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Let $(X_{n},d_{n})_{n \in \mathbb{N}}$ be a sequence of intrinsic metric spaces verifying :

  • $X_{n}$ have topological dimension $n$.
  • $X_{n+1}$ is n-connected.
  • $X_{n} \subset X_{n+1}$
  • The distance $d_{n}$ and $d_{n+1}$ generate the same topology on $X_{n}$.
  • $\forall x,y \in X_{n}$ : $d_{n+1}(x,y) \le d_{n}(x,y)$.
  • $(X_{n},d_{n})$ is quasi-isometric to $(X_{n+1},d_{n+1})$

Definition : Let $d$ be a distance on $ \bigcup{X_{n}}$, defined as follows : $d(x,y) = lim_{n \to \infty} d_{n}(x,y)$.

Remark : There is a small abuse in the previous definition because $d_{n}(x,y)$ is defined only for $x, y \in X_{n}$. But because we take $n \to \infty$, there is no problem.

Definition : Let $X:=\overline{\bigcup{X_{n}}}$, the complete metric space obtained as a completion of $\bigcup{X_{n}}$ for $d$.

Question : Is $X$ weakly contractible ?

Remark : If yes, perhaps some of these conditions are useless in the proof, and perhaps the useful conditions can be highly generalized.

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2  
I don't know this field at all, but this seems like a legitimate question; why the vote to close? –  Noah S Aug 20 '13 at 3:36
    
As the question concerns only the (weak) homotopy type of X, you don't need the metric, but only the topology it defines. Is this the limit (union) topology of the X_n's? –  Konrad Voelkel Aug 20 '13 at 14:04
    
@KonradVoelkel : What do you mean by limit topology ? The distance $d$ induces a topology on $X$ and $\bigcup X_{n}$ is a dense subset of $X$. Does this answer your question ? –  Sébastien Palcoux Aug 20 '13 at 14:22
    
I was merely hoping that your $X$ might be a limit object of the $X_n$ (in the categorical sense) in some category of metric spaces. Other than that, I would try to deform a map $S^n \to X$ to $\bigcup X_n$ ... –  Konrad Voelkel Aug 20 '13 at 16:07
    
Yes it's exactly the problem : is a continuous map $m:\mathbb{S}^{n} \to X$, homotopic to a continuous map $m′:\mathbb{S}^{n} \to \bigcup{X_{n}}$ and then to a continuous map $m′′:\mathbb{S}^{n} \to X_{n+1}$ ? (Keep in mind that in less specific cases, there are counter-examples : see the answer of Dan Petersen in my first post here). –  Sébastien Palcoux Aug 20 '13 at 20:25

2 Answers 2

up vote 2 down vote accepted

The intristic metric assumption seems to be added to exclude $S^1$ from the previous post. But what about $S^d$? It can be approximated with bigger and bigger disks $\{x\in S^d\ |\ x_1\leq 1-\frac{1}{n} \}$ (with the natural intristic metric) and in the limit you get the sphere. (Of course dimension of such a disk is $d$ so one needs some obvious bypasses for $n<d$.)

As it was already noticed quasi-isometry requirement is trivial for bounded spaces.

Similarly with the dimension assumption: we can always take cartesian product with some nice space like $\prod_{i=1}^n [0,\frac{1}{2^i}]$ with the $\ell_2$-metric ($\frac{1}{2^i}$ is unnecessary but this way all the bounds are uniform).

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You're right, thank you ! Do you know a slight additive assumption excluding $\mathbb{S}^{n}$. The CAT(0) assumption does the job (as suggested by Paul Fabel), but I think we can find a more general assumption, do you have an idea ? –  Sébastien Palcoux Aug 22 '13 at 8:44
    
I don't know where you are coming from, but CAT(0) looks like a very strong assumption to me - even a plane without a ball fails to be CAT(0). But indeed, it may work. –  savick01 Aug 22 '13 at 9:15
    
yes you're right, it's a strong assumption, that's why I ask you about a more general assumption, do you have idea ? (perhaps "uniquely geodesic" or "without conjugate points" or...). –  Sébastien Palcoux Aug 22 '13 at 9:22
    
There is also the "without boundary" assumption, but you can certainly modify your construction for absorbing this new assumption, right ? –  Sébastien Palcoux Aug 22 '13 at 11:01
    
What do you mean by "without boundary"? Did you add "manifold" assumption in the meantime? If you want a manifold without boundary, then you should specify what you mean by "intristic metric". If it is enough that the space is a length space, then there is no problem. But if you insist on a geodesic space, then there might be... –  savick01 Aug 22 '13 at 13:49

The following does not settle the original question.

The answer is `no' if we ignore the intrinsic metric requirement and merely demand metric.

For a counterexample first observe that the inverse limit $X$ of nested closed intervals $X_{n}$ (with retract bonding maps) might not be path connected.

(For example if $X$ is the closure of the graph of y= sin(1/x) with x in [0,1], and for $X_{n}$ take the subspace with x in [1/n,1]).

Now to arrange the correct dimensions multiply each $X_{n}$ by an n-1 cube (canonically embedded in Hilbert space) and adjust the bonding maps to collapse the terminal coordinates of the thickened $X_{n}$ factors.


The category of compact metric spaces X, each of which is the inverse limit of nested finite dimensional contractible retracts $X_{n}$, provides a more general source of examples.

Most of the conditions in the questions are satisfied automatically for such X, since any two compact metric spaces are quasi isometric, and the inverse limit topology is compatible with the metric as described in the question.

Multiplying factors by n-cubes arranges for dimension going to infinity.

As sketched in the comments, we can arrange a counterexample X with nontrivial fundamental group, by compactifying the open unit disk with an annulus, while remaining in the category at hand.

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Thank you for your answer. I'm not sure to understand your example. First, the closure of the graph $G$ of $y= sin(1/x)$ with $x \in [0,1]$, is weakly contractible, right ? Next, for simplicity, without the dimensional assumption, how do you deform $G$ for it becomes non-"weakly contractible" ? –  Sébastien Palcoux Aug 21 '13 at 11:50
    
No, the 0th homotopy group of a non path connected space is nontrivial. If this doesn't count, spin the original graph about the line x=1 in (x,y,z) space. Then X will have nontrivial fundamental group/ –  Paul Fabel Aug 21 '13 at 11:59
    
Yes, this doesn't count, and no, the spin you describe is called the Polish circle and it has the unexpected property to have a trivial fundamental group (see here). –  Sébastien Palcoux Aug 21 '13 at 12:11
1  
No, the spin I describe yields a compactification of the open unit disk by attaching a cylinder. –  Paul Fabel Aug 21 '13 at 12:23
    
I don't visualize your construction, let me think a little (clarifications are welcome). Note that $x=1$ is not a line but a plane in $(x,y,z)$ space. What's your line ? –  Sébastien Palcoux Aug 21 '13 at 12:50

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