7.4.4 The Right Skew Associator

The skew associator of the right tensor product of pointed sets is the natural transformation

\[ \alpha ^{\mathsf{Sets}_{*},\rhd } \colon {\rhd }\circ {(\operatorname {\mathrm{id}}_{\mathsf{Sets}_{*}}\times {\rhd })} \Longrightarrow {\rhd }\circ {({\rhd }\times \operatorname {\mathrm{id}}_{\mathsf{Sets}_{*}})}\circ {\mathbf{\alpha }^{\mathsf{Cats},-1}_{\mathsf{Sets}_{*},\mathsf{Sets}_{*},\mathsf{Sets}_{*}}} \]

as in the diagram

whose component

\[ \alpha ^{\mathsf{Sets}_{*},\rhd }_{X,Y,Z} \colon X\rhd (Y\rhd Z) \to (X\rhd Y)\rhd Z \]

at $(X,x_{0}),(Y,y_{0}),(Z,z_{0})\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$ is given by

\begin{align*} X\rhd (Y\rhd Z) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}|X|\odot (Y\rhd Z)\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}|X|\odot (|Y|\odot Z)\\ & \cong \bigvee _{x\in X}(|Y|\odot Z)\\ & \cong \bigvee _{x\in X}(\bigvee _{y\in Y}Z)\\ & \to \bigvee _{[(x,y)]\in \bigvee _{x\in X}Y}Z\\ & \cong \bigvee _{[(x,y)]\in |X|\odot Y}Z\\ & \cong ||X|\odot Y|\odot Z\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}|X\rhd Y|\odot Z\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}(X\rhd Y)\rhd Z,\end{align*}

where the map

\[ \bigvee _{x\in X}(\bigvee _{y\in Y}Z)\to \bigvee _{[(x,y)]\in \bigvee _{x\in X}Y}Z \]

is given by $[(x,[(y,z)])]\mapsto [([(x,y)],z)]$.

(Proven below in a bit.)

Unwinding the notation for elements, we have

\begin{align*} [(x,[(y,z)])] & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[(x,y\rhd z)]\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}x\rhd (y\rhd z) \end{align*}

and

\begin{align*} [([(x,y)],z)] & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[(x\rhd y,z)]\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}(x\rhd y)\rhd z. \end{align*}

So, in other words, $\alpha ^{\mathsf{Sets}_{*},\rhd }_{X,Y,Z}$ acts on elements via

\[ \alpha ^{\mathsf{Sets}_{*},\rhd }_{X,Y,Z}(x\rhd (y\rhd z)) \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}(x\rhd y)\rhd z \]

for each $x\rhd (y\rhd z)\in X\rhd (Y\rhd Z)$.

Taking $y=y_{0}$, we see that the morphism $\smash {\alpha ^{\mathsf{Sets}_{*},\rhd }_{X,Y,Z}}$ acts on elements as

\[ \alpha ^{\mathsf{Sets}_{*},\rhd }_{X,Y,Z}(x\rhd (y_{0}\rhd z))\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}(x\rhd y_{0})\rhd z. \]

However, by the definition of $\rhd $, we have $x\rhd y_{0}=x'\rhd y_{0}$ for all $x,x'\in X$, preventing $\alpha ^{\mathsf{Sets}_{*},\rhd }_{X,Y,Z}$ from being non-invertible.

Firstly, note that, given $(X,x_{0}),(Y,y_{0}),(Z,z_{0})\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$, the map

\[ \alpha ^{\mathsf{Sets}_{*},\rhd }_{X,Y,Z} \colon X\rhd (Y\rhd Z) \to (X\rhd Y)\rhd Z \]

is indeed a morphism of pointed sets, as we have

\[ \alpha ^{\mathsf{Sets}_{*},\rhd }_{X,Y,Z}(x_{0}\rhd (y_{0}\rhd z_{0}))=(x_{0}\rhd y_{0})\rhd z_{0}. \]

Next, we claim that $\alpha ^{\mathsf{Sets}_{*},\rhd }$ is a natural transformation. 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 $\alpha ^{\mathsf{Sets}_{*},\rhd }$ to be a natural transformation. This finishes the proof.


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