Can a perturbation of a matrix product always be represented as product of perturbations of its factor matrices? Given $A\in\mathbb{R}^{m\times n}$ with $A=BC$ for some $B\in\mathbb{R}^{m\times k}, C\in\mathbb{R}^{k\times n}$. Assume that $k\geq\min(m,n)$ so that this decomposition always exists for every $A\in\mathbb{R}^{m\times n}.$
Can we prove that for any perturbation $A'$ of $A$, there always exist some perturbations $B',C'$ of $B,C$ such that $A'=B'C'?$ 
More precisely, let $\epsilon>0$ and $A'$ satisfying $||A-A'||<\epsilon$ for some suitable norm.
Show that there always exist two matrices $B',C'$ which satisfy the following conditions:
a) $A'=B'C'$
b) $||B'-B||$ and $||C'-C||$ can be upper-bounded by $\epsilon$ and other constant terms.
The second condition implies that $B'$ and $C'$ can be chosen to be sufficiently close to $B$ and $C$ as long as $A'$ is sufficiently close to $A.$ Moreover, note that there is no constraint on $B$ and $C$, except that $k\geq\min(m,n).$
Intuitively i think it should be possible but i cannot prove/disprove it. If this is not possible then what conditions are needed for such existence of $B',C'$?
Thanks,

Some thoughts, but not a complete answer: 
The function $(B, C) \mapsto A$ is continuous and (by your assumption on $k$, surjective). It looks as if the answer might therefore be "yes". You take some target $A$, and look at the preimage of a neighborhood of $A$, and get an open set containing your particular $B$ and $C$. So you can pick a ball around $(B, C)$ and that ball is sent within the target neighborhood. But does it COVER an open ball in the target neighborhood? Maybe not. What if the map is degenerate, and the image of the domain ball is simply a point or a line in $A$-space? 
One way to avoid that is to use the derivative: if the map from $(B, C)$ to $BC$ had full-rank derivative, then in some small-enough neighborhood, it would be a homeomorphism onto the image of this neighborhood (or something like that -- I can never recall the exact statement of the inverse function theorem, or theorems on invariance of domain, etc.)
The problem is that the derivative of $(B, C) \mapsto BC$ with respect to $C$, looks like multiplication by $B$ (after all: $C \mapsto BC$ is a linear map!). If $B$ has rank $1$, then the rank of this derivative map is far lower than it needs to be to make the full-rank argument I proposed. 
I therefore would look closely at the case
$$
B = \begin{bmatrix}
1 & 1 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0 
\end{bmatrix},
C = \begin{bmatrix}
1 & 0 & 0 \\
1 & 0 & 0 \\
1 & 0 & 0 
\end{bmatrix}\\
A = \begin{bmatrix}
3 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 
\end{bmatrix}
$$
and see what happens when you perturb something like the $(2,3)$ element of $A$. 
Indeed, calling the multiplication function $f$, we have
$$
f(B + B', C + C') = BC + B'C + BC' + B'C'
$$
so that 
$$
df(B', C') = B'C + BC'
$$
which (for the $B$ and $C$ above) has the form
$$
\begin{bmatrix}
* & * & * \\
* & 0 & 0 \\
* & 0 & 0 
\end{bmatrix}
$$
so that to first order, at least, the $(2,3)$ element of the product doesn't change.
But then again, the domain here has dimension $9 + 9$, and the codomain has dimension $9$, so maybe looking at first-order only isn't enough...
