How to show the kernel of map $F: \pi_1(X,p)\to H_1(X)$ is just $[\pi_1,\pi_1]$? It's well known that for a path connected topological space $X$, $H_1(X)$ is isomorphic to the abelianization of $\pi_1(X,p)$, i.e. $\pi_1/[\pi_1,\pi_1]$.
For the proof, consider the map $F: [\alpha]_{\pi}\in \pi_1(X,p) \to [\alpha]_h \in H_1(X)$.  It's a homomorphism. It's easy to see that $[\pi_1,\pi_1]\subset Ker F$, how to prove that $ker F=[\pi_1,\pi_1]$?
For a singular 2-simplex $\sigma$,  $\partial \sigma=\sigma(e_1e_2)-\sigma(e_0e_2)+\sigma(e_0e_1)$, let $x=\sigma(e_0),y=\sigma(e_1),z=\sigma(e_2)$. In the book "Introduction to topological manfiold" by John Lee, Page 354. It's shown that the loop formed by $pxyppyzppzxp$ is equivalent to constant loop $c_p$ modulo $[\pi_1,\pi_1]$.
However, it seems to me that $pxyppyzppzxp$  is homotopic to the constant loop $c_p$.  First the loop $xyz$ based at x is homotopy to the constant map $c_x$ since $xyz$ is  the boundary of $\sigma$. Just choose $y'$ on $xy$, $z'$ on $xz$.  For $t=0$, $y'=y$, $z'=z$, as $y',z'$ go to $x$, $xy'z'$ go to $c_x$. So $pxy'ppy'z'ppz'xp$ go to $pxxppxxppxxp$, then go to $c_x$.
Can one explain where I am wrong?
 A: Your notation is not understandable. Only if one has a look into Lee's book, one can guess what you mean. The ominous loop $pxyppyzppzxp$ seems to  stand for the path in $X$ obtained by travelling from $p$ to $x$ to $y$ etc. (Fig. 13.8). But why did you replace $v_0$ by $x$, $v_1$ by $y$  and $v_2$ by $z$? And it is not explicit in $pxyppyzppzxp$ which precise path we take from one point to the next. My general advice is to write questions in such a manner that readers can understand them without doing "research". Otherwise you save some time, but readers (in general more than one) have to invest much more time because they are not familiar with the context. That said, let us come to the answer.
You are right, $pxyppyzppzxp$ is homotopic to the constant loop $c_p$. Lee actually proves that - you can replace $[\dots]_\Pi$ by $[\dots]_\pi$.
$[pxyppyzppzxp]_\pi =[c_p]_\pi$ in $\pi_1(X)$ is of course stronger than $[pxyppyzppzxp]_\Pi =[c_p]_\Pi$ in $\Pi = \pi_1/[\pi_1,\pi_1]$, but where is the problem?
The point is that Lee needs $\Pi = \pi_1/[\pi_1,\pi_1]$ to get a well-defined map $\beta : C_1(X) \to \Pi$. He defines $\beta$ on the generators of the free abelian group $C_1(X)$, and since $\Pi$ is abelian he gets a unique homomorphism $\beta : C_1(X) \to \Pi$.
However, on the generators $\sigma$ he can also define $\beta'(\sigma) = [\tilde \sigma]_\pi \in \pi_1(X)$, but if $\pi_1(X)$ is not abelian, there is no chance to get an extension to a  homomorphism $\beta' : C_1(X) \to \pi_1(X)$. If you know  the concept of the fundamental groupoid of a space $X$ (which we may denote by $\Pi(X)$) you will see that we get a homomorphism $\beta' : \Pi(X) \to \pi_1(X)$. Just define $\beta'([\sigma]) = [\tilde \sigma]_\pi$ for each path homotopy class in $\Pi(X)$. Then we get $\beta'([\sigma_1] \cdot \ldots \cdot [\sigma_n]) = \beta'([\sigma_1]) \cdot \ldots \cdot \beta'( [\sigma_n])$. Note that this equation only makes sense if $[\sigma_1] \cdot \ldots \cdot [\sigma_n]$ is defined, i.e. if the end point of $\sigma_i$ agrees with the initial  point of $\sigma_{i+1}$. This is what Lee really uses in his proof.
More conceptually, let $P(X)$ denote the set of all paths in $X$. This is the canonical basis of the free abelian group $C_1(X)$. We get a commutative diagram
$\require{AMScd}$
\begin{CD}
P(X)  @>{}>> \Pi(X) @>{\beta'}>> \pi_1(X)\\
@V{}VV @. @V{}VV \\
C_1(X) @>{\beta}>> @. \pi_1(X)/[\pi_1(X),\pi_1(X)]
\end{CD}
