@Alessandro Given a section $s : Y \to E$, take $sf : X \to E$, and upgrade that to $s' : X \to f^* E$ by defining $s'(x) = (sf(x), x)$. Does that work?

Hmmm sanity check: since $f^\ast E\subseteq E\times X$ the map $f^\ast E\to X$ is the restriction of the projection $E\times X\to X$?

Yup

@BalarkaSen Alright, looks like this works then. If $\pi:E\to Y$ then $\pi(sf(x))=f(x)$ since $s$ is a section, so $s'(x)$ lands in the right place, and it is also a section of $f^\ast E$, just because it does nothing to the second coordinate

Great

Thanks!

So now you can argue as follows: $E$ is trivial over $Y$ iff there are sections $s_1, \cdots, s_n$ such that for every $y \in Y$, $s_1(y), \cdots, s_n(y)$ spans $E_y$. Pull these sections back to $X$ to obtain a basis of sections for $f^* E$ as well.

That proves pullback of trivial bundle is trivial no problem

Ah, that's a cool argument

Very clean

Given any vector bundle $E$ over $Y$, choose a trivializing atlas $(U_i, \varphi_i)$ for this. Then for every pair $i, j$ of indices, denote $U_{ij} = U_i \cap U_j$. You have the various maps $\varphi_j \varphi_i^{-1} : U_{ij} \times \Bbb R^k \to U_{ij} \times \Bbb R^k$ which is of the form $x \mapsto (x, g_{ij}(x))$ where $g_{ij}(x) \in \text{GL}_k(\Bbb R)$

$g_{ij} : U_{ij} \to \text{GL}_k(\Bbb R)$ are called the transition functions of this atlas

Oh, right, we talked about this in AG

The data of the vector bundle $E$ over $Y$ and the data $(U_i, \varphi_i, g_{ij})$ of the atlas and the transition functions are equivalent in the following sense; $E = \bigsqcup U_i \times \Bbb R^k/\sim$ where $(x, v) \sim (x, g_{ij}(v))$ whenever $x \in U_i$ and $j$ is such that $U_i \cap U_j \neq \emptyset$, and the projection $E \to Y$ is given by forgetting about the $\Bbb R^k$ factor

$g_{ij}$ encodes the twists for $E$, so to speak

Right

We went through this to see the analogy between vector bundles and locally free sheaves

Aha

In both cases locally they look trivial, but there can be twists hidden in how they patch together

Right

So I claim that for any map $f : X \to Y$, a vector bundle $E$ over $Y$ with a trivializing atlas $(U_i, \varphi_i)$ and transition functions $g_{ij}$, $f^* E$ has a natural trivializing atlas and transition functions

Do you want to try to figure it out?

Not right now to be honest, but I want to understand $f^\ast E$ better so I'll try to work it out and come back to you!

For sure!

if $f_n:\mathbb{R}\to\mathbb{R}$ are Lebesgue integrable functions, the integrals are uniformly bounded, i.e. there is such $M$ that $|\int f_n|\leq M$ for all $n$, and the pointwise limit $f$ of the sequence $f_n$ exists and is finite almost everywhere, then is $f$ necessarily integrable?

@happyEddie Yes, because $f = \limsup f_n$

why is $\limsup f_n$ integrable?

Depends on how you construct Lebesgue integrals

for nonnegative functions, integrable functions are ones that can be approximated almost everywhere by nondecreasing sequence of simple functions

well i guess i need to see if it follows from the definition that $\limsup$ of integrable functions is integrable

at least the assumption on uniform bound is needed because the limit of $\chi_{[-n,n]}$ is not integrable

Wait I mistook integrable for measurable

But it's still true I think

yeah the pointwise limit $f$ will be measurable, that i know

fatou's lemma says that lim inf (which is lim in this case) is integrable if $f_n$ are nonnegative, let's see if i can decompose $f_n$ into positive and negative part and use fatou's lemma twice

or maybe look at the sequence $|f_n|$ instead... anyway i'm off for now, thanks

actually it doesn't hold with those assumptions, counterexample: $\chi_{[-n,0)}+\chi_{[0,n)}$, i think it needs $\int |f_n|\leq M$ instead... ok now i'm off for real :)

@happyEddie, $\int f = \lim \sup_{n \in \mathbb{N}} \int f_n$

assuming $f_n \nearrow f$ if i remember correctly

Does this differential equation make sense?

$f(x)=f'(x)+g'(x)$

Depends what you’re asking. As an equation which involves derivatives, it’s perfectly sound. However, the typical context in which you’d see that DE is where g(x) is known and f(x) is unknown

(You can solve that ODE to get g(x) in terms of f(x) easily. It’s harder if you want to get f(x) in terms of g(x), which is the more interesting case)

hey guys

if its okay with you, can someone give me feedback on the following proof

@Semiclassical ah okay good to know