User guillaume brunerie - MathOverflow

Orientation preserving self-homotopy equivalences of the 2-sphere

2013-03-23T05:16:13Z

According to this question, Hansen proved that the space $\mathrm{Aut}_0(\mathbb{S}^2)$ of orientation preserving self-homotopy equivalences of the 2-sphere is homotopy equivalent to $\mathrm{SO}(3)\times\mathbf{\Omega}$, where $\mathbf{\Omega}$ is the universal cover of the connected component of $\Omega^2(\mathbb{S}^2)$ containing the generator of $\pi_2(\mathbb{S}^2)$ (or any other component, they are all homotopy equivalent).

Let’s now consider the space $\mathrm{Aut}^{\mathrm{pr}}_0(\mathbb{S}^2)$ of pairs $(f,H)$ where $f$ is a self-homotopy equivalence of the 2-sphere and $H$ is ~~an homotopy between f and the identity function~~ a connected component of the space of homotopies between $f$ and the identity function. You can see $H$ as a "proof" that $f$ is orientation preserving and $\mathrm{Aut}^{\mathrm{pr}}_0(\mathbb{S}^2)$ is the space of orientation preserving self-homotopy equivalences $f$ of the 2-sphere where you do not throw away the proof that $f$ is orientation preserving (the "pr" stands for "proof relevant").

What is the homotopy type of $\mathrm{Aut}^{\mathrm{pr}}_0(\mathbb{S}^2)$?

One can reasonably expect that the answer is $\mathrm{SO}(3)$, the noise in $\mathrm{Aut}_0(\mathbb{S}^2)$ looks very similar to the missing homotopy, but I don’t know how to prove it.

Also, thanks to Springer I can’t find an online version of Hansen’s paper, so maybe this is already in his paper.

Homology theory constructed in a homotopy-invariant way

2011-10-18T12:33:32Z

Singular homology sends homotopic morphisms on equal morphisms and weakly equivalent spaces on isomorphic objects. So singular homology is in fact defined on the homotopy category of topological spaces.

But the usual definition of singular homology is on the category of topological spaces, and you can show that it is homotopy-invariant only after having defined it on the category of topological spaces.
For example, the definition uses the free abelian group on the underlying set of the space of singular $n$-simplexes, and taking the underlying set of a space do not make sense in the homotopy category.

I would like to have a construction of singular homology that can be entirely carried out in the homotopy category.

I was thinking of categorifying the usual construction:
Take the free spectrum ($(\infty,1)$ equivalent of abelian group?) on the space of singular n-simplexes, show that this is an "$(\infty,1)$ chain complex" and compute its "$(\infty,1)$ homology".

But I have the impression it will not work as is, because simplexes are only interesting as topological spaces, not as homotopy types.

Is there a way to construct singular homology (or in fact any homology theory) directly in the homotopy category without using the category of topological spaces?

Category-theoretic characterization of locally constant sheaves

2012-03-13T13:12:39Z

Let's consider $X$ a (locally connected) topological space and $\mathcal{Sh}(X)$ the topos of sheaves over $X$. If you see sheaves as étale spaces, locally constants sheaves correspond to covering spaces.

Is there an internal (topos-theoretic) characterization of the locally constant sheaves?

I've searched in Sheaves in Geometry and Logic, but I haven't found anything, and in the nLab there seems to be something but I do not understand it (and it does not really look like an internal characterization).

Answer by Guillaume Brunerie for Simple question about products in categories

2012-03-11T23:06:39Z

When first coming to the concept from the book days ago, I was confused if whether or not the auxilary object too was a product but I was confident that I understood that it too is a product,

If I've understood this sentence, this is what you are not understanding.

The definition of a product is that for any object $C$ and arrows $f:C\to A$ and $g:C\to B$, there is a unique arrow $(f,g):C\to A\times B$ making the diagram commute. In particular, the object $C$ has no reason to be a product. When he proves that products are unique up to isomorphism, he just applies the universal property taking for $C$ another product of $A$ and $B$, but in general you can apply it for any object $C$ (and to prove that something is a product, you have to verify it for every $C$)

In your example, the fact that you find that there is a unique arrow $(f,g)$ is reassuring, this is just a consequence of the fact that $Z_2\times{}Z_4$ is indeed a product of $Z_2$ and $Z_4$. If you want to prove that $Z_8$ is not such a product, you have to prove that for every pair of maps $pr_1:Z_8\to Z_2$ and $pr_2:Z_8\to Z_4$, there exists a group $C$ and maps $f:C\to Z_2$ and $g:C\to Z_4$ such that there is no arrow (or more than one) $C\to Z_8$ making the diagram commute.

Answer by Guillaume Brunerie for Is there a right adjoint to the contravariant functor Hom(-,B) in the category of Sets

2012-02-19T20:12:57Z

Left adjoints preserve colimits, but $\mathrm{Hom}(-,A)$ does not because $\mathrm{Hom}(\varnothing,A)\neq\varnothing$.

Right adjoints preserve limits, but if $A$ is not terminal, $\mathrm{Hom}(-,A)$ does not because $\mathrm{Hom}(\{*\},A)\neq\{*\}$ . (if $A$ is terminal, the functor is constant equal to $\{*\}$ and has a left adjoint which is the constant functor equal to $\varnothing$).

You can do the same reasoning to deduce that $-\oplus{}A$ does not have adjoints, unless $A=\varnothing$.

Answer by Guillaume Brunerie for Which functors between multicategories that come from monoidal categories are monoidal?

2012-02-13T00:23:27Z

I guess that you can take the image under $F$ of the identity in $M_C(;I)$ (the arity 0 is allowed).

Answer by Guillaume Brunerie for What do we mean by "Proving an algorithm"?

2012-02-12T17:23:49Z

It means to prove

that the algorithm terminates
that the answer given by the algorithm is correct

It is/seems sometimes obvious, but in general it isn’t.

Defining topological spaces with the notion of continuous path

2012-01-27T00:42:53Z

Let’s consider the following pseudo-definition of (nice) topological spaces : a space is a set $X$ together with distinguished paths $[0,1]\to{}X$ called continuous paths, distinguished maps $[0,1]\times[0,1]\to{}X$ called homotopies, and so on in every dimension (in a globular style), satisfying a bunch of properties. For example, the constant path should be continuous, the composition of two continuous paths is continuous, the slices of an homotopy are continuous paths, etc.

Is there a way to formalize precisely this definition, and how?

I apologize for the vagueness of the question.
And this is just a curiosity, I had to talk about topology to computer scientists a few days ago and the most intuitive definition of topological space I found is to say that a topological space is a set with a well-behaved notion of continuous path. And now I’m wondering whether this could be an honest definition of topological spaces.

Answer by Guillaume Brunerie for Algorithm on winning strategy of Winner (Simplified card game)

2012-01-28T07:37:40Z

What do you mean by best strategy? Given that the players don’t know the hand of the other player, it is impossible to know what will be a winning strategy.

For example, let’s suppose that I’m the first player and that my hand is {1, 4}.
If the hand of the other player is {4}, I win if I play 4 and I lose if I play 1.
But if the hand of the other player is {2, 3, 5}, I lose if I play 4 and I win if I play 1.

So if my hand is {1, 4}, there is no winning strategy, it depends on the hand of the other player.

(if I understood the rules correctly)

Edit: If you consider that both player know both hands, then the game is finite with perfect information, so you can draw the graph of all possible outcomes and determine recursively the winning and losing positions (as with all finite games with perfect information).

But perhaps there is a winning strategy more easily computable, I don’t know, I will try to see if there is one.

Fibration of Batanin/Leinster $\omega$-groupoids

2012-01-04T23:31:59Z

Is there (defined somewhere) a notion of fibration between two weak $\omega$-groupoids in the sense of Batanin/Leinster? I tried to search on Google and in Higher Operads, Higher Categories of Tom Leinster, but I haven't found anything.
This would probably be very useful for interpreting Martin-Löf type theory in the category of Batanin/Leinster weak $\omega$-groupoids.

Reduction rules for inductive types

2012-01-02T13:31:39Z

(I'm not sure if I should post this here rather than at Theoretical Computer Science, I've found a lot of type theory related questions on MathOverflow)

I'm working in Martin-Löf type theory with inductive types. Everything I say below for booleans should be understood with the type of booleans replaced by any inductive type, but for simplicity I'll do it in the case of boolean with match with written as if then else.

My question is about two reduction rules that I've never seen studied anywhere but that seems rather natural to me. I'd like to know if those rules have an "official" name and where I could read about them.

The first one (that I'm calling "lazy match") has the form

b : bool $\vdash$ if b then u else u $\rightarrow$ u : A

where A is a type, b : bool $\vdash$ u : A and "$\rightarrow$" is reduction (definitional equality, if you prefer).
I know there is a problem with this rule if b does not terminate, but I'm interested here in type theory as a logical system, so everything is supposed to terminate.

The second one (that I'm calling "deep match") is about exchanging two match and has the form

b : bool $\vdash$ if (if b then s else t) then u else v $\rightarrow$ if b then (if s then u else v) else (if t then u else v) : A

where A is a type, b : bool $\vdash$ s, t : bool and b : bool $\vdash$ u, v : A
Intuitively, when you match a match expression, the outer match can be distributed in the branches of the inner match.

Where can I read about these rules? Or are there obvious problems that I haven't seen? (there are perhaps problems with inductive predicates (as opposed to inductive types), but you can restrict it to inductive types)

Find weak equivalences from fibrations and cofibrations

2011-12-22T12:20:16Z

Let's suppose I have a category $\mathcal C$ with two weak factorisation systems $(C,F_W)$ and $(C_W,F)$, where $C_W\subset C$ and $F_W\subset F$.

I would like to have a model structure on $\mathcal C$ such that $C$ (resp. $C_W$) are the cofibrations (resp. the acyclic cofibrations) and $F$ (resp. $F_W$) are the fibrations (resp. the acyclic fibrations).

1. Is there always at most one possible choice for the class of weak equivalences? (I've heard that in a model structure, having two classes of arrows determines uniquely the third one, but I don't know how to do this when it's the weak equivalences that I want)

2. Under what conditions does a model structure indeed exists.

Answer by Guillaume Brunerie for undecidable sentences of first-order arithmetic whose truth values are unknown

2011-12-22T16:33:55Z

I think there is a little misunderstanding.

Paris and Harrington showed the strengthened finite Ramsey theorem is true but unprovable in first-order arithmetic; I don't know if there's a proof that extends the result to full-on undecidability rather than just unprovability.

Indeed, the Wikipedia page is only talking about unprovability, but the negation of the strengthened finite Ramsey theorem is also unprovable in Peano arithmetic for "trivial" reasons: if you can prove this negation, then second order arithmetic can also prove this negation (because second order arithmetic is stronger than Peano arithmetic), so this would mean that second order arithmetic is inconsistent (because second order arithmetic proves the strengthened finite Ramsey theorem).

So if you take for granted that second order arithmetic is consistent, then your example is actually undecidable in Peano arithmetic. And there are other examples like Goodstein's theorem.

Answer by Guillaume Brunerie for Characterization of Measureable Sets

2011-12-07T00:27:19Z

No, see here for a counter-example (it's a variant of the Cantor set which has non null Lebesgue measure and does not contain any interval).

And if you want a counter-example for $\mathbb{R}^2$ instead of $\mathbb{R}$, just cross it with an interval.

Functorial choice of pullbacks in a locally cartesian closed $(\infty,1)$-category

2011-12-03T20:23:33Z

In a locally cartesian closed category $\mathcal C$, for every map $f:A\to B$, there is an associated pullback functor $f^* : \mathcal C/B \to\mathcal C/A$. Moreover, if $g:B\to C$, the two functors $(g\circ f)^*$ and $f^*\circ g^*$ are canonically isomorphic, but they have no reason to be "equal". Even in the category of sets, the usual choice of pullbacks is only functorial up to isomorphism (see this paper of Hofmann)

This gives a pseudo functor from $\mathcal C^\mathrm{op}$ to $\mathcal{Cat}$. The fact that this is only a pseudo functor and not a functor causes some problems when trying to interpret dependent type theory in locally cartesian closed categories (see for example the previous paper of Hofmann, or this paper of Curien).

I was wondering what happens when you pass to $(\infty,1)$-categories. Intuitively, the fact that pullback are only functorial up to isomorphism should not be a problem anymore, because $(\infty,1)$-functors are also only functorial up to isomorphism anyway.

So my question is, if $\mathcal C$ is a locally cartesian closed $(\infty,1)$-category, does there always exists a functorial choice of pullbacks? (by which I mean an $(\infty,1)$-functor from $\mathcal C^\mathrm{op}$ to $(\infty,1)\mathcal{Cat}$ sending objects to slice categories and morphisms to pullback functors)

Local cartesian closedness in the category of compactly generated spaces

2011-12-02T21:07:18Z

According the the nLab, the category of compactly generated (CG) spaces is not locally cartesian closed. So if $A$ is a CG space and $C$ a CG space above $A$, $C$ may not be exponentiable.

What if we require that $C$ is fibrant over $A$?
If $C\to A$ is a (Serre) fibration, is $C$ exponentiable in the category of CG spaces above $A$?

Knots that turn around an axis

2011-11-25T22:56:02Z

Take a thick cord (the alim cord of your laptop or your mouse cord for example) and wrap it around your hand (or finger) turning always in the same direction but possibly knotting it. Then try to deform it using only the "obviously allowed moves", for exemple the first Reidemeister move is not allowed because it is difficult to perform it without "cheating".

More precisely, let's call a round knot a knot diagram which does not pass through $0$ and where the angle (in polar coordinates) of a point moving along the knot is always increasing.

I'm not sure of what should be the precise condition for two knots to be equivalent. For example I want to allow the second Reidemeister move, even if the result is not round anymore (but at the end, we must finish on a round knot of course). The idea is that only more or less planar moves are allowed, and Reidemeister 1 is not allowed because it involves untwisting a loop which is not really planar. Perhaps there are also new moves to be added, I'm not sure.

For example the torus knots are obviously round knots (and the torus knots with one parameter being 1 are trivial as usual knots but not at round knots), and the figure eight knot is also a round knot (and I haven't found a knot that cannot be represented as a round knot yet). There should also be an invariant which is the number of times the knot turns around 0.

Is there anything that can be said about these knots?

Do you have a more precise definition? Have they been studied? Are they classified?

Bijection between maps from a nice space to weakly homotopy equivalent spaces

2011-11-19T00:13:58Z

Let $S$ be a pointed topological space, that you can suppose nice (CW-complex, manifold, …), $X$, $Y$ be two (arbitrary) pointed topological spaces and $f:X\to{}Y$ a weak homotopy equivalence.

Let $[S, X]$ be the set of (pointed) maps between $S$ and $X$ up to homotopy, we have a canonical map $\phi:[S,X]\to[S,Y]$ (composition with $f$)

Is this map $\phi$ always a bijection? (and for what niceness of $S$?)

If $f$ is a homotopy equivalence (with a pointed inverse), then the answer is yes. If $S$ is a sphere, then the answer is yes by definition of weak homotopy equivalence, and if $S$ is the Hawaiian earring, then the answer can be no (for example take for $X$ the cone on the Hawaiian earring, see the introduction of http://arxiv.org/abs/1111.0731)

I’m asking this because I do not find very intuitive the definition of weak homotopy equivalence, we ask that $f$ is a bijection on the homotopy groups, but it would be equally natural to ask that $f$ is a bijection on maps from a torus, a lens space, or whatever.

Large cardinal axiom: everything that happen once must happen an unbounded number of times

2011-11-08T20:05:35Z

I remember reading something about a large cardinal axiom saying something like

If some cardinal $\kappa$ has some property $P$, then there should be a proper class of cardinals with the property $P$.

Of course this is inconsistent if you allow any property $P$, but this was used to justify for example the existence of uncountable inaccessible cardinals, because $\aleph_0$ is inaccessible, so there should be a proper class of inaccessible cardinals (because having only one inaccessible cardinal is not very homogeneous, and the ordinal hierarchy should be homogeneous).

Unfortunately I don't remember where I read that.

My questions are: Is there a precise formalisation of this axiom, and (if there is one) how high is it in the large cardinal hierarchy?

Answer by Guillaume Brunerie for 3-manifolds, cubes with handles

2011-11-01T20:43:23Z

This is not really an answer, but not really a comment either.

You can start from the classification theorem of compact surfaces (every compact connected 2-manifold is homeomorphic to either a connected sum of $g$ tori or a connected sum of $k$ projective planes, with a finite number of disks removed). To prove this you need the triangulation theorem, which is proved in Moise's Geometric Topology in Dimensions 2 and 3, and once you have a triangulation you can prove by hand that you will always get a connected sum of tori or projective planes.

After that, you can prove the result you want by hand by drawing nice pictures :)

Answer by Guillaume Brunerie for Is the category of rings co-well-powered?

2011-11-01T12:57:32Z

A category is locally small if for all objects $A$ and $B$, $\mathrm{Hom}(A,B)$ is a set. In the category of rings, $\mathrm{Hom}(A,B)$ is a subset of the set of all functions between the underlying sets of $A$ and $B$, so it's a set.

I don't understand what you are trying to do with your categories of monomorphisms and epimorphisms (perhaps you meant something else than "locally small"?)

Limits in an $(\infty,1)$-category

2011-10-30T19:37:57Z

In ordinary category theory, the notion of limit in a category $C$ is usually formulated with a category (of indices) $J$ and a functor $F:J\to C$ (a diagram in $C$), and a limit of this diagram is something satisfying some universal property.

In the context of quasi-categories (I looked in the nLab and in Higher Topos Theory), the category $J$ is replaced by a simplicial set $K$ and the functor $F$ is replaced by a map of simplicial sets $K\to C$ (a quasi-category is a particular simplicial set). But this definition is specific of quasi-categories, if you take another model for $(\infty,1)$-categories, it does not make sense to talk about a map between a simplicial set and an $(\infty,1)$-category.

Is there a definition of limits for $(\infty,1)$-categories independent of the model?

In particular, can we replace $K$ by an $(\infty,1)$-category and $F$ with an $(\infty,1)$-functor? If we can, why does everybody take a simplicial set instead (for quasi-categories)?

Homotopy groups of spheres in a $(\infty, 1)$-topos

2011-10-24T18:00:05Z

Let $H$ be an $(\infty,1)$-topos (seen as a generalization of the homotopy category of spaces).

You can define the suspension of an object $X$ as the (homotopy) pushout of $*\leftarrow X \to *$, hence you can define inductively the spheres $\mathbb{S}^n$ (the sphere of dimension $-1$ is the initial object of $H$ and the sphere of dimension $n+1$ is the suspension of the sphere of dimension $n$).

You can also define the loop spaces of a pointed object as the (homotopy) pullback of $*\to X \leftarrow *$. It will be itself pointed (because there is an obvious commutative diagram with a $1$ instead of $\Omega{}X$, so there is (I think) an arrow between this $1$ and $\Omega{}X$).

Then, given two integers $n, k$, you can define $\pi_k(\mathbb{S}^n)$ as the set of connected components (global elements up to homotopy) of the $k$-fold loop space of the $n$-sphere (I don’t know if this definition is one of the two described in the nlab)

Is there a natural group structure on $\pi_k(\mathbb{S}^n)$?

Is there something known about these groups in general?

For example,

Are they completely known for some $H$?
Is it always true that $\pi_k(\mathbb{S}^n)$ is trivial for $k<n$ and isomorphic to $\mathbb{Z}$ for $k=n$?
Are they isomorphic (or related in some way) to the usual homotopy groups of spheres?

Addition:

What if you assume that $H$ is a cohesive $(\infty,1)$-topos? (see here for the nLab page)

Answer by Guillaume Brunerie for topology by closed intervals on real line

2011-10-26T22:32:55Z

If you allow $[a,a]$ as a neighbourhood of $a$, then the topology is discrete and there is nothing more to say (in particular it is metrizable).

If you only allow non-trivial closed intervals, then you have a finer topology, because every $]a,b[$ is still open. But every open set for this new topology is also open for the usual topology (because there must be a non trivial closed interval around every point, so there is also a non trivial open interval), so the topology you get is in fact exactly the usual topology of $\mathbb{R}$.

Answer by Guillaume Brunerie for How has modern algebraic geometry affected other areas of math?

2011-10-06T00:25:00Z

The notion of (Grothendieck) topos is coming right from algebraic geometry, yet they are very useful in homotopy theory, see for example Lurie’s book "Higher Topos Theory".
In particular, the homotopy category is the archetypal example of $(\infty,1)$-topos.

Homotopic monoids and $A_\infty$ spaces

2011-08-17T15:33:55Z

Informally, an $A_\infty$-space is a monoid whose laws are only satisfied up to homotopy.

Let’s define now what I will call a "homotopic monoid" to be a space $M$ together with a point $e\in{}M$ and a multiplication $m:M\times {}M\to{}M$ with the monoid laws satisfied "up to a path". More precisely, if $F$ is the fibration over $M^3$ such that the fiber over $(x,y,z)$ is the space of paths (in $M$) going from $m(m(x,y),z)$ to $m(x,m(y,z))$, then I want a section of this fibration (the section is part of the structure of homotopic monoid). And of course I want the same thing for the laws with $e$.
At least from the point of view of homotopy type theory, this is a very natural homotopy-theoretic generalization of the notion of monoid (we just replaced equality by existence of a path).

My questions are:

Do those "homotopic monoids" have already been studied somewhere?
What is the relationship between an $A_\infty$-space structure and a homotopic monoid structure? I think it is easy to prove that every $A_\infty$ space has a homotopic monoid structure, but I’m not sure of the converse.

Is every closed set of Q² the intersection of some connected closed set of R² with Q²

2010-11-15T21:52:24Z

Let $F\subset\mathbb{Q}^2$ a closed set. Does there exists some closed and connected set $G\subset\mathbb{R}^2$ such that $F=G\cap\mathbb{Q}^2$?

For example if $F=\{a,b\}$ , you can take $G$ the reunion of two lines of different irrational slopes passing through $a$ and $b$. This is a connected set and the intersection with $\mathbb{Q}^2$ is $\{a,b\}$ because the slopes are irrationnals.

But I don’t know how to prove it in general (and I don’t know if it’s true). When there are many connected components this is not clear how to connect them without adding new rational points.

Answer by Guillaume Brunerie for Non-inversible monoids

2011-01-09T19:44:54Z

Let $M$ be a totally ordered set of cardinal $\ge3$ with a least element $\bot$, and we define a binary operation $x\cdot{}y=\max(x,y)$.

This is obviously a commutative monoid with $\bot$ as the identity element, but it is not invertible because if $c\lt{}b\lt{}a$ you have $a\cdot{}b=a\cdot{}c$ but $b\neq{}c$.

Forcing over an arbitrary model of ZFC

2010-12-07T00:01:51Z

I’m learning set theory and forcing in the (french) book from Jean-Louis Krivine “Théorie des ensembles”.

Given a countable transitive model $\mathscr{M}$ of ZFC together with a poset $P$, he constructs the model $\mathscr{M}[G]$ where $G$ is a $P$-generic in the ambient universe $\mathscr{U}$. The countability of $\mathscr{M}$ is essential in order for such a generic $G$ to exist.

But I saw several times in answers on MO that forcing could be defined over any model of ZFC and that, for example, CH and ~CH can both be forced over any model of ZFC.

My questions are:

How does this kind of forcing work? I guess that we cannot anymore suppose that there exists a generic $G$, so how is the new universe constructed? And what does the truth lemma becomes?
When do we need to use this kind of forcing? For example if I want to prove that some proposition $P$ is independant of ZFC, I can always assume that my initial model of ZFC is countable and “usual” forcing will probably be sufficient.

Any reference about this will be more than welcome.

Thanks.

Answer by Guillaume Brunerie for Problems where we can't make a canonical choice, solved by looking at all choices at once

2010-12-21T11:54:32Z

I don’t know if this is what you are looking for, but in linear algebra, if you have a finite dimensional vector space $V$ and $f\in\mathrm{End}(V)$, then in order to define the trace of $f$, you choose a basis $(e_1,\dots,e_n)$ of $V$ and define $\mathrm{Tr}(f)=\mathrm{Tr}(M_{(e_1,\dots,e_n)}(f))$

You have to make a choice but this definition does not depend on the basis.

Comment by Guillaume Brunerie

2013-03-23T05:35:15Z

Indeed, thanks, I changed the question to something less stupid.

Comment by Guillaume Brunerie

2012-12-09T21:56:51Z

Mathoverflow is not the place for homework problems (hint: take $y$ to be a big negative number)

Comment by Guillaume Brunerie

2012-11-05T00:28:46Z

@Najdorf Do you consider the law of excluded middle as being constructive?

Comment by Guillaume Brunerie

2012-11-04T16:14:22Z

@Andrej: I was assuming that if someone believe that GHC either holds or not then he probably also believe in excluded middle anyway. But I see your point, one can know that excluded middle is constructively problematic but not recognize a particular instance of it.

Comment by Guillaume Brunerie

2012-11-04T14:52:52Z

@Najdorf What is unreasonable about the fact that a finite set might have non-finite subsets? A finite set is a set $A$ such that there exists (constructively!) a natural number $n$ and a bijection $[n]\simeq A$. Now consider the subset $X$ of the singleton $\{0\}$ such that $X$ contains $0$ if and only if GHC holds. Classically $X$ is either empty or equal to $\{0\}$, but constructively there is no way to find a bijection between $X$ and some standard finite set $[n]$. A set which is not finite does not need to be infinite either.

Comment by Guillaume Brunerie

2012-04-29T17:33:27Z

Can the $(\infty,1)$-category of $\infty$-groupoids be presented by a model category in which all objects are fibrant-cofibrant?

Comment by Guillaume Brunerie

2012-04-22T02:25:50Z

If you allow a category to have a class of objects, you cannot even speak about the "class" of all categories, so you already have a problem with the objects of the category of categories.

Comment by Guillaume Brunerie

2012-04-09T11:36:03Z

"in ZFC, you have a concept of functions and relations as first order objects." What do you mean by that? The only first order objects in ZFC are the sets and nothing else.

Comment by Guillaume Brunerie

2012-04-08T02:46:31Z

Do you have an example of such a formal system?

Comment by Guillaume Brunerie

2012-04-07T23:57:33Z

The formula you want is $f(x)=1+\frac{14(x-m)}{M-m}$ where $m$ is the minimum value in the list and $M$ the maximum.

Comment by Guillaume Brunerie

2012-04-06T20:41:19Z

Well, I don't know, but it seems more plausible

Comment by Guillaume Brunerie

2012-04-06T20:23:21Z

$X$ will lie between some parallel lines through $a$ and $b$, not necessarily the lines orthogonal to $[a,b]$

Comment by Guillaume Brunerie

2012-03-29T19:41:07Z

I have often seen "$\alpha$-equivalence" instead.

Comment by Guillaume Brunerie

2012-03-26T21:42:05Z

@Martin @Tom: I do not understand, if you take any non trivial bipointed set there are a lot of different maps from it to $([0,1],0,1)$. Am I missing something?

Comment by Guillaume Brunerie

2012-03-26T04:10:28Z

I think you mean "terminal", not "initial".