TITLE: Given a space $X$, construct a CW complex $L(X)$ s.t. they have the same fundamental group
QUESTION [5 upvotes]: 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 $\pi_1(L(X))\cong \pi_1(X,x_0)$

REPLY [3 votes]: Your mistake is where you say that $QR$ corresponds to the loop $a^{-1}b$. Usually it does not correspond to $a^{-1}b$ but rather to another loop in $X$ that is homotopic to $a^{-1}b$ in $X$, but not in the wedge of circles constructed in step 1 of your question. The $2$-cell $e^2_\tau$ is added to $L(X)$ precisely for the purpose of producing a corresponding homotopy in $L(X)$.