How can we prove that $g = O(f) $ (Big-Oh notation) Problem description
I am trying to understand the first example of Big-Oh notation in Computational Complexity:A Modern Approach page 9.
If $f(n) = 100nlogn$ and $g(n) = n^2$ then we have the relations $f = O(g)$, $g = Ω(f)$, $f = o(g)$, $g = ω(f)$.

I understood why $f = O(g)$ and $f = o(g)$ . As for the rest I didn't. For instance in order for $g$ to be equal to $Ω(f)$ we must have $g = O(f)$.

Question
How can we prove that $g = O(f)$ and $g=o(f)$?
I think you have the wrong definition for $\Omega$. When we write $g=Ω(f)$, informally this means that $g$ is bounded below by ${\displaystyle f}$  asymptotically. So, it is basically just the reverse way of saying $f=O(g)$, which means that $|f|$ is bounded above by ${\displaystyle g}$ (up to constant factor) asymptotically. 

What you are describing is closest to the definition of Big theta $\Theta$.