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It is probably a trivial question. But I don't see the answer.

Is there any Hurewicz fibration $\mathbb{R}^n\to \mathbb{S}^n$ ?

Is there any fibration $X\to \mathbb{S}^n$, when $X\subset \mathbb{R}^n $?

I appreciate any help. Thank you very much!

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 Well, certainly there is when $n=1$... – Daniel Litt Sep 23 2011 at 7:38
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 My approach would be similar to Dylan's: If such a fibration existed for $n>1$, the fibre would be a closed subset $F\subset \mathbb{R}^n$ with the weak homotopy type of $\Omega S^n$, and would therefore have nonzero (co)homology in arbitrarily high degrees. In some sense you have an infinite dimensional space embedded in a finite dimensional Euclidean space, which feels wrong. I can't right now see how to rule it out though. – Mark Grant Sep 23 2011 at 9:18
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 Along this line, it's worth mentioning that Barratt-Milnor constructed examples (higher analogues of the Hawaiian earrings) that embed in $\mathbb{R}^n$ but have cohomology in arbitrarily high degrees. User BS discussed them here: mathoverflow.net/questions/4478/… – Tyler Lawson Sep 23 2011 at 12:39
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 However, $\Omega S^n$ has the homotopy type of a CW complex, and therefore (iirc?) its singular homology is the same as its \check{C}ech homology! – some guy on the street Sep 23 2011 at 14:36
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 A small note, the answer to the first question is clearly no for $n=2$ by the classification of non-compact $2$-manifolds + Mark's comment. I imagine there's a reasonable proof the answer is always no for $n>1$ but off the top of my head I'm not seeing it. – Ryan Budney Sep 23 2011 at 21:51
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2 Answers

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Edit: The following simplifies the original answer (which unnecessarily used singular cohomology).

If $f:\Bbb R^n\to S^n$ is a fibration, then as Mark noted, a fiber $F$ of $f$ is weak homotopy equivalent to $\Omega S^n$ (using the 5-lemma, see Prop. 4.66 in Hatcher). I claim that $F$ is in fact homotopy equivalent to $\Omega S^n$.

Indeed, $F$ is homotopy equivalent to the corresponding homotopy fiber of $f$ (Prop. 4.65 in Hatcher). The homotopy fiber consists of pairs $(x,p)$ where $x\in\Bbb R^n$ and $p$ is a path in $S^n$ connecting $f(x)$ and the basepoint. It is homotopy equivalent to the space $X$ of maps $[0,1]\to MC(f)$ (the mapping cylinder) sending $0$ into $\Bbb R^n$ and $1$ into the basepoint of $S^n$. Milnor showed that $X$ and $\Omega S^n$ are homotopy equivalent to CW-complexes. Hence, being weak homotopy equivalent to each other, by Whitehead's theorem they are homotopy equivalent to each other.

Now $F$ is finite-dimensional, so its Cech cohomology is eventually zero. Cech cohomology is a homotopy invariant, so we get that the cohomology of $\Omega S^n$ is eventually zero, contradicting Mark's comment. (It does not matter which cohomology of $\Omega S^n$, they are all isomorphic since $\Omega S^n$ is homotopically a CW-complex.)

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Hi Sergey, when you say that $F$ is finite-dimensional, are you talking about its covering dimension (I see no reason why $F$ should be something nice like a manifold or complex)? Do closed subsets of $\mathbb{R}^n$ necessarily have covering dimension less than or equal to $n$? – Mark Grant Sep 24 2011 at 7:55
3 
 Mark, yes, $F$ is of covering dimension $\le n$, that is, every open cover of $F$ can be refined by a cover whose nerve is of dimension $\le n$. Indeed, by the definition of subspace topology, every open cover of $F$ is the intersection with $F$ of an open cover of an open neighborhood of $F$; hence also of an open cover $C$ of $\Bbb R^n$. This $C$ has a refinement $D$ with a nerve $N$ of dimension $\le n$ since $\Bbb R^n$ is $n$-dimensional (see nice pictures at en.wikipedia.org/wiki/Lebesgue_covering_dimension). Finally, the nerve of $D$ intersected with $F$ is a subcomplex of $N$. – Sergey Melikhov Sep 24 2011 at 16:14
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 I should add that perhaps some confusion arises from Tyler Lawson's remark that the Barratt-Milnor examples of a subset of $\Bbb R^3$ "have cohomology in arbitrarily high degrees". This is true only of singular cohomology, which is not a geometrically interesting cohomology theory, in particular because it is not Brown representable (see mathoverflow.net/questions/47544). – Sergey Melikhov Sep 24 2011 at 16:49
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 Assume that $ p: E\to B $ is a fibration and $ E $ is contractible. We have that the homotopy fiber is homotopy equivalent to the fiber. But the homotopy fiber is the pullback of $ p: E\to B $ along $ v_0: PB\to B $. Since $ v_ 0 $ is a fibration, the homotopy fiber is, indeed, the homotopy pullback. Hence, the pullback of $ \ast\to B $ along $ v_0 $ is homotopy equivalent to the homotopy fiber (since $ E $ is contractible). This new pullback is $ \Omega B $. So, in general, if $ p: E\to B $ is a fibration with contractible total space, the fiber is homotopy equivalent to $ \Omega B $. – Fernando Lucatelli Feb 25 at 17:19
This makes it much simpler, thanks! – Sergey Melikhov Mar 1 at 0:40
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I think there's a pretty simple answer, patching together everything that's been said in the comments.

By Mark's answer, the fibers have $H_{n-1}$ isomorphic to $\mathbb Z$. And if $n>1$ these fibers are connected. Alexander duality tells you the Cech cohomology of the fiber in dimension $n-1$ is isomorphic to the reduced $0$-dimensional homology of the complement of the fiber in $\mathbb R^n$. And some guy on the street's comment tells you that Cech cohomology is regular cohomology.

So this is saying that the fibers separate $\mathbb R^n$, but since the base is $S^n$ with $n \geq 2$, that's impossible.

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1 
 Ryan, I'm not getting this. Mark has a weak homotopy equivalence. Cech cohomology is not an invariant of weak equivalence. – Sergey Melikhov Sep 24 2011 at 0:43
Since $\Omega S^n$ is a CW-space and so is the fiber, then the weak equivalence is an actual equivalence. – Dylan Wilson Sep 24 2011 at 0:50
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 Dylan, the fiber is not a CW complex: the original fibration is not assumed PL or smooth. – Sergey Melikhov Sep 24 2011 at 1:05
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 @Sergey: as you've pointed out in your answer, the weak equivalence can be made into a homotopy equivalence. Sorry I abandoned the question -- my calculus class had an exam and I had to run out of the room. Your answer is the more complete so I'd like to suggest Fernando accept yours. – Ryan Budney Sep 24 2011 at 3:48
1 
 Ryan, thanks. I'm still a bit puzzled by your comment on the $n=2$ case, can you do this case without using Milnor's theorem on functional spaces? In this connection the following example is worth mentioning. The pseudo-arc is a compact subset of $\Bbb R^2$ which does not have any nontrivial paths in it. So it is weakly equivalent to a discrete set of cardinality continuum (which I guess can be seen as CW-complex), but is not homotopy equivalent to any CW-complex. – Sergey Melikhov Sep 24 2011 at 17:32
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