Homology and Fundamental Group (Algebraic Topology - Allen Hatcher) I have some questions regarding 2 parts of the theorem of this section:
1) Having $f = \sum_{i,j}(-1)^jn_i\tau_{ij}$, it pairs $\tau_{ij}$ with opposite signs (in any way I assume), and says that the non paired $\tau_{ij}$'s are $f$.
I don't get why necessarily only some of the $\tau_{ij}$'s are $f$, and why it's ok to just pair any $\tau_{ij}$'s with opposite sign, because identifying the edges of the paired $\tau_{ij}$'s it then forms a $\Delta$-complex $K$, and then extends the maps $\sigma_i$ to a map $\sigma : K \to X$, but for $\sigma$ to be well defined, shouldn't the paired $\tau_{ij}$'s be exactly the same?
2)In the last paragraph of the proof it writes $[f] = \sum_{i,j}(-1)^jn_i[\tau_{ij}] \in \pi_1(X)_{ab}$ in additive notation, so the sums mean composition, but that means that $f$ is (or is homotopic to) $\prod_{i,j}\tau_{ij}^{(-1)^jn_i}$ as a loop, and I don't know where do you get that information, the information you have is $f = \sum_{i,j}(-1)^jn_i\tau_{ij}$, but as I see it that doesn't mean $f = \prod_{i,j}\tau_{ij}^{(-1)^jn_i}$ as a loop, since we have $f\simeq g \Rightarrow f \sim g$ but not $f\simeq g \Leftrightarrow f \sim g$.
Sorry for the extended "question" and thank you in advance.

1) This information is not directly for gluing, rather it is information on the orientation of the simplex. The goal is to obtain a coherently oriented two-dimensional $\Delta$ complex with boundary equal to $[f]$. Look back at the beginning of the chapter when Hatcher writes down the formula for the boundary of a 3-simplex. Notice that each of the 1-simplices appears twice, each time in an opposite direction. This means that they don't contribute to the boundary of the 2-complex.
2) The misconception here is that the sum represents loop composition, which it doesn't. Rather, the sum is to be interpreted in the abelianization of the fundamental group. As an example of what I mean here, let $X = S^1 \vee S^1$, which has a fundamental group which is free on two generators we will call $a$ and $b$. In particular, $f = aba^{-1}b^{-1}$ is not trivial. However, $[f] = [ aba^{-1}b^{-1}] = [a] + [b] - [a] -[b] = 0$, since homology is abelian. What the theorem says is that anything in the kernel of the homomorphism $\pi_1 \to H_1$ can be decomposed into loops which can be reordered into a null-homotopic loop, just as $f = aba^{-1}b^{-1}$ can be reordered to get $g = aa^{-1}bb^{-1}$. The claim isn't that $f$ and $g$ are homotopic, but that $[f] = [g]$. 
