Given a space $X$, construct a CW complex $L(X)$ s.t. they have the same fundamental group This is exercise 1.2.15 in Hatcher's Algebraic topology

Given a space $X$ with basepoint $x_0∈X$, we may construct a CW complex $L(X)$ having a single $0$-cell, a $1$-cell $e^1_γ$ for each loop $γ$ in $X$ based at $x_0$, and a $2$-cell $e^2_τ$ for each map $τ$ of a standard triangle $PQR$ into $X$ taking the three vertices $P,Q$ and $R$ of the triangle to $x_0$.
The $2$-cell $e^2_τ$ is attached to the three $1$-cells that are the loops obtained by restricting $τ$ to the three oriented edges $PQ,PR$ and $QR$.

Show that the natural map $L(X)→X$ induces an isomorphism $π_1(L(X))\cong π_1(X,x_0)$.
I met some problem in constructing the CW complex $L(X)$, here are my thoughts:
$1$. $L(X)$ have a single $0$-cell, and for each loop $γ$ in $X$ based at $x_0$, is has a $1$-cell $e^1_γ$.
At this time, $L(X)$ is a wedge sum of circles, each circle $S^1$ represents a loop in $X$ based at $x_0$.
$2$. For map $\tau:\text{triangle } PQR\to X$, $\tau$ maps $P,Q,R$ to $x_0$ and maps $PQ$, $PR$, $QR$ to loops in $X$ based at $x_0$.
Let $\overrightarrow{PQ}$ correspond to loop $a$ in $X$ (also a loop in $L(X))$, $\overrightarrow{PR}$ correspond to loop $b$ , then $\overrightarrow{QR}$ correspond to $a^{-1}b$.
$3$. $2$-cell $e^2_\tau$ is attached to the three $1$ cells $a,b,a^{-1}b$, we obtain relation $a\cdot a^{-1}b \cdot b^{-1}=1$, which is trivial.
There's something wrong here since I don't fully understand the construction of $L(X)$. So how does the 2-cells $e^2_\tau$ attached to $1$-skeleton of $L(X)$?

Update:
I now realized that triangles in $L(X)$ is just triangulation of $2$-cells in $X$,
and it doesn't change homotopy type, so $\pi_1(L(X))\cong \pi_1(X,x_0)$

$L(X)$ is attempting to "model" $X$ as a CW complex with the same fundamental group. To do this, we will consider all possible loops in $X$, then specify which ones should be considered homotopic. 


*

*As you note, we have a wedge sum of circles, one for each loop in $X$. You can think of each summand as "representing" it's corresponding loop in $X$. At the moment, this is a poor model of $X$, as homotopic loops in $X$ are not homotopic in $L(X)$. We will fix this in part 2.

*Convince yourself that if I have a map $\tau : PQR \to X$ where the vertices are sent to $x_0$, each edge is a loop which is homotopic to the product of the other two (possibly with inverses, depending on orientation). Also convince yourself that any two homotopic loops in $X$ have a $\tau$ which shows this homotopy. We want loops in $L(X)$ to behave the same as in $X$, so we will attach a "triangle" (ie 2-cell) along the loops in $L(X)$ corresponding to the images of the edges of $PQR$. After attaching this triangle, homotopic loops in $X$ will correspond to homotopic loops in $L(X)$.
Hopefully, I've given some motivation for this construction. 
