at.algebraic topology – Spaces homotopy equivalent over the topologist’s sine curve

Consider $$T=left{ left( x, sin tfrac{1}{x} right ) : x in (0,1) right} cup {(0,0)}subset mathbb{R}^2$$
with its subspace topology.

Denote $p_0=(0, 0)in T, p_1=(1, sin 1)in T$.

A TSC-homotopy between continuous maps $f, g:Xto Y$ is a continuous map $H:Xtimes Tto Y$ such that $H(x, p_0)=f, H(x, p_1)=g$.

What can be said about pairs of spaces $X, Y$ such that there exist continuous maps $f:Xto Y, g:Yto X$ and TSC-homotopies $fcirc gsim mathrm{id}_Y, gcirc fsimmathrm{id}_X$?

gt.geometric topology – Are there simplicial spheres with “non-geometric symmetries”?

Let $Delta$ be a simplicial sphere, that is, a finite (abstract) simplicial complex whose canonical geometric realization $|Delta|$ is homeomorphic to a sphere $mathbf S^dsubsetBbb R^{d+1}$.

Question: Can the homeomorphism $phi :|Delta|tomathbf S^d$ be chosen in a way, so that all combinatorial symmetries of $Delta$ are realized geometrically?

That is, if $sigma :DeltatoDelta$ is a combinatorial automorphism of $Delta$ (a bijective simplicial map) I want there to be an isometry $smash{f_sigma:mathbf S^dtomathbf S^d}$ so that
$$phicirc sigma = f_sigmacirc phi.$$

You can think of this as a subdivision of $mathbf S^d$ that has the same symmetries as the abstract simplicial complex $Delta$. If we consider the sphere embedded in $smash{Bbb R^{d+1}}$, the isometries are exactly the orthogonal transformations restricted to $smash{mathbf S^d}$.

at.algebraic topology – Construction of equivariant Steenrod algebra

I am reading through the calculations in Hu-Kriz “Real-oriented homotopy theory and an analogue of the Adams-Novikov spectral sequence” and I’ve got a small problem in understanding the computations of coefficients of the “first version” of $C_2$-equivariant Steenrod algebra.

Notation:

  • $H$ is a $C_2$-spectrum constructed as follows: we take the naive Eilenberg-MacLane spectrum associated to the constant Mackey functor $underline{mathbb{Z}/2}$ and extend it to a complete universe.
  • $A_star=(Hwedge H)_star$ is a dual Steenrod algebra associated to $H$ ($RO(C_2)$-graded) and $A_ast$ is its part graded over the integers.
  • $H^f_star=(Hwedge EQ_+)_star$ – the coefficients of homotopy orbits.

Now in Proposition 6.6 the authors compute $A_star$ by smashing $Hwedge H$ with the isotropy separation sequence. This yields the exact sequence
$$
ldotsto (Hwedge Hwedge E_{C_2+})_starto A_starto (Hwedge Hwedgewidetilde{EC_2})toldots.
$$

The authors claim that $(Hwedge Hwedge E_{C_2+})_starcong A_astotimes H^f_star$. I cannot see why this is true; I suppose it might come from applying homotopy orbits spectral sequence, but it seems quite complicated to me.

general topology – Prove that $forall xin E, exists zin Fsubset E, inf_{yin F}|x-y|=|x-z|$, where $F$ is compact and $E$ is a normed vector space.

Let $(E,|cdot|)$ be a normed vector space and let $Fsubseteq E$ be a non-empty compact subset. We define the distance from $x$ to $F$ as
$$begin{align}
dcolon &Etimesmathcal{P}(E)setminus emptysettomathbb{R}\
&(x,F)mapstoinf_{yin F}|x-y|.
end{align}$$

Prove that
$$forall xin E,exists zin F,; d(x,F)=|x-z|.$$

Definition of compact set: A set $F$ is said to be compact if and only if every sequence of elements of $F$ has a subsequence converging on $F$.

From this, it follow that $d(x,F)=0iff xin F$ (this was a previous problem and it only relies on $F$ being closed).
I don’t really know where to start with the proof, any good hints (or incomplete work) would be great.

general topology – Existence of a Jordan path in a path-connected set.

A path in set $E$ is a continuous function
$$gamma:(a,b)longrightarrow E, $$
with $A:=gamma(a)$ its starting point and $B:=gamma(b)$ its ending point.
Any two points $A, B$ in a path-connected set can be connected by a path, with $A, B$ as its starting and ending points.
A Jordan path $gamma$ is a injective path.
Any two points $P, Q$ in a Jordan-path connected set can be connected by a Jordan path.


Question: For any path-connected set $E$, is $E$ always Jordan-path connected?


When there are at most countably many self-intersecting points, it’s possible to ‘cut’ the loop and make the path a Jordan one. But I got problems when there are uncountably many self-intersecting points on the path.

gn.general topology – prove that a sequence whose range is relatively compact admits a convergent partial sequence

gn.general topology – prove that a sequence whose range is relatively compact admits a convergent partial sequence – MathOverflow

at.algebraic topology – Proofs that the classifying space of Connes’ cycle category $Lambda$ is $mathbb C mathbb P^infty$

Connes showed in Cohomologie cyclique et foncteurs $Ext^n$ (1983) that the classifying space of his cycle category $Lambda$ is $mathbb C mathbb P^infty = B(S^1) = K(mathbb Z,2)$.

Connes’ proof is not quite as conceptual as one might like. He shows that $|Lambda|$ is simply-connected, and then computes its cohomology via an explicit resolution of the constant functor $Lambda to Ab$, $(n) mapsto mathbb Z$ via a double complex which is cooked-up by gluing together the usual way of resolving cyclic groups and the usual way of resolving simplicial objects, with a few modifications. He’s able to get the ring structure on the cohomology using an endomorphism of this double complex. The result follows because $mathbb Cmathbb P^infty$ is characterized among simply-connected spaces by its cohomology ring.

Surely in the last nearly 40 years new proofs that $|Lambda| simeq mathbb C mathbb P^infty$ have been found.

Question 1: What are some alternate proofs that $|Lambda| simeq mathbb C mathbb P^infty$?

Question 2: (somewhat vague) In particular, is there a proof which somehow uses an explicit $S^1$ action on some category or simplicial set related to $Lambda$?

general topology – Checking Existence of a Surjective Continuous Map between Spaces

I am studying topology and recently tried to solve these two questions concerning the existence of a surjective continuous map between spaces. But I am not sure how to construct one, or to claim that there is no such surjective continuous map. The questions(in the form of T/F question) are as follows:

  1. There exists a surjective continuous map from $S^2$ to $mathbb{R^2}$ where $S^2$ is the unit sphere $S^2 := {(x,y,z) in mathbb{R^3}: x^2+y^2+z^2=1}.$
  1. There exists a surjective continuous map from $S^1$ to $S^1 times S^1$, where $S^1$ is the unit circle $S^1 := {(x,y) in mathbb{R^2}: x^2+y^2=1}.$

From what I think, I am guessing question 2 is False. (One of my friend told me it seems false due to topological group… But as I did not learn the concept of topological group, I felt there should be some other way to solve this.) That is, there is no surjective continuous map between spaces. But overall, I am not sure whether these statements are true or not. Also, I am not sure of a way to construct a surjective continuous map for them.

Is there some method or way of constructing surjective continuous map in this kind of situations? (Or some criteria that can tell whether these kinds of map exist or not.)

Thank you.

at.algebraic topology – Isotopies, Fiber Bundles and Selection Theorems

The following problem is a culmination of a few questions I’ve asked the last two months, and it’s still giving me some issues. I think I know the right way to solve it, but I’m having trouble with the details; my idea can be formulated in terms of selection theorems or fiber bundles.

Let $X subset mathbb{R}^n$ be any subspace, and let $I = (0,1)$. By a proper isotopy of $X$ I mean a continuous function $F: X times I rightarrow mathbb{R}^n$ such that for each $t in I$, $f_t := F|_{X times lbrace t rbrace}$ is an embedding and $f_0 = text{id}_X$. By an ambient isotopy I mean an isotopy on all of $mathbb{R}^n$.

If $F$ is a proper isotopy of a tamely embedded copy $X$ of $mathbb{S}^{n-1}$ in $mathbb{R}^n$, does $F$ extend to an ambient isotopy?

As noted in the answer here, this will be true as long as $F$ can be extended in some neighborhood of $X$.

I’m mostly interested in the case for the plane. In fact, for the plane it’s already known to be true, but the proof is very difficult. There was a follow-up paper where they defined a notion of length for plane curves that extended to unrectifiable curves, and which behaved continuously with respect to uniform convergence of compact sets (so basically, they didn’t require 0-regular convergence); between the preprint, the Annals paper and the follow-up they gave three different arguments, but all hinged on analytically controlling crosscuts using geometric function theory. My idea is also to use cross-cuts, but not to construct an explicit isotopy.

Using results from the Kirby-Edwards paper linked in the previous MO thread, and a ‘canonical’ Alexander-Pontryagin Duality Theorem in the plane, you can prove the isotopy extension theorem for compact, connected subsets of $mathbb{R}^2$ in a different way from the case for the circle (though it’s also fairly complicated, and is ill-suited to the generalizations they obtained in the follow-up).

What I’d like to do is get the case for the circle using some selection theorem, esp. the Michael Selection Theorem (or even better, a selection theorem whose proof is actually reasonable). To do this, let $D$ be a large, closed ball around the trace of $X$ under $F$. By the Annulus Theorem, each region between $partial(D)$ and $f_t(X) := X_t$ is a closed annulus, call it $A_t$.

For any $A_t$ there are many ways to partition it into crosscuts, so that each crosscut has one endpoint on each boundary component. By a crosscut, I mean an embedded copy of $I$. Let $mathcal{C}_t$ denote the collection of such partitions on $A_t$, and let $mathcal{C} = cup mathcal{C}_t$. Then what we want is a continuous selection of cross-cut decompositions, one for each $A_t$. To be precise, we should probably consider a family of parameterized cross-cut decompositions, so that each has a time parameterization (that will be our way of getting around $0$-regular convergence issues).

This is equivalently a fiber bundle problem on $mathbb{A}^n times I$ in the following sense. We have two cylinders, one the usual smooth cylinder, and the other one just some Jordan mess, as the boundaries. Can we warp it into the standard, smooth representative in a way that’s slice?

Long story short, the problem for the selection method is:

How do you topologize $mathcal{C}$ to apply the Michael Selection Theorem?

The problem for the fiber bundle method is:

If $mathbb{A}^n times I subset mathbb{R}^{n+1}$ is a bundle over $I$ whose slices $mathbb{A}^n_t$ are contained in hyperplaces orthogonal to the $(n+1)$-axis, is there an isotopy to the standard smooth (thickened) cylinder that’s slice? In other words, $f_s(mathbb{A}^n_t) subset mathbb{A}^n_t$ for all $s$ and $t$.

Thanks, appreciate any help at all!

algebraic topology – Free Groups – representation of 1 by empty word is unique

I am studying Free Products of Groups from Munkres’ Topology book. The definition of free products is:

Let $G$ group, ${G_a}_{a in J}$ family of groups that generates $G$ and $G_a cap G_b$ consists of the idenity element alone whenever $a neq b$.
We say that $G$ is the free product of $G_a$, if for each $x in G$ there is only one reduced word that represents $x$.

After that it says and proves, that it suffices to know that the representation of $1$ by the empty word is unique.

What does exactly the latter means?
Isn’t the empty word unique and doesn’t always represents $1$?
By the previous argument it looks to me like the condition of the uniqueness of the reduced word to be redundant.

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