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!



is there any fibration $\mathbb{R}^n\to \mathbb{S}^n$?
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.)


