A continuous map from $\mathbb S(\mathbb C^{n})$ to $U(n)$ Let $a$ in $\mathbb S(\mathbb C^{n})$, the unit sphere in $\mathbb C^n$.
Does there exists a continuous map $x\mapsto u_x$, from $\mathbb S(\mathbb C^{n})$ to $U(n)$, the group of unitary endomorphisms of $\mathbb C^{n}$, such that
 $u_x(a)$ equals $x$ for all $x$?
As far as I can tell (that is not very far), the fact that the composition of $x\mapsto u_x$ and $v\in U(n) \mapsto v(a)$ is the identity on $\mathbb S(\mathbb C^{n})$ does not seem to lead to an obvious contraction about the cohomology spaces.
But still, I am unable to define such a map.
And what if we consider the analogous question but with $\mathbb P^{n-1} \to PU(n)$ rather than $\mathbb S(\mathbb C^{n}) \to U(n)$?
 A: Note that $U(n)$ acts smoothly on $S(\mathbb{C}^n) = S^{2n-1}$ by left multiplication. The action is transitive with isotropy subgroup $U(n-1)$. So $p : U(n) \to S^{2n-1}$ given by $p(M) = Ma$ is a smooth principal $U(n-1)$-bundle. 
If $u : S(\mathbb{C}^n) \to U(n)$, $x \mapsto u_x$, then $p(u_x) = u_xa$, so a map $u$ such that $u_xa = x$ for all $x \in S(\mathbb{C}^n)$ is a section of $p$. Note that a principal bundle admits a section if and only if it is trivial.
If the bundle $p : U(n) \to S^{2n-1}$ where trivial, then $U(n)$ would be diffeomorphic to $U(n-1)\times S^{2n-1}$. If this were true, then 
$$\pi_{2n-2}(U(n)) = \pi_{2n-2}(U(n-1))\oplus\pi_{2n-2}(S^{2n-1}) = \pi_{2n-2}(U(n-1)).$$
By Bott periodicity, the left hand side is zero, while the right hand side is isomorphic to $\mathbb{Z}_{(n-1)!}$ (this was also proved by Bott). So we see that the bundle can only be trivial if $n =1$ or $2$.
For $n = 1$, $a \in S^1 = U(1)$,  while for $n = 2$, $a \in S^3 = SU(2) \subset U(2)$. In both cases, we can take $a^{-1}$ and therefore define the section $u(x) = xa^{-1}$.
