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12:28 AM
off topic - complex analysis. What is $2^i$ where $i$ is the square root of $-1$? In particular, how can one show that it has absolute value $2$?
 
Can you show that? I know that $i^i$ has infinitely many values, all real, and thus of varying sizes
 
Wolfram claims $2^i$ has absolute value 2, and a step in my text claimed $2^{x+iy}$ has absolute value $2^x$.
 
@TedShifrin I have an unburstable bubble :P This is lots of fun.
Largely cuz I'm learning a lot along the way
 
@TheSubstitute $2^i = e^{\ln(2^i)} = e^{i\ln 2}$
 
Well, @Stan, I'm glad to hear it. I don't want to say it's good when it's not. :)
 
12:42 AM
@JMoravitz I don't see why the rightmost expression has magnitude 2
 
@TedShifrin I wouldn't want you to. The point is to learn, make mistakes, and get better.
 
@TheSubstitute continuing (since I haven't seen the lightbulb of realization go off yet), $e^{i\ln 2} = 1\cdot e^{i\ln 2}$. This is then the complex number with $r=1$ and $\ln 2 = \theta$, or equivalently we can see this as $(\cos(\ln 2) + i\sin(\ln 2))$
@TheSubstitute thats because it doesn't have magnitude two., it has magnitude one
 
It doesn't, @TheSubstitute. But, @JMoravitz, remember that you need all the possible values of $\ln 2$. So you get $e^{i(\ln 2 + 2\pi i k)}$.
 
Fair enough, I get lazy and use Log instead of log too often i suppose.
 
user147690
@Balarka "Any quotient of a ring by a maximal ideal is a simple ring. In particular, a field is a simple ring" - Wiki. But I need to understand the quotient of a ring first
 
12:45 AM
@jm324354, @JMoravitz thanks for clarification!
 
Hello Math SErs
Starting tomorrow afternoon I will no longer be an actuary!!!!!!!!!!!!!!!!!
 
We'll still call you Sir Actuary Clarinetist just to bug you, @Clarinet.
 
LOL @TedShifrin
 
I'm not familiar with all of the best runtime algorithms for various situations. Can anyone tell me if a more efficient algorithm exists for this?
0
A: What's the least number of combinations you need to determine who the most efficient members are?

JMoravitzI will assume the following: You are able to keep a log of productivity and don't forget information once it is learned You are only able to test groups of three for productivity, not individuals Individual productivity of each applepicker remains constant each day regardless of who they are p...

 
user147690
@Clarinetist Do actuaries make a ridiculous amount of money where you live?
 
12:56 AM
hi @AlexC
 
Sir Actuary @Clarinetist, what will you be doing next instead?
 
user147690
Hey @TedS, how is life?
 
:) @JMoravitz
 
morning
 
@AlexClark I make less than the median income in this suburb of Des Moines, Iowa (I make $67k/year). My impression is that it depends on the company
 
12:57 AM
Getting a bit frazzling, @AlexC, but I'll endure :P
I'm not saying goodnight again, @MikeM.
 
@JMoravitz Lol, I will be working for Epic Systems. Chances are at least some of you have heard of that place
 
user147690
@TedShifrin Oh in general? Or because of us haha?
 
@AlexClark Considering that your pay is mainly based on your exams, it is a ridiculous amount of money
 
No, not you personally, @AlexC, although I'd love to blame you :D
 
12:58 AM
It has a magnitude of 1
 
user147690
@TedShifrin Hahahaha
 
Only the principal value, @columbus8myhw
 
user147690
@Clarinetist Something like this? payscale.com/research/AU/Job=Actuary/Salary
 
Well, alright @Ted
 
@AlexClark That's assuming you pass all of your exams (all 10ish of them)
@AlexClark I am stopping at 5.
Not interested in studying life insurance regulations.
 
user147690
1:00 AM
@Clarinetist Noone is I am sure, they are just interesting in earning more money ;)
 
I think some really are interested, @AlexC.
 
Different people have different tastes.
I have friends who have a hell of a time with excel.
 
user147690
Oh really? Wow that is a surprise
 
Well, I've enjoyed mastering LaTeX for writing books ... :P
 
user147690
(not the excel part: I enjoy excel myself)
 
1:01 AM
@AlexClark True, but I think this whole money rat race has driven me nuts. Hence another reason for me to get out
 
I'm looking to get away from Excel since I've used it daily for the last 9 months...
 
Absolutely, @Clarinetist. You should enjoy what you do for work.
Well, you will be doing all sorts of SAS and R programming, probably.
 
@TedShifrin Yep, of course :) Those will be in my first two M.S. classes
 
user147690
Does Artin cover rings? I can't recall(and I suspect not)
 
1:02 AM
Of course, @AlexC.
 
@AlexClark Surely there's some subject you've studied and not enjoyed in university. Numerous majors in that subject will be able to say they love it. This is analagous.
 
Rings, modules, and field theory are the latter half of the book.
 
@MikeMiller You don't really learn what an actuary is until you've worked as one. The major has nothing to do with the actual job.
 
user147690
@MikeMiller True, but the high salary jobs that require you to learn about regulations are a little different in my expectations
 
user147690
@MikeMiller But yes, I found operations research immensely boring unfortunately
 
user147690
1:04 AM
@TedShifrin Okay thanks
 
user147690
@TedS Balarka said Artin's algebra is the best algebra text he has found. What do you think, or no opinion?
 
@AlexClark They are quite different and no act sci major learns that stuff at a university.
 
user147690
@Clarinetist Where do you live again?
 
@AlexClark Des Moines, Iowa
 
user147690
@Clarinetist I thought you were in Australia for some reason
 
1:06 AM
Lol nope
 
user147690
@Clarinetist Sydney, I wonder who lives there, I know Benlim is Australian
 
user147690
Ahhhh
 
user147690
The Artist
 
@AlexClark I know I've met an Aussie (is that what you call them?) on this site
@AlexClark But yes, 6 figures is quite common when you hit management
 
user147690
@Clarinetist In Australia 6 figures is quite common for lecturers in math
 
1:08 AM
@AlexClark Dang, definitely not the case in the U.S.
 
user147690
@Clarinetist Yeah I was talking to Ted about that
 
@AlexClark I make more than a lot of my professors.
 
@Clarinetist: I'm claiming nothing about the job or the major or whatever. Just that there are undoubtedly people who enjoy it, and that this should not be surprising.
 
Point taken
 
user147690
@MikeMiller Are there any jobs you think noone enjoys?
 
1:09 AM
Retail.
 
Janitorial
 
user147690
@MikeMiller Retail? I think there are people who like retail, in fact I know someone who does
 
user147690
lmao
 
user147690
@Clarinetist I have a friend who does janitorial and he enjoys talking to the client(assuming house cleaning counts)
 
Interesting
 
user147690
1:10 AM
But I suppose it isn't the job he enjoys, just something he can do on the job(that isn't work)
 
I've wanted to get a job being a janitor at a university - the plan being to "Good Will Hunting" those unsuspecting undergrads, and butt in when they're talking math
2
 
I'm hoping to have normal human interaction where I will be relocating...
 
user147690
@pjs36 Hahaha, that would be amazing
 
If I weren't living with my gf, I would be insane by now
 
Morning @AlexClark.
 
1:12 AM
I know, it's a great plan! But with my luck, it wouldn't work :(
 
@pjs36: Odds are there won't be enough interesting conversations to butt in on.
 
user147690
@Clarinetist Yeah I have been there haha. Atleast you don't worry about isolation with that
 
user147690
Hey @Soham
 
@pjs36 I did most of the first chapter of Aluffi yesterday. It was fun.
Hey @Mike
 
user147690
1:13 AM
@SohamChowdhury You mean almost a 9th of the 0th chapter :P
 
@MikeMiller I know, but it would be worth it if I could just catch one unsuspecting person, telling them to integrate by parts, or remind them of the first isomorphism theorem...
 
@AlexClark Haha
@pjs36 Start a film franchise called Janitor From Hell perhaps?
 
@Soham Something like that, more or less. I'm glad Aluffi is working out, I never end up making it too terribly far
 
@pjs36 Universal properties today. Where did you get to?
 
Just bumming around usually, somewhere in the ring section
 
1:44 AM
@MikeMiller Thanks.
Does anyone know how to get a sans-serif font for Set and the like?
 
1:55 AM
what's to thank
 
The inverses thing. I was drawing a blank (sleep dep etc)
 
Let me test if {\sf Set} works: $\sf{Set}$
Whoops, I think I actually used \sf{Set} there
Actually either works, but one probably leads to unexpected consequences
 
2:13 AM
$\sf{Set}$, ${\sf Set}$
Thanks!
 
No problem, I actually had to look that one up myself, not too long ago
 
sure
 
2:39 AM
So I was looking back at my proof that epi $\equiv$ surjective function in $\sf{Set}$.
For one "direction" of the proof (epi $\implies$ surjective), does this work?
Let $f:A\rightarrow B$ be an epi in $\sf{Set}$.
Choose $a_1,a_2:B\rightarrow Z$.
If $f$ is not surjective, $\exists b\notin \text{im} f$. So $a_1$ doesn't need to be the same as $a_2$ at $b$.
For them to be equal everywhere, $f$ must take on all values in $B$.
 
Which bit are you talking about?
 
$a_1$ doesn't need to equal $a_2$ at $b$
you have to spell that out
 
Because $a_1\circ f = a_2\circ f \implies a_1 = a_2$ is only true if we check the first part for every $b$.
They have to agree everywhere.
So $f$ must take on all the values in $B$.
Am I missing a detail?
 
Because I am playing devil's advocate, I am not convinced that there exist $a_1$ and $a_2$ which agree everywhere else but $b$ (and disagree at $b$)
 
2:48 AM
But there may. The implication won't hold.
 
In other words, a better strategy is to construct explicit examples of $a_1$ and $a_2$
 
I found a proof very analogous to mine. Is this one correct?
 
Correct, with the same gap.
 
Hm.
So an explicit example mends the gap?
 
That's the only way I see of doing it.
 
2:50 AM
A counterexample of sorts?
 
There's a chance that the link is a proof in a category like the category of rings, since you should not have to use $\mathbf Z$ in $\sf{Set}$
 
$Z$ is a kind of variable, not $\mathbb{Z}$
I'm working from the same book.
 
Yeah, not the category of rings
Proof doesn't work if $Z$ is singleton or, even worse, empty!
 
2:53 AM
Yes, I just noticed that.
Devil's advocate, indeed! :)
So the proof won't be complete if I simply add a counterexample, I think. It's insufficient.
 
Well, pick any epimorphism $f:A\to B$. Assume $b$ is an element of $B$ but not the image of $f$.
You need to define a set $Z$ and a pair $a_1,a_2:B\to Z$ which contradict the assumption $f$ is epi.
This is sufficient to show that epi implies surjective.
 
Oh, an explicit construction. I got you wrong before, then.
 
Ah, sorry.
 
Then $Z$ must have at least two elements. Right?
And then $a_1(b) \neq a_2(b)$.
Does this work for all $Z$ in $\sf{Set}$?
 
3:00 AM
Oh, ok.
 
So, go with two elements
 
Mm. Thanks a bunch.
 
Give them fancy names like charles and lucas or just $x$ and $y$.
2
So, you have your set $Z$. Think of an explicit function $B\to Z$.
That'll be your $a_1$.
 
$a_1(b) = x$ for all $b\in B$
$a_2(b) = y$ for all $b\in B$
 
Hm, remember the context. $a_2\circ f=a_1\circ f$ still has to hold.
 
3:06 AM
Right. So $a_1(t) = x$ everywhere, $a_2(t) = y$ except at $b$.
Hey @Alex
 
Ok, check to see if it works.
 
user147690
@SohamChowdhury Hi
 
@Alex: You might like this. (Needs a little music though)
@KarlKronenfeld Do you know German? (I'm guessing from your name, don't mind)
 
@SohamChowdhury lol no
 
Is there anyone online who does?
That first quote in Artin's book.
It's an Euler quote.
 
3:10 AM
@SohamChowdhury Does it work?
 
I think so.
I don't solve exercises in Coq, so I may never know for sure.
Replace $y$ with $x$ in that, sorry.
 
There you go
 
It should be $a_1(t) = x$ everywhere, $a_2(t) = x$ everywhere except for $a_2(b) = y$
I edited it absentmindedly. It was right to begin with. :(
 
@SohamChowdhury What!? Everybody reformats their work and puts it in a proof engine to see if it is correct. :D
At any rate, you have a much better proof than what was given in that link now.
 
Thanks to you, I guess. "Devil's advocate"
Are you a prof?
 
3:14 AM
Not a prof. Just a "Devil's advocate"
 
Grad student?
 
Nah, undergrad
 
Ah, cool.
In the States?
 
Cool. BBL.
 
user147690
3:16 AM
@KarlKronenfeld Could you perhaps explain to me the correspondance theorem for groups?
 
user147690
Actually I'll read this different definition it seems better
 
@AlexClark Ok, if you have any questions, I'd be happy to answer.
It should seem like a completely natural statement once you get it.
 
@AlexClark consider the following analogy with numbers: Suppose h is a divisor of g. Then the integers {k: h|k|g} are in bijection with the integers {k/h:1|k/h|g/h}. Note that 1|k/h|g/h is equivalent to h|k|g.
if you can visualize this with numbers, you should be good
Let me put that in latex: if $h\mid g$ then there's a bijection $\{k\in\Bbb N: h\mid k\mid g\}\cong\{\frac{k}{h}\in\Bbb N :1\mid\frac{k}{h}\mid\frac{g}{h}\}$.
(it's like you're dividing everything by $h$)
 
Can someone explain a thing in real analysis?
 
@Karl, are you familiar with how one can make a category out of a set with some ref. and trans. relation on it?
 
3:27 AM
regarding exterior points, limit points isolated points etc?
 
Just shoot, Paradox
 
@SohamChowdhury Yeah
 
Unless anon wants to chime in with the search query for "just ask" again :)
 
0
Q: Find the limit points and exterior points of the following

Paradox 101 Let $X=\mathbb R$, with the usual metric on $\mathbb R$ and $A=((0,1)\cap \mathbb Q)\cup$ {$2,3$}. Find the limit points of $A$, exterior points of $A$, $A^o$, $\overline A$ and $\partial A$. Can anyone please explain how to answer this as I don't understand how to go about this? Any help wo...

In the second answer a user has stated that the limit point set is [0,1]
How did we get this?
 
Did you have any guesses as to what it would be, @Paradox101?
 
user147690
3:29 AM
May 16 at 12:21, by Alex Clark
in The Bridge, Sep 2 '10 at 22:25, by radp
:151729 Don't ask to ask, just ask.
 
So there's a question that asks what conditions a relation must satisfy for the category to be a groupoid.
I believe reflexive and symmetric. Is that right?
@AlexClark So the operation of "putting a message in the sidebar" reverses the order of composition?
 
The claim is saying that, for any number in $(0, 1)$ (not necessarily rational), you can find a sequence of rational points that tends to that number. That doesn't seem believable? Give me a number, and we'll find a sequence
 
user147690
@SohamChowdhury Well permalinking it, yes
 
@KarlKronenfeld that it would be (0,1)? I'm not sure I didn't get the concepts clearly
 
@SohamChowdhury Symmetric is not strong enough, since it says that $x\le y$ implies $y\le x$.
Oops, it actually is
 
3:31 AM
?
 
@pjs36 why not rational? I mean isn't (0,1) intersecting with the rationals in A?
 
Yeah, you're right @SohamChowdhury. You still need the transitive property of course.
 
So equivalence relations only?
 
The statement that if $A = \big(\Bbb Q \cap (0, 1)\big) \cup \{2,3\}$ then the set of limit points is $[0, 1]$: You can find a sequence of points in $A$ whose limit is any real number between $0$ and $1$
 
Yeah, I haven't thought about it in this way before @SohamChowdhury, but that does seem right.
 
3:33 AM
@KarlKronenfeld Oh, transitivity is required for composition/associativity?
 
@SohamChowdhury Yes.
 
Hmm, thanks.
Makes sense.
I do hope you're TAing.
 
user147690
@anon Thanks this helped
 
@pjs36 yes the set of limit points is (0,1). It shouldn't be a closed interval right? It should be an open one?
 
Ah, yes, I did mean square brackets there, the limit points are $[0, 1]$, thank you
 
3:35 AM
@pjs36 how can $\mathbb Q$ have no open set within it?
 
every open set of R has irrational numbers. hence no open set of R is contained in Q.
 
@anon Oh I got it. Thanks!
@pjs36 thanks a lot for your help!
 
You're very welcome, it was just a definition/right viewpoint thing that was tripping you up, from the sounds of things
 
I guess so.I still can't process the simple stuff in topology and metric spaces yet. I need a lot of practice.
 
does jasper loy still come here anymore
 
3:40 AM
He will soon enough..
 
It does take practice, and finding another reference that uses a different kind of language can help a lot; sometimes certain notations/descriptions just "click" more than others
 
Karl when was the last time you saw him?
< a month ago?
 
@Ethan I believe his name was WillHunting last I saw
 
I'm not a good reference since I don't come in here consistently. < a month, yes
 
user147690
@Ethan Something like that, I emailed him a week ago I think and he said he was fine
 
user147690
3:42 AM
he was here 19days ago
 
Hey there @ALexW, have you jet-set across the country yet? :)
 
how do i private message on here again, i want to verify his email? im not sure if i have an old one
wait
here is the sha256 of his gmail address: a3495b6913f1de0ffa4853526a5aaae7b791a337c812a0aee4a4fef3d43a3e2f
can someone confirm that for me?
 
user147690
That's correct Ethan
 
thx
 
Hey @pjs36! Haha, not yet, but soon. :) How are you?
 
3:46 AM
@AlexWertheim Good, thanks! The Amish are rebuilding stuff around my apartment complex, and it's quite the spectacle. How about you?
 
Haha, that must be something to see indeed! Can't say I've seen too many Amish-folk in my time. I'm well, thanks! Just packing a lot these days, which is a drag. But slowly and surely coming back to life (math-wise, anyway) ;)
 
user147690
I have never seen an Amish person before
 
morning @AlexW
 
You're not missing much, @AlexClark, except for some superior carpentry skills.
 
3:52 AM
Evening, @MikeM. How's it hanging?
 
Packing is the worst, but at least it'll be over soon!
 
Lol, yeah! It's actually not so bad. Happily enough, I actually didn't unpack much to begin with, and I don't have too many things, so it's coming pretty well.
Courtesy of Zev Chonoles' blog, have you all seen this? It's great: theproofistrivial.com
3
 
haha, @AlexClark, that's about right, except it's just a rag-tag band of 3, with one or two non-Amish mixed in for fun, I guess
 
@AlexWertheim Just biject it to a complete ultrafilter whose elements are dihedral complexity classes
Though "complete ultrafilter" is a little redundant
 
Can anyone please explain the following question?
-1
Q: Show that subspace metric induces subspace topology

Paradox 101 Let $(X,d)$ be a metric space, let $\tau$ be the topology on $X$ induced by $d$ and $A \subset X$. Define $d_A: A \times A \to \mathbb R$ as $d_A(a,b)=d(a,b) \forall a,b \in A$ . Show that $d_A$ induces the topology $\tau_A$ on $A$ where $\tau_A$={$A \cap U: U \in \tau$}. I don't understan...

 
3:57 AM
No @Karl, it's much simpler than that. Just view the problem as a thrice-differentiable field whose elements are combinatorial fields.
 
@Paradox101 what don't you understand specifically?
 
@KarlKronenfeld I guess that works, but wouldn't it be simpler if you just biject it to a Noetherian monoid whose elements are trivial groups?
 
ah noetherian monoids. Never would have thought of that.
 
@AlexWertheim: It's alright. Busy. Done for the night, though.
 
@pjs36 Wait, so I just have to show it is singleton?!
 
3:59 AM
Uh... are you not reading what I just said??
 
Hm, I guess I could just view the problem as a finite poset whose elements are nondeterministic linear transformations
 
Ah, I think you found the sweet spot there
 
You could view it as a moduli space whose elements are geometric objects.
 
@anon I don't understand subspace topology or how a metric induces a topology
 
Glad to hear it @MikeM. All work and no play makes Homer something something.
 
4:01 AM
@Paradox101 do you know what an open set in R (the reals) is?
 
@anon but that actually parses.
 
@MikeMiller that's because I didn't get it from the website :-)
 
Wait, people were using the website?
 
@anon any subset in which an element belonging to the set has an open ball around it?
 
@Paradox101 right. same definition in any metric space. a subset U is open if every element in it has an open ball around it (contained in U)
 
4:07 AM
@anon what does that have to do with inducing a topology though?Or the subspace topology?
 
@Paradox101 in order to know the subspace topology, you have to know what the original topology is don't you?
 
Ok. But how do I use that?
 
Do you know what your task is?
Well, first let's say this: what do the open sets in A look like?
 
Not sure
 
the metric d is defined on A. so write down the definition of an open ball in A. then write down what it means for a subset of A to be open in A.
 
4:14 AM
@KarlKronenfeld. Help me out with a hint. I need to find an example of a category which has epis that have no right-inverses.
No answers, please.
 
think of a category that's not set
this is true for most categories you think of
 
hint: use the picture on wikipedia's article on epimorphisms
:-)
 
i think that's way less instructive...
if you do that it feels more like you're building a category where it's not true, rather than seeing that you're just very lucky it's true (with choice) in Set
 
@anon Oh lol yes. But I need an actual example to grok it.
 
4:19 AM
@anon the same definition for open sets would be used on A?
 
@SohamChowdhury I was telling you how to make an actual example. use the picture...
 
@MikeMiller I don't really know all that many categories.
@anon Okay.
 
and reflect on the word "literally"
 
@SohamChowdhury: Do you know any mathematical objects? They usually form a category.
 
Can somebody go and explain to Arnie Dris that $\sigma(x)=\frac{x}{3}$ is not a Diophantine equation. He simply won't believe me. math.stackexchange.com/questions/1293668/…
 
4:21 AM
Groups, abelian groups, rings...
 
@MikeMiller true
 
@MikeMiller Not all that well, unfortunately. I have a sort of okay notion of how groups work.
 
You could also make an example with one object, but I am not sure how Mike would respond. :P
 
I'm actually working with an algebra book that introduces categories in the first chapter.
 
i don't really understand the point of learning categories and category theory without having a big class of examples that help motivate all the definitions being built, but oh well
you should try to understand the categories of groups and abelian groups and use them as some of your main examples.
 
4:23 AM
@MikeM: the category of semi-symplectic topological quantum paramonoids of Rice-Paddy type, satisfying the Mussolini-Rostropovich equations at infinity?
3
 
@MikeMiller That's the second chapter.
 
@AlexWertheim hehehe
 
misread Soham as responding to Alex for a second there
4
 
+star for Mussolini-Rostropovich
@anon lol
 
4:26 AM
on that note, @AlexW, how's studying going
 
Packing has interfered a bit recently, @MikeM (I know, only the latest in a long line of excuses, lol). But it goes. Still hammering away at A&M, and putting Lang in the mix now.
Embarrassingly, I'm yet unable to answer the problem I just referenced. In due time, I suppose. :)
 
well, of course, you first need to symplectic topological quantum paramonoids before generalizing
 
@Mike, is the category defined by $\mathbb{Z}$ and $\leq$ an example?
For any two integers there's only one morphism.
So everything is epi. (vacuously)
But there are no inverses.
 
right (no inverses except the identities)
 
Hahaha, ah yes. =P I'll start knocking out large chunks soon, I expect. Once I get rolling, I think I'll make much more evident progress.
 
4:31 AM
Ah, finally.
Everything is better once you have an example, but I don't have many yet. :)
Yes, except identities.
 
category theory was built to deal with examples like the category of sets, groups, abelian groups, rings, fields, ... while these might be decent examples for questions like this, i still find them very artificial. you might like to come back later, after learning about groups and so on, and do this with those in mind.
 
@SohamChowdhury Does the book have a section on universal objects/arrows before talking about any algebra?
 
@MikeMiller I get your point, but I'm sort of enjoying Aluffi. I'll drop it and switch to some other book if I get really stuck. I've almost done the first chapter.
@Karl: Yes, the first chapter is basic NST and a little about what a category is, what monos and epis are etc. With a few non-algebraic examples.
Second chapter is groups. And so on.
 
Can anyone explain how to show that $d$ and $\overline d$ induce the same topology on$X$?
 
sure, i understand. i'm not saying it'll be hard to understand or anything, just that these sorts of examples kind of miss the point. (this is also why, say, it's best to look at maclane's book after knowing a little algebraic topology, some algebra, etc... because this is where the examples come from, and tells you what they really mean)
 
4:36 AM
I had seen a very fair amount of algebra before looking into categories, but it wasn't until I took a topology class that the language really seemed motivated. I believe it was homotopies, but I'm almost certainly wrong about that, because I don't know topology well.
 
@KarlKronenfeld I just hit the last section of the first chapter. Universal properties. :)
 
Ah, homotopies of maps, that's what it was.
 
@SohamChowdhury Universal properties only make sense to me once you have those motivating examples mike was talking about. They go into applying category theory. I don't mind the fact that you learn what a category is without those examples, but introducing universals like this is a bit iffy to me.
 
@KarlKronenfeld The idea is that he then uses universal properties for everything after that. Modules and whatnot.
I see your point, though.
Ever since I did a wee bit of Haskell, I had to know a little category theory. To sort of find out what the fuss about lenses and monads was all about. :/
 
I can't say for sure, but my guess is Aluffi doesn't really expect the categorical language he introduces early on to be fully absorbed initially.
 
4:40 AM
@AlexWertheim Yes, he says as much.
 
@SohamChowdhury That's good. You'll be understanding the difference between product and coproduct of modules right off the bat.
 
@KarlKronenfeld There's an example I'm reading right now about the universals that products and coproducts satisfy in $\sf{Set}$.
Similar things later on too, then, I guess.
And how they're 'mirror constructions' of each other.
 
The thing is in the category of modules over a commutative ring they are the same object for finite products/coproducts, but they differ for infinite families.
That tends to throw people off when they don't know that products/coproducts are objects equipped with accompanying morphisms (etc.).
 
Hm.
I remember, when I bought my guitar, I didn't play it at all until I did some research on posture and efficient picking mechanics. That took almost a month. :P
You can chalk it up to that.
 
If you have the time, go ahead.
 
4:47 AM
Time for?
 
To do that research on the foundations without needing to know the stuff you are actually intending to learn.
 
I know, I know.
It was more about not getting any bad habits which would be hard to break later. (in that context)
 
Time for bed for me. Good night, friends!
 
Night, @AlexWertheim.
 
Fare well, @AlexWertheim !
 
4:49 AM
see ya @AlexW
 
@SohamChowdhury Yeah, with instruments that is a good point.
 
Here, it's about understanding this from the start as generally as possible.
Back to work now. BBL.
 
user147690
Can someone explain quotient rings to me? I am having trouble understanding this from wiki/Artin/D&F
 
Can anyone explain how to show that $d$ and $\overline d$ induce the same topology on$X$?
 
how can I use $\Gamma (n+\frac{1}{2})$ ? can someone show for instance $\Gamma (\frac{1}{2}+1)=\Gamma(\frac{3}{2})$ ?
 
4:58 AM
@AlexClark Is there something in particular bothering you about quotient rings, or a particular quotient ring?
 
user147690
@pjs36 Hmm I just don't understand the definition
 
Its basically modular arithmetic, except in different rings @AlexClark
 
user147690
So equivalence classes are just the additive cosets(via the ideal rather than a normal subgroup)
 
@AlexClark Pick an ideal of elements, set them equal to 0, see what happens to the rest of the ring
 

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