9
$\begingroup$

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 $1$):

enter image description here enter image description here

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.

$\endgroup$

1 Answer 1

4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.