Existence of open sets $U, V$ of matrices such that for every $A \in U$ there exists $B \in V$ such that $B^4 = A$ Prove there exist non-empty open sets U and V of $n \times n$ matrices over $\mathbb{R}$ such that for every matrix $A \in U$ there exists exactly one matrix $B \in V$ such that $B^4 = A$.
I've tried to approach this problem in several ways, using characteristic polynomials, Jordan Canonical Form and facts from calculus but failed to get anything useful.
 A: Hint: take $U$ to be a sufficiently small neighbourhood of the identity.  Recall the inverse function theorem.
A: Using the implicit function theorem, we can obtain a general result.
Let $B_0\in M_n(\mathbb{R})$, $spectrum(B_0)=(\lambda_i)$, ${B_0}^4=A_0$, $f:X\rightarrow X^4$.
$\textbf{Proposition}$. If, for every $(i,j)$, $\lambda_i^3+\lambda_j^3+\lambda_i\lambda_j^2+\lambda_i^2\lambda_j\not= 0$, then there are open neighborhoods $V,U$ of $B_0,A_0$ s.t. $f$ is a diffeomorphism from $V$ onto $U$.
$\textbf{Proof}$. It suffices to show that the derivative $Df_{B_0}:H\in M_n\mapsto HB_0^3+B_0HB_0^2+B_0^2HB_0+B_0^3H$ is one to one.
If we stack row by row the matrices into vectors, then 
$Df_{B_0}=I\otimes {B_0^3}^T+B_0\otimes {B_0^2}^T+B_0^2\otimes B_0^T+B_0^3\otimes I$. cf.
https://en.wikipedia.org/wiki/Kronecker_product
Since the $4$ above linear applications pairwise commute,  $spectrum(Df_{B_0})=(\lambda_i^3+\lambda_j^3+\lambda_i\lambda_j^2+\lambda_i^2\lambda_j)_{i,j}$ and we are done.
Note that, when $B_0=I$, all considered eigenvalues are equal to $4$.
