The Clowder Project

1. 1. We have $x=x^{\prime }$ and $y=y^{\prime }$.
2. 2. Both of the following conditions are satisfied:
   1. (a) We have $x=x_0$ or $y=0$.
   2. (b) We have $x^{\prime }=x_0$ or $y^{\prime }=0$.

In the first case, ${\rho }_X^{{\mathsf{Sets}}_{\ast }}$ clearly sends both elements to the same element in $X$. Meanwhile, in the latter case both elements are equal to the basepoint $x_0\wedge 0$ of $X\wedge S^0$, which gets sent to the basepoint $x_0$ of $X$.

and thus ${\rho }_X^{{\mathsf{Sets}}_{\ast }}$ is a morphism of pointed sets.

defined by

for each $x\in X$. Indeed:

• Invertibility I. We have
  $$\begin{array}{rl}[{\rho }_X^{{\mathsf{Sets}}_{\ast },-1}\circ {\rho }_X^{{\mathsf{Sets}}_{\ast }}](x\wedge 0)& ={\rho }_X^{{\mathsf{Sets}}_{\ast },-1}\left({\rho }_X^{{\mathsf{Sets}}_{\ast }}(x\wedge 0)\right)\\ & ={\rho }_X^{{\mathsf{Sets}}_{\ast },-1}\left(x_0\right)\\ & =x_0\wedge 1\\ & =x\wedge 0,\end{array}$$

  and

  $$\begin{array}{rl}[{\rho }_X^{{\mathsf{Sets}}_{\ast },-1}\circ {\rho }_X^{{\mathsf{Sets}}_{\ast }}](x\wedge 1)& ={\rho }_X^{{\mathsf{Sets}}_{\ast },-1}\left({\rho }_X^{{\mathsf{Sets}}_{\ast }}(x\wedge 1)\right)\\ & ={\rho }_X^{{\mathsf{Sets}}_{\ast },-1}\left(x\right)\\ & =x\wedge 1\end{array}$$

  for each $x\in X$, and thus we have

  $${\rho }_X^{{\mathsf{Sets}}_{\ast },-1}\circ {\rho }_X^{{\mathsf{Sets}}_{\ast }}={\text{id}}_{X\wedge S^0}.$$

• Invertibility II. We have
  $$\begin{array}{rl}[{\rho }_X^{{\mathsf{Sets}}_{\ast }}\circ {\rho }_X^{{\mathsf{Sets}}_{\ast },-1}]\left(x\right)& ={\rho }_X^{{\mathsf{Sets}}_{\ast }}\left({\rho }_X^{{\mathsf{Sets}}_{\ast },-1}\left(x\right)\right)\\ & ={\rho }_X^{{\mathsf{Sets}}_{\ast },-1}(x\wedge 1)\\ & =x\end{array}$$

  for each $x\in X$, and thus we have

  $${\rho }_X^{{\mathsf{Sets}}_{\ast }}\circ {\rho }_X^{{\mathsf{Sets}}_{\ast },-1}={\text{id}}_X.$$

the diagram

commutes. Indeed, this diagram acts on elements as

and hence indeed commutes, showing ${\rho }^{{\mathsf{Sets}}_{\ast }}$ to be a natural transformation.