Thursday, November 19, 2009

"Easy" Isomorphisms

So, if you read Hatcher (or anyone half-decent) on homotopy theory, you run into a long exact sequence of groups depending on an inclusion $j:U\rightarrow X$.

$\displaystyle{ \cdots \longrightarrow \pi_2(U)\longrightarrow \pi_2(X)\longrightarrow\pi_1(X|U) \longrightarrow \pi_1(U)\longrightarrow\pi_1(X)}$

where I'm indexing the relative homotopy functors $\pi_n(X|U)$ off by one from Hatcher's notation. This makes it easier to remember when they start being abelian, when they don't have a group structure anymore ... it's also handy to note that $\pi_n(X|U)$ actually is $pi_n$ of a functorial construct on the inclusion $j$.

Anyways, we have this long exact sequence, and it comes in handy whenever $U$ (or $X$) and $(X|U)$ --- whatever that means --- have easy homotopy groups. For instance, the 2-sphere and the 1-sphere both have easy homotopy groups; it just so happens that a long slog of an argument gives us that any Jordan Curve is a topological 1-sphere. Specializing the long exact sequence to a Jordan curve $J$ in a 2-sphere, we find

$\displaystyle{ \cdots 0 \longrightarrow \mathbb{Z} \longrightarrow \pi_1(S^2|J) \longrightarrow \mathbb{Z} \longrightarrow 0 }$

at the end of our sequence. Now, from algebra, there are only two possible groups that can sit in for $\pi_1(S^2|J)$; one of them is abelian, one has an abelian subgroup of index $2$; in any case, the long exact sequence says that $\pi_1(S^2|J)$ has an element $x$ represented by a disc whose boundary is mapped to the Jordan curve $J$ --- that's surjectivity of the last nonzero map --- and has another generator $y$ mapping the boundary of the disc to a single point --- that's by injectivity of the first nonzero map. We may just as well instead take as generators $x,xy$, so they both wrap the disk's edge around $J$ once.

Now, We're not quite at the Jordan curve theorem yet, but we're pretty-darn close!


Anonymous said...

I don't know how you speak this language.



Post a Comment