I have as an assignment problem that is Hatcher Exercise 1.2.15. The exercise is basically this. Given a space $X,x_0$, we construct a space $L(X)$ built from a one 0-cell, for each loop at $x_0$ in $X$ we have a 1-cell in $L(X)$, and for every map $T$ of the standard triangle into $X$, we have a 2-cell in $L(X)$ attached to three loops in $L(X)$ by the restriction of $T$ to the edge of a triangle.
I am asked to prove that from the natural map $L(X) \to X$, we get an induced map $\pi_1(L(X),z_0)\to \pi_1(X,x_0)$
that is a group isomorphism, where $z_0$ is the 0-cell that we have in $L(X)$.
Now I have spoken to my lecturer and he says that we can think of $L(X)$ as basically being built out of triangles, whereby for each standard triangle $\Delta$ in $L(X)$ we have a map $f_\Delta$ to $X$. We can then extend this to the whole of $L(X)$. The extension is well defined and our only problem is what happens if two triangles "touch". But the edge of a triangle is just one loop in $X$ so there is no problem. So this is my map $f : L(X) \to X$.
Now the surjectivity of $f_\ast$ is clear because for each loop in $X$ I have an edge of a triangle that maps to it. The problem now of course is injectivity. Suppose that $f_\ast([\phi]) = [\phi'] = f_\ast([\psi]) = [\psi'].$
Then this means that $\phi'\psi'^{-1}$ is homotopic to the constant map at $x_0$. We can think of this as a loop at $x_0$ so that $\phi'\psi'^{-1}$ determines the edge $E$ of a triangle in $L(X)$. My idea to show now that
$\phi\psi \simeq c_{z_0}$
is to say that every point on my edge $E$ must indeed be $z_0$. However I have tried for some time now to show this but nothing has worked out. Furthermore, it seems to me that when Hatcher says "for each loop at $x_0$ we have a $1$ - cell in $L(X)$" he surely means for each homotopy class of a loop we have a $1$ -cell in $L(X)$ yes? How do I go from here? Thanks.
Edit: I just had an idea to do the problem. I think Hatcher for each homotopy class of loops at $x_0$ we get a zero-cell in $L(X)$. Now my idea is this. For each class of loop $[g]$ at $x_0$ we know it is $f([l])$ for some class of loops $[l]$ at $z_0$ in $L(X)$. Now define $p : \pi_1(X,x_0) \to \pi_1(L(X),z_0)$ by
$p([f([l])]) = [l].$
This is well defined once we know that $[f([l])] = [f([m])]$ implies that $[l] = [m]$. Once we know this it is easy to check that $p$ is a group homomorphism and is the desired inverse of $f$ and I'm done. I am in the process of trying to prove that fact above in order to conclude that $p$ is well-defined.