7.5.4 The Associator
The associator of the smash product of pointed sets is the natural isomorphism
\[ \alpha ^{\mathsf{Sets}_{*}} \colon {\wedge }\circ {({\wedge }\times \operatorname {\mathrm{id}}_{\mathsf{Sets}_{*}})} \mathbin {\overset {\mathord {\sim }}{\Longrightarrow }}{\wedge }\circ {(\operatorname {\mathrm{id}}_{\mathsf{Sets}_{*}}\times {\wedge })}\circ {\mathbf{\alpha }^{\mathsf{Cats}}_{\mathsf{Sets}_{*},\mathsf{Sets}_{*},\mathsf{Sets}_{*}}}, \]
as in the diagram
whose component
\[ \alpha ^{\mathsf{Sets}_{*}}_{X,Y,Z} \colon (X\wedge Y)\wedge Z \overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }X\wedge (Y\wedge Z) \]
at $(X,x_{0}),(Y,y_{0}),(Z,z_{0})\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$ is given by
\[ \alpha ^{\mathsf{Sets}_{*}}_{X,Y,Z}((x\wedge y)\wedge z)\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}x\wedge (y\wedge z) \]
for each $(x\wedge y)\wedge z\in (X\wedge Y)\wedge Z$.
Well-Definedness
Let $[((x,y),z)]=[((x',y'),z')]$ be an element in $(X\wedge Y)\wedge Z$. Then either:
-
1.
We have $x=x'$, $y=y'$, and $z=z'$.
-
2.
Both of the following conditions are satisfied:
-
(a)
We have $x=x_{0}$ or $y=y_{0}$ or $z=z_{0}$.
In the first case, $\alpha ^{\mathsf{Sets}_{*}}_{X,Y,Z}$ clearly sends both elements to the same element in $X\wedge (Y\wedge Z)$. Meanwhile, in the latter case both elements are equal to the basepoint $(x_{0}\wedge y_{0})\wedge z_{0}$ of $(X\wedge Y)\wedge Z$, which gets sent to the basepoint $x_{0}\wedge (y_{0}\wedge z_{0})$ of $X\wedge (Y\wedge Z)$.
Being a Morphism of Pointed Sets
As just mentioned, we have
\[ \alpha ^{\mathsf{Sets}_{*}}_{X,Y,Z}((x_{0}\wedge y_{0})\wedge z_{0})\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}x_{0}\wedge (y_{0}\wedge z_{0}), \]
and thus $\alpha ^{\mathsf{Sets}_{*}}_{X,Y,Z}$ is a morphism of pointed sets.
Invertibility
The inverse of $\alpha ^{\mathsf{Sets}_{*}}_{X,Y,Z}$ is given by the morphism
\[ \alpha ^{\mathsf{Sets}_{*},-1}_{X,Y,Z} \colon X\wedge (Y\wedge Z) \overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }(X\wedge Y)\wedge Z \]
defined by
\[ \alpha ^{\mathsf{Sets}_{*},-1}_{X,Y,Z}(x\wedge (y\wedge z))\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}(x\wedge y)\wedge z \]
for each $x\wedge (y\wedge z)\in X\wedge (Y\wedge Z)$.
Naturality
We need to show that, given morphisms of pointed sets
\begin{align*} f & \colon (X,x_{0}) \to (X',x'_{0}),\\ g & \colon (Y,y_{0}) \to (Y',y'_{0}),\\ h & \colon (Z,z_{0}) \to (Z’,z’_{0}) \end{align*}
the diagram
commutes. Indeed, this diagram acts on elements as
and hence indeed commutes, showing $\alpha ^{\mathsf{Sets}_{*}}$ to be a natural transformation.
Being a Natural Isomorphism
Since $\alpha ^{\mathsf{Sets}_{*}}$ is natural and $\alpha ^{\mathsf{Sets}_{*},-1}$ is a componentwise inverse to $\alpha ^{\mathsf{Sets}_{*}}$, it follows from
Chapter 11: Categories,
Item 2 of
Proposition 11.9.7.1.2 that $\alpha ^{\mathsf{Sets}_{*},-1}$ is also natural. Thus $\alpha ^{\mathsf{Sets}_{*}}$ is a natural isomorphism.
