7.5.5 The Left Unitor

    Definition 7.5.5.1.1 ▶ The Left Unitor of $\wedge $

    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 $x\in X$.

    Proof of Definition 7.5.5.1.1.

    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:

      1. (a)

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

      2. (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 $X$.

    Being a Morphism of Pointed Sets
    As just mentioned, we have

    \[ \lambda ^{\mathsf{Sets}_{*}}_{X}(0\wedge x_{0})\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}x_{0}, \]

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

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

    \[ \lambda ^{\mathsf{Sets}_{*},-1}_{X}\colon X\overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }S^{0}\wedge X \]

    defined by

    \[ \lambda ^{\mathsf{Sets}_{*},-1}_{X}(x)\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}1\wedge x \]

    for each $x\in X$. Indeed:

  • 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.

    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.

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

    \[ f\colon (X,x_{0})\to (Y,y_{0}), \]

    the diagram

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

    Being a Natural Isomorphism
    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.

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