30 votes
Accepted

If homotopy groups of spaces are identical, then stable ones are also identical?

No, a counterexample is the rational sphere $S^{2n}_\mathbb{Q}$ and $K(\mathbb{Q},2n) \times K(\mathbb{Q},4n-1)$. By the work of Serre these have the same homotopy groups, though it is easy to see ...
Connor Malin's user avatar
  • 5,191
26 votes

What are some toy models for the stable homotopy groups of spheres?

My favorite warmup example to the stable homotopy groups of spheres is the following differential graded algebra. Let $A$ have the underlying ring $$ \Bbb Z[y] \otimes \Lambda[x], $$ a ring with a ...
Tyler Lawson's user avatar
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25 votes

Can a simply connected manifold satisfy $𝑀\simeq 𝑀\times 𝑀$?

Lemma. If $A$ is an abelian group satisfying $A\otimes A=0$ and $\mathop{\rm Tor}(A,A)=0$ then $A=0$. Proof. Since $\mathop{\rm Tor}$ is left exact on abelian groups, an inclusion of a finite cyclic ...
Dave Benson's user avatar
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22 votes
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Useful ideas in category theory which violate the principle of equivalence

I would think about this question in this way: If you have a construction that violates the equivalence principle then either (A) it is a strictified or simplified version of something that is ...
20 votes
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On the connections between condensed mathematics and homotopy theory

The way in which "condensed sets are similar to topological spaces" is very different from the way in which "$\infty$-groupoids are similar to topological spaces". In fact, ...
Maxime Ramzi's user avatar
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19 votes
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What is known about the homotopy type of the classifier of subobjects of simplicial sets?

It’s not hard to check that the subobject classifier $\Omega$ is trivially fibrant, and so its homotopy type is trivial. Orthogonality of $\Omega \to 1$ against a map $i : A \to B$ corresponds to the ...
Peter LeFanu Lumsdaine's user avatar
17 votes
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Where does the definition of ($\infty$-)groupoid cardinality come from?

I'll restrict to $\pi$-finite spaces (where the definition is guaranteed to make sense). Then homotopy cardinality is multiplicative in fiber sequences: if $E \to B$ is a fibration with connected base ...
Phil Tosteson's user avatar
16 votes

Why the stable module category?

To question 1: One big motivation for me is that two Frobenius algebras can be stable equivalent but not Morita equivalent and a classification up to stable equivalence can be very nice. For example a ...
Mare's user avatar
  • 25.8k
15 votes

What are some toy models for the stable homotopy groups of spheres?

You could say that I've made a living out of looking at the stable module category of a finite group (or rather its slight enlargement, the homotopy category of complexes of injective modules, $\...
Dave Benson's user avatar
  • 11.6k
14 votes

For which spaces $S^n$ ($n\geq 2$) is a universal covering space?

The answer is quite complicated. To begin with, the universal cover of your space $X$ is a sphere $S^n$ with a free action of a finite group $G=\pi_1(X)$. The group $G$ has to have periodic cohomology....
Dave Benson's user avatar
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13 votes
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Homotopy groups of categories of elements as higher colimits

To answer these questions, the best is to note that $|\int_C D|$, the geometric realization of this total category, is equivalently the colimit of $D$, viewed as a functor with values in the $\infty$-...
Maxime Ramzi's user avatar
  • 13.3k
13 votes

Homotopy type of the geometric realization of a poset

Once more, with feeling. Thanks to comments from Tyler Lawson and Neil Strickland. Let me use $B_n$ for the finite Boolean lattice of subsets of $[n] := \{1,2,\ldots,n\}$ (your $\mathcal{P}(S)$), and ...
Sam Hopkins's user avatar
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13 votes
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Categorical equivalences vs. categories of simplices

No it is not. But there is a replacement for this. There is another model of $\infty$-categories: marked simplicial sets. If $X$ is a simplicial set with a set of marked $1$-simplices $S$, the fibrant ...
D.-C. Cisinski's user avatar
13 votes
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Homotopy groups of finite CW complex finitely generated as Lie algebra

I think Ian Leary's answer to this question gives a counterexample. His construction shows that for every $k\ge 2$ there exists a group $G_k$, and a finite $k$-dimensional CW complex $X$ such that $\...
Gregory Arone's user avatar
13 votes

Categories on which one can determine all model structures?

If I remember correctly, about ten years ago I calculated the following: if $A$ is an Artinian commutative ring in which the square of each maximal ideal is zero, then the category of $A$-modules has ...
user509184's user avatar
13 votes
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Plus construction on Simplicial Sets?

The answer is yes. This is spelled out in the book The local structure of algebraic K-theory by Bjørn Ian Dundas, Thomas G. Goodwillie and Randy McCarthy. Check out Section 1.6.1 on page 26, where ...
David White's user avatar
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12 votes

For which spaces $S^n$ ($n\geq 2$) is a universal covering space?

A nice and quick survey on the groups acting freely on the sphere is given in chapter 3 of https://webusers.imj-prg.fr/~bernhard.keller/ictp2006/lecturenotes/skowronski.pdf . Theorem 3.26 gives a nice ...
Mare's user avatar
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12 votes

The image of the J-homomorphism of the tangent bundle of the sphere

There is probably more than one good way to look at this. I'll point out that $\tau_{S^n}$ is the boundary of a generator of $$ \pi_n(SO(n+1),SO(n))\cong \pi_n S^n. $$ The map $$ \pi_jS^n\cong \pi_j(...
Tom Goodwillie's user avatar
12 votes

Is a positive degree self map on a Riemann surface homotopic to a holomorphic self map?

The answer is "yes" if $S$ is the Riemann sphere. This is because a map $f$ of degree $d$ from the sphere to itself is homotopic to $z \mapsto z^d$. The answer is "basically no" ...
Sam Nead's user avatar
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12 votes

Categories on which one can determine all model structures?

Yes, this has been done in other settings. For example, Scott Balchin, Kyle Ormsby, Angélica M. Osorno, and Constanze Roitzheim wrote a paper, Model structures on finite total orders, that enumerates ...
David White's user avatar
  • 29.4k
11 votes

Why the stable module category?

One reason is just that $\text{stab}_{kG}(k,M)_*=\widehat{H}^{-*}(g;M)$ (the Tate cohomology of $G$ with coefficients in $M$). I think that Tate invented Tate cohomology for applications in Class ...
Neil Strickland's user avatar
11 votes

Big list: barycentric subdivision of simplicial sets

An important theoretical application is Kan's fibrant replacement functor $\def\Ex{{\sf Ex}}\def\Exi{\Ex^{\sf\infty}}\Exi$, defined as the filtered colimit of functors $\Ex^n$ ($n≥0$), where $\Ex$ is ...
11 votes
Accepted

Homotopic but not equivariantly homotopic maps

For any $G$-space the $G$-equivariant maps $[EG,X]_G$, also known as the homotopy fixed points $X^{hG}$, are a Borel homotopy invariant of $X$ meaning that it is an invariant of $G$-equivariant maps ...
Connor Malin's user avatar
  • 5,191
11 votes
Accepted

Is an exponentiable fibration with contractible fibers a homotopy equivalence?

The answer is no in general. A counterexample is the codiagonal functor $\Delta^1 \cup_{\partial\Delta^1} \Delta^1 \to \Delta^1$. This is an exponentiable fibration (as is every functor with codomain $...
Alexander Campbell's user avatar
11 votes
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Model categories as a tool to resolve size issues for localizing categories

I guess I'm the canonical person to answer this question. I wrote those notes as a PhD student, a long time ago, to go along with a talk I was giving at a grad student conference. They were basically ...
David White's user avatar
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10 votes

Besides $F_q$, for which rings $R$ is $K_i(R)$ completely known?

In the ten years since this question was asked, there has been a lot of progress in algebraic $K$-theory. For example, Achim Krause, Ben Antieau, and Thomas Nikolaus came up with an algorithm to ...
David White's user avatar
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10 votes
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Can we prove that any number of algebraic data cannot be a complete invariant of a homotopy type?

I'll answer the corresponding question for the homotopy category $\mathcal{S}$ of spectra. I doubt that this makes much difference, but I have not checked the details. We can choose a list $X_0,X_1,\...
Neil Strickland's user avatar
10 votes
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Non-triviality of Whitehead products in wedges of CW-complexes

Suppose $\pi_m(X)\cong \mathbb Z/p$ and $\pi_n(Y)\cong \mathbb Z/q$, where $p$ and $q$ are distinct primes. Then $[a, b]=0$, by bilinearity of the Whitehead product. On the other hand, if $a$ and $b$ ...
Gregory Arone's user avatar
9 votes
Accepted

Who introduced the notion of 2-categories?

It appears that the definition of 2-category was introduced independently by two authors, both of whom independently introduced the modern notion of enriched category, for which 2-categories appeared ...
varkor's user avatar
  • 8,675
9 votes
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Koszul duals of n-fold loop spaces

I am not sure if this gives what you want, but maybe it is: I went in the other direction in a paper The McCord model for the tensor product of a space and a commutative ring spectrum, in Progress in ...
Nicholas Kuhn's user avatar

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