8
$\begingroup$

The following question is an attempt to revise this one into what I intended. Important revisions are shown in bold.

Are there any known examples of a compact Riemannian manifold $M$ with (possibly empty) boundary and a smooth manifold $N$ such that $\dim M = \dim N$ and no topological embeddings $M\hookrightarrow N$ exist, but for each $\varepsilon > 0$ there is a continuous map $f:M\to N$ whose fibers $f^{-1}(n)$ all have diameter $< \varepsilon$?

For example, take $N=\mathbb{R}^2$ and $M$ to be a 2-torus with an open disk removed. Does there exist a sequence $f_k: M\to \mathbb{R}^2$ of continuous maps such that $$\sup_{x\in \mathbb{R}^2} \text{diam} f_k^{-1}(x) \to 0$$ as $k\to \infty$ with respect to some fixed metric on $M$?

Improve this question
$\endgroup$
9
  • 1
    $\begingroup$ For closed manifolds $M$ of dimension $\ge 5$ this is a consequence of Ferry's theorem. It will be also true in dimension 2 and maybe in dimension 3. Dimension 4 is, as usual, anybody's guess. $\endgroup$
    – Moishe Kohan
    Commented Dec 9 at 22:01
  • 2
    $\begingroup$ @MoisheKohan thank you. Is the consequence here that the answer to my question is “no” in those situations? If $M$ is closed and $N$ is open, I think I can prove your statements in all dimensions by an argument involving top degree homology. So while I’m interested in the case that $M$ is closed, I am especially interested in the case that $M$ has nonempty boundary. But I would also like to understand how you arrived at your conclusions, and in any case I am not sure that I know how to verify your conclusions if $M$ and $N$ are both closed manifolds. $\endgroup$
    – Matthew Kvalheim
    Commented Dec 9 at 23:00
  • 1
    $\begingroup$ If $M, N$ are both closed then it is a direct consequence of Theorem 1 in Ferry's paper "Homotoping $\epsilon$-maps to homeomorphisms," AJM, 1979. You have to assume $n\ge 5$, as I said. He also proves (in all dimensions) that the map is a homotopy-equivalence (with "short" tracks of the homotopies), which also does the job in dimension 2 and, mostly, in dimension 3. $\endgroup$
    – Moishe Kohan
    Commented Dec 10 at 15:44
  • 1
    $\begingroup$ Surjectivity is quite trivial to verify by looking at the top-dimensional homology. In dimension 3 I would not use geometrization but Heegaard splittings. (Geometrization would do the job in the aspherical case. However, even the case of maps between lens spaces is far from clear.) But details look complicated. $\endgroup$
    – Moishe Kohan
    Commented Dec 10 at 18:51
  • 1
    $\begingroup$ If $f: M\to N$ is not surjective, you set $Y:= f(M), X=M$. (This is OK, since all what ferry needs is a metric space $Y$ at this point.) Then construct $g: Y\to X$ using Corollary 2.3, so that $g\circ f: M\to M$ is homotopic to the identity map. But $g^*: H^n(M)\to H^n(Y)=0$ is zero (where $n= dim(M)=dim(N)$ and I use Chech cohomology with $Z_2$-coefficients), hence, $(g\circ f)^*=0$, which is a contradiction. $\endgroup$
    – Moishe Kohan
    Commented Dec 10 at 19:44

0

Reset to default

You must log in to answer this question.