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S Dec 13 at 18:29 history bounty started Matthew Kvalheim
S Dec 13 at 18:29 history notice added Matthew Kvalheim Draw attention
Dec 10 at 20:01 comment added Matthew Kvalheim @MoisheKohan thanks for explaining. I see. My “top-dimensional homology argument” is basically the same.
Dec 10 at 19:44 comment added Moishe Kohan 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.
Dec 10 at 19:27 comment added Matthew Kvalheim @MoisheKohan yes, I used top-dimensional homology in my mentioned verification of surjectivity. Regarding instead getting surjectivity from Ferry's Corollary 2.3, I am not sure what you mean since I think surjectivity is one of the hypotheses of that result. Thank you for your dimension 3 comments. I will need to think more about those.
Dec 10 at 18:59 comment added Moishe Kohan To get surjectivity, apply Ferry's Corollary 2.3, which is very general and does not even need the manifold assumption.
Dec 10 at 18:51 comment added Moishe Kohan 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.
Dec 10 at 16:10 comment added Matthew Kvalheim @MoisheKohan thanks, I think I now see how to prove what you said for $n\geq 5$ for $M$, $N$ closed. To do so, I think some work is required to verify Ferry’s hypothesis that small fiber maps must give surjections between connected components (surjectivity is mentioned in the first paragraph of Ferry 1979 and I suspect is implicitly assumed in his Theorem 1) in this case, but I think I see why that’s true now. I also think I see how to prove what you said in dimension $2$ using the classification of surfaces. Are you using the geometrization conjecture for dimension $3$?
Dec 10 at 15:44 comment added Moishe Kohan 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.
Dec 10 at 13:27 history edited Matthew Kvalheim CC BY-SA 4.0
Removed “compact” from the title since I do not want to assume that the codomain is necessarily compact
Dec 9 at 23:00 comment added Matthew Kvalheim @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.
Dec 9 at 22:01 comment added Moishe Kohan 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.
Dec 9 at 20:48 history edited Matthew Kvalheim CC BY-SA 4.0
Added more bold to indicate revisions from previous (linked) question
Dec 9 at 20:40 history asked Matthew Kvalheim CC BY-SA 4.0