$k$-invariant of Picard groupoid

Question

I am referring to Definition 17.27 of Dan Freed's notes. I have some understanding but not sure whether it is completely correct. Could somebody help me to spell out fully the definition filling up the details tha are left by Freed? I don't know why Freed wants to leave this central and important concept for students to work out rather than making sure the students are 100% affirmative about the concept in view of the fact that this is a pretty advanced concept that is not covered in standard textbooks.

Answer 1

to each $y$ we can associate an isomorphism $- \otimes \text{id}_y: \text{Aut}(1_D) \stackrel{\cong}{\to} \text{Aut}(y)$ defined by applying the monoidal product (as a functor) to any morphism $a \to b$ to give a morphism $a \otimes y \to b \otimes y$ and fixing $a = b = 1_D$ to make $(a \to b) \in \text{Aut}(1_D)$ and $(a \otimes y \to b \otimes y) \in \text{Aut}(y)$, which obviously has inverse $\text{id}_{y^{-1}}$, denote this isomorphism by $m_y: \text{Aut}(1_D) \stackrel{\cong}{\to} \text{Aut}(y)$

fixing $a = b = y$ we get a map $(y \to y) \to ((y \otimes y) \to (y \otimes y))$ which is also an isomorphism with inverse $- \otimes \text{id}_{y^{-1}}$, hence $\text{Aut}(y \otimes y) \cong \text{Aut}(y) \cong \text{Aut}(1_D)$, denote this isomorphism by $m_{y\otimes y}: \text{Aut}(1_D) \stackrel{\cong}{\to} \text{Aut}(y \otimes y)$, let $\sigma: y \otimes y \to y \otimes y$ be the symmetry transformation (ref. lecture 13), we can define a map $D \to \pi_1(D)$ by $y \mapsto m_{y\otimes y}^{-1}(\sigma)$

mimicking the association and isomorphism above we can define a map $\pi_0(D) \to \pi_1(D)$ by sending $[y] \mapsto m_{y_0 \otimes y_1}^{-1}(\sigma)$ where $y_0,y_1$ are any two representatives of $[y]$ and the isomorphism $m_{y_0 \otimes y_1}$ is defined as the following composition $$\left(1_D \to 1_D\right) \xrightarrow{- \otimes \text{id}_{y_0}} \left((1_D \otimes y_0) \to (1_D \otimes y_0)\right) = \left((y_0 \otimes 1_D) \to (y_0 \otimes 1_D)\right) \xrightarrow{- \otimes \text{id}_{y_1}} \left((y_0 \otimes y_1) \to (y_0 \otimes y_1)\right)$$ it is well-defined because let $y_0',y_1'$ be any other two representatives of $[y]$ then $\left((y_0 \otimes y_1) \to (y_0 \otimes y_1)\right)$ is isomorphic to $\left((y_0' \otimes y_1') \to (y_0' \otimes y_1')\right)$ as $D$ is a groupoid thus all morphisms are invertible, the definition of $m_{y_0\otimes y_1}$ also shows that $m_{y_0\otimes y_1} = m_{y_0} m_{y_1}$ which implies $[x] \otimes [y] \mapsto m_{(x_0\otimes x_1)\otimes(y_0\otimes y_1)}^{-1}(\sigma) = m_{y_0\otimes y_1}^{-1}(\sigma) \circ m_{x_0\otimes x_1}^{-1}(\sigma)$ hence $\pi_0(D) \to \pi_1(D)$ is a group homomorphism

now since $\sigma^2 = \text{id}$ we can extend $\pi_0(D) \to \pi_1(D)$ to $\pi_0(D) \otimes \mathbb{Z}_2 \to \pi_1(D)$...

(Waiting for Mike Miller's edit)