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!
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! 


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 5lemma, 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 CWcomplexes. Hence, being weak homotopy equivalent to each other, by Whitehead's theorem they are homotopy equivalent to each other. Now $F$ is finitedimensional, 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 CWcomplex.) 


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 $H_{n1}$ isomorphic to $\mathbb Z$. And if $n>1$ these fibers are connected. Alexander duality tells you the Cech cohomology of the fiber in dimension $n1$ 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. 

