Fiber product formulae for surgery obstructions

Is there a formula for the surgery obstruction of a fiber product of maps assuming the fiber product is also a homotopy equivalence?

In more detail, suppose that $X to Y$ and $Z to Y$ are homotopy equivalences of (topological, say, compact oriented) manifolds. There is a long history of product formulae for surgery obstructions which includes a surgery obstruction for the product map $X times Z to Y times Y$ to be homotopic to a homeomorphism. See for example Section 8 in Ranicki, Algebraic Theory of Surgery II. Applications to Topology.

Suppose I know that the fiber product $X times_Y Z to Y$ is also a homotopy equivalence. (Often it won’t be, but in my case I got lucky.) Is there a formula for the surgery obstruction of this map in terms of those of $X to Y$ and $Z to Y$? I am actually just interested in the case $X times_Y X to Y$, perturbed to make the fiber product transverse and would like to know what are the possibilities for the surgery obstruction of this map. (Could it be anything?)

werewolf the apocalypse – On Werewolves and Plastic Surgery

A peculiar situation arose on a Werewolf: The Apocalypse game I’m taking part as secondary storyteller.

During a session conducted by our main storyteller, one of our players brought up the idea of her character – a female glass walker – getting a boob job in the hopes of squeezing out an extra appearance point or a positive quality out of her excess cash. While that was made as a joke, it soon gave birth to a rather long discussion of what would and would not happen to an urban garou that decided to get something “done” at the local doctor.

Since the pack is mostly urban-themed with a bunch of cash (almost a literal case of “Wolves of Wall Street”), this now is a thing that a few players are now seriously considering.

Thus, the multifold question.

  • Is it possible for a garou to get regular plastic surgery done at his or her local doctor?
  • If regular plastic surgery won’t cut it, would a garou or kin doctor with the proper gifts/rites be able to get the job done?
  • Are implant-based procedures possible at all?
  • If plastic surgery isn’t possible, is there a way of getting “cosmetic procedures” done in a Gaia-friendly way?

Is someone else’s surgery a legally permitted reason for international travel from the UK?

Your need to see a person before a surgery, or to bring some people to see that person, is not a medical emergency. Her need for the surgery, may be, but it is not her who wants to travel. I doubt you will get an exception.

Where you might have a chance is, it says on your link that you can travel for the same reasons you can leave home, and you can leave home

to visit someone who is dying or to visit someone receiving treatment in a hospital, hospice or care home, or to accompany a family member or friend to a medical appointment

Now, it also says you can leave home to exercise, and I’m sure you couldn’t get permission to fly to another country to exercise, so this isn’t a guarantee of success, especially since this “major surgery” is one that can be postponed a few times already.

I can’t see a way for you to apply for permission to travel; it looks like you just do it and perhaps argue with someone at the airport about it? This looks too risky to me.

My personal definition of emergency travel (under which I did travel internationally in November) is: does any cost, quarantine or other issue mean I won’t go, or will I go no matter what? I will go no matter what, then I go. In my case this involved two separate two week quarantines, one of which cost me over $5000, and Covid testing. Didn’t matter, I had to go. If your need is not that urgent, then you know you should not go.

eye tracking – How can we protect form Cataract Eye surgery?

Cataract when left untreated can lead to permanent blindness. We at the The Eye Foundation pave way for perfect vision through our efficient Cataract Eye Surgery. Cost of cataract eye surgery at The Eye Foundation is affordable and deliver great results. Pave way to clear vision with Cataract eye surgery at The Eye Foundation. Cataract is characterized by a painless, progressive decline in vision. Apart from decreasing vision patients with cataract even in the early stages may experience increased glare while driving at night or reading in dim illumination, difficulty in differentiating objects at a distance and the need for frequent change of glasses.

Will US Customs and Border Protection allow my girlfriend (UK Citizen) to come help me recover from surgery (covid-19)?

Earlier this year she did a 14-day layover in Bermuda to travel from the UK to the US and that worked fine. She wants to come back in November to help me recover from surgery. I have a letter from the doctor’s office.

Anyone heard of people doing this in past few months?

usa – Will US Customs and Border Patrol allow my girlfriend (UK Citizen) to come help me recover from surgery (covid-19)?

Earlier this year she did a 14-day layover in Bermuda to travel from the UK to the US and that worked fine. She wants to come back in November to help me recover from surgery. I have a letter from the doctor’s office.

Anyone heard of people doing this in past few months?

at.algebraic topology – Nullity of the linking matrix of a framed link $L$ equals the first betti number of the manifold obtained by surgery on $L$

I have asked this on mathstackexchange as well. I’m not necessarily asking for a proof, just a hint or a point to the right direction (although I’m not saying that a proof isn’t welcome). I’m studying Witten invariants and I have come across a statement that most books agree that it’s not hard to see. Anyway this is the question and my progress so far.

Let $L$ be a framed link of $m$ components and $(L_i,f_i)$ be the component-framing pair. The linking matrix $A=(a_{i,j})$ where $a_{i,i}=f_i$ and $a_{i,j}=lk(L_i,L_j)$ when $ine j$. I have read that this matrix is a presentation matrix for the first homology group $H_1(M; Bbb Z)$ and that the nullity of $A$ is equal to the first betti number of $M$, where $M$ is the 3-manifold obtained from $S^3$ by surgery on $L$. Can someone explain this fact to me or point me to a proof or a reference?

EDIT: I am reading Nicolaescu’s Reidemeister Torsion of 3-manifolds p.77 and there I made some progress but I have some question’s as well. Suppose the $c_i$ are the meridians of the solid tori we glue back on the link complement $E$ to obtain $M$. This means that we send $c_i to f_iμ_i + λ_i$, where $μ_i, λ_i$ are the meridians and longitudes in the regular neighbourhood of $L$. It is clear to me that $μ_i$ form a basis for $H_1(E)$ and taking $K$ to be the free abelian generated by the $c_i$‘s and sending them to $f_iμ_i + λ_i$ we have a morphism $K to H_1(E)$ with matrix the linking matrix mentioned above. There Nicolaescu says that by taking the natural map $a:H_1(E) to H_1(M)$ we complete the presentation as this map is onto. Here is when it’s unclear to me. By $a$ I guess he means the map that sends $μ_i $ in $E$ to $μ_i $ in $M$. If this is the case then I don’t understand why it’s onto and while it’s clear to me that the $f_iμ_i + λ_i$‘s will be in the kernel because now they bound a disk in $M$ I don’t see why they should generate it.

differential graded algebras – How does the knot contact homology augmentation polynomial change under a surgery on the base manifold?

So for every knot $K in S^3$, there is a knot contact homology of $K$, and we can find the augmentation variety for this homology. The defining polynomial is known as the augmentation polynomial $A(K)$, which, as Aganagic et al. conjectured, corresponds to the classical A-polynomial when setting $Q=1$.

Now my question is that, how does this $A(K)$ change when I do surgery on $S^3$. As a concrete example, I want to find $K=text{unknot}$ and the surgery is the contact surgery manifold obtained on a Legendrian unknots. It has Thurston-Bennequin number $mathtt{tb} = -2$ and surgery coefficient $-1$.

What I knew was that “contact” surgery with coefficient $-1$ is equivalent to attaching a Legendrian handle on the manifold, so in that case is simply a good old Legendrian surgery.

My thought was that this surgery changes the symplectization from simply $mathbb Rtimes S^3$ to $mathbb Rtimes M$, but now the picture looks like this:

surgery effect
In that first term what I need consider is to consider, instead of pseudoholomorphic discs, pseudoholomorphic curves with genus $g = 1$, which Ekholm recently discussed. This then would give the contact homology and eventually (somehow) this would give me the augmentation polynomial.

Is this correct? Or am I completely wrong? Thank you everyone!

language barrier – Planning to travel to South Korea for voice surgery, do you know of any good phrase cards?

So, once this COVID-19 situation dies down, I’m hoping to travel to South Korea to get voice surgery. Now, I remember hearing of these laminated cards that you could use to point to a phrase or something in order to help communicate. Do you know what these are called? (I called it a phrase card in the title of this post, but since NOTHING is coming up when I look for Korean phrase cards, I’m thinking it might be called something else). It might look something like this (but include a bit more information):

A picture of a page including phrases in English and Korean. On this particular page, it is a picture of phrases to use at restaurants and at the airport

dnd 5e – Creating an animal companion for my UA ranger. To surgery?

So my ranger asked me the other day if I could make her a snapping turtle for an animal companion (she uses the UA Revised Ranger Beast Conclave). After looking in the Monster Manual and online, I found nothing that really worked for me. I have created the following statistics block for a possible turtle.

Snapping turtle

Medium tier, any orientation


Armor class 14 (natural armor)
Hit points 30 (4W8 + 12)
speed 15 feet, swim 30 feet.


begin {array} {cccccc}
text {STR} & text {DEX} & text {CON} & text {INT} & text {WIS} & text {CHA} \
16 "(+ 3) & 8" (- 1) & 16 "(+ 3) & 2" (- 4) & 12 "(+ 1) & 6" (- 2)
end {array}


Save throws Str +5
Senses passive perception 11
languages – –
Challenge 1/2 (100 XP)


Hold your breath. The snapping turtle can hold your breath for 1 hour.

Actions

Snap. Melee weapon attack: Hitting +5 will reach 5 feet, a goal. Beat: 6 (1d6 + 3) piercing damage.

Headbutt. Melee weapon attack: Hitting +5 will reach 5 feet, a goal. Beat: 8 (1d10 + 3) stick damage. Deals 2 (1d4) recoil damage to the turtle. A critical bug results in 5 (2d4) damage to the turtle.

Shell Defense. Snapping turtle can retreat to its shell as an action. Until it appears, it gets a +4 bonus on AC and has advantages in throws that save strength and constitution. While the turtle in its shell is vulnerable, its speed is 0, and cannot increase, it has a disadvantage with litters that save skill and cannot react. The only action it can take is a bonus action that comes out of its shell.

Is it overwhelmed?