This is a reult in algebraic topology, where we study the structure of topological spaces $X$. One early way to do this is to calculate a thing called $H_*(X)$, the ordinary homology of $X$. Later people discovered various "extraordinary homologies", which give more precise information. There are very many different extraordinary homologies, including those called $P(n)_*(X)$, $B(n)_*(X)$ and $K(n)_*(X)$. Of these, $P(0)$ (also called $BP$, or Brown-Peterson homology) is the most powerful, but it is often very hard to calculate. At the other end of the scale, $K(0)$ is the weakest and easiest to calculate. In general $K(n)$ is reasonably easy. Roughly speaking, the information in $P(n+1)$ is the information in $P(n)$ minus the information in $K(n)$, so all the $P(n)$'s are often hard to calculate.

From the definitions, the obvious guess would be that $B(n)$ is only a little easier than $P(n)$. However, this turns out to be wrong: $B(n)$ contains exactly the same information as $K(n)$ (and so is much easier than $P(n)$). If you know $K(n)_*(X)$ then Würgler's theorem allows you to calculate $B(n)_*(X)$, and a different but easier theorem lets you go in the opposite direction.

This is a result in algebraic topology, where we study the structure of topological spaces $X$. One early way to do this is to calculate a thing called $H_*(X)$, the ordinary homology of $X$. Later people discovered various "extraordinary homologies", which give more precise information. There are very many different extraordinary homologies, including those called $P(n)_*(X)$, $B(n)_*(X)$ and $K(n)_*(X)$. Of these, $P(0)$ (also called $BP$, or Brown-Peterson homology) is the most powerful, but it is often very hard to calculate. At the other end of the scale, $K(0)$ is the weakest and easiest to calculate. In general $K(n)$ is reasonably easy. Roughly speaking, the information in $P(n+1)$ is the information in $P(n)$ minus the information in $K(n)$, so all the $P(n)$'s are often hard to calculate.

From the definitions, the obvious guess would be that $B(n)$ is only a little easier than $P(n)$. However, this turns out to be wrong: $B(n)$ contains exactly the same information as $K(n)$ (and so is much easier than $P(n)$). If you know $K(n)_*(X)$ then Würgler's theorem allows you to calculate $B(n)_*(X)$, and a different but easier theorem lets you go in the opposite direction.