Isomorphism between $H_1(X)$ and $\pi_1(X)^{ab}$ I'd like to understand the proof of the following theorem, born from the conciseness of the proof given. It's also called Hurewicz's theorem in the special case $n=1$. The essential doubts about the proof, which doesn't go in detail about, are :
$\textbf{1}.$ Is $l'$ well defined ? It seems to me that the $u(x)$ is a generic path, which I think is not what we want since doesn't give the property $lh = id$. For example given a loop $\sigma = h([\gamma])$ with basis in $x_0$, we could consider instead of $u(x_0) = c_{x_{0}}$ the costant loop in $x_0$, a random loop so that we have a "$3$-loop" which doesn't return in $[\gamma]$ with $l'$.
$\textbf{2}.$ I don't how $\triangle^2$ being contractible implies that $\tau_2 \star \tau_1 \sim \tau_1$. I think I can prove the statement using that the simplex is convex but I don't see how the contractbile property of the space implies such homotopy
$\textbf{3}.$ I think the question three is linked to question one, I'd like to understand why the last equality holds : $[\sigma] = [u(\sigma_0)]+[\sigma] - [u(\sigma_1)] =  [(u(\sigma_0)\star \sigma)\star u(\sigma_1)^-]$
I'm currently reading the proof on Algebraic Topology of Tammo Tom Dieck, I add below a picture of the proof of the theorem.

 A: For question 1, $l'$ is well-defined because $C_1(X)$ is a free abelian group generated by the 1-simplices of $X$, so you get a well-defined map from $C_1(X)$ to any other abelian group by specifying where the generators go. If you have a loop based at $x_0$, then the key is that you have to use the same path $u$ from $x_0$ to $x_0$ at both ends in the definition of $l'$, traveling along $u$ once in each direction, so $l h$ will indeed be the identity map.
For question 2, the statement is accurate, but maybe to make it easier, the author should have said that $\Delta^2$ strongly deformation retracts to any of its vertices (for example because of convexity). But see Path homotopy in contractible space if you just want to use contractibility.
For question 3, it is not the case that $[\sigma] = \dots$. Rather, the extra terms $\sum a_{\sigma}([u(\sigma_0)] - [u(\sigma_1)])$ taken together give precisely the boundary of $\sum a_{\sigma} \sigma$ and therefore add up to zero, since the original sum is assumed to be a cycle. For the last equality, $u(\sigma_0) + \sigma - u(\sigma_1)$ should be homologous to $(u(\sigma_0)*\sigma)*u(\sigma_1)^{-}$, so the sums are equal when working in $H_1$.
