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1:06 AM
@PaulPlummer: Do you know if there's an algorithm to calculate the outer automorphism group of a group G? Like if I have a finite presentation for G, is there an algorithm to give me a presentation of Out(G)?
 
Well, I'm certainly no PaulPlummer, but that seems unlikely. Given the word problem, and all that.
 
Sure, but (as far as I know!) I'm not asking to be able to recognize the trivial group or anything like this.
(I agree that there are good odds it might be false - but I am not ready to throw out the possibilties that it's either unknown or true, either.)
 
@MikeMiller I am unsure, I highly doubt it, given what @pjs36 what said, and that there is a whole (large) chunk of geometric group theory just studying $Out(F_n)$
 
Ah, that's a good point.
Well, forget Out. What about Aut?
 
And I know you're probably only interested in infinite groups, and doing things theoretically rather than computationally?
 
1:22 AM
They're infinite, but I definitely do care about the computations. I guess I don't really want a presentation. Here's the real story. I have a finitely presented group $\pi$. What I really want to know is what the automorphisms of $\pi$ do on the level of $H = \pi/[\pi,\pi]$. In particular, if I have my favorite ten thousand groups $\pi$, it'd be nice if there was a computationally feasible way of checking whether or not every automorphism of $\pi$ descends to a trivial automorphism of $H$.
In every case I care about $H$ has rank at least 1.
 
Sorry, for some reason my internet on my computer is bugging out
On tablet right now
 
No worries, I know the feeling.
 
@MikeMiller I was just trying to see if Sage happened to have any "automorphism" group stuff... I know they have a class for finitely-presented groups, and they can handle commutators, etc
 
@pjs36 I'm just hopping in, but I noticed you were talking about sage. Do you use sage much?
how do you like it?
 
@pjs36 Ah, that's interesting. I wish I knew more about the algorithmic bits of group theory.
 
1:37 AM
I used it quite a lot around two years ago, @mixedmath working with abstract polytopes, but I haven't done much else. I really do like a lot, but I'm not super proficient with many areas
 
@MikeMiller Still on tablet, but looking around (I don't really know much about automorphisms of groups) but this seems to say no:math.stackexchange.com/questions/825580/…
 
Mm, yes, if I'm not even finitely generated I guess I don't have much hope of an algorithm.
 
@mixedmath Are you wondering about giving it a shot, or have you used it as well?
 
I am sure if you have a restricted class of groups, for example surface groups, there is a lot known about them
 
@pjs36 at one time, I used it quite a bit. I suppose this was about 4 years ago
I contributed for a bit as well, since I generally believe in open-source things, and especially open source math software
but I've also fallen off the bandwagon
 
1:41 AM
@PaulPlummer: In practice I want to test this on the fundamental groups of surface bundles over a circle. It's not too terrible to write down a presentation of these groups.
 
Basically you have to go by the "any statement about all finitely generated groups is either trivial or wrong"
 
Yes, I know what you mean, @mixedmath. I've wanted to contribute, but I never fully made up my mind to ("didn't know how" and all that). What did you work on, when you contributed?
 
Hello friends.
 
1:49 AM
@pjs36 some of the number theory, some of the documentation, and some of the matplotlib graphing interface
(which I still use, and which I still find problems in the documentation all the time about)
 
I believe it! Well, thanks for your work on everything, hopefully it'll motivate me to do the same, some time.
And hello there, @AlexWertheim
 
Hello @pjs36, @MikeM. :)
 
2:07 AM
@PaulPlummer: Google says these are HNN extensions, maybe that helps
How's things @AlexWertheim?
 
Hello all!
 
Not bad, @MikeMiller. I had a very productive day or two this past week. Then I got caught up in some more errands, and squirreled away some time watching the Pre Classic. You?
 
Ah, not much of a golf(?) guy myself. Gave a talk yesterday that has been contributing to much stress lately, so now things are much more peaceful. Today has been relaxing; time to be productive again tomorrow.
 
Ah, you finally gave the talk, nice!
 
If you're thinking of the high-dimensional topology talk, different talks :) That was for the graduate seminar, which is an informal thing where people just talk about cool math; that was pre-blog post. This was a talk for the topology group's seminar, and much more time consuming to prepare.
(But thanks!)
 
2:22 AM
Very nice @MikeMiller. Same here; I hope tomorrow is fruitful for both of us. (Track, by the way... not much of a golf guy myself!)
 
do you know guys where I can read about operators,stein spring dilation,hilbert spaces,W* algebra or C* algebra?
 
Ah, you give so many talks @MikeMiller, hopefully you'll forgive my confusion :P
 
@pjs36 do you know?
I need to understand some paper prof gave to me but I never seen this stuff but I will understand once I can see where I can read the material
 
@KarimMansour: That's a lot of topics. My introduction to C* algebras was through "C* algebras by example"; its a fields institute book. You need to know some of the fundamentals of functional analysis first.
 
I don't even know what half of those things are, @Karim, so unfortunately I can't help.
 
2:25 AM
@pjs36 No, this day shall live on in infamy!
 
stupid prof want me to make like honours project on alot of stuff I don't understand...
The aim of this note is to prove the following version of
Stinespring's theorem for a one parameter semigroup of completely
positive maps.
 
The whole point of an honors project is to learn a lot of stuff you don't understand.
 
I don't even know what completely positive maps are?
yeah
I just want a starting point to start reading
 
Googling positive map should help.
Completely positive, rather.
 
Ugh, I'm out, who will stand guard to prevent this from happening in my absence? If that edit gets approved...
 
2:28 AM
yeah I guess I can do it like this keep going through the paper googling everything I don't understand
would you suggest that @MikeMiller?
because I don't think like reading book on functional analysis or taking a specific topic and reading about it
that would take a long time
and honours project is 6 month from now
 
Talk to your professor. He's supposed to guide you.
6 months is a lot of time
 
2:53 AM
@pjs36, I just rejected it. Why is this guy proposing so many tag edits?
 
3:17 AM
what's up, math people
 
very good document
I am reading on hilbert spaces
Hi @DavidWheeler
 
allo @Karim
 
3:36 AM
hahahaha
 
David
@DavidWheeler I have a question do you know a good book to study category theory?
ok there is something I don't quite understand
if we have a inner product <v,w> complex inner product
we know that $<v,w> = <w,v>^c$
where c denotes the conjugate
 
and...?
 
now let us compute $<v,$\lambda$w> = \lamda^c<v,w>$
shouldn't it be
like do the vectors v,w behave similarly then and the conjugate affect the scalars only?
 
complex inner-products are sesquilinear, not bilinear.
 
I don't understand that terminology
@DavidWheeler
 
3:44 AM
$\langle v, \lambda w\rangle = \overline{\langle \lambda w, v\rangle} = \overline{\lambda}\overline{\langle w,v\rangle} = \overline{\lambda}\langle v,w\rangle$
 
I see
ok
makes sense thank you @DavidWheeler
 
Sesquilinear means "one-and-a-half times" linear (literally), that is, linear in one variable, and conjugate-linear in the other.
 
I see
yeah I am learning these stuff from my honours seminar I am doing project on completely positive maps and some stuff related to it that is EPR paradox etc
 
There is a reason why it's defined this way: we want to be able to use complex inner-products to define a metric, which has to be real-valued. In other words we want $\langle v, v\rangle \in \Bbb R$.
 
but there is many of techanilities I don't understand so just going over them
yeah
 
yeah @DavidWheeler also we know that inner product induces a norm
and we can prove it by taking the norm || . || = sqrt(<v,v>)
and we can prove that this indeed gives us a norm
but not all inner product come from norms though
 
Well, additively, $\Bbb C^n$ isn't really different than $\Bbb R^{2n}$, and so there's some geometric impetus for wanting to preserve a similar notion of "distance".
 
yeah I guess C $\cong R^2$
hm no
isomorphic symbol is not that
 
iso symbol is \cong
and strangely enough, congruence mod symbol is \equiv
 
weird
 
3:55 AM
hey, chat
 
$\Bbb C$ is only iso to $\Bbb R^2$ as a real vector space.
 
i've given up on pens. pencils are love, pencils are life.
 
what do you mean @DavidWheeler?
 
When you take into account complex multiplication, as well, things start to behave differently.
 
oh
I guess yeah geometrically C is just a 4d space so it makes sense for it to be isomorphic to $R^2$
 
3:57 AM
In other words, we can consider $\Bbb C$ as a $\Bbb R$-algebra, which is not really the case with $\Bbb R^2$, per se.
 
oh I see
 
When one deals with complex vector spaces, one "visualizes" the entire field $\Bbb C$ as being "a line", but this is only a crude analogy. The formal definitions pretty much take over, at that point.
 
It turns out that functions, and in particular, power series, are much better behaved with complex variables.
 
yeah for example we will have have solutions
 
4:01 AM
If $\psi_G:G\to A$ and $\psi_H:H\to A$ are homs, how does one denote/define the corresponding hom from $G*H$ to $A$? @Paul, @anon
 
for any polynomials
GxH ?
 
Free product.
 
you can always take the canonical hom
 
I think he may be referring to the free product
 
oh not familiar with free product
 
4:02 AM
which is the co-product in the category of groups
 
What's the notation, @David?
(if there is any)
 
maybe it's $\psi_G \amalg \psi_H$...
 
How could I define it?
 
element-wise?
If we call it $\theta$ for now, I would think $\theta(g_1h_1g_2h_2\cdots g_kh_k) = \psi_G(g_1)\psi_H(h_1)\psi_G(g_2)\psi_H(h_2)\cdots\psi_G(g_k)\psi_H(h_k)$
 
4:26 AM
Hello@DavidWheeler@karim
 
Hi @Rememberme
 
hi Sayan
 
Is anyone here familiar with generalized inverses?
 
Well @DavidWheeler since you last talk about maths with me I have really advanced in mathematics...... Now I am thinking of taking a break any takes on how much long my break should be??
 
No longer than 2 months
 
N S
4:30 AM
Hi
Can some one tell me what this formula is used for?
 
2 months....!! I thought you would say some where around 2 weeks@DavidWheeler
 
N S
Its some kinda divisibilty thing.. I just cant recall where I've seen.. It suddenly came into my mind
like floor[1000/2] +floor[1000/2^2] + floor[1000/2^3]...
 
@Rememberme A day is usually enough for me.
@NS Largest power of 2 dividing $n!$?
 
@Clarinetist - What do you mean by generalized inverses?
 
@PerplexedGuest Let $A$ be a matrix. Then a generalized inverse of $A$ is any matrix $G$ such that $AGA = A$.
 
4:34 AM
@Rememberme lol, for me, 2 weeks is just a long nap.
 
@Clarinetist Can't we right-multiply by $A^{-1}$?
 
@SohamChowdhury That is assuming $A$ is invertible
 
@SohamChowdhury usually one does this in investigating singular $A$.
 
So a generalized inverse works with non-invertible matrices?
 
4:35 AM
Yes.
 
So I think I would go for two weeks from tomorrow
 
Interesting.
 
@Rememberme Where?
To study combi? :P
 
Combi never.... :p I just don't like combo.....@SohamChowdhury I think I would continue my path in topology after two weeks...
 
Haha.
Do you know what generating functions are?
 
4:37 AM
I wonder how many numbers $n$ there are such that $n! + 1$ is prime ...
 
0
Q: How is $X(X^{\prime}X)^{-}X^{\prime}$ symmetric?

ClarinetistFor a matrix $X$, a generalized inverse of $X$ is any matrix $Y$ such that $XYX = X$. We use $X^{-}$ to indicate a generalized inverse of $X$. Suppose $X$ is a matrix. $X^{\prime}$ denotes the transpose of $X$. Then clearly $X^{\prime}X$ is symmetric and has a generalized inverse that is symmetr...

 
@PerplexedGuest 3 or 4
Can't remember, standard olympiad problem, I believe
 
Functions which generate transforms ?
I don't know
 
@Clarinetist
hi
 
Wild guess
 
4:38 AM
Hi @KarimMansour
 
@Rememberme Nope. You must've learned to do polynomial long division in school, right?
 
Yes yes
 
good you been doing some linear algebra lately
are you taking that as your hobby @Clarinetist ?
 
Try to do this one: $$\frac{x}{1-x-x^2}$$
 
Looks like it's 4.
 
4:39 AM
Very boring
 
Just a few terms.
 
@KarimMansour Preparing for a M.S. statistics program
 
You'll get a shock when you realise what's happening, @Rem.
 
user147690
@Rememberme What's boring?
 
4:39 AM
Trust me.
 
which uni @Clarinetist?
 
@AlexClark Poly division.
 
@KarimMansour Iowa State University
 
Long division@AlexClark
 
user147690
4:40 AM
@Rememberme Indeed, quite boring
 
@Rememberme Korchis?
@AlexC try finding the first five terms from that, you'll jump when you realise what's up.
$$\frac{x}{1-x-x^2}$$
It's an infinite series. With very interesting coefficients.
 
Omg. I think my question was dumber than it was supposed to be.
 
@Clarinetist The wind player in you will love this.
 
@Clarinetist, ya just take the transpose.
 
We have $$\left[X(X^{\prime}X)^{-}X^{\prime}\right]^{\prime} = X(X^{\prime}X)^{-}X^{\prime}$$ by symmetry
 
4:42 AM
@AlexC have you tried it?
 
user147690
@SohamChowdhury Not yet, just trying to learn something atm
 
The more I learn about $e$, the more fascinating it becomes.
 
@AlexC, @Rem: Behold.
$$\frac{x}{1-x-x^2} = 1 + x + 2x^2 + 3x^3 + 5x^4 + 8x^5 + 13x^6 + \cdots$$
Fibonacci!
@AlexClark What?
 
user147690
@SohamChowdhury Polynomial lifting
 
is sup sub-additive that is sup(A + B) $\leq$ sup(A) + sup(B) that should be the case intuitively right?
 
user147690
4:44 AM
I'm kind of run down tbh
 
For example: $$\int_{-\infty}^{x} e^{t} dt = e^{x}$$
 
Woooo. Cool stuff, I guess?
 
Ugh, my LaTeX skill is weak.
 
@PerplexedGuest \infty
@AlexClark Why?
 
user147690
@SohamChowdhury Haven't taken a break in days really
 
4:45 AM
Me neither, apart from the quiz that day.
Please grab one if you can.
 
user147690
@SohamChowdhury I feel like sleeping for a full day
 
@KarimMansour Have you ever seen the Dedekind-cut construction of the real numbers?
 
why @AlexClark
yeah @DavidWheeler
 
Man, fix your sleep schedule. When do you go to sleep, @AlexC?
 
user147690
I will get to fix all in a few days, I have two assignments due tomorrow, one on tues and one on wed
 
4:47 AM
@KarimMansour and do you recall how addition of real numbers is defined in that construction?
 
@AlexClark Get a good 7 hours of sleep, man. Less sleep $\neq$ more work done always.
$\hskip -1.4in \text{wise old man}\uparrow $:)
 
user147690
@SohamChowdhury I have been getting 8 a day since math requires sleep to even work at all, but my motivation is dead
 
yeah addition is that we for example $\alpha + \beta$ we add the components
for example
$\alpha$ = { c $\leq q_1$}
 
Hey, @Paul.
 
user147690
Pretty much this isn't how I like to learn. I prefer to have enough time to sit down and go through the text instead of learning to solve specifically these problems
 
4:49 AM
and $\beta$ = {d $\leq q_2$}
so now
 
Hello @SohamChowdhury
 
user147690
Hey @Paul
 
Now here's another question.

A lemma:

Suppose $G$, $H$ are generalized inverses of $X^{\prime}X$. Then (i) $XGX^{\prime}X = XHX^{\prime}X = X$ and (ii) $XGX^{\prime} = XHX^{\prime}$.

I have shown that for a particular generalized inverse of $X^{\prime}X$ (say $G$) that $XGX^{\prime}$ is symmetric, but how does this lemma demonstrate that for **any** $G$, $XGX^{\prime}$ is symmetric?
 
@AlexClark I had to literally learn a certain page of physics for 10th grade finals. I will never do that again.
 
$\alpha + \beta$ = {a + b | a $\in \alpha$,b $\in \beta$}
 
4:50 AM
@AlexClark Hello
 
@DavidWheeler
 
@Clarinetist how "generic" was $G$?
 
Yes thats the result you will get@SohamChowdhury
 
@MikeMiller Them being HNN extentions should make the problem more manageable, although the Baumslag-Solitar group is an HNN extention (which had the infinitely generated automorphism group)
 
@DavidWheeler From the book: since $X^{\prime}X$ is symmetric, there exists a generalized inverse $G$ that is symmetric (this was from a theorem). For this generalized inverse, $XGX^{\prime}$ is symmetric; so, by the lemma, $XGX^{\prime}$ must be symmetric for any choice of $G$
 
4:52 AM
@Rememberme Did you not find the Fibonacci generating function interesting?
 
that is why when you have holiday you study like me @AlexClark
 
user147690
@KarimMansour I love studying in the holidays :)
 
its good to do it that way I learn material not actually solving problems
yeah
 
@Clarinetist Is there a question in there?
Look at part (ii) of your lemma
 
so @DavidWheeler why was there some connection to what I said about sup ?
 
4:55 AM
I dont like Fibonacci @Soham
 
I don't like you.
 
As balarka quotes "Big names"
 
the real number $\alpha$ IS the sup of the set $\{x \in \Bbb Q: x < \alpha\}$
 
@DavidWheeler So basically, since $XGX^{\prime}$ is symmetric for a symmetric $G$, since $XGX^{\prime} = XHX^{\prime}$ for any generalized inverse $H$ of $X$, $H$ must be symmetric?
 
4:55 AM
oh ok ok I see now
 
Oh boy, that looks like that definition from that one section of Spivak I haven't read yet...
 
@SohamChowdhury Why on earth ? :p
 
If sup wasn't sub-additive, real number addition would not work.
 
how come
 
40
Q: Eyebrow-raising pattern of number of primes between terms of the Fibonacci number sequence?

HyperLuminalSo, $$1,1,2,3,5,8,13,21...$$ Any connection to primes?...it appears not. However, in between the Fibonacci numbers are how much primes? Let's see: 1 and 1 ZERO 1 and 2 NADA 2 and 3 ZILCH 3 and 5 ZIP 5 and 8 1 8 and 13 1 13 and 21 2 21 and 34 3 34 and 55 5 55 and 89 8 89 and 144 13 Hu...

Look at this @Soham
 
5:01 AM
so lets see suppose that sup isn't subadditive
so let us see where does this lead us
 
$\sup(x+y) > \sup(x) + \sup(y)$, for starters
 
hm?
ok
 
2
Q: Prove that $\sup(A+B) = \sup(A) + \sup(B)$ and why does $\sup(A+B)$ exist?

user87274We want to show that $\sup(A)+\sup(B)$ is the least upper bound of the set $A + B$. First, we need to show that $\sup(A) + \sup(B)$ is an upper bound for the set $A + B$. Indeed, if $z\in A + B$, then there exists $a\in A$ and $b\in B$ such that $z = a + b$. But by definition of $\sup(A)$ and $\s...

 
@Rememberme Already had.
 
Because of that reason I dislike sometimes number theory You can pose infinitely many questions in it you don't require any effort to make questions Just find some combinations and Voila!! You get a conjecture @Soham
 
5:09 AM
anyways I am gonna go to sleep will continue reading that paper tomorrow
night guys
 
@Clarinetist was reading the comments on your post, shouldn't $A$ be symmetric? I mean in general, not every square matrix is diagonalizable.
 
@DavidWheeler Yeah, I changed that a few seconds after submitting the comment
 
Even so, that is a fascinating theorem (the diagonalization theorem). And, yes, it doesn't matter for your lemma, because your considering $(X'X)^{-}$ and $X'X$ is certainly symmetric.
 
@DavidWheeler I look forward to relearning linear algebra for my own sake sometime, rather than just for statistics
 
Anyone here reads novels??
 
5:15 AM
The point of a lot of the "basis-less" versions is to produce theorems which reduce actual computations when we get to matrices.
 
Ahh.... when I was doing linear algebra I remember the computations very less proofs
 
Computation is tedious, and rarely very enlightening.
 
Very true
Thats why i like topology No computations at all!!@DavidW
 
Metric spaces have computations, especially if you want to find the distance between two regions.
 
I have never encountered any computations in metric spaces@DavidW
 
5:19 AM
A large part of altop is computing homotopy groups and whatnot.
 
Technically, you probably never have to have computations in anything....but....professors and authors like to give examples, which often necessitate them.
 
(I think)
 
@DavidWheeler Let me see if I'm following this logic correctly.

Since $X^{\prime}X$ is symmetric, then it has a generalized inverse that is symmetric. For this particular generalized inverse, say $\left(X^{\prime}X\right)^{-}$, $X\left(X^{\prime}X\right)^{-}X^{\prime}$ is symmetric [verified by simply taking the transpose], so that $X\left(X^{\prime}X\right)^{-}X^{\prime} = XGX^{\prime}$ is symmetric for any generalized inverse $G$ of $X^{\prime}X$. We can say even more here: if $XGX^{\prime} = XG^{\prime}X^{\prime}$ (because we have shown $XGX^{\prime}$ is symmetric)... does it follow tha
 
I don't know if we can go that far.
 
@DavidWheeler Could you explain to me why the book I have says that $G$ is symmetric for any choice of $G$? Maybe I'm missing something else from that lemma
 
5:25 AM
I don't know...you've skipped something, somewhere
 
Hmm, here's the lemma again:

Suppose $G$, $H$ are generalized inverses of $X^{\prime}X$. Then (i) $XGX^{\prime}X = XHX^{\prime}X = X$ and (ii) $XGX^{\prime} = XHX^{\prime}$.
 
@Paul, are you online? I'm confused about the answer to [this](math.stackexchange.com/questions/1039988/showing-that-the-product-group-of-g-and-h-satisfies-the-universal-property-f). The accepted answer says:
$$\sigma\left(g,h\right)=\sigma\left(\left(g,e_{H}\right)\left(e_{G},h\right)\right)\overset{\text{how?}}{=}\sigma\left(g,e_{H}\right)\sigma\left(e_{G},h\right)=\sigma\circ\psi_{G}\left(g\right)\sigma\circ\psi_{H}\left(h\right)=\varphi_{G}\left(g\right)\varphi_{H}\left(h\right)$$.
Isn't the marked step essentially assuming that $\sigma$ is a homomorphism?
 
The claim in the text is that $X(X^{\prime}X)^{-}X^{\prime}$ is symmetric for any choice of $(X^{\prime}X)^{-}$.
 
yes, but the claim is not that any two such gen. inverses are equal.
 
Okay, let me try this again.

Since $X^{\prime}X$ is symmetric, then it has a generalized inverse that is symmetric. For this particular generalized inverse, say $\left(X^{\prime}X\right)^{-}$, $X\left(X^{\prime}X\right)^{-}X^{\prime}$ is symmetric [verified by simply taking the transpose], so that $X\left(X^{\prime}X\right)^{-}X^{\prime} = XGX^{\prime}$ is symmetric for any generalized inverse $G$ of $X^{\prime}X$.
Wait
That's it
Isn't it?
-_-
 
5:29 AM
Yes
 
facepalm
 
Now, I haven't seen the proof of part (ii) of the lemma, but that is where the "meat" is buried.
Everything else is straight-forward.
 
How many users on MO have won a fields medal ??
 
@SohamChowdhury In category theorem, arrows are assumed to be homs for the category you're working in.
 
terrance tao is one of them
 
5:32 AM
Yes any more
 
Go ask.
 
where did you learn category @DavidWheeler?
 
@KarimMansour Nowhere. Everything I know about it I picked up by osmosis.
 
I see
 
@DavidWheeler never mind, I figured it out. It's part of the definition.
 
5:33 AM
I told ya @DavidWheeler is the unsung Balarka of the mathematics chat
 
but yeah arrows are the respective maps of your category
for example groups we take homomorphism
 
@SohamChowdhury Universal mapping properties assert the existence of such a homomorphism, see?
 
and for topological spaces we take homeomorphism
 
so (trick question), what do we take for a poset, considered as a category?
 
@DavidWheeler Do they? I believe they only assert that any such hom, if it exists, is unique up to a unique iso.
Existence has to be proven separately.
(AFAIK)
 
5:36 AM
@SohamChowdhury What has to be proven, is that product/co-products exist.
 
In some categories, they may not.
 
Once that's done, the hom exists.
Right.
The product is unique up to a unique iso, right.
 
However, if one has an explicit construction that satisfies the UMP, then the question of existence is settled a priori.
 
@PaulPlummer: Thanks. Do you know where I should be looking for references to this?
 
5:39 AM
It's instructive to look at the case $\bf{Set}$, first.
 
I have, already, @David.
It was in the first chapter.
 
Well, we are used to thinking of $A \times B$ as a "pairing" of $A$ and $B$
In the UMP view, we see it as something that acts as a "go-between" for pairs of maps from a common source, that allows us to consider the pair of maps as a single map.
 
I thought of a very interesting question.
At least to me.
 
What is it@PerplexedGuest
 
Does there exist a function $f(x)$ such that for any $a,b \in \mathbb{R}$ the following holds true? $$\int_{a}^{b} f(x)dx = f(b-a)$$
 
5:46 AM
@MikeMiller I have no idea what is going on with my computer... well I have been flipping through Trees by Serre and Combinatorial Group Theory by Lyndon and Schupp to see if there is anything that pops out
 
There is the trivial solution $f(x) = 0$, but I am curious as to whether or not there are more.
 
Thanks
 
The difference being: in $A\times B$ in the usual view, we have things like $(a,b)$...in the UMP view, we have $f: C \to A$ and $g: C \to B$, and we produce (note the etymological similarity with product) the map $f \times g$.
 
Actually, it can be shown that no such function exists for $a=0$ besides $f(x) = 0$ ...
 
@PerplexedGuest If any such functions exist, they must be non-linear.
 
5:50 AM
They must also look quite strange.
 
@PerplexedGuest $f(x) = x$?
Oh, no.
 
Yeah, it's tricky.
 
Nice question@PerplexedGuest
 
Most likely there is no such function, but how would one go about proving that ... ?
Thank you!
 
As @DavidW said It will be non linear
 
5:56 AM
@PerplexedGuest Things like this teach me to be cautious.
 
Oh wait - suppose I could prove that there exists no such function that satisfies these conditions when $a = 0$. Would that be enough to ascertain that no such functions exist in the general case?
 
Well certainly not for ALL $a,b$ since you have a counter-example
 
True.
Although I suppose my "proof" relies upon several assumptions.
 

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