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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.

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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 $\pi_1(X,x)$.

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  • $\begingroup$ 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? $\endgroup$ – Juan Mar 5 '14 at 2:08
  • $\begingroup$ It's an equivalence of categories. $\endgroup$ – Joe Johnson 126 Mar 5 '14 at 2:50
  • $\begingroup$ So, one can only assume that there is a natural iso between the composition and the identity functor? It is just that a few lines later in the proof he (May) seems to use the equality not the equivalence, i'm talking about the proof of the lemma in page 18. Just to make sure, do you agree that the composition FJ is constant on the morphisms. Thanks for your replies. $\endgroup$ – Juan Mar 5 '14 at 15:38

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