the opposite of projection
Please feel free to edit this question - I am decidedly rusty and probably inaccurate on my mathematical terminology :-)
I am familiar with the notion of projecting points from some space $\mathbb{R} ^n$ to another $\mathbb{R}^m$ where $m<n$, but am interested in what holds if $m>n$.
Suppose I have two non-intersecting and simply connected sets of points $C_1$ and $C_2$ defined over $\mathbb{R}^n$. Now suppose I "extend" the vector defining any point in either set into $\mathbb{R}^m$ by injecting an appropriate number of arbitrary scalars into the vector. Will the set of all possible extensions of this kind of $C_1$ remain non-intersecting with all possible extensions of $C_2$ so long as the "extension" is made on the same basis - scalars are injected at the same positions in each case? That is my intution from $\mathbb{R}^1$, $\mathbb{R}^2$ and $\mathbb{R}^3$, but I'd like to be sure this holds as one moves into higher dimensions.
Also, am I correct that in a simply connected set in $\mathbb{R}^n$, there exists at least one path which will connect any point in that set to any other point in that set via a path comprising points only from that set? And will this property of simple connectedness potentially be lost if one "extends" to higher dimensions? It seems to me that one can't say anything at all about the connectedness after this kind of extension.
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$\begingroup$The letter $\pi$ is often used to denote a projection map, so let $\pi:\mathbb{R}^m \rightarrow \mathbb{R}^n$ be projection onto the first $n$ of $m$ coordinates. The situation would be equivalent if some other subset of size $n$ of $m$ coordinates were chosen.
Consider a "lifting" of two (path-)connected subsets $C_1,C_2 \subseteq \mathbb{R}^n$ to m-dimensional subsets that project back to $C_1,C_2$ respectively. That is, suppose there are functions $f_i:C_i \rightarrow \mathbb{R}^m$ such that $\pi \circ f_i$ is the identity on $C_i$.
Now we are requiring a 1-to-1 correspondence between $C_i$ and its lift $f_i(C_i)$, though without requiring continuity of the functions $f_i$. In this situation $f_i(C_i)$ need not be connected even if we require $C_i$ to be connected, or path-connected, or even simply connected (the strongest of the three conditions).
For example, let $C_i$ be an interval restricted to the first component of $\mathbb{R}^n$, say $[i,i+\frac{1}{2}] \times \{0\}^{n-1}$. Define $f_i(r,0,\ldots,0)$ by inserting $m-n$ trailing coordinates using $1$s if $r$ is rational and $0$s otherwise. Then $f_i(C_i)$ is not connected, even though $C_i$ is as nicely connected as one can ask.
The other issue raised by the Question is whether disjointness of $C_1,C_2$ is necessarily preserved by their lifts $f_1(C_1),f_2(C_2)$. The answer is yes, and there's a simple argument involved. Suppose for contradiction that $x \in f_1(C_1) \cap f_2(C_2)$. Then what of $\pi(x)$? As $x \in f_i(C_i)$, we must have $\pi(x) \in C_i$ for both $i=1,2$. But this contradicts the disjointness of $C_1,C_2$.
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