inverse equivalence $\Pi(X) \to \pi_1(X,x)$ in proof of Van Kampen theorem

I would like to ask for help understanding May's (concise course of algebraic topology) proof of the Van Kampen theorem through colimits. Explicitly, I don't understand how to construct the inverse equivalence $F:\Pi(X) \to \pi_1(X,x)$ in such a way that $FJ=Id$, when $J:\pi_1(X,x)\to \Pi(X)$ is the inclusion. In the book says that one should make a choice of path classes $x\to y$ and whenever $y=x$ one should take $c_x$. I think that this last choise does not make sense, for it losses all the information of the classes in $\pi_1(X,x)$. Certainly, for $FJ$ the function on objects is the identity but the function on the morphisms is constant, $c_x$.

Even more, for the case $x\neq y$ the choise of classes is more troublesome if one wants to respect the composition of paths in $\Pi(X)$ and the group structure in $\pi_1(X,x)$.

Any help would be appreciated.

You're forgetting about all the other morphisms in $\Pi(X)$. Once you've selected a path representative $p_y:x\rightarrow y$ for each $y\in X$, you have to worry about all the morphisms $y\rightarrow x$. By choosing $p_y$, concatenation with any class of paths $y\rightarrow x$ will give you a morphism $x\rightarrow x$. This morphism will map, under $F$, to an element of $\pi_1(X,x)$. In fact, since $\pi_1(X,x)$ has only one object, $F(y)=x$ for all $y$. So any morphism $x\rightarrow y$ needs to map to an element of $\pi_1(X,x)$.
• I agree, but the claim that the composition gives the identity functor is not true, sending every morphism in $\pi_1(X,x)$ to the identity, right? – Juan Mar 5 '14 at 2:08