User some guy on the street - MathOverflow

Discrete disjoint covering of integer lattices

2013-05-07T16:20:44Z

Is there a covering of $\mathbb{Z}^n$ by disjoint translates of the ~~basis-and-origin~~ minimal integer $n$-simplex? By haphazard I have such coverings for $\mathbb{Z}$, $\mathbb{Z}^2$ and $\mathbb{Z}^3$, where the wanted translations are lattices spanned by $\{2\}$, $\{(2,-1),(-1,2)\}$, and $\{(1,1,-1),(1,-1,1),(-1,1,1)\}$, but rhyme nor reason can I see in this sequence of families to extend.

Status of the Isomorphism problem for automatic groups?

2012-12-17T00:39:43Z

I only ask because I don't know how to look for the answer.

Answer by some guy on the street for Conventional names for finite categories

2012-08-07T18:44:54Z

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 $\mathbb{N}$), a composable sequence, parallel arrows, equalized arrows... Not that diagrams in these categories aren't also interesting (a composable sequence in a composable sequence category, e.g., is well-worth half-an-hour's thought), but as diagram domains is where they all come up first for most of us; so why not call them what they are?

$\fbox{$\phantom{X}$}$, the trivial diagram domain/the shape of the empty diagram
$\fbox{$\bullet$}$, the object diagram domain/the shape of an object
$\fbox{$\overset\bullet\circlearrowleft$}$, the endomorphism diagram domain/the shape of an endomorphism...

Of course, to establish a convention, one must write a famous textbook. Good luck with that!

Answer by some guy on the street for What is the precise relationship between "prodsimplicial sets" and rooted trees?

2012-05-23T05:40:09Z

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{Prism} $

There are a short list of operations needed to generate the family of rooted trees:

$ T : \mathrm{Tree}\ \vdash \mathrm{Stem}\ T : \mathrm{Tree} $
$ F : \mathrm{list}\ \mathrm{Tree}\ \vdash \mathrm{graft}\ F : \mathrm{Tree}$

That makes the correspondence obvious; the distinction is in the semantics of prisms vs. trees.

Now, the generators I've given would seem to distinguish between $ \Pi (\Pi (A, B), C) $ and $ \Pi (A, B, C) $, just as they seem to distinguish between ... , well, I'd have to draw pictures, and I don't really want to. I think that's quite alright, I don't mind many isomorphic polyhedra having distinct descriptions. But if that worries you, you can argue that the strange plurality of trees is isomorphic to the strange plurality of products in a natural way. And that should be enough.

Answer by some guy on the street for Tracking spectral sequence differentials

2012-04-20T02:57:28Z

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 computational tool isn't that it tells you how to calculate things --- particularly, not how to calculate the things in it. Rather, it gives you a way to organize things you already know how to calculate; maybe it suggests a helpful summary of your calculations; and it will certainly remind you of what things you don't already know --- it will provide helpful questions to consider --- if you try to squeeze modules out of it.

As a small example, the usual snake-shapped diagram chase proving the isomorphism of, e.g. de-Rham and real Čech cohomology, is neatly summarized by noting that there's a related double complex which is very nearly exact in every way, and thus all the later pages are mostly zero, while the transgression is an isomorphism. We get a proof, because we already know what enough of the modules and differentials are.

If you really want to ask what the pages and differentials of some SpSeq are, there are algorithmic tools (i.e., programmes --- usually not suitable for human consumption!), provided you can properly specify what your particular spectral sequence is about. See e.g. (Ana Romero's thesis)1, building on homological perturbation lemmata (start at 2).

The most general context of Mather's Cube Theorems

2012-03-02T07:13:04Z

Quite simply, I'd like to know what is the broadest or most natural context in which either (or both) of Mather's cube theorems hold. If you like, this may mean any of

What properties of $Top$ or $Top^*$ are essential to the proofs?
(where) are model/homotopical categories verifying Mather's theorems studied as such in the literature?
Are there more examples known verifying Mather's theorems?

I ask because Mather's proof strikes me as fairly gritty and seems to rely on explicit cellular constructions.

For reference, the cube theorems concern a cubical diagram whose faces commute up to homotopy in a coherent way, and assert

If one pair of opposite faces are homotopy push-outs and the two remaining faces adjecent the source vertex are homotopy pull-backs, then the final two faces are also homotopy pull-backs
If two pairs of opposite faces are homotopy pull-backs, and the remaining face adjacent the target vertex is a homotopy push-out, then the remaining face is a homotopy push-out.

Answer by some guy on the street for A programming language that can only create algorithms with polynomial runtime?

2012-02-04T02:58:59Z

If I understand the paper's abstract, Yes.

Answer by some guy on the street for Thom's Principle: rich structures are more numerous in low dimension

2011-09-13T15:44:22Z

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 character from high-dimensional specializations.

Con

An example of increasing richness indicating Thom was thinking about something else:

"The" connected 0-manifold is a point
"The" connected compact 1-manifold is a circle
Connected compact 2-manifolds are connect sums of tori or of projective planes; they are uniformizable.
Connected compact smooth 3-manifolds are piecewise geometrizable, where the joints are among spheres and tori.
Connected compact $(3+n)$-manifolds "solve" the word problem for groups; particularly weird: there are smooth examples that have contractible stable closed periodic geodesics in any smooth metric.

This is sort-of what I'd call rigidity in low dimensions, and richness in high dimensions. Perhaps the theories in low dimension are richer in the sense that there are more universal statements we can prove, but there seems to be a greater wealth of useful examples in higher dimension.

Whether this is an instance of Thom's principle as quoted or an exception, it is still an instance of Guy's law, in that the low-dimensional behaviour isn't representative of high-dimensional behaviour.

Answer by some guy on the street for Examples where it's useful to know that a mathematical object belongs to some family of objects

2011-09-02T01:29:04Z

I have two related sorts of example to suggest, probably exhibiting my categorical bias vs. the analytic/geometric-topology weight of the preceding examples.

Galois-theoretic

Let $P\in K[x]$ be an irreducible (separable) polynomial of degree $n$ over a field $K$, and consider the extension field $L = F[x]/(P)$. It is fruitful (i.e., it's the whole categorical view of Galois theory) to view $L$ as one of serveral isomorphic extensions, with possibly non-trivial automorphism group relative to $F$. In one sense, everything you might wish to know about $L$ and $K$ is contained in your ability to understand $K$ and $P$, but it's evidently fruitful to look at commutative diagrams $L//F\to L'//F$

Topological-algebraic

So you want to understand some algebraic gadget? A vector space/lie algebra/group? Then study bundles of such things!

For example, it can be a bit stifling to think of a Lie algebra of a Lie group as the tangent space at the identity. Sometimes it's better to think of it as either space of half-invariant vector fields; and then the Lie bracket is given by the Lie bracket! (ha-hah!) But there is plenty of room to study other things, exploting the bundle of isomorphic Lie algebras.

Other times it's helpful to study the universal bundle $EG\to BG$ of some discrete group $G$. These realise the group both as a particular fiber and as the fundamental group of $BG$ at your favourite basepoint, and so you learn things about the group (e.g. related homology functors) by studying the whole collection of groups $\pi_1(BG,b)$ and torsors $EG_b$ for $b\in BG$.

Connecting group ring, abelianization

2011-08-15T13:55:30Z

For reasons arising in algebraic topology, I'm wanting to better understand the relations between two functors from groups to abelian groups, $\mathbb{Z}[\cdot]$ and $\operatorname{ab}$; group ring and abelianization.

Specifically, we can extend the homomorphism $G\to \operatorname{ab}(G)$ to an additive map $\mathbb{Z}[G]\to \operatorname{ab}(G)$, and this map is clearly natural. It seems too vague to ask "what should I do with this next?", so instead the question is

What references are there that consider this natural transformation?

For instance, I'd be immensely pleased if this is a good way to start a free resolution of $\operatorname{ab}$ without forgetting too much about $G$, and to learn what's the natural way to continue... but I'm not picky! You'll understand when I complain that automatic search-engines aren't too helpful in looking for anonymous natural transformations.

Thanks.

What's so difficult about $\pi_{15}(SO)$?

2011-07-22T17:09:56Z

Regarding the table of $SO(n)$s-of-origin in Davis+Mahowald (if you can get MathSciNet), is there a good reason that it should take longer for $\pi_{15}(SO)$ to be representable than $\pi_{19}(SO)$, which is already fully representable in $SO(11)$?

Topological Localization of (the simply-connected cover of) SO or Spin

2011-05-24T15:33:57Z

This is essentially a (cw) reference request, because it seems like the sort of thing that should have been looked at already.

Setting aside, for now, how to think what the localization of a general space is really; what's the right way to think about localization of $SO(n)$ --- or, if that doesn't make sense, of $Spin(n)$ --- as a space? (whether Sulivan's construction or Bousfield-Kan or ...)

some points of curiosity: Is it still a homotopy group? Is it better known as something else? Is there any good interaction between the algebraic side of localization and the group structure?

Answer by some guy on the street for Homology or cohomology?

2011-05-02T05:00:05Z

(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 call the other cohomology and I'm sure it'll be fine. This is also probably why someone worked out the language "left-derived" and "right-derived".

Personally, I think the distinction between "homological" and "cohomological" chain complexes is artificial; what might make a chain complex "co"-chain is that it was initially zero, whereas "really-chain" complexes might be eventually zero. Of course, we're often interested in things that do both, and things that do neither --- e.g. either theory on manifolds, or geometric K theory and elliptic cohomology. Anyways, if you're playing with chain complexes, I'd always call the kernel-mod-image construction the "homology" of a chain complex. We'd reserve the right to write "cohomology of X" for the homology of a contravariant construction from X, but it's still a homology of something.

Historically, "Homology" comes from a relation that Poincaré described on submanifolds of a manifold; there are both covariant and contravariant ways to make this functorial and algebraic on the category of smooth manifolds; for eitehr one you probably have to deal with singular submanifolds eventually. The contravariant one is called cobordism[1] these days and is represented by a spectrum indexed by codimension. There is that theorem that all reasonable "cohomology" theories on spaces are corepresentable; there are reasonable representable homology theories (like homotopy), and there are homology theories that aren't representable (like singular homology of spaces, iirc). Perhaps if your things are always corepresenable it'd be reasonable to call them cohomologies.

[1] That's "co" meaning "together", not "dual". That is, two things may be "co-bordant"; not that "co-bordisms" pair with bordisms. Although, I suppose they might, anyways...

Limits are terminal objects in another category; (when) are they colimits of (another diagram)?

2011-04-07T04:42:50Z

Let $C$ be a category with finite limits; that is, for any finite category $D$ and functor $F:D\to C$ the category $\mathrm{Cone} F$ of cones over $F$ is inhabited and has terminal objects (we could turn the morphisms around and call the limit an initial obect, but never mind).

That is, define $$ (A,\phi): \mathrm{Cone} F \iff \phi:Nat(\Delta_A, F) $$ where $\Delta_A(d) = A$ and $\Delta_A(\delta:d\to d')= 1_A$; and for $(A,\phi),(B,\psi):\mathrm{Cone} F$ define $$ \mathrm{Cone}F((A,\phi),(B,\psi)) = \{ f:A\to B \mid \phi_d = \psi_d\circ f\} .$$ A limit for $F$ is a terminal object $(H,\eta)$ in $\mathrm{Cone}F$.

Now there is an obvious forgetful functor $F^\perp: \mathrm{Cone} F \to C$ with $F^\perp(A,\phi)=A$. More: for any limit $(H,\eta)$ there is a co-cone $(H,\chi)$ with $\chi_\phi:F^\perp(\phi)\to H$, the unique one showing that $(H,\eta)$ IS a limit. For any co-cone $(A,c)$ under $F^\perp$, there is by hypothesis a morphism $c_\eta:H\to A$, again by hypothesis making everything commute that should; thus $\chi$ ( $\eta$ (or $H$)) is versal for cocones under $F^\perp$.

I'd like to eventually conclude that $H$ is universal --- that $c_\eta$ is the only thing making everything commute where it should, so that $\chi$ makes $H$ a colimit for $F^\perp$, but from the diagrams, I'm afraid it probably doesn't.

Are there supplementary assumptions that'll make everything work out nicely? Does it work out nicely and I just don't see it?

Or maybe I'm wrong about something!? That'd be OK, too.

Answer by some guy on the street for Geometric meaning of torsion in homotopy groups

2011-04-07T03:55:56Z

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 $\varphi: D^{k+1}\to X$.

The slightly less silly answer --- the slightly deeper answer, that is --- is that, equivalently, $f$ itself extends to a map $\tilde f: P^{k+1}(p)\to X$. Here $P^{k+1}(p)$ is the CW complex built of a point, a $k$-cell and a $(k+1)$-cell, where the $(k+1)$-cell is attached by a degree-$p$ map ${"p}$; and it's called the $(k+1)$th (cyclic) Moore space of order $p$.

For any $p$ the Moore spaces form a suspension spectrum $P^{k+1}(p) \simeq \mathbb{S}^1\wedge P^k(p)$, and there is in fact a long cofibration sequence $$ \mathbb{S}^1 \overset{"p}{\to} \mathbb{S}^1 \to P^2(p) \to \mathbb{S}^2 \overset{"p}{\to} \dots $$ which gives rise to a long exact sequence of generalized homotopy groups $$ \pi_1(X) \leftarrow \pi_2(X,p) \leftarrow \pi_2(X) \overset{p}{\leftarrow} \pi_2(X) \leftarrow \pi_3(X,p)\leftarrow \cdots $$ which in turn, for $n$ large enough, breaks up into the homotopy Universal Coefficient Theorem $$ 0\leftarrow \mathrm{Tor}(\pi_n(X),\mathbb{Z}/(p)) \leftarrow \pi_{n+1}(X,p) \leftarrow \pi_{n+1}(X)\otimes \mathbb{Z}/(p) \leftarrow 0.$$

The letter $P$ is used here because, at least because they have also been called Peterson spaces, and maybe because $P^2(2)\simeq \mathbb{RP}^2$ is the real projective plane.

Note that I'm using the letter $p$ because when I worry about these things $p$ is usually prime, but that's not necessary in the above.

We give up continuing the sequence at the first $\pi_1(X)$ because, without more information, it's not clear that $({"p})_\sharp$ should be a group homomorphism. Someone else can remind me what we can still say about the underlying sets.

Contractability of Exotic R^4s

2010-04-14T20:43:04Z

Notation: $\mathbf{R}^4$ is a smooth manifold with underlying topology $(\mathbb{R})^4$; ${\mathbb{R}}^4$ is the standard smooth structure.

The two things I know best about $\mathbf{R}^4$ is that it is locally diffeomorphic to $\mathbb{R} ^4$, and that it's contractible. It's easy to see that the contraction can be acheived by a smooth map ${\mathbb{R}}^4\times I\rightarrow{\mathbb{R}}^4$.

Do I suppose correctly that the same contraction is not smooth as a map ${\mathbf{R}}^4\times I\rightarrow{\mathbf{R}}^4$?
Do the exotic smooth structures have any smooth contractions?
If not, are there continuous contractions $\mathbf{R}^4\times I\rightarrow\mathbf{R}^4$ within the smooth maps $\mathbf{R}^4\rightarrow\mathbf{R}^4$?

Answer by some guy on the street for Clearing misconceptions: Defining "is a model of ZFC" in ZFC

2011-01-11T15:24:18Z

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-orderness: First-order theories with infinite models have models of every larger size; but in any universe there is exactly one isomorphism class of complete ordered field --- being a complete ordered field is NOT a first-order property of the field, but a first-order statement about a thing in some universe.

Contrapositivewise, teaching clearly about these two sorts of phenomena ought to help us keep clear of trouble. Unless, of course, ZFC proves itself consistent...

A curious construction of a chain complex and its homology

2010-12-29T16:01:40Z

... curious to me, that is.

Suppose two module filtrations $$ \cdots < A_3 < A_2 < A_1 < \cdots $$ and $$ \cdots < B_3 < B_2 < B_1 < \cdots $$ are comparable in the sense that for all $j$, $ B_{j+1} < A_{j} < B_{j-1} $; then there are natural complexes $$ \cdots \to \frac{A_3}{B_4} \to \frac{A_2}{B_3} \to \frac{A_1}{B_2} \to \cdots $$ and $$ \cdots \to \frac{B_3}{A_4} \to \frac{B_2}{A_3} \to \frac{B_1}{A_2} \to \cdots $$ both of which have homology groups $$ \frac{A_i\cap B_i}{A_{i+1} + B_{i+1}} .$$

My question is in two parts:

this canonical isomorphism $H(A^+/B)\simeq H(B^+/A)$, has it got a name?
is it useful?

Answer by some guy on the street for Associativity with infinite nesting

2010-12-07T05:18:37Z

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 notation for the infinite connect sum as $+$ is the wrong notation for $\sum$. Similarly, a conditionally-convergent infinite sum of numbers is the limit of a particular sequence: the initial finite sums; whereas an absolutely convergent sum of real numbers $x_i$ is a (sum of) limit(s) for much bigger diagrams: $$ \sum x_i = \sup_{W\ \mbox{finite}} \sum_{i\in W} x_i + \inf_{V\ \mbox{finite}} \sum_{j\in V} x_j$$

Looking for a better description of our infinite connect sum, let's first note that $A\# B$ means a particular colimit $$\begin{array}{c} {} & & S^n & & \\ &\swarrow & & \searrow & \\ Q & & & & R \\ & \searrow & & \swarrow & \\ & & Q\sqcup_{S^n} R & & \end{array}$$ Note that $A$ and $B$ are also colimits, $A= Q \sqcup_{S^n} D$ and $B=D\sqcup_{S^n} R$. In fact, it's best to keep things open at both ends: for $X$ and two maps $S^n\rightrightarrows X$, let $Q = D\sqcup_{S^n}X$ and similarly define $R$ in terms of $Y$ and two maps.

Instead of the infiite connect sum, we have an infinite diagram $$\begin{array}{c} {} & & S^n & & & & S^n & & \\ &\swarrow & & \searrow & &\swarrow & & \searrow & & \swarrow \cdots \\ X & & & & Y & & & & X \\ \end{array}$$ and its colimit.

The two-ended construction you describe is (if I understand you correctly) a colimit for a different diagram. The fact that you'll get the homeomorphic spaces in the colimit spot for these two diagrams is a nontrivial fact, and is where you need to use the hypothesis $X\sqcup_{S^n} Y \simeq I\times S^n$.

In summary, to test your intuition, pay attention to what limit (or limits) you're considering, and then see if additional hypotheses let you decide on or against equivalence.

Answer by some guy on the street for A conceptual proof that local fibrations over paracompact spaces are global fibrations?

2010-11-13T20:44:14Z

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-finite cover $V\to B$ such that the pullback $$\begin{array}{ccc} E' & \rightarrow & E \\ \downarrow & & \downarrow \\ V & \rightarrow & B\end{array} $$ has $E'\rightarrow V$ a Hurrewicz fibration. Fix a lifting problem $X\to E$, $X\times I\to B$. Then consider also the pull-backs $$\begin{array}{cccccc} X' & \rightarrow & X & H &\to & X\times I \\ \downarrow & & \downarrow & \downarrow & & \downarrow \\ E' & \rightarrow & E & V & \to & B \end{array} $$ and note that $$ \begin{array}{ccc} X' & \to & X \\ \downarrow & & \downarrow \\ V & \to & B \end{array} $$ is again a pull-back. Since $V\to B$ is a locally-finite covering, so are all the horizontal maps in these pull-backs. Because $H$ is defined by a pull-back, the composites $X'\to X\to X\times I$ and $X'\to V$ give a unique map $X'\to H$ such that $$ \begin{array}{ccc} X' & \to & E' \\ \downarrow & & \downarrow \\ H & \to & V \end{array} $$ commutes. Now, the structure of $H$ is as a locally-finite cover of $X\times I$; in particular, every point of $X'\to H$ inhabits an open set in $H$ of the form $U\times [0,a)$; each of these defines a lifting problem, which is solvable by construction of $E'\to V$.

My instinct at this point is to appeal to local-finiteness again, using the compactness of $I$ to solve finitely-many lifting problems in $E'\to V$, hoping they glue together properly; which, admitedly, was the problem to start with; only now it looks more hopeful: in particular, I expect we can claim a lemma that for $E'\to V$ a Hurrewicz fibration, we can solve not only the initial-value lifting problems, but the initial-closed-interval lifting problems: for $\alpha:X\to [0,1]$ continuous, $$ \begin{array}{ccc} X\times[0,\alpha] & \to & E' \\ \downarrow & \nearrow & \downarrow \\ X\times I & \to & V \end{array} $$

It might be worth-while noting that, because $V\to B$ is locally-finite, the nerve of the cover has finite-depth on a neighborhood of every point. We might also save some work by using a universal lifting problem, as in Hurewicz Connection at the nCatLab.

Answer by some guy on the street for What is your favorite isomorphism?

2010-09-02T22:36:43Z

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

Answer by some guy on the street for A canonical and categorical construction for geometric realization

2010-08-25T21:18:49Z

I'd just like to point out that there is a monad on $Top$, (which in the homotopy category looks rather dull,) assigning to each space $X$ its cone $CX$, the mapping cylinder of $X\to * $. The unit map is the inclusion of $X \to CX$, and the composition $CCX\to CX$ may as well be the map $ [[x,s],t]\mapsto [x,s+t-ts].$ (This ugly formula is just a natural obfuscation of the heuristic description of $CX$ as the union of convex combinations of points $x\in X$ and the new point $*$.) Another way to think of it is that $CX$ is the underlying space of the free contraction of $X$.

The topological (realization of the) simplex category is just the orbit of a one-point space $\star$ under this monad, together with the maps derived from the monad and the maps $C\star\to \star$ and $\star\to *\to C\star$. I think this gets at what Grigory M means above by "most natural" contractible set on $n$ points. Somewhere in this nonsense I should say "Bar construction", but I can't remember precisely where.

Answer by some guy on the street for Efficiently sampling points uniformly from the surface of an n-sphere

2010-05-15T04:57:46Z

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 $y\in [-1,1]$ with density $k (1-y^2)^{(n-2)/2}$ for the appropriate constant $k$. Letting $\mathbf{x}'=(y,(1-y^2)^{1/2} \mathbf{x})$ is a uniform point on the $n$-sphere.

Alternatively, in the unlikely event that you don't mind funny correlations between consecutively generated points, you might prefer randomly generating elements of the orthogonal group, for instance letting $A_n$ be sufficiently random${}^1$ antisymmetric matrices, $O_n=(I-A_n)^{-1}(I+A_n)$ is orthonormal, and then the sequence of matrices $U_0=I$, $U_{n+1} = O_n U_n$ is uniformly distributed w.r.t. Haar measure. If later you want to kill consecutive correlations, generate a bunch of them and then sample the sequence at random.

${}^1$: "Sufficiently random": Leveraging the mean-value theorem for all it's worth, we might even take the $A_n$s concentrated on a "generic curve in general position". For sure this isn't efficient, but it gives an idea of how little is needed to get quantitative convergence to uniformity.

Answer by some guy on the street for What do people mean by "subcategory"?

2010-05-12T21:59:07Z

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 use "subcategory" to mean "a subset of $\operatorname{Hom}$, closed under $1_{-}$ and composition." Then you can suitably mimic Mac Lane's definition by specifying $$\operatorname{SubC}(A,X)=\operatorname{SubC}(X,A) = \left\{\begin{array}{c} \{1_A\} & A=X \\ \emptyset & A\neq X \end{array} \right. $$ for any objects $A$ you might want to ignore; they become isolated and trivial. That does feel a bit kludgy, though.

or do they also require full? containing all the automorphisms?

Probably not, in either case. That just looks weird to me... but what do I know?

Are there other useful intermediate notions?

Definitely.

Mac Lane's definition of subcategory given in the question corresponds to a functor which is strongly faithful in the sense that $F(g)=F(h)\implies g=h$ without hypotheses on the sources and targets of $g,h$; this strong property comes from the fact that functors don't generally have sensible image categories: what do you do if all objects are mapped to the same object!?

The ordinary sense of faithful is a slightly less strict condition: if $f,g:S\to T$ and $F(f)=F(g)$ then $f=g$.

Consider functors of groupoids $F:\mathcal{A}\to\mathcal{B}$. Every such functor factors in an essentially unique way as $$ F = G_3 \circ G_2 \circ G_1 $$ where $G_i$ omits only property $i$ among

Faithful
Full
Essentially Surjective

The same construction${}^1$ that gives this factorization makes good sense for general categories as well, although it's then complicated by other functor properties you might want to consider (reflects isomorphisms, reflects isomorphy, etc.). $G_3$ might be called "full objectwise-subcategory" (the reference calls it "forgets only property") and $G_2$ ("only structure"), I think, generalizes the notion I describe at the top of this answer.

(To be clear, "$G:\mathcal{X}\to \mathcal{Y}$ is essentially surjective" means that every morphism of $\mathcal{Y}$ factors as $i\circ G(\varphi) \circ j$ for isomorphisms $i,j$ in $\mathcal{Y}$. This implies a property of $G$ relative to objects which isn't worth spelling out here.)

${}^1$take a day to read this page

Answer by some guy on the street for What is the earliest definition given by a universal mapping property?

2010-05-04T18:08:18Z

I'm betting on Supremum.

Answer by some guy on the street for How should one present curl and divergence in an undergraduate multivariable calculus class?

2010-04-19T21:07:35Z

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:

Theorem (Stokes) Given a differentiable vector field $X$ on a region $U$ of $\mathbb{R}^3$ there is a unique continuous vector field $\operatorname{curl} X$ such that for any regularly parametrized surface $(u,v):D^2\rightarrow U$ with normal field $\hat{\mathbf{n}}$ and boundary tangent field $\mathbf{s}$, the integrals $$\iint (\operatorname{curl} X)\cdot \hat{\mathbf{n}} dA $$ and $$ \oint X \cdot \mathbf{s}\ dt $$ are equal

From there you can proceed by deducing properties of $\operatorname{curl} X$ sufficient to give its formula in coordinates and at the same time prove the theorem.

Note for instance that even stated only as an existence theorem, it already says there's a sufficient local criterion for local integrability of a vector field; the actual formula for curl then tells you what the criterion is.

This style of approach also gives you a quick proof that $\operatorname{div}\operatorname{curl}(X)=0$, because a sphere can be regularly parametrized by a disk such that $\mathbf{s} \equiv 0$.

(*) Here I mean roughly that if you show them sufficient variations of the mean value theorem to prove that itterated derivatives commute when they're continuous, then it should be feasible to give this construction with comparable rigor.

(**) none of these words used here with any standard mathematical sense.

Answer by some guy on the street for Subset of the plane that intersects every line exactly twice

2010-04-15T16:56:02Z

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, and let $B_l$ be the set of lines spanned by points of $A_l$, and let $C_l=\cup B_l$. Suppose further that $l'\prec l$ implies $l'\in B_l$. Note that $|A_l| \leq |B_l| < |l|$, so that $$|C_l\cap l| = |\cup_{l'\in B_l} l\cap l'| < |l| .$$

If $l\in B_l$, let $A_{Sl} = A_l$.
If $a\in l \cap A_l$, take the minimal point $b\in l\backslash C_l$ and let $A_{Sl} = A_l\cup \{b\} $.
If $A_l\cap l = \emptyset$, take the minimal two points $a,b\in l\backslash C_l$, and let $A_{Sl}=A_l\cup\{a,b\}$.
Otherwise if $l'\ $ is a limit ordinal, let $A_{l'} = \bigcup_{l\prec l'} A_l$.

It is easy to check that $A_{l'}$ has no three points colinear --- they'd have to all be in some $A_l$ for $l\prec l'$. The final union $\bigcup_l A_l$ has exactly two points on each line $l$.

Answer by some guy on the street for Is there a category-theoretic definition of the arithmetic Grothendieck group

2010-04-12T17:35:23Z

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 furthermore every complex is equivalent to a two-term complex, a.k.a. a bundle morphism; the graded tensor product of complexes also gives a ring structure on $K_0$. If you like, classical $K_0$ is a categorical interpolation between the Euler Characteristic and the homology of a chain complex. The present construction looks like a refinement of that idea, where instead of representing a trivial element, an exact sequence is equivalent to a particular (sum of) differential form(s).

Answer by some guy on the street for How to motivate the skein relations?

2010-04-05T18:31:50Z

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 gadget determined by a link-diagram, AND you know that it satisfies a skein relation, THEN that information may be enough to prove that your gadget is isotopy-invariant, because if you're lucky, you can relate the Reidemeister moves to a very few skein-relation applications. Of course, this may not always be the case, but it's handy when it is.

Answer by some guy on the street for Do the empty set AND the entire set really need to be open?

2010-03-25T05:52:39Z

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 neighborhoods of every point --- then a function $f:X\rightarrow Y$ is continuous iff for all $x\in X$ and every neighborhood $V$ of $f(x)$ there is a neighborhood $U$ of $x$ such that $f(U)\subset V$. You'll remember this as the "epsilon-delta"-style of continuity criterion from calculus, only without mentioning $ε$ or $δ$. The viewpoints are equivalent in the sense that one can define an open set as being a neighborhood of all its points, or contrariwise define a neighborhood of $x$ as including some open set containing $x$.

There's another alternative, to study presentations of topologies; usually a base for a topology is given. These have the advantage of being available as raw data, because the closure requirements for a base are much less stringent from a set-theoretic point of view than those of the whole system of open sets; one consequence is that you can study the topological space $(X,\langle B\rangle)$ in any universe including $X$, $B$ and $(X,B)$. This notion of a presented topology then lets you compare how the properties of a space-given-the-base change after a forcing extension of the universe --- if you're into that sort of thing.

The moral of this story is that open sets are not really the object of study in topology, and changing what you mean by "open set" is mostly going to distract you; continuous functions are one object of study --- as others have pointed out --- and there are many ways to define those.

Comment by some guy on the street

2013-05-09T18:26:38Z

I think that's what I was trying to say... but I'd have quibbled $\Sigma a_{ij} v_j$; and now I'm convinced that there are always exactly $n$ such $[a_{i\dot}]$; and so the interest is in how might this fail to give a basis for $\mathbb{Z}^n$... I'm now convinced this gives a list of all solutions in every case; I still like that Kevin's answer gives a specific solution in all dimensions, which is closer to what I was wondering about, but this is really nifty!

Comment by some guy on the street

2013-05-09T15:34:22Z

could you explain the "$\Sigma a_i v_i=u_i"? I can see that there are finitely many HNFs to check, and that they're enough... are we basically trying to see if the cube spanned by an HNF contains enough independent integer points?

Comment by some guy on the street

2013-05-09T12:06:40Z

And it turns out that the Cartan matrix for $A_n$ is always a solution; this explains the bases already in the $n=1,2$ case; the haphazard found base for $n=3$ I suspect has the noncyclic group of order $4$ as quotient.

Comment by some guy on the street

2013-05-09T01:50:54Z

I also want to say that this really is quite beautiful. Thanks!

Comment by some guy on the street

2013-05-09T01:44:12Z

OK, I think I believe you, but let me check this...

Comment by some guy on the street

2013-05-08T14:30:53Z

or else I could have omited "translates of"

Comment by some guy on the street

2013-05-08T14:30:29Z

The intent is, pick one such simplex, and cover $Z^n$ with translates of that.

Comment by some guy on the street

2013-05-08T13:46:45Z

@Gerry, Ben has it right; I should have said "minimal $n$-simplex", because they're all $SL_n$-the same... in fact, I think I will.

Comment by some guy on the street

2013-04-23T20:49:39Z

I think you are looking for Spectral Sequences (which thought makes me feel somewhat gloomy...) on which there is a thorough reference by McCleary. Good luck to you!

Comment by some guy on the street

2013-04-02T20:24:47Z

ahem, [nlab entry](ncatlab.org/nlab/show/bar+and+cobar+construction)

Comment by some guy on the street

2013-04-02T20:22:35Z

... are you wanting to update the nLab entry? or is it something else?

Comment by some guy on the street

2013-04-01T01:05:45Z

all spaces are $A_1$; H means $A_2$; $A_n$ is the playground Stasheff built, and where associahedra are from; he constructs a family of $H$ spaces that are $A_p$ but not $A_{p+1}$ for many (I think prime?) values of $p$ (or $p+1$, perhaps... this paper is sitting in my computer, why don't I look it up?!) Anyways, Stasheff, Transactions AMS Vol 108 No.2 pp 275-292

Comment by some guy on the street

2012-12-19T03:06:52Z

@André, you know, I mean it feels a bit like cheating...

Comment by some guy on the street

2012-12-19T03:05:42Z

@Jim: what I mean is, this is the sort of picky question that I'm sure, if settled, was only done recently and I furthermore don't know who would write about it, nor what keywords to put into mathscinet/arxiv/google scholar search etc. Eight years ago or so my first move would have been to ask Dani Wise, just because he taught me about Gromov hyperbolic groups and the Rips complex, but he's not handy right now and I can't get to a library.

Comment by some guy on the street

2012-08-13T15:54:24Z

True. Choosing one convention or the other is a matter of poetics; the question struck me as being one of poetics at the beginning, anyway. However: it seems that one shouldn't call any category a "free coequalizer", because the thing that comes to mind admits functors that aren't coequalizer of anything, but it nearly is "the shape of a coequalizer" in a sensible way. This distinction may be artificial, however, since being a coequalizer is a property, not just stuff.