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1:41 AM
To prove that $\lim_{x \to 0} \arcsin x =0$ using the definition is it enough to argue like this: let $0<\epsilon \le 1$ be arbitrary. Setting $\delta_\epsilon =\sin \epsilon$, if $|x|<\delta_\epsilon$ then $-\sin \epsilon <x<\sin \epsilon$ and so, being $\text{arcsin}$ increasing and odd, we have $-\epsilon<\arcsin x <\epsilon$. I assumed $0<\epsilon<1$ because $\arcsin \sin t = t$ if and only if $t \in [-1,1]$; I proved previously that assuming "small" $\epsilon$ doesn't affect the validity
of the proof. Is everything correct?
1 hour later…
2:45 AM
@Gwyn It’s good that you thought about domain/range issues, but you have it wrong. You need $|t|\le\pi/2$, no?
@TedShifrin Uuuh sure, I got confused with sin of arcsin instead of arcsin of sin. Thanks for pointing that out.
I am ashamed because I always use the example: "What is $\arcsin (\sin 2)$?" with my physicists friend to shock them because very often they automatically answer $2$. Karma is real :D
3:18 AM
No need to be ashamed :)
When there is sin there is shame.
does arcsin undo the shame?
only for a given range of sin
3:38 AM
Anyone know if there's a nice way to reduce this for an arbitrary $x\in \mathbb{R}$?
$$\frac{\sqrt{-(e^{ix} - 1)^2}}{e^{ix} - 1}$$
if not it's not a big deal
but I feel like there oughta be a way to decompose the square root of a complex number
The answer is $\pm i$.
i don’t know what “decompose” means.
Wait seriously
remember the square root has two values.
well I know that for complex numbers $a$ and $b$, you can't simply write $\sqrt{ab}=\sqrt{a}\sqrt{b}$
you can if you mean the multivalued function.
3:43 AM
or for any nonnegative number i mean
e.g. if you could generally use that, you could write $i^2 = \sqrt{-1}\cdot \sqrt{-1}=\sqrt{-1\cdot -1} = \sqrt{1} = 1$
you haven’t heard a word I typed.
might've been misunderstanding you, sorry
you didn’t even stop to think about it.
3:44 AM
I mean I am, and I'm still at a loss here
hello :D
why? Each square root has a $\pm$.
hi EM
how are you doing Ted?
sigh, yeah, that makes sense
i'm tired and I was reading that as a principal square root
I said multivalued three times.
3:47 AM
@TedShifrin hold on though, wouldn't $x=0$ yield 0?
The expression isn’t defined when $e^{ix}=1$.
yeah i'm a dumbass at this hour
thanks though
Go to sleep and leave us in peace.
What’re you up to, EM?
Just got home and I am happy.
I solved my first Fresnel integral.
Ah, residues?
3:52 AM
no sadly.
didn't use complex analysis method.
Oh, what did you use?
how do I explain this with words. This is hard haha.
I can try to send my work pdf by email.
do you still use your UGA email?
Differentiating with respect to a parameter?
I use u-sub.
then I "replace" function with integral.

I just don't know to write how I did it.
I have my work in pdf.
OK, you can email. I’m not at my computer now, however.
4:05 AM
I send it.
its long pdf of work.
I should learn the residue and Feynman technique on this integral.
Oops I just typed my password
haha my friend you are drunk LOL.
which may seem super strange to others
4:20 AM
Hello everyone
How might I show that $\forall n\in \mathbb{Z}$, $n$ and $n^2+1$ are relatively prime?

My first thought is this means $\gcd{(n,n^2+1)=1}\iff an+b(n^2+1)=1$ for some $a,b\in \mathbb{Z}$

Does this mean all I have to do is find a $a,b$ that satisfy that?
Is that correct?
You can make an easy argument, alternatively, that no prime divides both.
Yes, it’s correct.
4:26 AM

$an+b(n^2+1)=1\implies a=\frac{1-b}{n}-bn$. And if we let $b=nk+1$ for some $k\in \mathbb{Z}$ then $a=\frac{1-(nk+1)}{n}-bn=k-bn$ so $a\in \mathbb{Z}$. Does this look good? I'll also think about no prime dividing both
No, this looks terrible. You need to specific integers that work.
Yes. Tell me actual values of $a$ and $b$.
Think rather than writing formulas.
4:31 AM
Thank you as always, professor. How might I show no prime divides both?
In fact, show that if $d|n$ and $d|(n^2+1)$, then $d=\pm 1$.
I was just writing it down and ended up with $d|n\implies d|n^2$ but $d$ can't divide $n^2+1$
Why not?
I guess unless $d=\pm 1$
4:40 AM
That's pretty snazzy
5:23 AM
Something feels strange about this: If $v$ is a Lipschitz continuous function, then $\max(v,0)$ has gradient equal to $0$ almost everywhere on the level set $\{v = 0 \}$. Which means if $v \geq 0$ everywhere and is Lipschitz, then on $\{v = 0 \}$, $\nabla v = 0$ almost everywhere, which seems .. wrong? Am I misunderstanding something here?
So now you have: Since $\color{#C00}{n^2+1}-n\cdot\color{#C00}{n}=1$, if $d\mid \color{#C00}{n^2+1}$ and $d\mid\color{#C00}{n}$, then $d\mid1$.
wait nvm. That isn't weird at all, the only way $\nabla v = 0$ can fail is on the measure-theoretic boundary of $\{ v = 0\}$, and that always has lebesgue measure zero
If my name was Lipschitz, I don't think I'd name anything after myself
good thing standard practice is other people attributing things to your name rather than you attributing things to your own name
5:39 AM
Apparently he may have named it after himself
oh, interesting - I had no idea
in that case I agree, its a little weird
But yeah, I hadn't considered that things often get attributed to others
5:53 AM
Hi, anybody familiar with the concept of support of a measure?
6:05 AM
Sure but what's the question?
I'm reading a statistics paper that includes a proposition that states in part, "Suppose that the law of $X$ has a dense support on $\mathcal X$." In the paper, $(\mathcal X, d_{\mathcal X})$ is a metric space and $(\mathcal X, \Sigma, \mu_X)$ is a probability space, and I believe $X$ would be a random variable from $\mathcal X$ to $\mathbb R$.
My issue is that the quote discusses the support of the law on $\mathcal X$. But the law of a random variable is the image measure, so it's a function on $\mathcal B(\mathbb R)$. So how can a function whose domain is $\mathcal B(\mathbb R)$ have anything to do with $\mathcal X$? Can this be reconciled?
Sorry, not sure what happened with the edit
Hmm that looks more like an issue with probability terminology (which I know nothing about) than the support itself, I'm afraid I cannot help
Here's a quote from p. 20 of Kallenberg's probability theory textbook: "For any measure $\mu$ on a topological space $S$, we define the support $\text{supp } \mu$ as the set of points $s \in S$ such that $\mu B > 0$ for every neighborhood $B$ of $s$.
okay, fair enough, thanks
6:34 AM
Given an $n$-dimensional smooth manifold $M$, $N(M)$ is defined as the minimal cardinality of an open cover $\{U_i\}$
of $M$ such that each $U_i$ admits a
smooth embedding into $\Bbb{R}^n$. What will be $N(\Bbb{R}P^3)$?
I figured out that $N(\Bbb{R}P^2)$ will be $3$, but I don't have any idea for the above
Any idea how to start this one?
4 hours later…
10:44 AM
@Novice I have a feeling that they're considering $\mathcal{X}$-valued random variables
11:40 AM
@Thorgott Thanks, good to know I can ask. In the meantime I've made some progress, but if I continue to be stuck I'll try here
12:06 PM
This question is too soft to be posted on main, but are discontinuities of the Dirichlet function removable?
12:31 PM
@DannyuNDos To determine this, you just need to find if the limits exist at each discontinuity.
(i.e. at each rational)
But you might have troubles.
Depending on what function is the Dirichlet function to you.
1 on rationals, 0 on irrationals.
12:50 PM
So try taking a limit of this Dirichlet function as x goes to 0.
Does it exist, @DannyuNDos?
I mean... Of course, left limits and right limits exist nowhere, but I had something different in my mind.
What do you mean by "something different"?
Since the Dirichlet function differs from the constant function to zero on a null set, I thought the removal of discontinuities would/should result in the said constant function.
Okay, but in your version of the Dirichlet function, it's discontinuous everywhere, not just as at the rationals. Though there are variants which are continuous at all the irrationals.
Yeah, the Thomae function. I'd argue that discontinuities of Thomae function are also removable tho.
1:00 PM
How are you arguing that the discontinuities are removable in the original case??
I mean... For Dirichlet function, removing discontinuities on the rationals would "cost" less than removing discontinuities on the irrationals. 'Cause the set of rationals is a null set.
Sounds like you aren't using the definition of removable discontinuity.
Yeah, I'm not. Rather, I'm trying to make a more rigorous classification of discontinuities.
The usual removable/jump/infinity doesn't quite cut it.
Are the discontinuities of $$f(x) = \begin{cases}1,&x\in[0,1]\\0&x\notin[0,1]\end{cases}$$ removable? In the same way, it would "cost less" to set $f$ equal to 0 everywhere than to set it equal to 1!!!
I don't think your idea is rigorous.
That function has discontinuities only at 0 and 1.
1:11 PM
@DannyuNDos That isn't the "usual" classification, AFAIK. In an elementary calc class, I typically talk about removable discontinuities (essentially, we can change the value of the function at a removable discontinuity in order to make the function continuous at that point), jump discontinuities (left and right limits exist, but are not equal), and oscillatory discontinuities (the function is bounded, but the limit doesn't exist).
I don't generally consider "blow ups" to be discontinuities, since we typically don't define functions at such points. It doesn't make sense to talk about functions being discontinuous at points which aren't in their domain.
But, I suppose, you could define $f(x) = 1/x$ for $x \ne 0$ and $f(x) = 1$ for $x=0$, and talk about the "infinite" discontinuity at zero. Seems a little artificial, though.
@DannyuNDos No, they are oscillatory. At the points where the Thomae function is discontinuous, the limits don't exist. This is similar to $\sin(1/x)$ at zero (where you define the function however you like at zero).
1:30 PM
S^2 is 1-skeleton of CP^oo, right? @Thorgott
@soupless did you ever get the last two digits? The last two digits of $2^{3^{4^k}}$ are the same for $k\ge1$ since $3^4\equiv1\pmod{20}$.
An intuitive question - say $f$ is a real function which is differentiable n times at a real point $x_0$. Now suppose $Q$ is a polynomial s.t for each $m \leq n$, the $m$'th derivative of the function function $R(x) = Q(x) - f(x)$ at $x_0$ equals zero. Is it the same as saying that $Q$ is $f$'s Taylor polynomial around $x_0$ of at least order $n$?
I believe that follows by the formula for the Taylor series.
you might need $n$ times differentiability in a neighborhood, but otherwise, it should be true.
I see, thanks
2:38 PM
@Koro You mean 2?
yeah, should be 2
the $2n$-skeleton of $\mathbb{CP}^{\infty}$ is $\mathbb{CP}^n$ and $\mathbb{CP}^1=S^2$
the $2n+1$-skeleton is $\mathbb{CP}^n$ as well, because there are no odd-dimensional cells
Thor, I'd figured that out already. I was confused because I related skeleta to the definition $CP^n= S^{2n+1}/S^1$.
CP^1=S^2 so I thought CP^1= 1 skeleton.
but CP^1 is 2 skeleton of CP^oo and is also a 3 skeleton.
yeah, always pay attention to the dimensions when working with complex stuff
3:00 PM
@Thorgott: I want to show that a certain map induces a weak equivalence from S^1 to $\Omega CP^\infty$.
I know that $\pi_k (\Omega CP^\infty)=0$ for k not equal to 2 and is 0 otherwise.
the map that does this is: induced by the inclusion $S^2\to CP^\infty$.
in the following way: the groups $[S^2, CP^\infty]_*\simeq [S^1, \Omega CP^\infty]_*$
Here is what I did (please let me know in case you find my proof incomplete): I denote the inclusion $S^2\to CP^\infty$ by $i$. Since S^2 is 3 skeleton, it is clear that the pair (CP^oo, CP^1) is 3 connected, and therefore i induces isomorphisms between $\pi_k S^2$ and $\pi_k (CP^\infty)$ for k=1,2.
Let's consider only the non trivial case k=2.
i is a generator of the group $\pi_2(S^2)$ so its isomorphic image say $i'$ (under the isomorphism $[S^2, CP^\infty]_*= [S^1, \Omega CP^\infty]_*$) is also a generator of $\pi_1(\Omega CP^\infty)$, which is isomorphic to $\pi_2(CP^\infty)$.
So the induced map gives isomorphism of homotopy groups.
@Thorgott (pinging again to indicate-my message is complete now.)
3:41 PM
@冥王Hades Let me know when you're about to bring a total eclipse. I want to watch too. But for your own enjoyment, wouldn't it be more fun and pleasing to watch God does it than yourself? Sometimes I would rather listen to a world-class performance than hearing myself play a piece that I like.
@冥王Hades The next "God-scheduled" total eclipse will pass your hometown (!) on April 8, 2024. If you plan a sideshow of total eclipse to upstage this one, please let me know when/where since being in other part of Canada it's too far for me to watch. @XanderHenderson
3:55 PM
Timeanddate.com has a nice very graphics and detailed map showing where a Total Solar Eclipse can be seen between 1900 to 2199. Looks like my best time is not until Aug 22, 2044 and I may have to travel to Calgary for that. For you, @冥王Hades, DFW is close enough to smack center next year!
4:35 PM
Another error in my textbook I'm afraid, but I want to check in
Haven't we already determined your textbook has plenty of errors?
Oh, this is a different textbook.
Oh no, same text
This is Amann Escher
I thought it was analysis.
4:36 PM
But I am concerned that the claim (f) must include $\varphi(1_k) \neq 0$, right?
It is, but they have a whole algebraic preliminary
So in my algebra book, I define ring homomorphisms to take $1$ to $1$. But some books do not.
No, I think they're right here.
It seems AE don't, but if they don't then in this theorem they must state that $\varphi(1_k) \neq 0$, right? For otherwise, if it maps to 0 then this claim needn't be true?
They said $\varphi \ne 0$.
If $1$ maps to $0$, then everything maps to $0$.
Ah, of course!
This is a very backwards way of stating it but I guess correct? Thank you for pointing that out Prof. Shifrin
I will include a note that mentions ring homomorphisms often include the extra bit in the definition
which I guess saves one from writing "ring homomorphism which is nontrivial"
No, no. The question is whether you wish to allow $\varphi\colon\Bbb Z\to\Bbb Z$, $\varphi(n)=2n$, to be a ring homomorphism.
4:41 PM
authors define stuff differently.
@Koro that sounds about right
during my discrete class, the author started the natural numbers from zero.
$\mathbb{CP}^{\infty}$ is a $K(\mathbb{Z},2)$, so $\Omega\mathbb{CP}^1$ is a $K(\mathbb{Z},1)$, same as $S^1$
Parallel to what I just said, some authors do not require that a ring have a $1$. Some books call such things a rng. :P
@EM4 That seems extremely popular in France. Not so much in the US.
really, why would they count zero as natural number.
@TedShifrin I seen your email reply :) hahah.
4:43 PM
@TedShifrin Can you elaborate? That would seem to fail to be a ring homomorphism no matter what, right? That is, $\varphi(n^2) = 2n^2 \neq \varphi(n)\varphi(n) = 4n^2$?
My book defines a ring as the barest of definitions (without a multiplicative identity or commutativity)
So their ring homomorphism definition is only that it respect addition and multiplication between the rings
Hmmm ... You're right, EE18. .... What am I meaning to say? Oh, you never want to build in commutativity. Many interesting rings (like matrices) are non-commutative.
A mathematical physics text of mine did build in commutativity (Szekeres!) I guess matrices were "viewed" as an algebra over a field so maybe that was the reason, but it was weird...
For some reason I thought EE18 changed avatar and they're now EM4
maybe EE18 and I are twins.
@Jakobian Hi, I think you might be right about this
It seems unusually abstract (for statistics) to consider random variables taking values in a metric space
4:49 PM
You sometimes want to restrict the image
usually it will be $\mathcal{X}\subseteq\mathbb{R}^n$ either way
Thanks for your help
A question now that I've done the proof...I didn't understand AE's proof a priori. In particular, they suggested tacitly that we could restrict and corestrict $\varphi$ such that it was between $K^\times$ and $K^{'^{\times}}$. It was the corestriction which was not obvious to me. That is, how could we be sure that $\varphi(a) \neq 0$ for all $a \in K^\times$? Prof. Shifrin, I guess your earlier argument also leads us to $\varphi(a) \neq 0$ as required?
Well, the kernel of a ring homomorphism is an ideal and the only ideals in a field are $(0)$ and the whole field. So only $0$ can map to $0$. We need to prove this bare-hands in this case, I imagine.
Suppose $a\ne 0$ and $\varphi(a)=0$. Then for any $x\in F$, $\varphi(x)=\varphi(aa^{-1}x)=\varphi(a)\varphi(a^{-1}x)=0$.
5:09 PM
Is there some bounding relation between the number of prime ideals and the number of maximal ideals of a ring? of course the former is bigger than the later
Local rings have only one nontrivial maximal ideal. But they can have plenty of prime ideals, no?
:64490012 was it not true?
The link to the deleted comment didn't work
@TedShifrin yes
Which deleted comment, @robjohn? I've made several stupid comments lately. :)
So I don't know what you can be looking for with your question, @Soumik.
@TedShifrin "For sure."
in response to my comment about the Taylor series
5:17 PM
I’m not yet familiar with ideals (except the ones in an algebra?) so perhaps I’ll have to come back here after
There do need to be some constraints on the derivatives, too
Oh. I had missed that the original comment asserted repeated differentiability at $x_0$ and just saw your comment about differentiability on an interval. I don't think that's needed.
@EE18 That is the ideal in algebra :)
@TedShifrin Nothing :/
But I gave you an independent proof, EE18.
Can you give some hint about the manifold question I asked earlier?
5:19 PM
I have no idea about that.
That's really a general topology notion, nothing to do with geometry, as far as I know.
My only comment is that $\Bbb RP^3 = SO(3)$. Is there something special you can do with Lie groups?
why is O(n) a manifold?
Why is U(n) a manifold?
I was told that because A-->A^t A has regular value I that's why.
why why why??
What do you mean why why why??
This is an elementary exercise with the regular value theorem.
I don't understand why those spaces are manifolds.
I don't know regular value theorem yet.
5:25 PM
Well, learn it.
any other way to deduce?
will that also tell me dimension of U(n)?
It's just the implicit function theorem.
Yes, of course.
ok, I'll take a look at it.
ok, thanks.
5:31 PM
@TedShifrin If we don't care about the Taylor series converging to the function on a neighborhood, then we don't need differentiability in a neighborhood.
@Thorgott thanks.
@robjohn It's just important not to fall into the trap into which I've fallen recently that if $P$ is a polynomial of degree $\ge k$ that agrees with $f$ to order $k$ at $0$, then $f$ must be $k$ times diff at $0$.
@TedShifrin if you localize $k[x_i:i\in I]$ at the maximal ideal $(x_i:i\in I)$, you should get a local ring with at least $|\mathcal{P}(I)|$ many prime ideals
Awkwardly confusing choice of letter for index set :P
Right. That's the sort of thing I had in my brain when I made my comment.
@TedShifrin the differentiability at $x_0$ was already given, but the conditions in a neighborhood were meant to insure agreement. However, they were just asking about the Taylor series at that point, and for that, I think what they said is sufficient.
5:37 PM
@robjohn Do you have a slick answer to this? I want to just observe that $\partial c/\partial\bar t$ satisfies a linear differential equation along the curve?
I have a wacky conjecture
Recall that you can color the quadrants of the plane, the octants of space, and in general the orthants of n-space with two colors so that adjacent orthants have different colors. For example you can color a point black if an even number of coordinates is positive and white otherwise.
Draw a sphere surrounding, but not centered on, the origin. We are interested in how much of the sphere is in each color.
Conjecture: The amount of the sphere in each color is equal, regardless of the location of the sphere's center, in even dimensions greater than 2. It does not hold in general for odd dimensions or two dimensions.
This is a weird high-dimensional version of the pizza theorem.
Why the parity difference, DogAteMy?
'Cause I need opposite orthants to have the same color.
Also, this is true both for the ball and for the sphere.
Meaning, the coloring equally divides the surface as well as the interior, in these dimensions.
Ah, duh. What about a cube?
Instead of a sphere? Doubtful
You could push nearly the whole cube into one orthant and still have it surround the origin
5:50 PM
Why can't you do that with a ball/sphere?
I mean, curvature stops it to some extent, but ....
I'm worried by sphere thay barely contain the origins
Spheres also have weird concentration of measure business going on in higher dim
6:08 PM
Never mind… it does not seem to work numerically
6:22 PM
@Thorgott how do I show that i is a generator of $[S^2, CP^\infty]_*$?
There does seem to be a higher-dimensional equivalent of the pizza theorem, listed on the Wikipedia page, but this isn't it.
sandwich theorem and now pizza's theorem?
what's next? biryani theorem?
No, samosa.
6:35 PM
6:53 PM
@Thorgot: I have proven that [i] is a generator of $[S^2, CP^\infty]_*$. So by adjunction group isomorphism, [i] goes to a generator [i'] of $[S^1,\Omega CP^\infty]_*$. How do I show that $i': S^1\to \Omega CP^\infty$ induces an isomorphism on $\pi_1$?
7:10 PM
@Koro Do you really want to show that, or are you just trying to use that towards showing $[S^2,\Bbb{CP}^\infty]\cong \Bbb Z$? I have the feeling that your question there is harder than the original question
@F.White I got these as intermediary steps towards showing that an induced map $S^1\to \Omega CP^\infty$ is a weak equivalence.
Should that be $S^2$? I was suggesting that maybe your intermediary step is actually making a harder question, rather than an easier one
where the map is induced as follows: We have inclusion $S^2\to CP^\infty$. So by adjunction, this gives a map $S^1\to \Omega CP^\infty$.
@F.White I know. Do you have any alternative?
Just to be clear, are you saying your end goal is to know $S^1\to \Omega CP^\infty$ is a weak equivalence, or you are using that towards your actual end goal which is to show that $[S^2,\Bbb{CP}^\infty]\cong \Bbb Z$?
the former.
[S^2, CP^oo]= Z has been shown already.
(follows by noting that (CP^oo, S^2) is 3 connected.)
if I take a generator of S^2--> CP^oo and map this via the adjunction isomorphism say $\psi$, then I get a generator $\psi(i)$ of [S^1, \Omega CP^oo]. I'm stuck at showing that $\pi_1(\psi(i))$ is an isomorphism.
7:20 PM
@Koro that group is the same as $\pi_2$ and you know that $i$ induces an isomorphism on $\pi_2$
Ok, so you know $\pi_2(\Bbb CP^\infty)\cong \pi_1(\Omega\Bbb{CP}^\infty)\cong \Bbb Z$. So generator gets sent to generator, and maybe I'm being silly but doesn't the result follow?
$\pi_k$, when k is not equal to 1 is clearly an isomorphism.
@F.White but how do you know that this isomorphism was induced by that particular 'induced map'?
@Thorgott I used the fact that identity of S^2 generates \pi_2 (S^2) and then went from there.
Thor: you may see only this summary of the discussion since our last discussion:
4 mins ago, by Koro
if I take a generator of S^2--> CP^oo and map this via the adjunction isomorphism say $\psi$, then I get a generator $\psi(i)$ of [S^1, \Omega CP^oo]. I'm stuck at showing that $\pi_1(\psi(i))$ is an isomorphism.
hi chat
i have a question that i've google for some time and i didn't find anything
so we know that given any topological space $X$, the category $\mathrm{Sh}(X)$ is an abelian category
thus we have kernels, images, cokernels, etc. and thus we do also have quotient objects
that said, we can even define the (co)homology of chain complexes of objects in that category, and thus we can define (co)homology
so why do we actually care about the global section functor and derived functors, injective resolutions blablabla to define sheaf cohomology when we can simply quotient out chains of sheaves?
@Koro This is a fair question. I'll come back to it in a minute when I finish some computations on my algebraic geometry problem.
7:36 PM
@F.White sure, no problem.
@LucasHenrique Global sections functor is left exact, so it takes short exact sequences $0\to A\to B\to C\to 0$ to $0\to \Gamma A \to \Gamma B\to \Gamma C$ and we can extend this to the right to get a long exact sequence $0\to \Gamma A\to \Gamma B\to \Gamma C \to R^1\Gamma A \to R^1\Gamma B\to \dots$, maybe this helps you understand the map $\Gamma B\to \Gamma C$ better? I guess it matters what you are trying to do
@GratefulDisciple Watch God do it? I am Hades, I am God, I can and have summoned eclipses before.
@LucasHenrique You could ask why people care about derived hom or derived tensor too, and there are things people are trying to compute/understand via those
@Koro it's a morphism between two free cyclic groups, so it's an isomorphism because it sends a generator to a generator
@Thorgott I suppose the point s/he is objecting to, is that the adjunction only gives a bijection of sets, a priori, since this isn't an additive category, so you have to say something about the multiplicative structures
7:44 PM
the adjunction does give an isomorphism of groups
I know that the adjunction gives an isomorphism of groups.
@Thorgott generator goes to generator, fine. But why is the generator in the image a weak equivalence?
@LucasHenrique sheaf cohomology is not the same thing as the cohomology of random complexes of sheaves
@Koro it induces an isomorphism on $\pi_1$ and the other homotopy groups are trivial
@Thorgott if it were, then all the abstract nonsense effort would be pointless :p. what is sheaf cohomology is pretty much what i'm trying to understand
I'm talking about \pi_1. pi_k, k not equal to 1 are trivial.
i think that once i understand what is the i-th derived functor, things are gonna be clearer
7:47 PM
this amount of advanced discussion in chat is a rare sight
$\pi_1(\psi(i))$ is a map between two groups isomorphic to $\mathbb{Z}$ and it maps a generator to a generator
$\psi(i)$ is a generator of $[S^1, \Omega CP^\infty]_* =Z$ so it could happen that $\pi_1(\psi(i)): Z\to Z$ is $2\times $ or $3\times $ or even a zero map.
@LucasHenrique Left exact covariant functors have right derived functors - take an injective resolution (any will do equally well) and then compute cohomology in some degree. Right exact covariant functors have left derived functors - take a projective resolution (any will do equally well) and then compute cohomology in some degree. That's all there is to it. (The resolutions define isomorphic objects in the derived category to the objects you are resolving )
@Koro what does $\pi_1(\psi(i))$ map the generator of $\pi_1(S^1)$ (which as you know is represented by the identity map) to?
it should map it to [$\psi(i)\circ id]$
and that is $[\psi(i)]$
7:55 PM
Which you said is a generator
and you know that to be a generator of $[S^1,\Omega\mathbb{CP}^{\infty}]=\pi_1(\Omega\mathbb{CP}^{\infty})$
ok so $\pi_1(\psi(i))$ is an isomorphism as it sends generator to generator.
(both origin and source are cyclic hence an isomorphism)
Not a very convincing way to phrase that somehow
a homomorphism from cyclic group A to cyclic group B which sends a generator of A to that of B is an isomorphism.
if A and B are isomorphic
I am only saying the phrasing is suspect, it almost sounds like a question
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