There are many countries, for example, India, where I guess that English is still a widely used language, even though I think it may not be the official one
In any case, English is the de-facto international language
There's a long exact sequence $\cdots \to H_n(X) \to H_n(X) \to H_n(T_f) \to \cdots$ where $T_f$ is the mapping torus of $f$, and the map $H_n(X) \to H_n(X)$ is $1 - f_*$
Falls out of Mayer Vietoris
So your construction is sound
But I am 70% sure the loop has to be $[\ell_1, [\ell_2, [\ell_3, \cdots]\cdots]]$. It's homologous to $[\ell_2, [\ell_3, \cdots]\cdots]$
I just don't know how to prove it's indeed not nullhomologous. The "turtles to infinity" is what should prevent it to be zero in $H_1(\Bbb H)$
From what I've seen it's like "The Warsaw circle (sin(1/x) but connected to itself) should be like a circle but homotopy/homology groups can't see it! We need something that can"
> Very nice! Actually, [a_1,a_2][a_2,a_3][a_3,a_4]... works. As you found, homology fixed points are the key (for any mapping torus) and there are lots of them. There's a result of K. Eda called the 0-form lemma which tells us which "reduced" loops are null-homologous in 1d spaces.
> The proof of this lemma is really something, but it basically says a reduced loop (with no null-homotopic subloops) in a 1-d space is null-homologous if and only if it factors as the product of finitely many reduced subpaths each of which has an inverse pair in the factorization.
(I used a instead of $\ell$ because no fancy script l in ASCII)
I think it's a hard question in general to find what's the minimal number of simplicies needed to bound a nullhomologous curve
It's like asking what the minimal area of the disk which bounds a nullhomotopic embedded loop in a Riemannian manifold (you can always make the disk immersed by generic perturbations I believe)
Then I don't see the problem. When you cross a torus with time $1/2^n$, you are effectively constructing a homotopy from the outermost circle to the $n$-th inner circle, and since the times add up to $1$, you can prove it uniformly converges to a nullhomotopy
This is a classic telescoping trick
Let's see if I miss something. You have a map $D^2 \setminus 0 \to \Bbb H$ with boundary mapping to the outer circle
$D^2 \setminus 0$ is partitioned into annuli of radius $1/2^n$
Your complaint is that there are points in any of those annuli which are $1$ away from the basepoint?
That's not true, everything is shrinking
As you traverse inwards and inwards, the uniform distance of the loop from the basepoint during one full traversal becomes smaller and smaller
Oh no how is that true
The circle has some radius
So $D^2 \setminus 0 \to \Bbb H$ cannot possibly extend to $0$
> Meet the Hawaiian mapping torus - the mapping torus of the shift map. First singular homology is infinite cyclic generated by inner loop. Can you find the non-trivial elements of H_2? If you look at the image, it might feel H_2 is trivial since there is no "enclosed space!"
That's the actual quote (and what I said was nonsense)
