Oh, I think I have to prove something else...

suppose $f:X\to Z$ and $g:Y\to Z$ are continuous

suppose that $f=g$ on $X\cap Y$

then is the "combined" map on $X\cup Y$ continuous as well?

yes

gluing lemma

I should prove it first.

then move on to whitehead's paper

although I think it was used already in the last proof...

@0celo7 Oh? Are my ears working right?

You want to prove something when you can google it?

Wow.

I always try to prove things before I google

Sometimes I give up quicker than other times ;)

eww, do you need three cases?

one where $X,Y$ are open, $X$ is open and $Y$ is closed and $X,Y$ are closed?

oh, $X$ and $Y$ has to be either both closed or both open in $X \cup Y$. Otherwise you're going to have trouble.

Say $1$ on $[0, 1]$ and $0$ on $(1, 2]$. Vacuously equal on the intersection, but clearly not continuous.

So this is true?

Just making sure before I try to prove it

Yes, this is right.

k

I'd parse it a bit differently, though. $X \cup Y$ in itself is not determined by $X$ and $Y$. You have an ambient space $W$ which is $X \cup Y$ where $X, Y$ are both closed or both open in $W$.

There's no way to make arbitrary union out of two random topological spaces

But I am sure you're aware of this. Just a heads-up.

Really?

How does the construction on page 3 of Milnor work then

seems like $Y\cup_g e^k$ is ill-defined

if you're telling me $Y\cup e^k$ is ill-defined already

No, no, you have the information about the attaching map.

More information than the two topological spaces.

how is the disjoint union defined though

@0celo7 Balarka meant "union" in the sense that they have non-empty intersection (because you were talking about $X\cap Y$). Of course the disjoint union is defined.

@0celo7 Yes, that you can do.

But then they have trivial intersection, like ACM said.

...why aren't they defined in the other case

I've never heard that

You have to identify $X$ and $Y$ along something. I can make a million different topological spaces out of a cantor set and a 8-dimensional torus. More information is needed

Guys

Let's solve the mass gap problem

And win that million dollars

@Slereah On it.

MY THEORY : obviously there's a mass gap because particles have a mass

Just let me seclude myself for a few years and emerge a crazyman with a brilliant solution in a few years, as is custom.

The precise amount of information needed is exactly the gluing information. Ref: pushouts in category theory.

Build on that

Good old Perelman

Then you get to act all smug

"Oh it was really quite obvious when you think about it"

@ACuriousMind Hello Perelman

$X \cap Y \neq \emptyset$ is unnecessary, but oh well

@BalarkaSen right

For what I need it for I guess the intersection is nonempty...but it seems to work either way

ok, this is harder than I thought it would be

Use le definition of continuity.

Wow, I did not think of that.

Who would have thought I have to use the definition of continuity

I sure didn't

Apparently not

Well, we have $h^{-1}(U)=f^{-1}(U)\cup g^{-1}(U)$, which are both open by definition?

What is there to even prove here

Careful there. $h^{-1}(U) \cap X$ need not be $f^{-1}(U)$.

You're splitting up $h^{-1}(U)$ into $h^{-1}(U) \cap X$ and $h^{-1}(U) \cap Y$, right?

I just wrote $\{p\in X\cup Y\mid h(p)\in U\}=\{p\in X\mid f(p)\in U\}\cup \{p\in Y\mid g(p)\in U\}$

which is the above equation

It is unclear to me how you arrive at that conclusion. You $f$ and $g$'s should really be $h$.

$h$ is $f$ on $X$ and $g$ on $Y$

and it is either on the overlap

That's what's given. How do you prove what you just wrote? I think it's wrong.

Do we agree on $\{p\in X\cup Y\mid h(p)\in U\}=\{p\in X\mid h(p)\in U\}\cup \{p\in Y\mid h(p)\in U\}$, @BalarkaSen ?

Yeah.

I maintain that $\{p\in X\mid h(p)\in U\}= \{p\in X\mid f(p)\in U\}$

I can give a proof, but it seems trivial...

$h=f$ on $X$.

(by definition)

In fact, I don't think I need any restriction on the topology of $X$ and $Y$ if this is correct.

I mean, you may be overcounting.

@0celo7 Yes, which is clearly false.

Why would I be overcounting?

Let the above disputed equality be $A=B$. Suppose $p\in A$. Then $h(p)\in U$. But $f(p)=h(p)\in U$, so $f(p)\in U$, i.e. $p\in B$.

Suppose $p\in B$. Then $f(p)\in U$. But $h(p)=f(p)$, so $h(p)\in U$, i.e. $p\in A$.

We have shown inclusion both ways, so equality holds.

Take the thing I wrote: $[0, 2] = (0, 1) \cup [1, 2) \to \{0, 1\}$ which maps $[0, 1]$ to $0$ and $[1, 2)$ to $1$. Take preimage of $\{1\}$, which is open. You get all of $(1, 2]$, which is not open in $(0, 2)$.

It is upon you to figure out what you messed up. I'm off to sleep.

night

@ACuriousMind can you help?

Sorry, meant $(0, 2)$ there, not $[0, 2]$.

Redo: $(0, 2) = (0, 1) \cup [1, 2) \to \{0, 1\}$ which maps $(0, 1)$ to $0$ and $[1, 2)$ to $1$.

Ah!

That is right

What I did is right

{1} is open, preimage is [1, 2), not open in (0, 2).

The issue is that $f^{-1}(U)$ and the same for $g$ are open in the topology of $X$ and $Y$

not in the topology of the union

$h^{-1}(U)=f^{-1}(U)\cup g^{-1}(U)$ is definitely true

@0celo7 Great, there you go.

but it doesn't have to be open

need to think some more.

If $X$ is open in $X\cup Y$ and $A\subset X$ is open in $X$ it will be open in $X\cup Y$.

Good catch there. I was confused about the counterexample I wrong down, so didn't pay attention to your thing well enough. Your set theory is correct, sorry about that. Just not the topology.

@0celo7 Correct.

Now how to prove that...

So you have done case 1 (?) where X and Y are both open.

@BalarkaSen Yes.

...how the heck do you actually prove that

I hate topology

Preimage of closed sets by continuous maps is also closed, yeah? I think the same proof should push through.

@BalarkaSen Yup.

Hmm, I think you have to use the subspace topology on $X$ inherited from the union

Right.

Ah! $X$ is open in the union. So the open sets of $X$ viewed as a subspace are those which are intersections of subsets of $X$ with $X$ itself?

...no

Ugh, now I've confused myself

Open sets of $X$ are intersection of open subsets of $X \cup Y$ with $X$.

Right

So $A$ is the intersection of an open set of $X\cup Y$ with $X$.

But that open set of $X\cup Y$ has to be $A$ itself since $A\subset X$.

Thus $A$ is open in the union.

Well, that was an entirely unenlightening topology exercise :P

Important fact nonetheless

yup

it's why continuous piecewise functions made sense I guess

Right. The same proof works for gluing smooth functions if moreover you assume at the intersection the derivatives coincide.

yeah

Congrats, you have just proved that the presheaf of continuous functions is a sheaf. -nlab

Anyway, I hope you worked the details out right. I am not to be trusted in the middle of the night, so I should flee.

I think it's good now

I'll write the proof formally now, and that should allow me to catch potential errors

then on to algebraic topology, woot

@BalarkaSen ...what

It's what a sheaf is. Assigning some random object to each open set of your topological space so that they agree on the intersection in some sense.

E.g., the assignment $U \mapsto C^0(U; \Bbb R)$ is a sheaf.

Replace $C^0$ by $C^\infty$ or $C^\omega$ or whatever.

crap, I'm still not using the fact that $X$ is open.

You're looking at $f^{-1}(U)$. That's open in $X$, which is open in $X \cup Y$. So $f^{-1}(U)$ is open, right?

Same for $g^{-1}(U)$.

@BalarkaSen Why?

"open in $X$" means there is some open $A\subset X\cup Y$ s.t. $A\cap X=f^{-1}(U)$, right?

Yes.

Now $A$ is open in $X \cup Y$, $X$ is open in $X \cup Y$. Finite intersection of open sets is open, so $f^{-1}(U)$ is open.

There's going to be a subtle argument involving $X^c$ being closed, I think.

@BalarkaSen Or that.

Yes, perfect.

Alright, really gone now.

cheerio

0celo7 learns topology

i'm done

You're just one and a half year late posting that ;)

In other news, there are 34 hits for "Hitler" in the chat log of this chat

I wonder whether that is above or below the SE standard

Hm, the math chat has double that

lol, the ELU chat has 400 mentions of Hitler.

Seems we're comparatively Hitler-free here

Physics is a Jewish science, after all.

Nice work Sherlock :P

@ACuriousMind Wonder if I've gotten any better since then :P

Probably not

Hello, I have a question about finding a position vs time function of two objects under the influence of gravity. Is there anyone here that can help?

damn you whitehead and your 40s math

hides

I'm a mathematician, don't talk to me

@ACuriousMind This notation is making me cry

@user50213 I'm guessing you know Newton's law of gravitation? What problem do you have solving F=ma?

HOW are you supposed to come up with that???

@ACuriousMind I know the inverse square law and I also have some basic background in calculus. My question is here: physics.stackexchange.com/questions/118705/…

Is there a way to come up with a position vs time function for two objects falling towards each other with no other influence other than gravity?

I tried to solve it and came out with a non linear PDE which I do not know how to solve.

Sorry, a nonlinear DE, not a PDE.

what does it look like, perhaps it might be one of those known DEs?

@user50213 I don't know how you got a difficult PDE out of that. Choose radial coordinates and go to the frame where one of the bodies is at rest. Then you have (ignoring constants) $\frac{1}{r^2} = \ddot{r}$, which is not a hard equation to solve.

Wait

It can be a hard equation to solve, it's called the Kepler problem

Or two-body problem

Right I'm not actually sure how to solve that equation.

But you can find the solution to that in many places

In which cases can you find a solution?

@user50213 Here's the Wikipedia article with the solution.

What I wrote there was the case of two masses that start at rest, i.e. with zero angular momentum

Okay, I don't know what I was thinking

That's a horrible equation to solve

The solution for the Kepler problem isn't in terms of position-time, but just the shape of the orbit.

So basically it can't be solved explicitly?

@user50213 Yeah, you won't get any nice functional form for position as a function of time

But there won't be any orbit if two objects start from rest right?

No, they'll just collide in that case

Yes I saw that, the solution looks scary

Looking at that makes me wonder how wolfram alpha choose its anzat and get that answer

At least one can find the time when they collide from that

How?

@ACuriousMind I really wish I didn't have the need to verify this proof

I'm regretting it heavily

@user50213 set x(t) = 0 (multiply it out first so you don't divide by 0), then solve for t

If I see it correctly the l.h.s. just becomes 0

Not exactly sure how I would get rid of the 1/x(t) inside the log

@user50213 there's an x(t) in front of it, but inside the log

So t=-c_2?

Yes...which of course begs the question how c_2 is related to the initial conditions...

@ACuriousMind There.

@user50213 There's an easier way to get the time to collision, though: By energy conservation, you can express momentum as a function of radius. So you can express velocity as a function of radius, and then you use $T = \int\mathrm{d}t = \int\frac{\mathrm{d}t}{\mathrm{d}r}\mathrm{d}r$ where this integration should be easier than the differential equation (the expression for velocity/its inverse $\frac{\mathrm{d}t}{\mathrm{d}r}$ is not that ugly)

Is there a way for me to view the formatted version of your last message?

@user50213 Look in the upper right corner of the chat room, there is a link "For MathJaX see:"

@ACuriousMind I will try that out, thank you very much for your help.

@ACuriousMind did you read my wonderful proof

@0celo7 No, I was busy reactivating first-year mechanics memories :P

this next one isn't any nicer

holy crap you did you read this book

@0celo7 For your purposes, you can probably assume the answer is "no" :P

that's not a grammatically correct answer

I don't think my prof thinks I read like I do, either

he said it should be trivial to knock back the first 40 pages...

there's so many details though

@0celo7 I have to say that I don't think your thoroughness is a bad thing

@ACuriousMind I used to not read like this and I don't remember a single thing from QM, QFT, ST, etc.

@0celo7 Well, I'm not sure I'd remember much of them if I'd only read about them, either

what do you mean

how much of Milnor do you remember

Not much

I remember things from when I had to use them

I guess. I tend to remember analysis because it shows up everywhere

I also retain much more from lectures than from reading

I'm not sure why that is, exactly

Sadly we don't have a regular differential topology lecture

And Milnor has a lot of prerequisites

You'd be stuck for half a semester just trying to explain all the Riemannian geometry in the middle

@0celo7 There hasn't been one here either lately :(

@ACuriousMind the closest one we have is a Ricci flow lecture that happens whenever enough grad students get to that level

that's differential topology I guess?

The only topologist who usually gives that lecture did the intro analysis, which is three consecutive lecture, so he couldn't do it for one and a half year and now he's starting with algebraic topology again which I already took

@ACuriousMind The actual stuff I want to do, geometric analysis, is too hard for now

I need to take the advanced analysis and PDE courses :(

@ACuriousMind have you heard of the $h$-principle?

My advisor said it was a big deal when he was in school

Heard of it, but I couldn't tell you what it entails

Apparently it's geometric analysis via homotopy theory and symplectic topology

@ACuriousMind Crap! What is a "cluster point" on page 23 and why doesn't that sequence have one :/

All the critical points are isolated...

@0celo7 Pretty sure that's another word for "accumulation point"

@ACuriousMind is that just the same as a limit of the sequence in the standard topology?

@0celo7 No, it's a bit weaker. Every limit point is an accumulation point, but not every accumulation point is a limit point

E.g. the sequence 1,-1,1,-1,1,-1,... has no limit point, but both 1 and -1 are accumulation points

It just means that every neighbourhood of the accumulation point contains infinitely many elements of the sequence

Oh, right

Wait, let me see if I can prove it

Suppose $\{c_i\}$ had the accumulation point $c$. Then $M^c$ is compact by hypothesis. Since all critical points are nondegenerate, we can form an open cover of $M^c$ by taking neighborhoods of the separated critical points $c_i<c$, of which there are infinitely many. This open cover has no finite subcover, which is a contradiction.

I don't think that's right

No, I don't buy that. What if it accumulates from above?

Oh wait, it's an increasing sequence

...no, @ACuriousMind I don't get it

what the heck and why is $M^a$ empty for $a<c_1$

@0celo7 The open cover of $M^c$ you want are the interiors of the $M^{c_i}, c_i<c$, I think.

Interesting...and why does that show there cannot be an accumulation point?

@0celo7 Because that's an infinite cover without a finite subcover if $c$ is an accumulation point.

What if it accumulates "after $c$"

As you rightly noted, it's an increasing sequence

So?

Does that imply no $c_i>c$?

That means that if c is an accumulation point, then the infinitely many c_i must lie below it, because as soon as one $c_n$ lies above it, you have that no other $c_i$ after that can lie in $[c-\epsilon,c+\epsilon]$ for $\epsilon < c_n - c$.

So the only way every neighbourhood of c can contain infinitely many points is by them lying below it

how can you have an infinite number of points below $c$ and then keep going above it

@0celo7 I think you can't. That there is the proof that an accumulation point of an increasing sequence is already a limit point, I think.

If there is a $c_n$ above it, then there are only finitely many points in $[c-\epsilon,c+\epsilon]$, so it's not an accumulation point

Google does not turn up anything

@ACuriousMind Yes, basically

you can only have $n-1$ points below it

that was my point just now

Ok, so why is $M^a=\emptyset$ for $a<c_1$?

I might have to email my prof, this is too hard...

Ok, this proof is definitely out of my reach

Fuck

@0celo7 I'm not sure if this is shown at this point, but $M^a$ would be a compact manifold that has a function without critical points on it, which can't exist.

What the heck is a direct limit topology

Oh, it doesn't need any of the theory, just use that the function must have a minimum or a maximum on a compact set.

@ACuriousMind Oh, right

what is a limit map?

Oh dear

"direct limit" is an algebraic/categorial construction

I think no one understands those notions of limit the first time they read the definition, they take a bit getting used to (at least I took my time getting used to them, and most people I know)

I don't think any of this is explained in Lee's topology book

@ACuriousMind oh my god this is horrible

Well, this was fun while it lasted

@ACuriousMind Seriously, what do I do

You either admit defeat or you fight through it; I can't and won't tell you which is the right way.

How do I fight through it

I don't even know what to do

By...looking up the things you don't understand?

where

I find that Googling helps (not a joke)

And allowing yourself more time to dwell on arguments, possibly

Trying to do this on a schedule might be a bad idea

I don't know what these words mean

What will dwelling do

I'm not talking about the stuff in the first paragraph

I don't know what this direct limit business is about