Take the 2-minute tour ×  The title is probably a bit too broad.
I frequently encountered the following situation: suppose I need
to select a solution to a linear equation from a compact set.
Can I make this selection continuous?  Formally, let $S \subset
\mathbb{R}^n$ be a compact set. Let $A$ be a $k \times n$ matrix
($k < n$), which we view as a linear function $A: \mathbb{R}^n
\to \mathbb{R}^k$. Let $T = A(S)$ be the range of $A$ on $S$. Is
there a continuous function $g: T \to S$, such that $A g(y) =
y$?  To construct $g$, we only need to pick a value from the
solution set $A^{-1}(\{y\}) \cap S$, which is compact. The
question is: can we choose it in a continuous way? It is easy to
see that we can choose $g$ to be Borel measurable, say, choose
$g$ to be the one with the minimum Euclidean norm from the
solution set.  share|improve this question alright. I think this
is impossible in general. Example: $n = 2, k =1$. $S = [-1,0]
\times $ {1} $\cup$ $[0,1] \times $ {0}. Then T=[-1, 1] . There
is a problem at