7.3.4 The Left Skew Associator

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

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

as in the diagram

whose component

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

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

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

where the map

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

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

(Proven below in a bit.)

Unwinding the notation for elements, we have

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

and

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

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

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

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

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

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

However, by the definition of $\lhd $, we have $y_{0}\lhd z=y_{0}\lhd z'$ for all $z,z'\in Z$, preventing $\alpha ^{\mathsf{Sets}_{*},\lhd }_{X,Y,Z}$ 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

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

is indeed a morphism of pointed sets, as we have

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

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


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