Brushing up on Categories for the Fundy Group.

May 7, 2010

Last post I talked about fundamental groups and some of the stuff that we can do with them.  For example, we have that a disk is different from a circle, since the disk has a trivial fundamental group and the circle’s fundamental group is the integers.

$\pi_{1}(D^1) = 0$

$\pi_{1}(S^1) = {\mathbb Z}$

But is this enough to say anything about how truly different the circle and the disk are?  Yes, the circle has a hole, and the disk does not have a hole, but is there some kind of continuous deformation that let’s a circle become a disk?  Or one that let’s a disk become a circle?  Maybe we could pull at the disk and spin it around a lot until it becomes a circle, continuously!  We just don’t have the machinery at this point to be able to prove that the circle and the disk are not similar in the sense that there is no continuous deformation between them.  Upsetting, but true.

Before we can build up this machinery, and can truly think about spaces and fundamental groups as truly related species, we need to build up something called Category theory.

After the jump.

Let’s get some definitions out of the way here.

Defn: A category, ${\mathcal C}$ is defined in the following way: we have a section inside of it for objects (these are like the elements in a set) and morphisms (also called maps) between objects.  The morphisms have to follow certain composition rules (for each of these, let $f: A\rightarrow B$, $g: B\rightarrow C$, and $h: C\rightarrow D$):

1. Associativity: We must have that $h\circ ( g \circ f ) = (h\circ g) \circ f$.
2. Identity: We must have, for every object x in the category, we must have an identity morphism denoted $1_{x}$ such that $1_{A}(A) = A$ and such that for every map $f:A\rightarrow B$ we have $1_B \circ f = f = f \circ 1_A$.

That’s it.  A category is just a box full of objects that all have maps from one to another.  The maps have to be sufficiently nice (in that they must associate) and there has to be an identity map.  Notice, though, that there is no mention of inverses; this is no accident.  There does not have to be an inverse for a particular map in the category.

For now, we really only care about two categories: Top, the category with topological spaces as objects and continuous maps between them as the morphisms; and Grp, the category with groups as objects and group homomorphisms as morphisms.

Now, with regular ol’ sets, we can map them back and forth between each other with maps: for example, we can map the set $\{\mbox{possum, fox, orange, apple}\}$ to the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$ by sending the word in the first set to the number of letters it has in the second set.  So possum goes to 6, fox goes to 3, orange goes to 6, and apple goes to 5.  Similarly, we could send them to numbers based on some rank.  Or, we could send them all to the number 4.  It doesn’t matter, really, they’re just maps.

With categories, we have a bit more structure: we have objects and maps!  So if we want maps between two categories, we need to not only map objects to objects, but maps to maps.  We, for some reason, call maps between categories functors.

Defn:  A (covariant) functor $F: {\mathcal C}\rightarrow {\mathcal D}$ is a map between categories such that for each object $x\in {\mathcal C}$ there exists a corresponding object $F(x)\in {\mathcal D}$, and (this is the important part to remember!) for each set of morphisms $f, g \in {\mathcal C}$ we have corresponding morphisms $F(f), F(g) \in {\mathcal C}$ such that if we have $g\circ f \in {\mathcal C}$, this corresponds to $F(g\circ f) \in {\mathcal C}$ and $F(g \circ f) = F(g)\circ F(f))$.  Also, identity functions need to go to identity functions: $F(1_{a}) = 1_{F(a)}$.

That last part, that $F(g\circ f) = F(g)\circ F(f)$, is pretty important, damnit, so remember it.

Okay, yeah, now you’re probably saying to yourself, “Well, that’s all fine and good, but I’ve never seen a functor before, and I probably never will!”

But SURPRISE because $\pi_{1}$ is a functor from the category of topological spaces to the category of groups!  Yes, that’s right.  For each space, $\pi_{1}$ associates a group to this space.  Hence,

$\pi_{1}(S^1) = {\mathbb Z}$

where $S^1$ is a space and ${\mathbb Z}$ is a group.  Cute.  This all is nontrivial to prove (namely, we can’t just SAY that things are a functor and it becomes true somehow.  We’d need to PROVE this, but this is done in several places around the internet, and it’s not all that difficult to prove for yourself.)  So, now, what can we do with this?

Thm: If X and Y are spaces, and $h: X\rightarrow Y$ is a homeomorphism (that is, a bijective continuous map with continuous inverse) then $\pi_{1}(X)$ and $\pi_{1}(Y)$ are isomorphic.

This theorem, which is proved in any standard algebraic topology book, is extremely useful and it’s essentially a consequence of the functoriality.  How can we use this?  Well, here’s one thing.

Cor: $S^1$ isn’t homeomorphic to $D^1$.

Because if it were, then their fundamental groups would  be isomorphic.  But, in particular, $0$ has one element, and the group of integers has at least two.  Best contradiction ever.