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• Jul 19 2020 10:05

Add link to web page Update README.md Fix link to t… Merge pull request #2 from boll… (compare)

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• Apr 02 2020 05:40

• Mar 31 2020 07:17

poincare duality notes notes reformatted, pictures (compare)

• Mar 30 2020 10:12

note sketch for duality lecture. (compare)

• Mar 28 2020 10:59

• Mar 27 2020 14:44

notes on free resolutions start of lecture draft (compare)

• Mar 26 2020 07:00

• Mar 26 2020 06:56

• Mar 25 2020 11:28

first online lecture made notes (compare)

• Mar 24 2020 04:05

tree homotopy picture Merge branch 'master' of https:… (compare)

• Mar 24 2020 01:11

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• Mar 24 2020 01:11
• Mar 24 2020 01:08 Chinmaya Kausik
@Chinmaya-Kausik
Yes, that would make sense. In general though, this particular homeomorphism works, right? You should be able to check. Chinmaya Kausik
@Chinmaya-Kausik
I think one can actually use the same embeddings to show things for the triple product too Let me try to do it the other way, which should be equivalent and give the same formula. By the way I want to avoid embeddings as they should not be needed. Chinmaya Kausik
@Chinmaya-Kausik

You should be able to check.

Yes, sorry. AR-MA210
@AR-MA210
@Chinmaya-Kausik At $s=0$, (assuming $r$ is the join coordinate and $r+s=1$, all points of $X$ are identified, so the map $rx+sy \to x$ is well defined (as a map on the join), right? I might be missing the issue Chinmaya Kausik
@Chinmaya-Kausik
And speaking of the answer, does Ronnie Brown's initial topology make sense?

@Chinmaya-Kausik At $s=0$, (assuming $r$ is the join coordinate and $r+s=1$, all points of $X$ are identified, so the map $rx+sy \to x$ is well defined (as a map on the join), right? I might be missing the issue

But not to $y$ na

We need all maps to be defined everywhere if I'm not wrong
To define an initial topology The triple product:

• We start with three spaces $X$, $Y$ and $Z$.
• We construct a quotient of $X\times Y\times Z \times \Delta$, with $\Delta$ a $2$-simplex.
• We take a quotient where at each vertex two of the three spaces are identified to points.

This is probably the formal linear combination approach you mentioned.

The remaining task is to map to the separate joins. Chinmaya Kausik
@Chinmaya-Kausik

The triple product:

• We start with three spaces $X$, $Y$ and $Z$.
• We construct a quotient of $X\times Y\times Z \times \Delta$, with $\Delta$ a $2$-simplex.
• We take a quotient where at each vertex two of the three spaces are identified to points.

This is probably the formal linear combination approach you mentioned.

Added double  for latex

I see Here we take a point in the triangle $\Delta$ and map the fibre over each point to a join.
By the way I am working this out as I type.
Let us call the vertices $X$, $Y$ and $Z$ based on which space is not identified to a point at a vertex.
First let us consider $(X * Y) * Z)$.
Clearly the edge from $X$ to $Y$ should be mapped to $X * Y$, and the map is an obvious one. In general, any point in $\Delta$ other than $Z$ is on a line $l$\$ that is parallel to $XY$. Identify $l$ with $XY$ by projection from $Z$ to give an identification of the fibre over $l$ with $X * Y$.
I now see a mistake.
We also need to quotient along the sides, with the space $Z$ pinched to a point on the line $XY$ for example. Otherwise "Clearly the edge from XX to YY should be mapped to X * YX∗Y, and the map is an obvious one." has a problem - there may be a map but it is not a homeomorphism.
Now proceeding with this correction, we get $((x, y), s)$ by projecting radially from $Z$, and we get $t$ as the distance from the base $XY$, normalized so that $Z$ is at distance $1$.
The above is Wrong too. At any point in the interior, $x$, $y$ and $z$ are all given. All I should say is that we get $s$ by projecting radially from ZZ, and we get tt as the distance from the base XYXY, normalized so that ZZ is at distance $1$.
What is to be verified is this being well-defined at $Z$ and along the sides $XZ$ and $YZ$. I now see what is happening.
• A priori we have a map on
KaTeX parse error: Undefined control sequence: \Y at position 9: X \times\̲Y̲ ̲\times Z\times …: X \times\Y \times Z\times R
where $R = [0,1]\times [0, 1]$
• At $t = 1$ the value of $s$ "does not matter" in $((X * Y) * Z)$, as $X * Y$ has been collapsed to a point.
• This allows one to pass to the quotient of the rectangle, where the top edge is pinched. This edge is just a triangle.
I now see what is happening.
• A priori we have a map on $X \times Y \times Z\times R$ where $R = [0,1]\times [0, 1]$
• At $t = 1$ the value of $s$ "does not matter" in $((X * Y) * Z)$, as $X * Y$ has been collapsed to a point.
• This allows one to pass to the quotient of the rectangle, where the top edge is pinched. This edge is just a triangle.
What I have described, starting with a triple product, may be the cleanest way to formalize this.
If the above is correct, then to complete the proof we map to $X * (Y * Z)$ by projecting from $X$ (instead of $Z$) to get $t$ (instead of $s$). Further, $s = 0$ at $X$ and $s = 1$ on $YZ$. Chinmaya Kausik
@Chinmaya-Kausik
I see. So I guess the key insight is converting a rectangle to a simplex by passing to a quotient.
To achieve the formal linear combination picture. That's what it looks like. Indeed one converts a rectangle into a simplex in two different ways for the two different associations. Once this is done, the two coincide. Chinmaya Kausik
@Chinmaya-Kausik
I see, thank you so much. Thank You. I too understand this clearly now. If any of you is online can you go to teams.microsoft.com and log in with your iisc account. I want to test this as an alternative, since we have a paid subscription. Chinmaya Kausik
@Chinmaya-Kausik
Oh, I see. Hadn't read this. I am online.
As you may have seen on teams. AR-MA210
@AR-MA210
I saw another version of Whitehead's theorem somewhere, for homology groups, stated as "If a continuous map between simply connected CW complexes $f:X \to Y$ induces isomorphisms $H_*(X) \cong H_*(Y)$, $f$ is a homotopy equivalence.". Is it possible to prove this using the theorem from today's class + Hurewicz's theorem or some other tools we have? It is another theorem.
It is in Hatcher. It follows from the result that I stated and did not prove, plus Hurewicz theorem and a little more AR-MA210
@AR-MA210
Ok, I'll look into it. Thanks I have started the teams, but not yet the video. Sumanta
@dasumanta
Okay. When I went to the link, I found they even have a debian linux app (they will surely have windows and mac), so I downloaded this. It may be good for everyone to do the same. Shabarish CH
I am on the app for MacOS philipthomasarun
@philipthomasarun
I am on the app on Windows Chinmaya Kausik
@Chinmaya-Kausik
I'm trying to launch it on Chrome Shabarish CH