The monoidal structure on the fundamental groupoids of spectrum

Question

I am trying to understand Anderson duality and Picard categories from appendix B of Hopkins and Singer's paper, and I somehow get stuck on Example B.7 (Page 87).

For a spectrum $E$, they consider each of the fundamental groupoids $\pi_{\leq 1}E_n$. Then they equipped a tensor product $$\otimes: E_n\times E_n\rightarrow E_n$$ with these groupoids, making them Picard categories.

I have some confusion about this procedure:

1. The precise meaning of $\pi_{\leq 1}E_n$. To my understanding, $\pi_{1}E_n$ is the category in which objects are points in $E_n$ and morphisms are homotopy classes of paths in $E_n$, so it is a groupoid. But $\pi_0E_n$ is the set of path-connected components of the space $E_n$, so it is not a groupoid in which objects are points in $E_n$. Then it seems to me that the tensor product they constructed only gives rise to a monoidal structure on the groupoids $\pi_{1}E_n$, but not the sets $\pi_{0}E_n$.

2. They identify the space $E_n$ with the space of maps of spectra $S^0\rightarrow \Sigma^n E$. I believe here they use $S^0$ for the sphere spectrum $S$, and the latter set is the collection of functions of degree $-r$ between $S$ and $E$. But I didn't figure out how this identification works.

3. Then they claimed that a choice of "loop multiplication" amounts to a choice of deformation of the diagonal $S^0\rightarrow S^0\times S^0$ to a map $S^0\rightarrow S^0\vee S^0$. I am quite confused about how can we get a map $\otimes: E_n\times E_n\rightarrow E_n$ from such a deformation.

Corrections and comments are welcomed!

Answer 1

1. $\pi_{\leq 1}$ and your $\pi_1$ mean the same thing - $\pi_{\leq 1}$ is simply used to indicate that you take the groupoid and not just the fundamental group. If you think of groupoids as spaces, another reason to use $\pi_{\leq 1}$ is because you're seeing it as a $1$-truncation operation: you're chopping off everything $>1$, so you're left with $\leq 1$.

(note that $\pi_0$ does acquire the structure of an ordinary monoid/group because for any monoidal groupoid $\mathcal G$, $\pi_0\mathcal G$ has the structure of an ordinary monoid)

2. I'm unsure how to answer this because I'm unsure what you mean by "degree $-n$ maps" if not "maps $S^0\to \Sigma^n E$". $S^0$ is indeed the sphere spectrum here.

3. The construction goes : $E_n\times E_n = map(S^0,\Sigma^n E)\times map(S^0,\Sigma^n E)\simeq map(S^0\vee S^0, \Sigma^n E)$, then you precompose with your diagonal $S^0\to S^0\vee S^0$ to obtain something in $map(S^0, \Sigma^n E) = E_n$. All in all, this gives you $E_n\times E_n\to E_n$.

Answer 2

Maxime Ramzi has already answered questions 1 and 3. I'll answer question 2.

This is really more of just a topological fact, namely we have the following isomorphisms (assuming the notation that $\Sigma$ is defined via shifting the indices of sequential spectra and $\Omega^\infty$ returns the zeroth space). $$ E_n\simeq \mathbf{Top}(S^0,E_n) \simeq \mathbf{Top}(S^0,\Omega^\infty\Sigma^n E) \simeq \mathbf{Sp}(\Sigma^\infty S^0, \Sigma^nE) $$

Edit in response to request for clarification below:

For the last isomorphism, $\Sigma^\infty$ and $\Omega^\infty$ are adjoint functors, which is a fairly straightforward proof that I'll include below. As for a reference, any source that builds a category of spectra should have some sort of proof of this fact. I should note that by $\mathbf{Top}$ I mean the category of pointed topological spaces, specifically whichever category was used to build your notion of spectrum.

Now we want to prove that for $X\in\mathbf{Top}$, $E\in\mathbf{Sp}$, $$\newcommand\Top{\mathbf{Top}}\newcommand\Sp{\mathbf{Sp}} \Sp(\Sigma^\infty X, E)\simeq \Top(X,\Omega^\infty E). $$ Now conveniently for us, $\Omega^\infty \Sigma^\infty X=X$, so $\Omega^\infty$ induces a map $\Sp(\Sigma^\infty X, E)\to \Top(X,\Omega^\infty E)$, and we just need to show that this map is a bijection. Again, $\Omega^\infty$ evaluates a sequential spectrum at its zeroth space, so it sends a map $(f_i)_{i\in\mathbb{Z}}$ to $f_0$. (It doesn't matter whether your spectra are indexed by integers or natural numbers. I'm going to assume integers.)

Let's take a closer look at maps of spectra $f:\Sigma^\infty X\to E$. $f$ has components $f_i : (\Sigma^\infty X)_i \to E_i$. For $i < 0$, $(\Sigma^\infty X)_i = *$, so $f_i$ must be the inclusion of the basepoint into $E_i$ for $i < 0$. For $i\ge 0$ we use the fact that maps of spectra commute with the structure maps of the spectra. Consider such a commutative square: $$ \require{AMScd} \begin{CD} \Sigma \Sigma^i X @>1>> \Sigma^{i+1} X\\ @V\Sigma f_iVV @VVf_{i+1}V \\ \Sigma E_i @>\sigma_i>> E_{i+1}, \end{CD} $$ where $\sigma_i : \Sigma E_i\to E_{i+1}$ is the structure map in $E$.

This tells us that $f_{i+1} = \sigma_i\circ (\Sigma f_i)$ for all $i\ge 0$. In other words for $i>0$, $f_i$ is determined by $f_0$.

In fact, not only is $f_i$ determined by $f_0$ if we start with a morphism of spectra $f:\Sigma^\infty X \to E$, for any map $h:X\to E_0$ we can define a map $g : \Sigma^\infty X \to E$ by taking $g_0=h$, $g_i = *\to E_i$ for $i < 0$, and inductively defining $g_{i+1} = \sigma_i \circ (\Sigma g_i)$ for $i \ge 0$.

This gives us a map $\Top(X,\Omega^\infty E)\to \Sp(\Sigma^\infty X, E)$ that is inverse to the map induced by $\Omega^\infty$, proving the claim.