, 27 tweets, 8 min read
Follow-up thread with the other half of the solution to this problem: computing the homology of this glued ("adjunction") space!

1/n
This will mostly be an excuse to talk about one of my favorite theorems from introductory algebraic topology: the Mayer Vietoris sequence.

(Full disclosure, there is a *much* easier way to do this problem, namely finding a nice space homotopy equivalent to X.)

2/n
Recall that our adjunction space X was described as "2 spheres glued along their equators by a degree 2 map".

3/n To compute π_1, we then used a "standard trick": breaking X into two subspaces A, B which "fuzzed" into each other a bit, where A, B, A∩B were easier to understand.

The claim is that we can double our mileage, and also use this decomposition to compute the homology H_*(X).

4/n
If you haven't seen homology H_* before, the important fact here is that it's an algebraic invariant of our space -- for our purposes, just an indexed collection of groups. I.e. if two spaces output a different collection, they can't be homeomorphic.

5/n
The intuition here is that homotopy groups (like π_1) help measure embedded non-contractible spheres, which measure n-dimensional holes in your space.

These are generally hard to compute, and homology is like a "linearization" of homotopy that is easier to compute.

6/n A nice thing about the category of topological spaces is that we have many "models" to work with. In particular, we can take this linearization business seriously, and think about homology in terms of simplices.

(After all, a triangle is nothing but a pointy circle!)

7/n This works well for nice enough spaces: for a surface like the one here, we can think about breaking it up into a bunch of points (0-simplices), edges (1-simplices), triangles (2-simplices), and so on.

8/n The idea now is that we'll want to use "chains" to measure where the holes are. Abstractly, chains will be formal linear combinations of simplices with integer coefficients.

Here I've tried to show what a chain of 2-simplices might look like, wrapping like a bandana.

9/n How does this measure "holes"? In the previous picture, you can imagine that if we modified it by moving one triangle "up" at a time, we could eventually shrink the green band up. If they had a ponytail, we'd get stuck! Some missing triangles would cause the same problem.

10/n
And with a bit of algebraic machinery (taking certain equivalence classes of chains), homology will spit out a number that tells us about the missing regions.

Okay, so what is Mayer-Vietoris doing for us then?

11/n
If we've split up our space X into A and B, we can think about chains that just live in A, or B, or the intersection, or chains that break up nicely into a chain in A plus a chain in B.

Some investigation produces maps that yield a short exact sequence of chain complexes:

12/n The maps won't be important for us here, but they essentially come from including a chain in A∩B into either A or B (with some signs for orientation), and then taking a chain in A and a chain in B and adding them to get "a chain in A plus a chain in B" (the last term).

13/n
And now the beautiful machinery of homological algebra kicks in for us: by the Snake Lemma, any short exact sequence of chain complexes will induce a *long* exact sequence in homology.

We call this particular one the Mayer-Vietoris sequence.

14/n Okay, but what does that actually mean? How do we actually use it?

For me, anyway, it's easiest to throw all of the information into a little table. The rows measure which degree of homology you're in, and the columns tell you about the operation you're considering.

15/n Next, I'm just going to fill in the information we know from the original decomposition. Both A and B deformation retracted onto 2-spheres, and the intersection onto a 1-sphere.

The union was X by construction, and H_* X is what we're trying to find.

16/n But now our decomposition pays off! By Algebraic Topology 101, we know everything about the homology of S^n: Z in degrees 0 and n, zero elsewhere.

We'll also simplify further by taking "reduced" homology -- why is this a good thing to do?

17/n
What it buys us is zeroing out the bottom row, at the cost of giving no information about H_0 X. H

However, we can use some ad-hoc reasoning: it's a theorem that H_0 X is one copy of Z for each path component of X. We can argue that X is path-connected, and so H_0 X = Z.

18/n
So we bought a lot of zeros in our LES (which is always the situation you hope for!)

Switching to reduced homology and filling in the known homology of spheres gives us this:

19/n Now homological algebra does us even more favors: for any exact sequence of the form 0 -> A -> 0, we can immediately conclude A=0.

(Nothing fancy, this just says A admits an injective and surjective map to zero.)

So this just leaves us with the red bit:

20/n But great! This leaves us with the following SES, and a little Algebra finishes the job.

We have a short exact sequence of abelian groups, i.e. Z-modules, and the last term (Z) is a free Z-module and thus projective.

21/n But it's a theorem that 0->A->B->C->0 splits iff C is projective, which means we have an isomorphism B≅AxC.

For us, that means H_2 X ≅ (ZxZ)xZ = Z^3.

22/n
Collecting up all of this information, we have

H_* X = [Z, 0, 0, Z^3, 0, 0, ....]

i.e. Z in degree 0, Z^3 in degree 2, and zero elsewhere!

This gives the solution, but it's a bit unsatisfying: where did we use the fact that the gluing map was degree 2?

23/n
What happened here was that we got *too* lucky: in a generic situation, these attaching maps would show up in those chain maps from earlier, and we'd need to look at their induced maps and their images/kernels in homology.

We skate by because too many zeros appeared!

24/n
Some questions:

1. Can you construct a space that *forces* considering degrees of attaching maps in MV?
(Try the Klein bottle!)

2. Can you modify X to make H_1 X and/or H_1 A ⊕ H_1 B nontrivial?

25/26
3. There is a homotopy-theoretic argument that handles both π_1 and H_* simultaneously. What is it, and can the homotopy equivalence be easily "seen"? (e.g. drawn or visualized)
Does it also help explain why the degree of the map didn't play a role here?

26/26!
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