In continuation with previous question .
Let $\Omega=\{x_1,x_2\}$ be a two element set, and $f:\Omega\to\mathbb R$ be defined as
$$f(x_1)=1,f(x_2)=2$$
Consider $\Sigma_1=\{\varnothing ,\Omega\}$ the trivial $\sigma$-algebra, and $\Sigma_2 = P(\Omega)$ the powerset. Then, $f$ is $\Sigma_2$-$B(\mathbb R)$-measurable, because if the domain has the power set as the $\sigma$-algebra,then every function is measurable. One of the comments in answer states $f^{−1}(\{3\})=\varnothing\in\Sigma_1$.The empty set belongs to every $\sigma$-algebra.
Does this also mean that $f(\{\phi\}) = \{3\}$? Since a function should map every element from $\Omega$, is this function even defined for null set?