algebraic topology – show that the lifting of a constant map is constant.

Let $p : mathbb{R} to S^1$ be a covering map then show that the lifting of a constant map is constant.

My attempt : By using the lifting property , we have map $ widetilde{f} : (0,1) to E$ such that $pcirc widetilde{f} =f$

$f$ is constant $implies p$ is constant $implies widetilde{f}$ is constant

since $widetilde{f}=p^{-1}$ is constant (inverse image of an constant is constant).

at.algebraic topology – Available frameworks for homotopy type theory

I am thinking about trying to formalise some parts of classical unstable homotopy theory in homotopy type theory, especially the EHP and Toda fibrations, and some related work of Gray, Anick and Cohen-Moore-Neisendorfer. I am encouraged by the successful formalisation of the Blakers-Massey and Freudenthal theorems; I would expect to make extensive use of similar techniques. I would also expect to use the James construction, which I believe has also been formalised. Some version of localisation with respect to a prime will also be needed.

My question here is as follows: what is the current status of the various different libraries for working with HoTT? If possible, I would prefer Lean over Coq, and Coq over Agda. I am aware of, which seems moderately active. I am not clear whether that should be regarded as superseding all other attempts to do HoTT in Coq such as I am also unclear about how the state of the art in Lean or Agda compares with Coq.

gt.geometric topology – Philosophy behind Grigori Perelman’s proof of the Poincaré conjecture

The Poincaré conjecture states that:

Every simply connected, closed $3$-manifold is homeomorphic to the $3$-sphere.

Grigori Perelman presented a proof of the conjecture in three papers made available in 2002 and 2003 on 1,2,3.

It is far too difficult to understand his method for me. Can someone briefly explain the philosophy behind his work and comment on why it might be expected to shed light on the geometric topology and problems like the Poincaré conjecture?

at.algebraic topology – $mathbb Z$-formality of spheres

A topological space $X$ is $mathbb Z$-formal, if the singular cochain complex $C^*(X,mathbb Z)$ is
quasi-isomorphic to $H^*(X, mathbb Z)$ as an augmented differential graded ring.

It’s quite simple to write down specific quasi-isomorphisms to show that the Spheres $S^n$ are $mathbb Q$-formal spaces by fixing a volume form $v in Omega^n(S^n)$ and considering the maps $H^*(S^n)=operatorname{span}(1,(v)) to Omega^*(S^n)$ sending $1$ to the $1$-form and $(v)$ to $v$
and the canonical map $C^*(S^n) to Omega^*$.

Is it also possible to show the $mathbb Z$-formality of the Spheres $S^n$ by writing down specific quasi-isomorphisms, or is it easier to use another method for showing $mathbb Z$-formality?

general topology – why is deleted infinite broom not path connected

There are many posts in google about this. I am not from core mathematics background. Deleted infinite broom are lines connecting (0,0) to $(1,frac{1}{n})$ but point (1,0) is not included.

I know this is NOT path-connected but connected.

If i had to prove that it is not path connected, there should exist 2 points between which no path exists. But in proof why is that (1,0) is considered when it is not part of X. X = deleted infinite broom.

I understood defn of path connectedness as any 2 points x and y in X, there should be path P(which is continuous function from (0,1) to X)

algebraic topology – A question about retraction of the torus onto the cylinder.

Prove that the cylinder $S^1 times (0,1)$ is a retract of the torus $S^1 times S^1.$

I know that $S^1 times S^1$ is obtained from $S^1 times I$ by identifying $(x,0) sim (x,1),$ $x in S^1.$ From here how do I embed $S^1 times I$ inside $S^1 times S^1.$ My idea is to reflect the torus through $xz$-plane and then the corresponding orbit space is the cylinder. So cylinder can be thought of as embedded inside the torus if we identify it with one of the halves of the torus sitting in either side of the torus. Let the half be $A$ with which the cylinder is being identified. Let $A’$ be the portion of the torus obtained by reflecting $A$ through $xz$-plane. Then if let us take the map $r$ which is identity on $A$ and reflect every point of $A’$ to the corresponding point of $A.$ Then will $r$ work as the required retraction?

Any suggestion in thus regard will be greatly appreciated. Thanks!

general topology – How to prove the inverse limit $lim_{n>0} R/mathfrak{a}^{n}$ of a quotient of adic ring is Hausdorff?

I’m reading Bosch’s Lectures on Formal and Rigid Geometry. And I’m have some problems. Let $R$ be an adic ring. How to prove that the completion of $R$ constructed by dividing the ring of all Cauchy sequences in $R$ by the ideal of all zero sequences is Hausdorff? And consider the inverse limit $widehat{R}=lim_{n>0} R/mathfrak{a}^{n}$, where $mathfrak{a}$ is the ideal of definition of $R$, how to prove that it is Hausdorff?

reference request – The topology of the Deck group of covering map

I have never been interested in this before, and I have become interested in to find some answers and my teaching on the fundamental group has led me in this direction. Neither, I don’t know if it suits for MO. So, please down-vote or vote-to-close after you provided some references.

I think if $X$ is semi-locally simply connected, simply connnected, and locally path connected then for the universal cover $p:widetilde{X}to X$, its group of Deck transformations, say $mathrm{Deck}(p)$ acts properly discontinuously on $X$, hence it is equipped with the discrete topology. This then would imply that any covering over $X$ must have discrete group of Deck transformations. Indeed, this does not imply that it is finite, infinite or even countable. So, it could be $mathbb{R}^delta$ that is $mathbb{R}$ equipped with the discrete topology. Is this conclusion correct?

So, if I wish to find a covering whose group of Deck transformations has a non-discrete topology then either $X$, the base space of my covering map, should not be semi-locally simply connected ( like the infinite earring) or $X$ should not be either path connected nor locally path connected. I wonder if there is a place that I can look for examples of such coverings where the topology of the Deck group is determined. I know that in Munkres’s book there are some statements/exercises about this. But, anything more recent or some survey articles on this?

general topology – Prove that $Bbb{R}^2/Bbb{Z}^2approx S^1times S^1$

Here $Bbb{R}^2/Bbb{Z}^2$ is the quotient space obatined from $Bbb{R}^2$ by identifying points of $Bbb{Z}^2$ i.e. $(x,y)sim (x’,y’)iff (x,y),(x’,y’)inBbb{Z}^2$.

$S^1times S^1:={(z,w)| z,win S^1}$

I define $f:Bbb{R}^2to {S^1}times S^1$ by $f(x,y)=(e^{2pi i x},e^{2pi i y})$. $f$ is continuous and onto.

As $f(n,m)=(1,1) forall (n,m)in Bbb{Z}^2$ i.e. $f$ agrees on $Bbb{Z}^2$. By the property of quotient space, $f$ induces a continuous map $tilde{f}:mathbb{R}^2/Bbb{Z}^2to S^1times S^1$ such that $tilde{f}((x,y))=f(x,y)$. This map is onto as well. But this map is not injective. I couldn’t move forward from here.

Although I have observed one thing- instead of only identifying the points of $Bbb{Z}^2$ if we identify the points as follows-

$$(x,y)sim (x’,y’)iff x-x’,y-y’inBbb{Z}^2$$

Then we would have $Bbb{R}^2/simapprox S^1times S^1$, the same $f$ will give rise to this homomorphism.

Can anyone help me to solve the problem? Thanks for help in advance.

at.algebraic topology – Why does $iota_4^2 in H^8(K(mathbb Z/2,4);mathbb Z/2)$ not come from $H^8(K(mathbb Z/2,4);mathbb Z)$?

In Hatcher’s Chapter 5 ( on page 574 (page 57 in the pdf), he states that $iota_4^2 in H^8(K(mathbb Z/2,4);mathbb Z/2)$ is not in the image of $H^8(K(mathbb Z/2,4);mathbb Z)to H^8(K(mathbb Z/2,4);mathbb Z/2)$. The argument for the lower classes ($iota_4, Sq^2iota_4, Sq^2Sq^1iota_4$) which don’t come from $H^*(K(mathbb Z/2,4);mathbb Z)$ is that the Bockstein homomorphism $Sq^1colon H^*(K(mathbb Z/2,4);mathbb Z/2) to H^{*+1}(K(mathbb Z/2,4);mathbb Z/2)$ applied to those classes is nonzero, so it can’t come from $H^*(K(mathbb Z/2,4);mathbb Z)$. This argument doesn’t work for $iota_4^2 = Sq^4 iota_4$ because $Sq^1Sq^4 iota_4 = Sq^5 iota_4 = 0$.

How can one argue that $iota_4^2 in H^8(K(mathbb Z/2,4);mathbb Z/2)$ doesn’t come from $H^8(K(mathbb Z/2,4);mathbb Z)$? Is it even true?