Abelianization of the Fundamental Group

I guess I have no reason to offer explanation for lack of posting, but in general this has been one of the best weeks ever and at the same time one of the worst. The worst because I’ve been fairly ill and can’t seem to fully conquer it. It has been the best week for reasons I won’t mention, since I try to keep personal stuff out of this blog as much as possible (but if you know of my other blog which is purely my personal stuff, then you can read about it to your heart’s content, but I refuse to give any hints at all as to how to find that). Both of these factors has lead to a fairly unproductive week.

I may take a more algebraic topology approach for awhile. This is mainly since I’m doing a reading course on Hatcher (with two other students), and before I go present stuff to them and the prof I want to clarify my ideas.

Tomorrow I’m presenting the proof that $H_1(X)\cong \pi_1(X)^{ab}$ for path connected spaces. This is a pretty wonderful result if you think about it. We have exactly how first homology and the fundamental group relate. In fact, the first thing you’d think to do (granted, this might take a little while) is the thing that works.

We can naturally think of paths and singular 1-simplices as the same thing, since they are both just continuous maps to the space out of a closed interval. So after rescaling, a loop $f:[0,1]\to X$ is actually also a 1-cycle since $\partial f=f(1)-f(0)=0$.

The overall idea of this proof is then to show that $h: \pi_1(X, x_0)\to H_1(X)$ is a well-defined homomorphism with image all of $H_1(X)$ and kernel the commutator subgroup. Almost all of these facts are fairly straightforward.

First, we’ll need a few ways in which our different modes of thinking about loops versus 1-cycles correlate. If as a path $f\equiv c$, a constant, then $f\sim 0$ the cycle is homologous to 0. If two paths are homotopic (in the path homotopic and hence equivalence class of $\pi_1(X)$ sense), denoted $f\simeq g$ then they are homologous. Concatenation of paths (and hence the operation in the fundamental group) is homologous to addition of the cycles (the operation in first homology). Lastly, traversing a path backwards is homologous to negating the cycle: $\overline{f}\sim -f$.

So we’ll use these four facts without proof, since they are fairly standard and the proof is long enough as it is.

Recall the definition $h([f])=f$. The second fact, gives us that $h$ is well-defined since any other representative of the equivalence class will be homotopic to the original, and hence the outputs will be homologous.

The third fact gives us that $h$ is a homomorphism of groups.

Our first bit of effort comes from showing that $h$ is surjective. Here we will use the path-connected hypothesis (everything else so far is true without it). Let $\sum n_i\sigma_i$ be any 1-cycle. We must construct a loop that maps to it.

Since the $n_i$ are integers, we can assume each is $\pm 1$ by just repeating the $\sigma_i$ as many times as needed. But all the $\sigma_i$ with -1 in front can be replaced by $-\overline{\sigma_i}$ by the fourth property. This converts all the $n_i$ to 1. Thus $\sum n_i\sigma_i\sim \sum \sigma_k$.

But $\partial(\sum \sigma_k)=0$, so all the endpoints must cancel. So for any $\sigma_k$ that is not a loop, in order to cancel $\sigma_k(1)$, there must be a $\sigma_j$ such that $\sigma_j(0)=\sigma_k(1)$. i.e. there is some $\sigma_j$ that we can concatenate with to form $\sigma_k\cdot \sigma_j$. In order to cancel the $\sigma_k(0)$ some other $\sigma_j$ must exists with endpoint $\sigma_j(1)=\sigma_k(0)$.

So we can concatenate, then rescale, and group all of these cycles into a collection of loops by the third property. So the only remaining thing we must do is get it to be a single loop. But $X$ is path-connected, so pick some basepoint $x_0\in X$. For any of these possibly disjoint loops floating around, we can pick a basepoint at each and connect with a path $\gamma_i$ from $x_0$ to the basepoint of the i-th loop. By the third and fourth properties $\gamma_i\cdot \sigma_i\cdot \overline{\gamma_i}\sim \sigma_i$. So now all loops start and end at $x_0$ and we can combine into a single loop $\sigma$. Thus $h([\sigma])=\sum n_i\sigma_i$.

Now comes the hard part. We want $ker h=\pi_1(X, x_0)'$. The one containment is easy. Since $H_1(X)$ is abelian, by the universal property of the commutator subgroup, $\pi_1(X)'\subset ker h$. The method to get the other direction is to show that for any $h([f])=0$, we must have that $[f]$ is trivial in the abelianization.

Suppose $[f]\in\pi_1(X)$ such that $h([f])=0$. Since $f$ is a cycle, there is some 2-chain $\sum n_i\sigma_i$ such that $\partial (\sum n_i\sigma_i)=f$. So as before, we can assume each $n_i=\pm 1$. Now the goal is to associate a 2-dimensional $\Delta$-complex to $\sum n_i\sigma_i$ by taking for each $\sigma_i$ a $\Delta_i^2$ and identifying pairs of edges which we’ll call $K$.

So before writing this process down, we should examine what the process will be geometrically. It turns out that $K$ will be an orientable compact surface with boundary, since we are just fitting together a finite collection of disjoint 2-simplices (this is not meant to be obvious). The component containing the boundary is a closed orientable surface with an open disk removed. Since connected sums of tori can be expressed as a 2n-gon with pairs of edges identified in the manner $aba^{-1}b^{-1}, cdc^{-1}d^{-1}$ etc, we see that $f$ is homotopic to a product of commutators.

Writing this in detail algebraically is much trickier. Given any $\sigma_i$, we have $\partial \sigma_i=\tau_{i0}-\tau_{i1}+\tau_{i2}$, where $\tau_{ij}$ are singular 1-simplices. Thus $f=\partial(\sum n_i\sigma_i)=\sum (-1)^j n_i\tau_{ij}$.

Keep the picture of a triangle in your head. When we fit together the triangles we are getting pairs of edges. The signs on these pairs are opposite and so will cancel when we sum. The remaining (of the three sides) $\tau_{ij}$ is a copy of $f$. This forms our $\Delta$-complex $K$.

Now form $\sigma : K\to X$ by fitting together the $\sigma_i$ maps. Deform $\sigma$ relative the edges that correspond to $f$ by mapping each vertex to $x_0$. So we have a homotopy on the union of the 0-skeleton with edge $f$, so by the homotopy extension property we get a homotopy on all of $K$.

Now restrict $\sigma$ to the simplices $\Delta_i^2$ to get a new chain $\sum n_i\sigma_i$ with boundary $f$ and $\tau_{ij}$ loops at $x_0$.

Now we just need to check whether the class is trivial or not: $[f]=\sum (-1)n_i[\tau_{ij}]=\sum n_i [\partial \sigma_i]$ where $[\partial \sigma_i]=[\tau_{i0}]-[\tau_{i1}]+[\tau_{i2}]$. But $\sigma_i$ gives a nullhomotopy of $\tau_{i0}-\tau_{i1}+\tau_{i2}$ and we are done.

Thus $ker h=\pi_1(X, x_0)'$ and by the First Iso Theorem we have $H_1(X)\cong \pi_1(X, x_0)^{ab}$.

6 thoughts on “Abelianization of the Fundamental Group”

1. *gasp* another blog!?! No wonder you manage to keep this one focused on your work! đ

2. hilbertthm90 says:

That is definitely the trick. Otherwise I’d be venting all sorts of craziness for you all to see.

3. So you get everyone else’s Ranty McCrazy business, and all we get is math with the occasional bit of art thrown in? No fair!

Okay, just kidding. Sort of. You know I want to read the other stuff. đ

4. This is exactly the same as Hatcher’s book, with no further explanations!

5. Plutonium says:

Sorry for the necropost: would you please remind me, or give me a ref. of the definition of the map h from Pi_1(X) to H_1(X) ? Is it just the Abelianization functor?

6. hilbertthm90 says:

The map h is given by taking a representative for your loop and then considering it as a 1-cycle. It isn’t a priori obvious this is well-defined, but the post sketches how to check this (at least where H_1(X) is singular homology).