$\pi_1(S^1, 1)$ 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.
 A: 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.
Question: is this the best way of seeing this? The fact that Brown didn't include this argument in the proof makes me feel like there must be a more "immediate" way to see it.
