The following is regarding the book Algebraic Topology by said authour.
First, there is a theorem about pushout of fundamental groupoids:
To compute the fundamental groupoid of the circle, he proceeds a follows:
He constructs a groupoid $G$ with with object set $S^1$ and morphism set $S^1\times\mathbb{R}$. A domain of a morphism $(a,t)$ is $a$ and the codomain is $ae^{2\pi i t}$. One has an open cover of $S^1$ by $X_0 = S^1\setminus \{(0,1)\}$ and $X_1 = S^1\setminus \{(0,-1)\}$ with inclusions $i_k\colon X_0\cap X_1\to X_k$ and $j_k\colon X_k\to S^1$. The sets $X_0$ and $X_1$ are contractible (since they are homeomorphic to $\mathbb{R}$), thus there exists a single morphism $(a,b)_k\colon a\to b$ between any two objects of $a,b$ of $\Pi(X_k)$.
He considers bijective maps $f_0\colon (0,1)\to X_0$ and $f_1\colon (-1/2,1/2)\to X_1$ given by $t \mapsto e^{2\pi i t}$. The author then defines functors $\gamma_k\colon \Pi(X_k)\to G$ by the identity on objects and by $\gamma_k((a,b)_k) = (a, f^{-1}_k(b) – f^{-1}_k(a))$ on morphisms. Moreover, he consider a functor $\zeta\colon G\to \Pi(S^1)$ which is the identity on objects and which sends a morphism $(a,t)$ of $G$ to the homotopy class of the path $u_{a,t}\colon [0,1]\to S^1$ given by $u_{a,t}(s) = ae^{2\pi i t s}$. (1) However, I don't see why it is functorial. For example, consider a composition $(b,s)\circ (a,t)$ in $G$. Then $b = ae^{2\pi i t}$. $u_{a,s + t}(k) = ae^{2 \pi i (s + t)}$ and $(u_{a,t}*u_{b,s})(k)$ is $a e^{4\pi i t k}$ whenever $k \in [0,1/2]$ and $(u_{a,t}*u_{b,s})(k)$ is $ae^{2\pi i t + 4\pi i s (2k – 1)}$ whenever $k \in [1/2,1]$. Supposedly, there should be an obvious homotopy between these, but I can't see it.
tom Dieck then says that the diagram
commutes, and applies the aforementioned theorem about pushout of groupoids to $(\gamma_0,\gamma_1)$, obtaining a functor $\gamma\colon \Pi(S^1)\to G$. The uniqueness of the pushout property shows $\zeta\gamma = 1$. In order to show that $\gamma\zeta = 1$, he says it is enough to prove that the morphisms in $G$ are generated by the images of $\gamma_0$ and $\gamma_1$. (2) But why is it enough? Given $(a,t) \in G$, the authour chooses a decomposition $t = t_1 + … + t_m$ such that $|t_r| < 1/2$ for all $r$. (3) I see he encorporated a condition that the domain of $f_0$ is $(-1/2,1/2)$. But what about $(0,1)$, the domain of $f_1$? He sets $a_0 = a$ and $a_r = ae^{2\pi i (t_1 + … + t_r)}$. Then $(a,t) = (a_{m – 1},t_m)\circ … \circ (a_1,t_2)$ in $G$. Since $|t_r| < 1/2$, he says, there is $r(k) = 0,1$ for each $r$ such that $a_{r – 1}e^{2\pi i t_r s} \in X_{k(r)}$ for some $s \in [0,1]$. (4) Where did $s$ come from? Then, tom Dieck says, $(a_{r – 1}, t_r) = \gamma_{k(r)}((a_{r – 1}, a_r)_{k(r)})$. (5) Again, I don't see how we have this identity.
Best Answer
(1) The idea is the same as for proving that $z\mapsto z^{n+m}$ and $(z\mapsto z^n)*(z\mapsto z^m)$ are homotopic.
The general context as follows : let $G$ be an $H$-space (that is, it has a multiplication $\mu : G\times G\to G$, of which we ask nothing except that it has, for simplicity as in our case, a unit - in general it's enough to require a unit up to homotopy). Then $\Pi(G)$ is a "monoidal" (with the level of generality I gave, that's not quite true because it's not associative - we won't need that) groupoid, that is, it has a "tensor product" : $\otimes : \Pi(G)\times \Pi(G)\to \Pi(G)$. This is simply induced by $\mu$; in particular it is a functor.
Now in our situation, what do we have ? We have a path, $u_{a,t}$, which is of the form $id_a\otimes \gamma$ for some $\gamma : 1\to x$ (for $x$ such that $ax = b$), and $u_{b,s}$, which is of the form $\delta \otimes id_b$ for some $\delta : 1\to y$. Now we want to compose these and use the functoriality of $\otimes$. This does not work, because of course the codomain of $\gamma$ may be different from $b$. But we can mix things up a bit, and since $ax = b$, write $\omega\otimes id_x$ for some $\omega : a \to y$ (with $ay = c$ - think of it this way : $u_{b,s} = b\times$ something, so we may rewrite it as $a\times $something $\times x$)
With this set up, we get $u_{b,s}\circ u_{a,t} = (\omega\otimes id_x)\circ (id_a\otimes \gamma)$; and now we may use functoriality because the (co)domains match up : we get $(\omega \circ id_a )\otimes (id_x\circ\gamma) = \omega\otimes \gamma$. Now $\omega \otimes \gamma$, in our setting, is just a multiplication of paths : now you can remember how $\omega, \delta$ have been defined, and convince yourself that this is just $u_{a,s+t}$ (recall that $\otimes$ is defined by the multiplication on $S^1$ !)
I'll let you write a general statement if you're interested.
(2) It is enough for the following reason : suppose they do, then any two morphisms $f,g : G\to K$ such that $f\circ \gamma_i = g\circ \gamma_i$ ($i=0,1$) must be equal (*).
It follows that if you have $K$ and morphisms $f_i: \Pi(X_i)\to K$ making the appropriate diagram commute, you can factor them through $\Pi(S^1)$, and then, thanks to $\zeta$, through $G$.
So this yields the existence in the universal property of the pushout, and the property (*) yields the uniqueness : so $G$ is a pushout, just like $\Pi(S^1)$, they must be uniquely isomorphic.
(3)I don't understabd. Who's $f_1$ ?
(4) I think it's "for all $s\in [0,1]$", not "for some": you want the path between times $t_1 +...+t_{r-1}$ and $t_1+...+t_r$ to be totally contained in $X_{k(r)}$ : to achieve this, you subdivide your path in small pieces where it has to be contained in one of the two.
(5) The path is completely contained in $X_{k(r)}$, so it is in the image of $\gamma_{k(r)}$ (the precise value is not really necessary, but you can obtain it in any case by noting that $\Pi(X_{k(r)})$ has only one morphism between any two points), which is what we wanted to show.