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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.

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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 y=0 –  mr.gondolier Apr 6 '10 at 3:56
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I think what is lacking here is convexity. Otherwise, one can make use of this selection theorem: en.wikipedia.org/wiki/Michael_selection_theorem –  mr.gondolier Apr 6 '10 at 4:20

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