7.3.5 The Left Skew Left Unitor

Definition 7.3.5.1.1 ▶ The Left Skew Left Unitor of $\lhd $

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

whose component

\[ \lambda ^{\mathsf{Sets}_{*},\lhd }_{X} \colon S^{0}\lhd X \to X \]

at $(X,x_{0})\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$ is given by the composition

\begin{align*} S^{0}\lhd X & \cong |X|\odot S^{0}\\ & \cong \bigvee _{x\in X}S^{0}\\ & \rightarrow X, \end{align*}

where $\bigvee _{x\in X}S^{0}\to X$ is the map given by

\begin{align*} [(x,0)] & \mapsto x_{0},\\ [(x,1)] & \mapsto x \end{align*}

for each $x\in X$.

Proof of Definition 7.3.5.1.1.

(Proven below in a bit.)

Remark 7.3.5.1.2 ▶ Unwinding Definition 7.3.5.1.1

In other words, $\smash {\lambda ^{\mathsf{Sets}_{*},\lhd }_{X}}$ acts on elements as

\begin{align*} \lambda ^{\mathsf{Sets}_{*},\lhd }_{X}(0\lhd x) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}x_{0},\\ \lambda ^{\mathsf{Sets}_{*},\lhd }_{X}(1\lhd x) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}x\end{align*}

for each $1\lhd x\in S^{0}\lhd X$.

Remark 7.3.5.1.3 ▶ Non-Invertibility of the Skew Left Unitor of $\lhd $

The morphism $\smash {\lambda ^{\mathsf{Sets}_{*},\lhd }_{X}}$ is almost invertible, with its would-be-inverse

\[ \phi _{X}\colon X\to S^{0}\lhd X \]

given by

\[ \phi _{X}(x)\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}1\lhd x \]

for each $x\in X$. Indeed, we have

\begin{align*} [\lambda ^{\mathsf{Sets}_{*},\lhd }_{X}\circ \phi ](x) & = \lambda ^{\mathsf{Sets}_{*},\lhd }_{X}(\phi (x))\\ & = \lambda ^{\mathsf{Sets}_{*},\lhd }_{X}(1\lhd x)\\ & = x\\ & = [\operatorname {\mathrm{id}}_{X}](x) \end{align*}

so that

\[ \lambda ^{\mathsf{Sets}_{*},\lhd }_{X}\circ \phi =\operatorname {\mathrm{id}}_{X} \]

and

\begin{align*} [\phi \circ \lambda ^{\mathsf{Sets}_{*},\lhd }_{X}](1\lhd x) & = \phi (\lambda ^{\mathsf{Sets}_{*},\lhd }_{X}(1\lhd x))\\ & = \phi (x)\\ & = 1\lhd x\\ & = [\operatorname {\mathrm{id}}_{S^{0}\lhd X}](1\lhd x), \end{align*}

but

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

where $0\lhd x\neq 1\lhd x_{0}$. Thus

\[ \phi \circ \lambda ^{\mathsf{Sets}_{*},\lhd }_{X}\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{?}}}=}}\operatorname {\mathrm{id}}_{S^{0}\lhd X} \]

holds for all elements in $S^{0}\lhd X$ except one.

Proof of Definition 7.3.5.1.1.

Firstly, note that, given $(X,x_{0})\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$, the map

\[ \lambda ^{\mathsf{Sets}_{*},\lhd }_{X} \colon S^{0}\lhd X \to X \]

is indeed a morphism of pointed sets, as we have

\[ \lambda ^{\mathsf{Sets}_{*},\lhd }_{X}(0\lhd x_{0})=x_{0}. \]

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

Previous
Up
Next

View Remark 7.3.5.1.2 as PDF
Statistics Cite

Noticed something off, or have any comments? Feel free to reach out!

You can also use the contact form below: