Well, the silly answer is that $f:\mathbb{S}^k\to X$ represents a torsion element of order $p$ if $p \cdot f:\mathbb{S}^k\to X$ extends along $\mathbb{S}^k \hookrightarrow D^{k+1}$ to a map $\varphi: ... View answer 11 votes Depending (*) on the underlying degree of analyticity (**) in your calculus course, it might be just as well to start with the Stokes theorem, stating it as an existence and uniqueness theorem: ... View answer 10 votes Pro I think the examples given are instances of Guy's "strong law of small numbers". That seems at least poetic reason for low-dimensional specializations of your favorite theory to be different in ... View answer 8 votes If I understand the paper's abstract, Yes. View answer Accepted answer 8 votes By AC, choose a cardinal well-ordering of the lines in in the plane and any well-ordering of all the points. We proceed by transfinite induction. Suppose$A_l$is a set of points, no three colinear, ... View answer 8 votes If I understand you right, you're assuming that there is already a Riemann metric chosen for you. This of course integrates to a distance function ("metric" in the sense of metric spaces) whether or ... View answer 6 votes I suspect these sorts of problems arise in two ways: ignoring wrongness (e.g. Skolem's paradox, that there are countable models of set theory which believe in uncountable sets) and ignoring first-... View answer Accepted answer 5 votes There are a short list of operations described as generating the desired polyhedra:$ X : \mathrm{Prism} \vdash C X : \mathrm{Prism}  l : \mathrm{list}\ \mathrm{Prism} \vdash \Pi l : \mathrm{...

This is probably making a hash of the earlier answers, but bundles are special fibrations; specifically, they are fibrations with (not canonically) isomorphic fibers. And we all like fibrations, ...

On reflection, SAS tells me that Euclidean geometry has a strong semi-local homogeneity, in that every neighborhood of every point is isotropically isomorphic with some neighborhood of any other point ...

Oh! From uniqueness of the countable dense linear order without endpoints: take (for instance) a countable ordinal $\lambda$, and consider the anti-lex order on $\mathbb{Q}\times\lambda$. This is a ...

Do people tend to mean the official definition? I think "official" belongs in scare-quotes... I tend to think that "subcategory" is an evil notion. I'm not published anywhere, but in my notes I ...

Here's an irresponsible under-referenced response. On the one hand, this is hard! Indeed, I've read somewhere --- not handy --- that this question lurks among quantum gravity's many difficulties. ...

First, it's important that the infinite connect sum $A\# B\#A\#\cdots$ is not the limit of the finite connect sums $A,A\#B, A\#B\# A,\dots$; in fact, I'm sure the binary connect sum is as wrong a ...

I notice that the categories considered for naming here are all the domains, or shapes, of basic diagrams; an object, an arrow, an endomorphism (n.b., my instinct was just to call that $\mathbb{N}$), ...

Without meaning to be snarky, I think there's some confusion here about what spectral sequences are for. In particular, the sense in which a SpSeq (or even a long exact sequence!) serves as a ...

I should say I'm fond of the Thom isomorphism, but I still find the contents rather mysterious.

In case you don't want to worry about dividing by small norms (admittedly, less and less of an issue for higher $n$...) Generate a point $\mathbf{x}$ on the $(n-1)$-sphere; and generate a number $y\... View answer 2 votes The following runs out of steam towards the end; I may also be making important mistakes, so be on your guard --- but that's half the fun! Anyways, it was too long for a comment. Choose a locally-... View answer Accepted answer 2 votes (CW because it's more an over-long comment than a real answer.) I think there are too many competing normalizations to make a good choice. In lieu of sensible default, call one of them homology, and ... View answer 2 votes I'm betting on Supremum. View answer Accepted answer 2 votes The classical group$K_0$can also be thought of as consisting of equivalence classes of chain complexes of vector bundles, such that the exact sequences represent the zero of$K_0$--- and ... View answer 2 votes Why they are useful is related to one reason Polynomial invariants are useful themselves: they let you prove theorems, when they acutally do let you prove theorems. In particular, if you have a ... View answer 2 votes To address the question in a somewhat less-categorical way, I would point out that you can in fact do all of topology without using the expression "open set", by instead refering to the filter of ... View answer 2 votes Sorry if this is silly; but might it have something to do with needing in the sheaf category to consider the sheafified presheaf cokernel in order to talk about projections? that is, I (think I) can ... View answer 2 votes An incomplete answer on the subject of countable dense linear orders without endpoints; I left some other thoughts at the Cafe; on further reflection, one can think of the maps$\cdot\times \frac{p}{...

As I understand the word "tautology" from Mathematical Logic; a logic $L$ means a formal system of deductions $\Sigma \vdash_L \Phi$, satisfying some rules, where $\Sigma$ is a set of propositions and ...

In a pre-order $\prec$ (or a category) one can speak of initial objects $0$, or terminal objects $1$, meaning that $0\prec x$ for all $x$ --- (or $0\rightarrow^! x$ ) --- which also gives the notion ...

An incomplete answer; but perhaps it helps to rephrase the problem as below. The reason the round circle does have this property is that without loss of generality, the map $f$ fixes the origin; and ...