Remark 5.4.4.1.3 (00FC): Non-Invertibility of the Skew Associator of $\rhd $

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Taking $y = y_{0}$ , we see that the morphism $α_{X, Y, Z}^{{Sets}_{*}, ⊳}$ acts on elements as

α_{X, Y, Z}^{{Sets}_{*}, ⊳} (x ⊳ (y_{0} ⊳ z)) \overset{def}{=} (x ⊳ y_{0}) ⊳ z .

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

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

α_{X, Y, Z}^{{Sets}_{*}, ⊳} : X ⊳ (Y ⊳ Z) \to (X ⊳ Y) ⊳ Z

is indeed a morphism of pointed sets, as we have

α_{X, Y, Z}^{{Sets}_{*}, ⊳} (x_{0} ⊳ (y_{0} ⊳ z_{0})) = (x_{0} ⊳ y_{0}) ⊳ z_{0} .

Next, we claim that $α^{{Sets}_{*}, ⊳}$ is a natural transformation. We need to show that, given morphisms of pointed sets

\begin{aligned} f & : (X, x_{0}) \to (X^{'}, x_{0}^{'}), \\ g & : (Y, y_{0}) \to (Y^{'}, y_{0}^{'}), \\ h & : (Z, z_{0}) \to (Z^{'}, z_{0}^{'}) \end{aligned}

the diagram

commutes. Indeed, this diagram acts on elements as

and hence indeed commutes, showing $α^{{Sets}_{*}, ⊳}$ to be a natural transformation. This finishes the proof.