Hey @Mike

@loch my algebra prof actually sorta said that he views it as a kind of generalization of quadratic reciprocity (one proof he gave of it had to do with the fact that $\mathbb{Q}(\sqrt{p^*}) \subset \mathbb{Q}(\zeta_p)$, where $p^* = \pm p$ so as to be 1 mod 4)

@BalarkaSen connectivity of Y is irrelevant here, taking subcomplexes kills that

@Symposium Yeah it's a really good book! (in fact his grad level books are also really good - but not spelled out as much as his undergrad one)

@BalarkaSen what is $n$-connected?

@Daminark hmmm i see - im not sufficiently NT minded to say much about this though unfortunately :p

@MikeMiller Not sure what you mean.

Who knows what the homotopy groups of Y^k are

Compared to Y

I guess you're pointing out it doesn't change homotopy groups in degree up to that

So I'm being bad

Your proof is good

Oh yeah I just care about $\pi_i$ for $i \leq n$. Sorry for not being clear.

@LeakyNun $\pi_i$ trivial for $i\le n$

so just shrink everything lol

@loch The reason I'm actually trying to learn some basic algebraic geometry (from Kirwan's book on algebraic curves) is because I want to read The Arithmetic of Elliptic Curves so bad! (Apparently you don't need much as he introduces majority of the required material, but I still want to know some beforehand if that makes sense).

It does make sense but I think it might not be too reasonable

I guess you should feel comfortable with tangent spaces 1-forms and stuff on varieties

But the generality you're seeing might overcomplicate Silverman's comparatively simple (and beautiful) story

I guess your point was cellular approximation is stronger; it homotopes $f : X \to Y$ to a map such that $f(X)$ is a subcomplex of $Y$. But now use $Y^{n}$ is contractible, so snap.

Yeah what Mike said

Yeah that makes a lot of sense.

sorry stupid Q - why is $Y^n$ contractible?

Consider the constant map $Y^n \to pt$

That's isomorphism on all homotopy groups

So by Whitehead's theorem is a homotopy equivalence.

(That wasn't a stupid question)

aha i see

@BalarkaSen interesting

(Yeah, on retrospect, I don't know why I thought $[X, Y] \cong [X, Y^{n+1}]$ was relevant. I just needed to homotope it inside $Y^n$, regardless of knowing what the homotopy class of that homotopy was, and then compose with the nullhomotopy. I have forgotten a chunk of math, so writing out a note-to-self in this format)

$Y$ being $n-$connected means that $\pi_i(Y) =0$ for $0\le i\le n$. Cellular approximation gives us $\pi_i(Y^n) = \pi_i(Y)$ for $i=0<i<n$ - but i don't think we have $\pi_n(Y^n) = \pi_n(Y)$ in general?

Here's table of contents for Kirwan's book.

@Balarka I agree with loch's most recent comment and I think your original proof didn't work

Yeah, I see now. I was equating $n$-connectedness with $Y^n$ being contractible.

That's not right.

y^{n+1} plus higher cells is a wedge of spheres though

$Y^{n-1}$ is contractible.

Sorry if the images are blurry.

No

Y^1 is not contractiblw

but uh

You are ignoring the higher homotopy groups you don't have insight into. Addint a low d cell changes the high homotopy groups in crazy ways

you dont need to know every detail - i guess that's the main point :p

Add a low-d cell to the point!

@loch Thank you very much! That's neat!

like i think riemann-roch is only used once (but im not 100% sure) - to prove that you have weierstrass equations

and bezout is only used to prove the group law etc.

Riemann-Roch is used, I remember that

they are important for their own sake, but they are also things that you can assume and move on without worrying too much

yeah i think it's used to prove the existence of weierstrass equations by computing $h^i(E,\mathcal{O}(nP))$ for $n=1,2,3$, or something like that.

@MikeMiller OK, let's see. By cellular approximation, $[S^i, Y] \cong [S^i, Y^{i+1}]$. Now, $[S^i, Y] \cong 0$ for $i = 0, \cdots, n$. So $\pi_i Y^{i+1} \cong 0$ for $i = 0, \cdots, n$ should have been my conclusion. I can't tell anything about $\pi_2 Y_2$, say, eg.

I was wrongly applying cellular approximation altogather. Facepalm.

That's interesting. So it's not trivial that any $n$-connected space can be given the CW structure of a $n$-dimensional CW complex with trivial $n$-skeleton after all.

@Symposium now that i think of it again - i think a better advice would be learn ch 2 - and then read silverman and come back if you see something you're not familiar with (e.g. differentials) - you might find that you don't really need all that much lol

Now I want to know a proof of that fact without invoking machinery

@loch Sounds good. According to the appendix Riemann-Roch is used in the section on Effective Methods, and the section on Distance Functions as well.

not sure if typo - but a $n-$dimensional CW complex with trivial $n$-skeleton is trivial?

Yeah, sorry, CW complex with trivial $n$-skeleton.

The $n$-dimensional adjective was a typo.

hm what's the proof with machinery?

I think I can do it using obstruction theory.

Oops, turns out appendix and index sound similar.

Hmm. Let's see if I can stop myself from being confused. Consider the map $Y^{n+1} \to \bigvee S^{n+1}$ given by sending $Y^n$ to the wedge point and the $(n+1)$-cells to the $(n+1)$-spheres in the wedge. $0 = \pi_n Y \cong \pi_n Y^{n+1}$, so this is an isomorphism in $\pi_n$. This is an isomorphism in $\pi_i$ for $i \leq n-1$.

Is this ok, @loch?

yes i think so

Attach $(n+2)$-cells to $Y^{n+1}$ to make $Y^{n+2}$. Ahh, I see.

I was going to use the same attaching maps to build a $(n+2)$-complex from $\bigvee S^{n+1}$.

But the attaching maps might attaching to lower-dimensional spheres.

Because homotopy groups of spheres are fucking weird as fuck

That's why this approach fails. Interesting.

There should be a way to systematically build a bigger complex from $\bigvee S^{n+1}$ and inductively extend the map from $Y^{n+1}$ to a map from $Y$ thereof, though.

i think a proof is given in davis and kirk based on the proof of CW approximation

I'll have a look at it once I come up with a proof. :)

It seems to be an interesting exercise

Is there a Lebesgue measure on Hilbert's Cube i.e. $[0,1]^\omega$?

math.nagoya-u.ac.jp/~richard/teaching/s2017/Nelson_2015.pdf

I think this book is brilliant

from a quick skim it does look like a very friendly book

Title isn't a lie

That's always a good start

oh and the diagram on the cover is writing an open set as a countable union of closed rectangles

This is such a great song

Do some functions not have inverses??

or does every function have an inverse

@geocalc33 define inverse

what do you mean by an inverse of a function $f:X \to Y$?

a reflection across y=x

@geocalc33 but that might not be a function

then sure, every function has an inverse relation

I just found one that doesn't have an inverse relation in terms of standard mathematical functions

i'm going to claim it as my own

I swear this is the third time I'm seeing subdivision

the first time is in excision in algebraic topology

the second time is in Cauchy Integral Formula for Closed Triangles

what's subdivision

now I'm seeing it in additivity of Lebesgue outer measure for disjoint sets

@geocalc33 when you divide a shape into smaller parts

@leakynun

I'm asking stack exchange about the inverse thing

we can't answer it if you don't define it properly and precisely

and formally

My question is this: Is it possible for a function to not have an inverse relation in terms of standard mathematical functions?

In the usual sense of those words, it is certainly possible. $f(x)=x^2$ is an obvious one, since for instance $f(2)=f(-2)=4$.

So $2,-2$ are both mapped to $4$ and there's no unique inverse for $4$.

but the inverse of x^2 is plus or minus sqrt(x)

That's two functions. Not one.

so I guess what im asking is: is it possible for the inverse relation, to not be able to be defined in terms of any mathematical functions

like you can't even graph it using standard mathematical functions

If you can graph a function, then you can reflect that function across $y=x$. If what you get is the graph of another function, then that's the inverse function and you've plotted it

If that reflection isn't the graph of a function, then there's not an inverse function.

but it's an inverse relation

okay i see

morning

Hi thee.

Hi, in this example Weierstrass M-test is used I guess. But since $\sum_k k^{-1}$ diverges, how in this case is $g_k(x)$ said to be uniformly convergent?

@LeylaAlkan It's saying that $g_k \rightarrow 0$ uniformly on $\mathbb{R}$, nothing about $\sum g_k$!

Ooops, yes. I see now, thanks

@loch The solution to the problem was to do it the other way around. $Y$ be $n$-connected as before. Enumerate the generators $g_i$ of $\pi_{n+1}(Y)$ and consider the map $f : \bigvee_i S^{n+1}_i \to Y$ where $f|S_i^{n+1}$ is a representative of $g_i$.

This is an isomorphism on $\pi_k$ for all $0 \leq k \leq n$. Attach $n+2$-cells to $\bigvee_i S^{n+1}$ by attaching maps of the representatives of the generators of the kernel of $\pi_{n+1} f$ to promote it to a $n+2$-complex $K$. $f$ extends to $\tilde{f} : K \to Y$ by extending over the nullhomotopy of the attaching maps of our new cells.

Since $\pi_n K \cong \pi_n K^{n+1} = \pi_n Y^{n+1} \cong \pi_n Y = 0$, this is still an isomorphism on $\pi_k$ for $0 \leq k \leq n$, but now is also an isomorphism on $\pi_{n+1}$ by construction.

I think if you iterate this construction you'll arrive at a CW model of $Y$ which has trivial $n$-skeleton.

Hello, i want to prove that $x\in cl(A)\Longrightleftarrow \exists (x_n)\in A, \lim_{n\to\infty}d(x_n,x)=0$

By which I mean, at the next step, list the generators of the cokernel of $\pi_{n+2}(\tilde{f})$, and take $K \bigvee S^{n+2}$ for each of those generators, and define $g : K \bigvee S^{n+2} \to Y$ by $\tilde{f}$ on $K$, and the representatives of those generators on the $S^{n+2}$'s. Then take the kernel of $g$, and attach $n+3$-cells corresponding to them, and attach to this $K \bigvee S^{n+2}$ fellow to get a newer complex $K_1$ and $\tilde{g} : K_1 \to Y$ the extension of the map.

This is an isomorphism on $\pi_{n+2}$ by construction, and the homotopy groups below are unchanged.

Hmm, maybe I'm a little skeptic about what happens at $\pi_{n+1}$.

Oh, nothing, because $\pi_{n+1} K_1 \cong \pi_{n+1} K_1^{n+2}$ and $K_1^{n+2}$ is just $K \bigvee S^{n+2}$. $\pi_m(X \vee S^{m+1})$ is just $\pi_m(X)$, isn't it?

I can crunch and shrink the bit of the $m$-sphere that lands in the $S^{m+1}$ factor.

i know that $x\in cl(A)\Longrightleftarrow \forall \varepsilone>0, B(x,\varepsilone)\cap A\neq\emptyset \Longrightleftarrow \forall \varepsilone>0,\exists x_{\varepsilone}\in A, d(x,x_{\varepsilone})<\varepsilone$

can i directly change $\varepsilone$ into $\frac1n$ ?

someone here?

@BalarkaSen yeah I think you're right with the construction.

Nice, thanks a lot for cross-checking!

I think it works too

I need to think a little bit about $\pi_m(X\vee S^{m+1}) \isom \pi_m(X)$ - although I think what you said makes sense

Yeah I think it's, take a map $S^m \to X \vee S^{m+1} \to S^{m+1}$ (where the last map is the retract sending $X$ to the wedge point), which is nullhomotopic - call the nullhomotopy $h_t$. Then consider homotoping $S^m \to X \vee S^{m+1}$ by $\text{id} \vee h_t$ until it's constant on the second wedge factor.

hello @LeakyNun

@BalarkaSen Yeah!

I used to know this stuff but after 6 months of inertness I have forgotten pretty much everything. Slowly trying to get back there... the crowd in this chat has always been a great help. Thanks a lot, @loch.

I can feel the TeX in my soul

The red text

It gives me strength

you are with me ?

Yup

@PolineSandra you messed up your TeX which makes things hard to read..

That's better than the $Y^{n+1} \to \bigvee S^{n+1}$ thing because $(n+2)$-cells might be attached to $Y^{n+1}$ in terrifying ways; a bits of the boundary $S^{n+1}$ of such a cell might be mapped to $S^d$ for some $d < \!\! < n+1$, and I have no way to attach stuff accordingly to $\bigvee S^{n+1}$ now in accordance with that.

sorry

@BalarkaSen Yeah

@PolineSandra sure

I was reviewing the proof that a product of spheres is parallelisable if one of them is odd-dimensional. It's interesting how simple the proof is!

and i have directly an equivalence $\forall n\in\mathbb{N}, \exists (x_n)\in A, d(x_n,x)<\frac1n$

@PolineSandra you should probably put the there exists $(x_n) \in A$ part in front of $\forall n\in \mathbb{N}$ part - but otherwise yes!

i don't understand "part in front"

@BalarkaSen So this extends to - say - if $\pi_m(Y) = 0$, then $\pi_m(X\vee Y) = \pi_m(X)$ by the same argument then

I think so, yes.

@PolineSandra what i mean is you should say $\exists (x_n) \in A : \forall n\in \mathbb{N}, d(x_n,x) < \frac{1}{n}$

oh that's neat - I was wondering how to compute $\pi_2(S^2 \vee S^1)$ just now .. but now I know !

and the equivalence stay ?

@PolineSandra actually now that i think of it i don't think it matters

@loch You can actually compute that one by looking at the universal cover of $S^2 \vee S^1$, which is $S^2$ stuck to each integer point in $\Bbb R$. Higher homotopy groups of the base space and the universal covers coincide, so you're done.

And I think that's a counterexample to your claim.

Consider pinching the equator of $S^2$ to get $S^2 \vee S^2$, and stretching the two lobes so it becomes $S^2 \cup [-1, 1] \cup S^2$ (so two spheres with a stick between them like a dumbbell)

Now wrap that dude around $S^2 \vee S^1$ so that those two spheres map to the first copy of $S^2$, and the interval $[-1, 1]$ wraps around $S^1$.

I see the issue now. Call this map $f : S^2 \to S^2 \vee S^1$. Compose this with the retract to $S^1$ so it becomes $r \circ f: S^2 \to S^1$. The point is, even if this guy is nullhomotopic, I can't make it into a nullhomotopy of $f$ because $r \circ f$ hits the wedge point in $S^1$ twice.

If that makes sense

@BalarkaSen hmmm I see intuitively that this makes sense, but I'm missing some details, what does the preimage of the covering map from a point on $S^1$ far from the one joining it to $S^2$ look like?

I think I was expecting the universal cover to be $\Bbb R$ with countably many $S^2$ attached

That's what I said, no? An $S^2$ stuck to each integer in $\Bbb R$.

@AlessandroCodenotti It looks like a copy of $\Bbb N + t$ in $\Bbb R$ where $0 < t < 1$.

Ohhh, I completely missed the "integer"

I can't read

Sorry, I shouldn't do math before my cofffee in the morning :P

I'm basically talking math when I haven't slept all night, so we're close cousins, if not blood brothers at this point

@loch Let's see. So my corrected claim would be that any map $f : S^m \to X \vee_{p} Y$ ($p$ being basepoint of the wedge) such that $r \circ f: S^m \to Y$ admits a nullhomotopy $h$ rel $(r \circ f)^{-1}(p)$ is homotopic to a map entirely lying in $X$.

Does this always work (with some niceness assumptions on $Y$)? I have a space $X$ with its universal cover $\rho:C\to X$, is the universal cover of $X\vee Y$ the fundamental cover of $X$ with a copy of $Y$ attached to every point in the preimage through $\rho$ of the point in $X$ at which I attached $Y$?

Yes.

Well, no.

That works only if $Y$ is simply connected.

Okay what's my job

@TedShifrin, I have been watching your videoes for long time, but can't figure out what does this mean: I ask myself- self ...?

The correct idea is to take the bipartite graph with blue vertices of valence $|\pi_1(X)|$ and red vertices of valence $|\pi_1(Y)|$ and construct the space corresponding to this graph where to each blue vertex you assign a copy of $\widetilde{Y}$, to each red vertex you assign a copy of $\widetilde{X}$ and to each edge you assign a wedge operation.

That's the universal cover of $X \vee Y$.

Relevant shitty answer

Right, thinking about $S^1\vee S^1$ and its fundamental cover is the illuminating example for me here

Ah I considered taking a universal cover at first and for some reason I wasn't sure how to compute its $\pi_2$ - but now that I think of it I think i know it - this is a wedge of countably many $S^2$ and I can apply Hurewicz to compute its $\pi_2$

@MikeMiller Is this statement correct?

@loch Yup.

@Alessandro Yeah. That's literally the graph in that case.

Hmm, "valence" wasn't correct.

I'm too sleepy to write math. You can figure out what I meant :P

.

Nah, it was correct, lol. The vertex spaces for $S^1 \vee S^1$ are $\Bbb R$, was the confusing part for my sleepy brain.

@Balarka I believe that statement well enough. It seems lame though. That inverse image could be nasty.

Random request: I wish there was an encyclopedia of the most commonly used mathematical notation organized by field

@MikeMiller True. But it still proves that $\pi_m(X \vee S^{m+1}) \cong \pi_m(X)$ (What's a non-lame proof?)

Also, if you missed it, I proved that any $n$-connected CW complex admits a CW model with $n$-skeleton = trivial. That proves the statement that any map $K \to X$ from an $n$-dimensional complex $K$ to a $n$-connected complex $X$ is nullhomotopic.

Let $f:\Bbb R\to\Bbb R$ be differentiable function, and consider the following:

i: $|f(x)-f(y)|\le1$ for all $x,y\in \Bbb R$ with $|x-y|\le1$.

ii:$|f'(x)|\le1$ for all $x\in \Bbb R$.

Is it true that i holds iff ii does?

@LeakyNun

i dunno

it's difficult

@Silent can you prove either of the implications?

Hello @Silent my old friend!

Assuming $x<y$, $\displaystyle |f(y) - f(x)| = \left| \int_x^y f'(t) \ \mathrm dt \right| \le \int_x^y |f'(t)| \ \mathrm dt \le \int_x^y 1 \ \mathrm dt = y-x$

so 2 implies 1 as long as $f'$ is integrable

which suggests using MVT

wait, I haven't shown 2 implies 1

ah, I proved something stronger than 1

anyway, if $x<y$ and $y-x \le 1$, then there is $\xi \in (x,y)$ such that $f'(\xi) = \dfrac {f(y) - f(x)} {y-x}$

so $\left| \dfrac {f(y)-f(x)} {y-x} \right| \le 1$

so $|f(y) - f(x)| \le |y-x|$

but $y-x \le 1$

so $|f(y) - f(x)| \le 1$ as required

but 1 does not imply 2

take $f(x) = \sin^2(5x)$

then 1 is satisfied since $|f(x)-f(y)| \le 1$ for all $x,y \in \Bbb R$, even if $|x-y| > 1$

but $f'(\pi/20) = 5$

mornin'

$$ \Biggl (\sin 30 ^o . (\cos x)^\infty + (\frac{1}{\cos x} )^{(\cos x - \frac{1}{ \cos x }}) \Biggl ) = (\frac{3}{2})^{\frac{1}{\cos x }} $$

solve for x

Is the exponent on $\cos{x}$ really $\infty$?

@LeakyNun Thank you very much!

yes it's infinity

@LeakyNun hi! how are you?

JEE Advanced Question

@AbhasKumarSinha So what exactly does it mean? It must be a notation that I haven't met yet because as it's written now it doesn't make any sense to me!

@Symposium me too either

hi

@Symposium Find x where it's an real number

Oh no not the nerd patrol! :'(

Let $\nabla$ be a connection, why does $(\nabla_X Y)(p)$ only depend on the value of $X$ in $p$ ?

ugh

@Symposium You have any idea?

Regarding the infinity business, keep in mind that $|\cos(x)| \le 1$. So if for a given $x$, you have that it's equal to 1, then raising "to the infinity" gives 1, sort of as a limit of raising it to finite powers

@AbhasKumarSinha Nope! It doesn't make sense to write $\cos{x}^{\infty}$ so unless we find the actual question there really is nothing to solve.

If it's less than 1, then you die off

To me that's the only reasonably way a priori to interpret that

@Symposium Using some rough estimation, I got x =1, but I dunno how to prove it

@Daminark Yes that helps to prove x = 1

now, how to prove x = 1?

$\cos x = 1$, thus x = $n \pi$

Proved!

wat

take $\cos = n\pi$

simple

yeeeh!!!

I solved one IIT Advanced Question.

XD

@secret hii

where n is any integer

@Symposium any problem?

@AbhasKumarSinha yeeeeh!

Kidding, I really don't understand what you wrote.

I just wanted to say yeeeeh!

take $ x = n \pi$ where n is any natural number and pi is an angle = 180 degrees

that's it

no big deal

yeeeeeh!

for ex - take $x = | n * 180 |$ where n is an integer or $x = n\pi$ when n is a natural number

@Symposium you understood it?

But $x = \pi$ does not satisfy your equation (granted to we assume the infinity exponent to mean a limit of sorts).

you can prove it by taking x = 180 + y and x = 180 - y and then solving the question

@Symposium woops sorry, x must be lying in 1st or 4th quadrant

that I forgot to mention

XD

At this stage I'm just gonna go yeeeeeh!

@Symposium $x = | n2 \pi |$ where n is an integer

yeeeeeh!

got it?

it looks difficult but it's easy

Noooooope!

just substitute $x$ with $| n * 180| $ solve to get the problem

@Symposium what's the problem?

We just said $x=\pi$ doesn't work.

@Symposium no, not x = pi but $x = n * 2 \pi$

* stands for multiplication

$x=2\pi$ doesn't work either.

how?

x = 0, 360, -360, -720, 720 ...............

$\cos 2 n\pi = 1$

and if you'll substitute all $\cos x$ with $\cos 2 n\pi$ or simply 1, you'll get it

$$ \Biggl (\sin 30 ^o . (\cos x)^\infty + (\frac{1}{\cos x} )^{(\cos x - \frac{1}{ \cos x }}) \Biggl ) = (\frac{3}{2})^{\frac{1}{\cos x }} $$ refral

@Symposium now try

note that $\cos x - \frac {1}{\cos x }$ is the power raised on $\frac{1}{\cos x }$

@Symposium now you got it?

@Symposium Simply, replace all $\cos x $ with 1

@AbhasKumarSinha Oops sorry $\sin 30 ^o$ I took it to mean $\cos 30 ^o$ the whole time.

that happens sometimes

I fully thought you were trolling for a while now, whereas it was me that was reading $\cos $ everywhere.

Apologies.

So yeeeeeeh!

you can prove it by taking x = $| n\pi |$ and cancelling cos both sides and solving the linear equation

yeeeeeeeh!

no problem

Indeed you can!

Yeeeeeeh!

XD

you tooo

yeeeeeeeh

I'm liking this yeeeeeh thing! Feels great!

xD

yeeeeeh! feels great yeeeeeeeh!

XD

big one yeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee‌​eeeeeeeeeeeeeeeeeeeeeeh!

Ive got an issue with probability

@Symposium Are you a phd? or what?

Given that sort of question

Haha! Most decidedly not!

@JakeRose what's the problem?

@Symposium so what?

I'm undergrad.

me too

undergrad

@JakeRose there's nothing wrong in the question

Sorry I got distracted

yeeeeeh!

Why is P(pass 2nd time|fail first time)=4/5?

Why do I have to do 2/5 x 4/5?

what is |? do you mean :?

if your first statement?

given?

Thats the notation used at my uni sorry

@JakeRose because that's what is given

But doesnt it imply that if its the second time

(the phrase "second attempt" automatically implies that the first time is fail, or else you won't need a second attempt)

So doesnt that mean its given?

F

this is what the question says

it means

P(pass first attempt) = 3/5

Oops

P(pass second attempt | fail first attempt) = 4/5

I mean not equal

I agree with you

Sorry

bad question wording

hi @loch

What would P(passed in 3 or fewer)=?

I think it should be 3/5+4/5+7/8 But that is obviously wrong

yeeeeeeeeh!

hello

world

!

The answer is 5050

Is it wrong

Or am I wrong?

Halp

The factorisation looks wrong to me

The answer is 5050.

@Symposium yes

Teach me master

@AvnishKabaj Master?

@Symposium you got the correct answer

Right?

Yes.

One sec... I'm gonna write it up for you.

@LeakyNun, is there any other way for this, other than trial n error?

Cayley-Hamilton

@Symposium thaanks

Any idea where I messed up ?

@AvnishKabaj No worries. I'm not sure how you got the third step.

@LeakyNun thank you very much

@Symposium isn't $$\frac{x^m - a}{x-a}$$ when x tends to a $$m\cdot a^{m-1}$$

@AvnishKabaj I'm not sure where you used that.

@LeakyNun So, i have this follow up question: characteristic polynomial of $A$ will have constant term, iff $A$ invertible?

nonzero constant term

yes

over a field

yes

note that the constant term is just the determinant

oh! thanks for that!

Also note that you can't evaluate the denominator and numerator separately at that point since your denominator goes to 0 then @AvnishKabaj

@Symposium That cleared everything up

Thanks a lot

I'm glad.

Could someone take a look at my question about polyhedra?

I ask how one could count the number of polyhedra with $n$ vertices...

I think it seems quite interesting.

measure theory be like "to prove this result, it suffices to prove it for closed rectangles"

and then Lebesgue integration be like "to prove this result, it suffices to look at indicator functions of closed rectangles"

reductions upon reductions

to prove this, it suffices to assume that it is bounded

Can we create a model that captures all possible "Prove X => it's sufficient to prove Y". What will the morphisium => be?

I am guessing that will be the category of all implications Y => X