The First Homology Group is the Abelianization of the Fundamental Group.

I am trying to understand the proof of the following fact from Hatcher's Algebraic Topology, section 2.A.

Theorem. Let $X$ be a path connected space. Then the abelianization of $\pi_1(X, x_0)$ is isomorphic to $H_1(X)$.

I am having trouble understanding the last step of the proof.

Step 1. First we define a map $h:\pi_1(X, x_0)\to H_1(X)$ which sends the homotopy class $[f]$ of a loop $f$ based at $x_0$ to the homology class of the cycle $f$. One checks that this is a well-defined group homomorphism. Further $h$ is surjective.

Step 2. Now since $H_1(X)$ is abelian, the map $h$ factors through the abelianization $\pi_1(X, x_0)^{ab}$ of $\pi_1(X, x_0)$.

Step 3. The main part is to show that the induced map $\pi_1(X, x_0)^{ab}\to H_1(X)$ is injective. To do this we need to show that if a loop $f$ based at $x_0$ in $X$ is a boundary, then $f$ is in the commutator subgroup. Since $f$ is a boundary, we have singular $2$-simplices $\sigma_i$ such that $f=\partial(\sum_i n_i\sigma_i)$. We have $\partial \sigma_i = \tau_{i0} - \tau_{i1} + \tau_{i2}$, where $\tau_{ij}$'s are singular one simplices obtained by restricting $\sigma_i$ on the edges of the standard $2$-simplex.
Hatcher shows that one may assume that each $\tau_{ij}$ is a loop based at $x_0$.

Step 4. This step is where I am having problem. Hatcher writes "Using the additive notation in the abelian group $\pi_1(X, x_0)^{ab}$, we have the formula $[f]=\sum_{i, j}(-1)^jn_i[\tau_{ij}]$ because of the cancelling pairs of $\tau_{ij}$'s. We can rewrite the summation $\sum_{i, j} (-1)^j n_i[\tau_{ij}]$ as $\sum_i n_i [\partial \sigma_i]$ where $[\partial \sigma_i]=[\tau_{i0}]-[\tau_{i1}]+[\tau_{i2}]$. Since $\sigma_i$ gives nullhomotopy of the composed loop $\tau_{i0}-\tau_{i1}+\tau_{i2}$, we conclude that $[f]=0$ in $\pi_1(X, x_0)^{ab}$."

I do not seem to understand anything in the paragraph quoted above. Can somebody elaborate on that? The additive notation is especially confusing. So if possible please use the multiplicative notation. Thank you.

Solution 1:

I do not understand a lot of the last paragraph either, but I will give a slightly different proof of the statement based on Bredon, Topology and Geometry, p. 174.

For every point $x\in X$, fix a path $\lambda_x$ from $x_0$ to $x$. Let $\sigma$ be a singular 1-simplex in $X$, i.e. a continuous path $\sigma\colon[0,1]\to X$. Then $\Psi(\sigma):=\overline{\lambda_{\sigma(1)}}\cdot\sigma\cdot\lambda_{\sigma(0)}$ is a loop based at $x_0$, where the dot is juxtaposition of paths and $\overline\lambda(t)=\lambda(1-t)$ is the inverse path. I claim that this induces a well-defined map $\Psi_*\colon H_1(X)\to\pi_1(X,x_0)^{\text{ab}}$, such that $\Psi_*h\colon\pi_1(X,x_0)\to\pi_1(X,x_0)^{\text{ab}}$ is the canonical projection. From this it follows that $h$ induces a monomorphism $\pi_1(X,x_0)^{\text{ab}}\to H_1(X)$.

First well-definedness: Since $\pi_1(X,x_0)^{\text{ab}}$ is abelian, $\Psi$ extends to a homomorphism $\Psi\colon\Delta_1(X)\to\pi_1(X,x_0)^{\text{ab}}$ from the group of singular 1-chains. It is now straightforward to check that boundaries of singular 2-simplices go to zero in $\pi_1(X,x_0)^{\text{ab}}$ (I can elaborate on this if you like), so $\Psi$ induces a homomorphism $\Psi_*\colon H_1(X)\to\pi_1(X,x_0)^{\text{ab}}$ as claimed.

Now if $\sigma$ is a loop based on $x_0$ then $\Psi(h(\sigma))=\overline{\lambda_{x_0}}\cdot\sigma\cdot\lambda_{x_0}$, and the class of this in $\pi_1(X,x_0)^{\text{ab}}$ equals the class of $\sigma\cdot\overline{\lambda_{x_0}}\cdot\lambda_{x_0}\simeq\sigma$, so that $\Psi_*h$ is the canonical projection.

Solution 2:

I follow the proof by Hatcher that the OP outlines.

The first important point is the following, once you have proved that $f = \Sigma_{i,j}(-1)^jn_i\tau_{i,j}$ for your 1-cycle $f$, you need to remember that the group of 1-chains (which contains the 1-cycles as a subgroup) is the free $\mathbb{Z}$-module (ie the free abelian group) on the set of continuous maps from $\Delta_1$ to $X$. Hence, in particular, every element $f$ of this group as a unique expression of the form $n_1f_1+\ldots +n_pf_p$ with $n_i\in\mathbb{Z}$ and $f_i$ a continuous map from $\Delta_1$ to $X$. So, from the equality $f = \Sigma_{i,j}(-1)^jn_i\tau_{i,j}$ you conclude, as noted by Hatcher (third paragraph from the end), that $f$ is one of the $\tau_{i,j}$'s and the remaining $\tau_{i,j}$'s form canceling pairs. This allows you to state this equality between homotopy classes $[\Sigma_{i,j}(-1)^jn_i\tau_{i,j}] = \Sigma_{i,j}(-1)^jn_i[\tau_{i,j}]$. But $\Sigma_{i,j}(-1)^jn_i[\tau_{i,j}] = \Sigma_in_i[\partial\sigma_i]$.

The second important point consists in noting the fact that, $\sigma_i$ being a singular 2-simplex with boundary given by $\tau_{i0} -\tau_{i1} +\tau_{i2}$, you can continuously deform this boundary, through $\sigma_i$, into the constant loop at $x_0$. Hence, one has the following equalities between homotopy classes $[\partial \sigma_i] = [\tau_{i0}]-[\tau_{i1}]+[\tau_{i2}]=0$.

Now, third point, you need to realize this last equality means $[\tau_{i2}] = -([\tau_{i0}]-[\tau_{i1}])$, or in multiplicative notation as you wish $[\tau_{i2}]=([\tau_{i0}][\tau_{i1}]^{-1})^{-1}$. Thus $[\partial\sigma_i]=[\tau_{i0}] -[\tau_{i1}] +[\tau_{i2}]$, being an element followed by its inverse, belongs to the commutator subgroup and so $\Sigma_in_i[\partial\sigma_i]=[f]$ does, meaning that $[f]$ is trivial in $\pi(X,x_0)_{ab}$.