@DylanMoreland Well, as someone pointed out earlier in chat, there's a difference between zero, and nonzero but small. Zero either means "I do not know there's this thing called accepting answers" or it means "I couldn't figure out how to accept answers". On the other hand, 17% means...
@Srivatsan 17% means "I ask interesting questions that tend to accumulate too-long-for-comments half-answers which are interesting and illuminate the problem but do not resolve it completely".
(Vassili is the one who names bizarre Diophantine equation systems after himself, claims they will solve the world's political and technological problems, and complains about answers that don't find all solutions, right?)
A propos of naming things after oneself, last Monday was the anniversary of the discovery of the kalle-numbers. I think I must have missed the celebrations.
@QED Here's what I'm worried about on that $\mathbf Q(\zeta_n)$ question. At some point you have to show something like, "with $n = 3$, I can write $0$ as $\zeta^2 + \zeta + 1$. I have to send this to zero. So I also have to know that $\zeta^4 + \zeta^2 + 1 = 0$.
This is a dumb example, because you can just conjugate. But for arbitrary $n$ I think this is where the work is.
It seems like you need the minimal polynomial and the knowledge that its degree is $\varphi(n) = 2$ to do that. And if I take a product and the degree gets $\geq 2$ then how do I know how rewrite it?
@AsafKaragila I'll take a look. I'm pretty indifferent about voting but when things are slow I must admit that it's fun to point out these interactions. Probably a poor way to find entertainment.
"I am not obliged to show my appreciation" is something I disagree with strongly, though.
Interesting. He basically spells out what I've been thinking: Is it really necessary to accept answers? I mean, I want to but I don't care whether the OP of the question I'm about to answer has acceptance rate zero.
But then again I don't get the idea of badges either...
@robjohn I relate it either to the fact that I was in US where it was totally different climate, or that I was going by the plane with a person who had a cold and was sneezing all the way
he introduces the isomorphism between objects of the category and then gives theorem about the isomorphism between two categories
well, I guess he means them regarded as objects of Cat - but then how do we compare that two arrows are the same?
from what I understood, when we define a category $\mathbf C$, we put the objects and arrows by definition, claiming that there is a composition of arrows which satisfies associative property (includes the equality between arrows) and identity arrow for each element (includes the equality between arrows)
e.g. if $\mathbf C = \mathrm{Rel}$ then two arrows $f,g\in\mathrm{Hom}_\mathbf C(A,B)$ are equal if and only if $f=g$ as subsets of the set $A\times B$
but how can we now that $f=g$ if $\mathbf C$ is some abstract category?
for example, I need to prove for the isomorphism of two objects that $f\circ g = \mathrm{id}_B$ and vice versa
Composition is a map $\operatorname{Hom}{(B,C)} \times \operatorname{Hom}{(A,B)} \to \operatorname{Hom}{(A,C)}$ and associativity means that the two ways of composing three morphisms give the same result.
In each $\operatorname{Hom}{(A,A)}$ there is a distinguished element called the identity and to say that two objects are isomorphic is to say that there are morphisms $f: A \to B$ and $g: B \to A$ such that $gf = \operatorname{id}_A$ and $fg = \operatorname{id}_B$.
@tb: so the idea is to get rid of objects when thinking about equality of arrows - objects only appear to give source and target (which have to be the same for equal arrows), the main information if two arrows are equal or not is given by the composition map. right?
@tb Quote: Every category $\mathbf C$ with a set of arrows is isomorphic to one in which the objects are sets and the arrows are functions.
as a sketch of the prove he gives a functor from the original category to the on where objects are sets. I proved the rest yesterday but I don't have a feeling that I understand everything in the proof I wrote
So he maps an object to the set $\operatorname{Hom}{(A,A)}$ and a morphism to the function $\operatorname{Hom}{(A,A)} \to \operatorname{Hom}{(B,B)}$ it induces?
I'm a bit confused also with such notation: from what I understand, it means $$ \bar A = \bigcup\limits_{B\in\mathrm{Ob}(\mathbf C)}\mathrm{Hom}_\mathbf C(B,A)$$ I don't know if union is allowed here, though
@tb yes. so for each morphism $g\in \mathrm{Hom}_\mathbf C(A,D)$ we define the morphism $g\in\mathrm{Hom}_{\mathbf C'}(\bar A,\bar D)$
I don't know. he distinguishes sets and classes (without discussing what is the difference) and I am not sure if I can just directly take unions of any collection that I meet. However, in this particular case we have $\mathrm{Hom}_\mathbf C$ are all sets, so it should be allowed
@tb 300 pages is long? )) well, I'm reading a PhD thesis of a student which contains some CT and I tried to learn briefly the main concepts from there - but then I had a lot of question which he certainly didn't have to answer - and I asked Zhen for the reference
the student provides MacLane's "Categories for working mathematicians" as the main reference, but I don't have it in the library
Anyway, for people wanting a quick introduction to categories, I always recommend the second chapter of Hilton-Stammbach, A course in homological algebra, this gives you the main ideas quickly and concisely.
@tb I hope too, I didn't have an intention to read the whole book if there is a brief introduction. Maybe you can also tell what do you think of this: they introduce categories through multigrpahs which seems much more comfortable for me
But I think we shouldn't talk too much in Swiss German. This excludes pretty much everyone from this conversation and it's a bit impolite towards them...
@tb I have no idea. A natural transformation of adjoints like that gives rise to a correspondence of commutative diagrams, but I haven't used this fact much...
well, it gives you the three diagrams I used for free, but I don't see how to apply this observation to see the correspondence between maps from the push-out and maps to the pull-back.
@ZhenLin okay, I just wondered why three people upvoted that answer which feels to me like a (useful) general comment while my answer was left untouched. So it seemed to me like I was missing something obvious.
@Ilya: Conceptually, the proof has two steps: showing that the Yoneda embedding is full and faithful, and then showing that the presheaf topos on a small category is a concrete category.
@ZhenLin oh, maybe I was incorrect in titling the theorem. It is theorem 1.7 in the link I put in my previous post: each category $\mathbf C$ with a set of arrows is isomorphic to one in which the objects are sets and the arrows are functions
I have the following problem: given $\bar f,\bar g\in \mathrm{Hom}_{\mathbf C}(\bar A,\bar B)$ I want to prove that $\bar f(\bar a) = \bar g(\bar a)$ for all $\bar a\in \bar A$ implies that $f=g$
@Zhen here $\bar f(\bar a) = f\circ \bar a$ and $\bar g(\bar a) = g\circ \bar a$ and $\bar a$ is any morphism in $\mathbf C$ such that $\mathrm{cod}\bar a = A$
@tb: by the way, you advised an alternative way: to take objects being $\mathrm{Hom}(A,A)$ - but how should I define morphisms in a new category then?
I have a question: what's useful about Michael's answer there? It gives a notation and doesn't explain anything and yet it has 6 upvotes? Frankly, I find this bizarre.
@tb thanks. the definition of coslice I derived by myself, seem to fit the purposes of the book. Btw, what if in the definition of slice we will require an arrow $g:(f:X\to C)\to(f':X'\to C)$ to be an arrow $g:X'\to X$ such that $f\circ g = f'$ rather than an arrow $g:X\to X'$ such that $f'\circ g = f$?
@Ilya that's not it. Coslices are of the form $(f: C \to X)$ and a morphism $g: (f:C \to X) \to (f': C \to X')$ is given by a morphism $g: X \to X'$ such that $gf = f'$
@JM no idea. Would be nice if the tools gave access to these kinds of statistics. He crossed the 1000 and 2500 views just yesterday. He'll be having a shiny golden one in the next few hours, I assume.
By the way, your fern looks like a shark's tail when you're faded out in the chat window. :)
Also, if you're going to tell a joke, you need to be absolutely sure that you've gauged the audience correctly. It's tough to recover from a flattened joke.
Whats worse is one old friend went on to share it writing.." i do not get whats in it but am sharing it since i believe you must have shared something great "
