I am trying to understand the proof of the following fact from Hatcher'sAlgebraic 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.

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.Step 1.

Further $h$ is surjective.

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 2.

The main part is to show that the induced map $\pi_1(X, x_0)^{ab}\to H_1(X)$ is injective.Step 3.

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.