Let $G$ be a locally compact group, and let $N$ be a closed, normal subgroup, and let $\pi\colon G\to G/N$ be the quotient homomorphism. It is known that there exists a Borel cross section, i.e., a measurable map $\sigma\colon G/N\to G$ such that $\pi\circ\sigma=\mathrm{id}\_{G/N}$. Moreover, the Borel cross section can be chosen locally bounded, i.e., for every compact subset $K\subset G/N$, the set $\sigma(K)$ has compact closure in $G$.


Consider the $2$-cocycle $\omega\colon G\times G/N \to N$ defined by the equality
$$
\sigma(gx)\omega(g,x)=g\sigma(x)
$$
for all $g\in G$ and $x\in G/N$.



> 
> Question: Given a compact subset $K$ of $G$, can $\sigma$ be chosen such that the set $\omega(K\times G/N)$ has compact closure in $G$?
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> 
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Some remarks:


1. The answer is `yes' in any of the following three cases: If $N$ is compact, or if $G/N$ is compact, or if $G$ is a semidirect product of $N$ and $G/N$.
2. I would interpret a positive answer to the question as saying that $G$ is a `coarsely trivial' bundle over $G/N$.


Edit 03.03.2015: Logical error in question fixed. The section $\sigma$ may depend on the compact subset $K$.



Coarsely trivial Borel cross section for $G\to G/N$
No, contrary to what you said, it is not always possible, even if $G$ is a semidirect product. Let us fix some $g \in N$, so $\sigma(gx) = \sigma(x)$ for all $x$. Then $\sigma(x)^{-1} g \sigma(x) = \omega(g,x)$ is a bounded function of $x$. It is easy to construct a counterexample to this, by arranging for the conjugacy class of $g$ to be unbounded. For example, let $G = \mathbb{R}^\times \ltimes \mathbb{R}$ be the $ax + b$ group with $N = \{(1,\*)\}$ and take $g = (0,1)$. To eliminate this type of counterexample, you could replace $g$ on the right-hand side of your equation with $\sigma(g)$.




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Here is a different type of counterexample (not a semidirect product). Let $N$ be the center of the discrete Heisenberg group:
 $$ N = \begin{bmatrix} 1 & 0 & \mathbb{Z} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
 \subset
 \begin{bmatrix} 1 & \mathbb{Z} & \mathbb{Z} \\ 0 & 1 & \mathbb{Z} \\ 0 & 0 & 1 \end{bmatrix}
 = G .$$
Let $\{a,b\}$ be a generating set of $G$, let $K = \{a,b,a^{-1},b^{-1}\}$, and let $z = [a,b] = a^{-1} b^{-1} a b \in N$.
Then
 $$ z^{n^2} \sigma(1 \cdot N) 
 = a^{-n} b^{-n} a^n b^n \, \sigma(1 \cdot N) 
 = \sigma (a^{-n} b^{-n} a^n b^n \cdot N) \cdot \prod\_{i=1}^{4n} \omega( s\_i, x\_i )
 = \sigma (1 \cdot N) \cdot \prod\_{i=1}^{4n} \omega( s\_i, x\_i )
 ,$$
since $a^{-n} b^{-n} a^n b^n = z^{n^2} \in N$,
where each $s\_i$ is in $K$ and $x\_i = s\_{i-1} s\_{i-1} \cdots s\_1 N$. The exponent of $z$ on the left-hand side is a quadratic function of $n$, and the right-hand side has only linearly many terms in its product, so $\omega( s\_i, x\_i )$ cannot be a bounded element of $N$.


