Section to Skew-Symmetrization Map Let $A$ be an $n\times n$ matrix skew-symmetric matrix.  Define the map
$\mathbb{R}^{d^2}\to Skew_d$ by
$$
B\mapsto B^{\top} - B.
$$
Does this map have a continuous right inverse?
 A: The following addresses the question as first stated, which read,  if I recall correctly:
"Let $A$ be an $n\times n$ matrix skew-symmetric matrix.  Does there necessarily exist a unique $n\times n$ matrix $B$ such that
$A = B^T - B."   \tag 0$
To which I responded:
Pick any symmetric matrix
$C = C^T; \tag 1$
let
$B = -\dfrac{1}{2}(A + C);  \tag 2$
then
$B^T =  -\dfrac{1}{2}( A^T + C^T) = -\dfrac{1}{2}(-A + C), \tag 3$ 
since
$A^T = -A, \tag{3.5}$
and
$B^T - B = -\dfrac{1}{2}(-A + C) - (-\dfrac{1}{2}(A + C)) = A.  \tag 4$
These calculations indicate the existence of an infinite number of matrices $B$ such that (4) holds; thus no such $B$ is unique.
The preceding result may be used to address the present question of the existence of a continuous right inverse to the skew-symmetrization map, which we denote by 
$\Sigma(B) = B^T - B. \tag 5$
According to what has been presented in the above, every skew-symmetric matrix is of the form $\Sigma(B)$ for some $B$.  Thus a right inverse $\theta$ to $\Sigma$ will assign to every skew-symmetric $A$ a matrix $\theta(A)$ such that 
$\Sigma ( \theta(A)) = A. \tag 6$
Picking any symmetric matrix $C$ as in (1), we set
$\theta(A) = -\dfrac{1}{2}(A + C) \tag 7$
as in (2); it is easy to see that such a $\theta$ is continuous, since both matrix addition and division by $2$ are continuous operations.  Now (6) is simply the result of our preceding calculations, and $\theta$ is a right inverse of $\Sigma$.
