartz horizon]

It’s difficult to parse what you’re saying here

[fierce lily]

I mean, probably you can explain what is unique isomorphism in category thm( or if universal property can automatically let unique isomorphism hold)

[fierce lily]

because I think I don't really know what is unique up to isomorphism

[quartz horizon]

It’s a little difficult to do this in generality

At least the way I would do it is introducing naturality

Hmmm actually

[quartz horizon]

Ok I can try to do this for products of sets first, if that’s ok?

Given two sets A and B, you can form the Cartesian product A x B. What it “is” is the collection of ordered pairs (a, b) for a in A, b in B.

But you can also ask what it “does”, how it relates to other sets, and what you can use it for. In particular, you can notice the following.

If you have a function f : Z -> A x B, it has to take the form f(z) = (g(z), h(z)) for some functions g : Z -> A and h : Z -> B. In other words, you can “unpackage” a single function f : Z -> A x B into a pair of functions g : Z -> A and h : Z -> B

Conversely, given two functions g : Z -> A and h : Z -> B, you can “package” them together into a single f : Z -> A x B, by defining f(z) = (g(z), h(z)).

This is what the Cartesian product “does”, and is called its universal property, since it tells you how it relates to every other set in the “universe”. What it does is let you package and unpackage pairs of functions!

You can notice that there are other sets which also let you package and unpackage pairs of functions - B x A “does” the same thing as A x B. Category theory then guarantees there’s a unique isomorphism between A x B and B x A which “plays nicely” with packaging/unpackaging in a manner you can make precise. This isomorphism just swaps the elements in the ordered pair

I gotta get to sleep now but I hope that’s somewhat useful

[novel plank]

How would you solve exercise 2.5.2?

I don't see any straightforward way of proving this just using the definitions of induced topology

[heady skiff]

why are the U_y an open cover for A?

do they mean for each y in A?

[heady skiff]

my book states "every normal Hausdorff space is regular" without proof, but I'm struggling to see why - I know the intuition is that if we have a closed subset A of X and a point outside of A, then we can find disjoint open sets containing each, but who's to say that the singleton is closed?

or ig I haven't used the hausdorff condition anywhere

but still having trouble seeing

[unreal stratus]

There seem to be multiple typos

Lol

[heady skiff]

lol yeah i was able to kinda parse it

/the argument

[unreal stratus]

Like they say U_x and U_y

Should be U_y and V_y in some order

Honestly I would just look up another proof to avoid confusion lol

[heady skiff]

okay

[heady skiff]

lso why is this the case

are we assuming that singletons are closed

isn't that a separation axiom or whatever it's called

[unreal stratus]

Which is implied by Hausdorff

[unreal stratus]

So is being Hausdorff yeah

"Singletons are closed" is T1 which is implied by Hausdorff aka T2

[plush folio]

It should be a straightforward application of the induced topology, but a small difficulty is that you need the formula $f(V \cap f^{-1}(Z)) = f(V) \cap Z$. In general images don't preserve intersections, but for this we have equality. See if you can prove it

[gentle osprey]

sheddow

[scenic ridge]

so the solution that I read of SO_3 being homeomorphic to V(3, 2) sort of confuses me

it says that an element of SO_3 can be described as a pair of orthagonal vectors each of length 1 as the third column is the cross of the first 2

V(3, 2) is obviously also a pair of orthagonal vectors of length 1

and thus they are homeomorphic (this is the part i dont get)

[prime elbow]

I want to characterize the continuous function from R with co-countable topology to R with usual topology.

since the closed set maps to the closed set, so for x in R, the preimage set of x must be countable.

Any hint?

[prime elbow]

To prove lower limit topology on R is not metrizable.

We know that if X is metrizable and it is separable then it is second countable.

But the lower limit topology on R is not second countable and it is separable so lower limit topology cannot be metrizable.

[prime elbow]

let A be infinite subset of A such that Z\A is also infinite. Define a Hausdroff topology on Z such that A is open and singleton set are not open.
any hint?

[red yoke]

Do you know of a countably infinite Hausdorff topological space with no isolated points (open singletons)

[raw stag]

Q?

Under the induced topology from R?

[red yoke]

You can take disjoint copies of this to get A and Z\A

[raw stag]

Yeah so cofinite topology or sth?

[red yoke]

Nah that's not hausdorff

[raw stag]

Sorry I meant coproduct topology

I always mess up the names

Apparently Z has a topology that is homeomorphic Q with this topology

It's the Furstenberg topology i think

[prime elbow]

how? disjoint copies?

[red yoke]

Product of Q with some discrete space

Note that for any infinite cardinal κ, we have κ·|Q| = κ

[heady skiff]

does anybody know if this argument is wrong, since Y doesn't necessarily contain 0?

[languid patrol]

what are X and Y

can they be any ordered set?

[alpine nest]

Subtraction also won't necessarily make sense in Y

[heady skiff]

well i guess only Y is ordered

true

[bronze wadi]

I get why f induces the bijection between open sets but can we go the other way?

I wanted to map x to a point contained in the intersection of the image of open sets containing x in X but I have no guarantee it will not be empty

[heady skiff]

here could we maybe fix a in Y and consider the intersection of the preimages of (a, oo) under f and (-oo, a) under g?

well i guess this wouldn't work since if f(x) > g(x) then we don't necessarily have f(x) > a and a > g(x)...

hm

[dire dove]

X is open

Actually yes i see what you mean

Any surjective function from X to Y both with the indiscrete topology induces a bijection of open sets

I guess it should be "equivalently, f is a bijection and ... "

[prime elbow]

Not necessarily, I think

[raw stag]

wdym

[unreal stratus]

As in, from a bijection on opens to a bijection on points?

Unfortunately not in general, but under some conditions, yes

[raw stag]

i was just saying that Z is homeo to Q with the frustenberg topology

[prime elbow]

When Q is subspace topology with respect to usual topology on R?

[raw stag]

obv

[prime elbow]

How?

But Furstenberg topology doesn't satisfy the given topology condition

May be

[raw stag]

i was not talking about your condition

[prime elbow]

Okay

But now I am thinking about how to prove that Furstenberg topology is homeomorphic to Q

[raw stag]

well its metrizable but not completely

so that should ring some bells

[prime elbow]

What?🥲

[raw stag]

you know what metrizable is right?

[prime elbow]

A topology such that there exists a metric function on it such that it metrics topology same as given topology

[raw stag]

yeah

[prime elbow]

So how does it help me to prove that ?

[raw stag]

help you prove what?

the homeo thing?

[prime elbow]

Yes

[raw stag]

claim that any countable metric space without isolated points is homeo to Q

[prime elbow]

Q×N, work?

[prime elbow]

No idea

How bijection and continuous plays here ? Like i know there is a bijection function between that countable set to Q but do I show that the function is continuous or any different function is continuous?

[raw stag]

well the metric space thing is needed is what i would like to say

[raw stag]

heres a reference

http://at.yorku.ca/p/a/c/a/25.pdf

[prime elbow]

Thank you ❤️

[heady skiff]

yo can someone let me know if they've ever heard of this pick up line before

are you the discrete topology? because man you are fine...

if X is a compact topological space, is X/~ also compact since it's the surjective image of X

[sonic crane]

Nice

[rancid mantle]

Hi! I'm having issues understanding pullbacks in topology. The ones that have the fiber product and the commutative square. I don't get why would they be useful nor what do they even mean. Also I'm still getting used to categories, universal properties and that sort of stuff, so please don't have that as a given with me.

The sort of things that I have to prove are like "prove homeomorphisms are stable under base change" or "composition of proper functions is proper".

[gentle crypt]

They help you glue spaces together, yeah?

[rancid mantle]

Isn't that the pushout?

I have a little less of an issue with that because gluing things together does feel intuitive

It would be nice to have something more useful than raw intuition though, but my main question is about pullbacks

[paper wedge]

#diff-geo-diff-top

@rancid mantle

i wish i could help but i only know the regular pullbacks of functions under forms

ig in a way they are the same thing

[rancid mantle]

Are you sure? This deals a lot with category stuff. No differentials, no difeomorphism and no geometry.

[paper wedge]

then yeah ig this is way out of my level, i thought fiber bundles in geometry context.

ig maybe #advanced-algebra or #category-theory

[rancid mantle]

I found some stuff on internet about pullbacks in more general contexts but I really didn't understand a thing.

[paper wedge]

all what i can say is the pullback of a section is like a section on the pull back bundle

such that it is like a natural transformation in a way

[rancid mantle]

What is a bundle

[paper wedge]

ie the square diagram

[paper wedge]

do u know about smooth manifolds

[rancid mantle]

No

[paper wedge]

ig i won't be able to help you much as this is only the context i have ever seen this in

[rancid mantle]

This is my first topology course

[red yoke]

Pullbacks of topological spaces are just "product parametrized over a space"

[quartz horizon]

Did you manage to understand pullbacks

[prime elbow]

Can someone tell me why we are always intersected in separate the points in topology?

[tender halo]

what?

i dont understand the question

[prime elbow]

They define regular topological space as the space, for any x in X and closed C such that x not in C then there exists disjoint open sets such that x in U and C contained in V.

So they say that we separate the points

[tender halo]

sure, ok

whats the question then

[prime elbow]

Why are we interested in separation?

[heady skiff]

isn't this just bc the restriction of F to X x {y_0} is still continuous and similarly for k?

[tiny obsidian]

Essentially

[keen seal]

this seems wrong

damn

guess I'm wrong

[raw stag]

It's weird enough

But it's true

[tender halo]

well thats kinda what we expect good spaces to look like

we want things to be "far away" from each other

[prime elbow]

thank you

[tender halo]

its a measure of how "fine" a space is

[prime elbow]

i am not sure but i want to show that if X is uncountable with co-finite topology then X cannot be second countable.

fix x in X. then intersection of all open sets containing x is {x}.

but how can i say that there exists countable basis B_i such that intersection B_i is {x}

if i assumed X is second countable

and yes {x}\subset \bigcap B_i

B_i contains {x]

[rancid mantle]

Nope

[rancid mantle]

That seems better but I still don't understand why would that be true or where does that idea come from. Where's that parameterization?

[quartz horizon]

Ooh ok I can try to explain if you like?

[rancid mantle]

Yeah sure

[quartz horizon]

So have you met the universal property of product, for starters?

[rancid mantle]

Yes. I can't formulate it right now but I got to use some universal properties successfully

Ah it's the one about the projections right?

[quartz horizon]

The way I like to think of it is in terms of “packaging” and “unpackaging”

For sets, if you have a function $f : Z \to A \times B$, it must take the form $f(z) = (g(z), h(z))$ for some $g : Z \to A$ and $h : Z \to B$

[gentle osprey]

Pseudonium

[quartz horizon]

And conversely, if you have $g : Z \to A, h : Z \to B$, you can “package” them together into a single function $f : Z \to A \times B$ defined by $f(z) = (g(z), h(z))$

[gentle osprey]

Pseudonium

[rancid mantle]

And if you are in some category other than set then you have the type of morphism from the category right?

[quartz horizon]

So this is what the product “does” - it lets you package and unpackage functions (or, in general, morphisms) with a common domain

[quartz horizon]

Yep

So the product of topological spaces lets you package and unpackage continuous functions

[rancid mantle]

And that is a consequence of the definition of product topology, right?

[quartz horizon]

https://media.discordapp.net/attachments/834837272707727372/1263928509978640454/image.png?ex=66ef1327&is=66edc1a7&hm=e12711bbf4056a85acf46021d28f0bce934ea2aae44bc967e34328f554e0479c&

https://media.discordapp.net/attachments/834837272707727372/1263928868088315945/image.png?ex=66ef137c&is=66edc1fc&hm=0cc2e949dfa5d48f0b9982c93d48af9e1d4ba3dc785ee8223719fd8c45d98dea&

[quartz horizon]

Yep yep

So you have two perspectives here

What the product topology “is”, in terms of the specific construction and topology

And what the product topology “does”, in terms of letting you package and unpackage continuous functions

A big part of categorical thinking is being comfortable with interconverting between these two

What’s perhaps surprising is that knowing what something “does” determines what it “is” up to unique isomorphism (assuming that something satisfying the property actually exists)

All fine so far?

[rancid mantle]

Yes

[quartz horizon]

Now, what happens if you “unpackage” the identity A x B -> A x B?

[rancid mantle]

You get the projections

[quartz horizon]

Exactly! That’s essentially (half of) applying the yoneda lemma

[rancid mantle]

Or inclusions, I don't know which direction are unpacking

[quartz horizon]

Following where the identity goes usually leads you to useful places

[rancid mantle]

Oh wait the inclusions don't make sense

[quartz horizon]

Its this diagram for unpackaging

We’ve just set C = A x B, and f to be the identity

[rancid mantle]

Yes

[quartz horizon]

For pullbacks, the other universal property you want to know is what I like to call “the universal property of subset”

Or subspace in this case

[rancid mantle]

I'm not understanding this

What do you mean by the identity going somewhere

[quartz horizon]

We followed where the identity function went under “unpackaging”

And this got us the projections

[rancid mantle]

Right

[quartz horizon]

That’s all I really mean

We have some kind of way to transform morphisms into pairs of morphisms

And it’s useful to see how this transformation acts on the identity morphism

It often gives you a “natural” or “special” result

[rancid mantle]

Just so I can keep this chip in my head until it shows up again (I suspect it will, from the way you phrased this), are there any other processes to apply to identity that I can follow to see what happens?

[quartz horizon]

Oh yeah lots and lots and lots

This is half of the yoneda lemma

Just following where the identity goes is a surprisingly useful trick

[rancid mantle]

I'm not familiar with that but I'll just keep it somewhere 😅

[quartz horizon]

Well you’re about to meet another example of it

[quartz horizon]

With this

So say S is a subset of X

Then functions Z -> S naturally correspond to functions Z -> X whose image is contained in S

You can interconvert between the two

Does that make sense?

[rancid mantle]

Yes

[quartz horizon]

What’s useful is that this also holds for continuous functions

If S is a subspace of X (with the subspace topology)

Then continuous functions Z -> S naturally correspond to continuous functions Z -> X whose image is contained in S

You can check this using the definition of the subspace topology

[rancid mantle]

Yes

[quartz horizon]

This is what the subspace topology “does”, how it relates to other spaces

It lets you shrink the codomain of a continuous function

To a subspace containing the image

Now, what happens when you follow the identity S -> S?

[rancid mantle]

Wait I want to try to work it out

What I want to do is to take the identity on S and expand the codomain to X

So I get the inclusion

[quartz horizon]

Yep!

[quartz horizon]

So here’s your second example, haha

In this case, following where the identity goes gets you the inclusion map

You might also notice that in both cases, following where the identity goes describes one direction of the interconversion

“Unpackaging” for the product is just composing with the projections

And “expanding the codomain” for the subspace is just composing with the inclusion map

[rancid mantle]

Okay

[quartz horizon]

Ok so now for pullbacks

You wanna understand what a pullback “is”

But I think it’s useful to try and see what it “does” too

It’s similar to the product in that it “packages” things

Specifically, if you have A -> C and B -> C, then what the pullback P “does” is the following

Morphisms Z -> P naturally correspond to pairs of morphisms Z -> A, Z -> B which compose to the same morphism Z -> C

So it “packages” together the whole diagram A -> C, B -> C into a single object, P, such that morphisms into P naturally correspond to “morphisms into the diagram”

[quartz horizon]

It might be instructive to follow where the identity P -> P goes here

[rancid mantle]

Oh

[quartz horizon]

You can also construct the pullback as a product followed by a subset

Pairs of morphisms Z -> A, Z -> B correspond to a single morphism Z -> A x B

But you don’t want to consider all morphisms Z -> A x B. Only the subset of those which end up with equal composites to C after unpackaging

So the pullback will be a subset/space of A x B

[quartz horizon]

But this isn’t the only way to do it - often in special cases, there’s a nice description of what “morphisms Z -> A, Z -> B with equal composites to C” mean

[rancid mantle]

This is by definition? I mean, this is what you are defining the pullbacks to do right now, right?

[quartz horizon]

Yes

[quartz horizon]

E.g. preimages can be described by pullbacks

[rancid mantle]

Like you have the "whole" object in P and how it relates to A and B and you can then do whatever you want because you have the whole information encoded there?

And you also ask the information in P to not conflict when you go to C through A or B

[quartz horizon]

Yeah

[rancid mantle]

And why is this useful? This is something my course did not succeed in yet. We just proved properties about these things and used it to get a very strange definition of compact spaces, but it wasn't really useful up to now.

[quartz horizon]

It’s what I said here

[quartz horizon]

In the same way that a product is useful by “packaging” two spaces into one, a pullback is useful by “packaging” the data of 3 spaces and how they relate into a single space

They also come up in preimage and pullback bundles

[rancid mantle]

It will give morphisms P->A and P->B that don't conflict when composed with the other ones, but now I don't know what those morphisms look like. They are not projections or inclusions (I mean they will but I don't want to give answers that are not deduced by what is being said here)

[rancid mantle]

Okay I think something clicked here