The Clowder Project

7.5.5 The Left Unitor

The left unitor of the smash product of pointed sets is the natural isomorphism

whose component

\[ \lambda ^{\mathsf{Sets}_{*}}_{X} \colon S^{0}\wedge X \overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }X \]

at $X\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$ is given by

\begin{align*} 0\wedge x & \mapsto x_{0},\\ 1\wedge x & \mapsto x \end{align*}

for each $x\in X$.

Well-Definedness

Let $[(x,y)]=[(x',y')]$ be an element in $S^{0}\wedge X$. Then either:

1. 1.

   We have $x=x'$ and $y=y'$.

2. 2.

   Both of the following conditions are satisfied:

(b)

We have $x'=0$ or $y'=x_{0}$.

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

As just mentioned, we have

and thus $\lambda ^{\mathsf{Sets}_{*}}_{X}$ is a morphism of pointed sets.

The inverse of $\lambda ^{\mathsf{Sets}_{*}}_{X}$ is the morphism

defined by

for each $x\in X$. Indeed:

1. 1.

   Invertibility I. We have

   \begin{align*} [\lambda ^{\mathsf{Sets}_{*},-1}_{X}\circ \lambda ^{\mathsf{Sets}_{*}}_{X}](0\wedge x) & = \lambda ^{\mathsf{Sets}_{*},-1}_{X}(\lambda ^{\mathsf{Sets}_{*}}_{X}(0\wedge x))\\ & = \lambda ^{\mathsf{Sets}_{*},-1}_{X}(x_{0})\\ & = 1\wedge x_{0}\\ & = 0\wedge x, \end{align*}

   and

   \begin{align*} [\lambda ^{\mathsf{Sets}_{*},-1}_{X}\circ \lambda ^{\mathsf{Sets}_{*}}_{X}](1\wedge x) & = \lambda ^{\mathsf{Sets}_{*},-1}_{X}(\lambda ^{\mathsf{Sets}_{*}}_{X}(1\wedge x))\\ & = \lambda ^{\mathsf{Sets}_{*},-1}_{X}(x)\\ & = 1\wedge x \end{align*}

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

   \[ \lambda ^{\mathsf{Sets}_{*},-1}_{X}\circ \lambda ^{\mathsf{Sets}_{*}}_{X}=\operatorname {\mathrm{id}}_{S^{0}\wedge X}. \]
2. 2.

   Invertibility II. We have

   \begin{align*} [\lambda ^{\mathsf{Sets}_{*}}_{X}\circ \lambda ^{\mathsf{Sets}_{*},-1}_{X}](x) & = \lambda ^{\mathsf{Sets}_{*}}_{X}(\lambda ^{\mathsf{Sets}_{*},-1}_{X}(x))\\ & = \lambda ^{\mathsf{Sets}_{*},-1}_{X}(1\wedge x)\\ & = x \end{align*}

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

   \[ \lambda ^{\mathsf{Sets}_{*}}_{X}\circ \lambda ^{\mathsf{Sets}_{*},-1}_{X}=\operatorname {\mathrm{id}}_{X}. \]

This shows $\lambda ^{\mathsf{Sets}_{*}}_{X}$ to be invertible.

We need to show that, given a morphism of pointed sets

the diagram

commutes. Indeed, this diagram acts on elements as and and hence indeed commutes, showing $\lambda ^{\mathsf{Sets}_{*}}$ to be a natural transformation.

Since $\lambda ^{\mathsf{Sets}_{*}}$ is natural and $\lambda ^{\mathsf{Sets}_{*},-1}$ is a componentwise inverse to $\lambda ^{\mathsf{Sets}_{*}}$, it follows from Chapter 11: Categories, Item 2 of Proposition 11.9.7.1.2 that $\lambda ^{\mathsf{Sets}_{*},-1}$ is also natural. Thus $\lambda ^{\mathsf{Sets}_{*}}$ is a natural isomorphism.