# $\pi_1(S^1, 1)$ via the fundamental groupoid

I'm currently reading Ronnie Brown's Topology and Groupoids and am stuck on a small detail of his computation of the fundamental group of the circle (in particular his computation of the group's generator). Recall that there is a functor $$\pi:\mathbf{Top}\rightarrow \mathbf{Grpd}$$ from the category of the topological spaces to the category of groupoids taking a space $$X$$ to its "fundamental groupoid" $$\pi X$$, the category with objects the elements of $$X$$ and morphisms the homotopy classes of paths in $$X$$. For any set $$A$$, we define $$\pi XA$$ to be the full subcategory of $$\pi X$$ with objects $$X\cap A$$. Chapter 6 gives a proof of a kind of baby van-Kampen theorem for fundamental groupoids: if $$X$$ is a topological space with subspaces $$X_0, X_1, X_2$$ such that $$X_0=X_1\cap X_2$$ and $$X=\mathrm{Int}(X_1)\cup\mathrm{Int}{X_2}$$, and $$A$$ is a set (for convenience assume contained in $$X$$) such that $$A$$ meets every path component of $$X_0$$, $$X_1$$, and $$X_2$$, then the following commutative square induced by inclusion maps is a pushout in $$\mathbf{Grpd}$$: $$\require{AMScd}$$ $$\begin{CD} \pi X_0A @>i_1>> \pi X_1A\\ @Vi_2VV @VVu_1V\\ \pi X_2A @>u_2>> \pi XA \end{CD}$$ A further theorem is that, if $$A'$$ is a subset of $$A\cap X_1$$ that meets every path component of $$X_1$$, then letting $$A_1=A'\cup (A\setminus X_1)$$ we can extend this square to another groupoid pushout $$\require{AMScd}$$ $$\begin{CD} \pi X_0A @>i_1>> \pi X_1A @>r>> \pi X_1A_1\\ @Vi_2VV @VVV @VVu_1V\\ \pi X_2A @>u_2>> \pi XA @>r'>> \pi XA_1 \end{CD}$$ where $$r$$ and $$r'$$ are deformation retractions – i.e. their composition with the natural inclusion functors on the left are homotopic as functors to the respective identity on $$\pi X_1 A$$ and $$\pi X A$$ ($$\mathrm{rel}(\pi X_1 A_1)$$ and $$\mathrm{rel}(\pi X A_1)$$ respectively). (The use of $$u_1$$ here as the induced inclusion map is a slight abuse of notation.)

I've read and understood the proofs of both of these results. Now, the final result of the chapter is to apply the latter theorem to a computation of the fundamental group of the circle (which is isomorphic to the "object group" $$\pi(S^1, 1):=\pi S^1(1, 1)$$). Here is the proof (Brown uses $$+$$ to denote composition of paths and $$\mathbf{I}$$ to denote the unique tree groupoid on two objects $$0$$ and $$1$$):

Everything is clear to me except the very last argument that $$\varphi$$ is a generator, and in particular the claim that the retraction $$r'$$ satisfies the identity $$r'=-\varphi_1+\varphi_2$$. As best as I can see, an argument for this might run by taking $$F:\pi S^1 A \times \mathbf{I}\rightarrow\pi S^1 A_1$$ to be a functor homotopy $$ir'\simeq \mathrm{id}_{\pi S^1 A} \space\mathrm{rel}(\pi S^1 A_1)$$ where $$i:\pi S^1 A_1 \rightarrow \pi S^1 A$$ is the map induced by inclusion. Then, if $$\iota$$ is the unique element of $$\pi I(0, 1)$$, we have by definition of functor homotopy a commutative square in $$\pi S^1 A$$ $$\require{AMScd}$$ $$\begin{CD} F(1, 0)=ir'(1)=1 @>ir'\varphi_2=r'\varphi_2>> F(-1, 0)=ir'(-1)=1\\ @VF(\mathrm{id}_{1}, \iota)=\mathrm{id}_{1}VV @VVF(\mathrm{id}_{-1}, \iota)V\\ F(1, 1)=\mathrm{id}_{\pi S^1 A}(1)=1 @>\mathrm{id}_{\pi S^1 A}(\varphi_2)=\varphi_2>> F(-1, 1)=\mathrm{id}_{\pi S^1 A}(-1)=-1 \end{CD}$$ (where the arrow on the left side of the square is the identity by the $$\mathrm{rel} (\pi S^1 A_1)$$ condition). Clearly if we had $$F(\mathrm{id}_{-1}, \iota)=\varphi_1$$ then we would be done, but I don't see anywhere in the construction of $$r'$$ why this has to be the case. Does anyone have any insight? Sorry for the overly long post; if notation is unclear check out Chapter 6 of the attached pdf above. Thank you so much in advance.

Aha, I've figured it out! The construction of $$r'$$ and $$r$$ in the composed commutative square in the question statement also gives that $$\require{AMScd}$$ $$\begin{CD} \pi X_1A @>r>> \pi X_1A_1\\ @Vu_1VV @VVu_1V\\ \pi XA @>r'>> \pi XA_1 \end{CD}$$ commutes (and is in fact a pushout as well, though we do not need this); again we abuse notation with respect to $$u_1$$. In the context of $$S^1$$ and $$X, X_1, A, A_1$$ as given, this is a square $$\require{AMScd}$$ $$\begin{CD} \pi X_1\{-1, 1\}\cong\mathbf{I} @>r>> \pi X_1\{1\}\cong\mathbf{0}\\ @Vu_1VV @VVu_1V\\ \pi S^1\{-1, 1\} @>r'>> \pi S^1\{1\}\cong\pi_1(S^1, 1) \end{CD}$$ where $$\mathbf{0}$$ is the tree groupoid on the singleton $$\{1\}$$ and $$\mathbf{I}$$ is as above. In particular, $$\varphi_1$$ is the unique element of $$\pi X_1(1, -1)$$, so the commutativity of the square gives $$r'(\varphi_1)=r(\varphi_1)=id_1$$. Using the same functor homotopy square as in the question text but substituting $$\varphi_1$$ for $$\varphi_2$$ in the horizontal morphisms gives also that $$\varphi_1 id_1=F(id_{-1}, \iota)r'(\varphi_1)=F(id_{-1}, \iota)id_1$$, whence $$F(id_{-1}, \iota)=\varphi_1$$ as desired.