The symmetric monoidal structure on the category $\mathsf{Sets}_{*}$ of Proposition 7.5.9.1.1 is uniquely determined by the following requirements:
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1.
Existence of an Internal Hom. The tensor product
\[ \otimes _{\mathsf{Sets}_{*}}\colon \mathsf{Sets}_{*}\times \mathsf{Sets}_{*} \to \mathsf{Sets}_{*} \]of $\mathsf{Sets}_{*}$ admits an internal Hom $[-_{1},-_{2}]_{\mathsf{Sets}_{*}}$.
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The Unit Object Is $S^{0}$. We have $\mathbb {1}_{\mathsf{Sets}_{*}}\cong S^{0}$.
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Let $X\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$.
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Since $\otimes _{\mathsf{Sets}_{*}}$ is part of a closed monoidal structure, it preserves colimits in each variable by
.
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Since $X\cong \bigvee _{x\in X^{-}}S^{0}$ and $\otimes _{\mathsf{Sets}_{*}}$ preserves colimits in each variable, we have
\begin{align*} X\otimes _{\mathsf{Sets}_{*}}Y & \cong (\bigvee _{x\in X^{-}}S^{0})\otimes _{\mathsf{Sets}_{*}}Y\\ & \cong \bigvee _{x\in X^{-}}(S^{0}\otimes _{\mathsf{Sets}_{*}}Y)\\ & \cong \bigvee _{x\in X^{-}}Y\\ & \cong \bigvee _{x\in X^{-}}S^{0}\wedge Y\\ & \cong (\bigvee _{x\in X^{-}}S^{0})\wedge Y\\ & \cong X\wedge Y, \end{align*}naturally in $Y\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$, where we have used that $S^{0}$ is the monoidal unit for $\otimes _{\mathsf{Sets}_{*}}$. Thus $X\otimes _{\mathsf{Sets}_{*}}-\cong X\wedge -$ for each $X\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$.
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Similarly, $-\otimes _{\mathsf{Sets}_{*}}Y\cong -\wedge Y$ for each $Y\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$.
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By
, we then have $\mathord {\otimes _{\mathsf{Sets}_{*}}}\cong \mathord {\wedge }$.
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Subdiagrams $(1)$, $(2)$, and $(3)$ commute by the naturality of $\operatorname {\mathrm{id}}^{\otimes }_{\mathsf{Sets}_{*}}$.
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Subdiagram $(4)$ commutes by
.
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Subdiagram $(5)$ commutes by the naturality of $\rho ^{\mathsf{Sets}_{*},-1}$.
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Subdiagram $(6)$ commutes trivially.
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Subdiagram $(7)$ commutes by the naturality of $\rho ^{\mathsf{Sets}_{*}}$, where the equality $\rho ^{\mathsf{Sets}_{*}}_{S^{0}}=\lambda ^{\mathsf{Sets}_{*}}_{S^{0}}$ comes from
.
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Subdiagram $(5)$ commutes by the naturality of $\lambda ^{\prime ,-1}$.
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Subdiagram $(\ddagger )$ commutes, as proved above.
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Subdiagram $(4)$ commutes by the naturality of $\operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathbb {1}|\mathsf{Sets}_{*}}$.
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Subdiagram $(1)$ commutes by the naturality of $\operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}}$.
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Subdiagram $(3)$ commutes by the naturality of $\lambda ^{\mathsf{Sets}_{*},-1}$.
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Subdiagram $(1)$ commutes by the naturality of $\operatorname {\mathrm{id}}^{\otimes }_{\mathsf{Sets}_{*}}$;
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Subdiagrams $(2)$ and $(3)$ commute by the functoriality of $\otimes $;
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Subdiagram $(4)$ commutes by the left monoidal unity of $(\operatorname {\mathrm{id}}^{\otimes },\operatorname {\mathrm{id}}^{\otimes }_{\mathbb {1}})$, which we proved above;
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Subdiagram $(5)$ commutes by the naturality of $\lambda '$;
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Subdiagram $(6)$ commutes by the naturality of $\rho '$, where the equality $\rho '_{\mathbb {1}_{\mathsf{Sets}_{*}}}=\lambda '_{\mathbb {1}_{\mathsf{Sets}_{*}}}$ comes from
;
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Subdiagrams $(1)$ and $(6)$ commute by
;
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Subdiagram $(2)$ commutes by the naturality of $\operatorname {\mathrm{id}}^{\otimes }_{\mathsf{Sets}_{*}}$;
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Subdiagram $(\S )$ commutes, as was shown above;
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Subdiagram $(3)$ commutes by the naturality of $\lambda ^{\mathsf{Sets}_{*}}$;
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Subdiagram $(4)$ commutes trivially;
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Subdiagram $(5)$ commutes by Chapter 13: Constructions With Monoidal Categories, Item 2c of Item 2 of Proposition 13.1.1.1.4, whose proof uses only the left monoidal unity of $(\operatorname {\mathrm{id}}^{\otimes },\operatorname {\mathrm{id}}^{\otimes }_{\mathbb {1}})$, which has been proven above;
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Subdiagram $(1)$ commutes by the naturality of $\operatorname {\mathrm{id}}^{\otimes }_{\mathsf{Sets}_{*}}$;
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Subdiagram $(2)$ commutes by the naturality of $\sigma '$ and the fact that $\operatorname {\mathrm{id}}^{\otimes }_{\mathbb {1}}$ is invertible;
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Subdiagram $(\ddagger )$ commutes as proved above;
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Subdiagram $(3)$ commutes by the naturality of $\sigma ^{\mathsf{Sets}_{*}}$ and the fact that $\operatorname {\mathrm{id}}^{\otimes }_{\mathbb {1}}$ is invertible;
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Subdiagram $(4)$ commutes by the naturality of $\operatorname {\mathrm{id}}^{\otimes }_{\mathsf{Sets}_{*}}$;
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Subdiagram $(2)$ commutes by the naturality of $\sigma ^{\mathsf{Sets}_{*},-1}$.
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Subdiagram $(5)$ commutes by the naturality of $\operatorname {\mathrm{id}}^{\otimes ,-1}$.
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Subdiagram $(¶)$ commutes, as proved above.
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Subdiagram $(4)$ commutes by the naturality of $\sigma ^{\prime ,-1}$.
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Subdiagram $(1)$ commutes by the naturality of $\operatorname {\mathrm{id}}^{\otimes ,-1}$.
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Subdiagram $(1)$ commutes by the braidedness of $\operatorname {\mathrm{id}}^{\otimes }$, as proved above.
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Subdiagram $(2)$ commutes by
.
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Subdiagrams $(1)$, $(2)$, and $(3)$ commute by the naturality of $\operatorname {\mathrm{id}}^{\otimes }_{\mathsf{Sets}_{*}}$.
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Subdiagrams $(4)$, $(5)$, and $(6)$ commute by the naturality of $\lambda ^{\mathsf{Sets}_{*}}$, where the equality $\lambda ^{\mathsf{Sets}_{*}}_{S^{0}}=\rho ^{\mathsf{Sets}_{*}}_{S^{0}}$ comes from
.
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Subdiagram $(5)$ commutes by the naturality of $\rho ^{\prime ,-1}$.
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Subdiagram $(\dagger )$ commutes, as proved above.
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Subdiagram $(4)$ commutes by the naturality of $\operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathbb {1}|\mathsf{Sets}_{*}}$.
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Subdiagram $(1)$ commutes by the naturality of $\operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}}$.
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Subdiagram $(3)$ commutes by the naturality of $\rho ^{\mathsf{Sets}_{*},-1}$.
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Subdiagrams $(1)$, $(4)$, $(5)$, $(8)$, and $(11)$ commute by the naturality of $\operatorname {\mathrm{id}}^{\otimes }_{\mathsf{Sets}_{*}}$;
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Subdiagram $(2)$ commutes by the right monoidal unity of $(\operatorname {\mathrm{id}}^{\otimes }_{\mathsf{Sets}_{*}},\operatorname {\mathrm{id}}^{\otimes }_{\mathbb {1}|\mathsf{Sets}_{*}})$;
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Subdiagram $(3)$ commutes by the triangle identity for $(\alpha ',\lambda ',\rho ')$;
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Subdiagram $(6)$ commutes by
;
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Subdiagram $(7)$ commutes by the naturality of $\rho ^{\mathsf{Sets}_{*}}$;
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Subdiagram $(9)$ commutes by
;
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Subdiagram $(10)$ commutes by
;
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Subdiagrams $(1)$, $(3)$, $(4)$, and $(6)$ commute by the naturality of $\operatorname {\mathrm{id}}^{\otimes }_{Sets_{*}}$;
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Subdiagram $(\ddagger )$ commutes, as proved above;
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Subdiagram $(2)$ commutes by the naturality of $\alpha '$;
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Subdiagram $(5)$ commutes by the naturality of $\alpha ^{\mathsf{Sets}_{*}}$;
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Subdiagram $(1)$ commutes by the naturality of $\alpha ^{\mathsf{Sets}_{*},-1}$.
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Subdiagram $(2)$ commutes by the naturality of $\operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}}$.
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Subdiagram $(3)$ commutes by the naturality of $\operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}}$.
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Subdiagram $(\dagger )$ commutes, as proved above.
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Subdiagram $(4)$ commutes by the naturality of $\operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}}$.
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Subdiagram $(5)$ commutes by the naturality of $\operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}}$.
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Subdiagram $(6)$ commutes by the naturality of $\alpha ^{\prime ,-1}$.
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1.
Two-Sided Preservation of Colimits. The tensor product
\[ \otimes _{\mathsf{Sets}_{*}}\colon \mathsf{Sets}_{*}\times \mathsf{Sets}_{*} \to \mathsf{Sets}_{*} \]of $\mathsf{Sets}_{*}$ preserves colimits separately in each variable.
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The Unit Object Is $S^{0}$. We have $\mathbb {1}_{\mathsf{Sets}_{*}}\cong S^{0}$.
7.5.10 The Universal Property of $(\mathsf{Sets}_{*},\wedge ,S^{0})$
More precisely, the full subcategory of the category $\mathcal{M}^{\mathrm{cld}}_{\mathbb {E}_{\infty }}(\mathsf{Sets}_{*})$ of spanned by the closed symmetric monoidal categories $\left(\phantom{\mathrlap {\lambda ^{\mathsf{Sets}_{*}}}}\mathsf{Sets}_{*}\right.$, $\otimes _{\mathsf{Sets}_{*}}$, $[-_{1},-_{2}]_{\mathsf{Sets}_{*}}$, $\mathbb {1}_{\mathsf{Sets}_{*}}$, $\lambda ^{\mathsf{Sets}_{*}}$, $\rho ^{\mathsf{Sets}_{*}}$, $\left.\sigma ^{\mathsf{Sets}_{*}}\right)$ satisfying Item 1 and Item 2 is contractible (i.e. equivalent to the punctual category).
Proof of Theorem 7.5.10.1.1.
Unwinding the Statement
Let $(\mathsf{Sets}_{*},\otimes _{\mathsf{Sets}_{*}},[-_{1},-_{2}]_{\mathsf{Sets}_{*}},\mathbb {1}_{\mathsf{Sets}_{*}},\lambda ',\rho ',\sigma ')$ be a closed symmetric monoidal category satisfying Item 1 and Item 2. We need to show that the identity functor
\[ \operatorname {\mathrm{id}}_{\mathsf{Sets}_{*}}\colon \mathsf{Sets}_{*}\to \mathsf{Sets}_{*} \]
admits a unique closed symmetric monoidal functor structure
\[ \begin{array}{cccc} \phantom{\operatorname {\mathrm{id}}^{\otimes }_{\mathsf{Sets}_{*}}}\mathllap {\operatorname {\mathrm{id}}^{\otimes }_{\mathsf{Sets}_{*}}} \colon \mkern -10mu & X\otimes _{\mathsf{Sets}_{*}}Y \mkern -10mu& {}\mathbin {\overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }}& \mkern -10mu{}X\wedge Y,\\ \phantom{\operatorname {\mathrm{id}}^{\otimes }_{\mathsf{Sets}_{*}}}\mathllap {\operatorname {\mathrm{id}}^{\operatorname {\mathrm{Hom}}}_{\mathsf{Sets}_{*}}} \colon \mkern -10mu & [X,Y]_{\mathsf{Sets}_{*}} \mkern -10mu& {}\mathbin {\overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }}& \mkern -10mu{}\mathsf{Sets}_{*}(X,Y),\\ \phantom{\operatorname {\mathrm{id}}^{\otimes }_{\mathsf{Sets}_{*}}}\mathllap {\operatorname {\mathrm{id}}^{\otimes }_{\mathbb {1}|\mathsf{Sets}_{*}}} \colon \mkern -10mu & \mathbb {1}_{\mathsf{Sets}_{*}} \mkern -10mu& {}\mathbin {\overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }}& \mkern -10mu{}S^{0}, \end{array} \]
making it into a symmetric monoidal strongly closed isomorphism of categories from $\left(\phantom{\mathrlap {\lambda '}}\mathsf{Sets}_{*}\right.$, $\otimes _{\mathsf{Sets}_{*}}$, $[-_{1},-_{2}]_{\mathsf{Sets}_{*}}$, $\mathbb {1}_{\mathsf{Sets}_{*}}$, $\lambda '$, $\rho '$, $\left.\sigma '\right)$ to the closed symmetric monoidal category $\left(\phantom{\mathrlap {\lambda ^{\mathsf{Sets}_{*}}}}\mathsf{Sets}_{*}\right.$, $\times $, $\mathsf{Sets}_{*}(-_{1},-_{2})$, $\mathbb {1}_{\mathsf{Sets}_{*}}$, $\lambda ^{\mathsf{Sets}_{*}}$, $\rho ^{\mathsf{Sets}_{*}}$, $\left.\sigma ^{\mathsf{Sets}_{*}}\right)$ of Proposition 7.5.9.1.1.
Constructing an Isomorphism $[-_{1},-_{2}]_{\mathsf{Sets}_{*}}\cong \mathsf{Sets}_{*}(-_{1},-_{2})$
By , we have a natural isomorphism
\[ \mathsf{Sets}_{*}(S^{0},[-_{1},-_{2}]_{\mathsf{Sets}_{*}})\cong \mathsf{Sets}_{*}(-_{1},-_{2}). \]
By Chapter 6: Pointed Sets, Item 4 of Proposition 6.1.4.1.1, we also have a natural isomorphism
\[ \mathsf{Sets}_{*}(S^{0},[-_{1},-_{2}]_{\mathsf{Sets}_{*}})\cong [-_{1},-_{2}]_{\mathsf{Sets}_{*}}. \]
Composing both natural isomorphisms, we obtain a natural isomorphism
\[ \mathsf{Sets}_{*}(-_{1},-_{2})\cong [-_{1},-_{2}]_{\mathsf{Sets}_{*}}. \]
Given $X,Y\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$, we will write
\[ \operatorname {\mathrm{id}}^{\operatorname {\mathrm{Hom}}}_{X,Y}\colon \mathsf{Sets}_{*}(X,Y)\overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }[X,Y]_{\mathsf{Sets}_{*}} \]
for the component of this isomorphism at $(X,Y)$.
Constructing an Isomorphism $\mathord {\otimes _{\mathsf{Sets}_{*}}}\cong \mathord {\wedge }$
Since $\otimes _{\mathsf{Sets}_{*}}$ is adjoint in each variable to $[-_{1},-_{2}]_{\mathsf{Sets}_{*}}$ by assumption and $\wedge $ is adjoint in each variable to $\mathsf{Sets}_{*}(-_{1},-_{2})$ by Chapter 4: Constructions With Sets, Item 2 of Proposition 4.3.5.1.2, uniqueness of adjoints () gives us natural isomorphisms
\begin{align*} X\otimes _{\mathsf{Sets}_{*}}- & \cong X\wedge -,\\ -\otimes _{\mathsf{Sets}_{*}}Y & \cong Y\wedge -. \end{align*}
By , we then have $\mathord {\otimes _{\mathsf{Sets}_{*}}}\cong \mathord {\wedge }$. We will write
\[ \operatorname {\mathrm{id}}^{\otimes }_{\mathsf{Sets}_{*}|X,Y}\colon X\otimes _{\mathsf{Sets}_{*}}Y\overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }X\wedge Y \]
for the component of this isomorphism at $(X,Y)$.
Alternative Construction of an Isomorphism $\mathord {\otimes _{\mathsf{Sets}_{*}}}\cong \mathord {\wedge }$
Alternatively, we may construct a natural isomorphism $\mathord {\otimes _{\mathsf{Sets}_{*}}}\cong \mathord {\wedge }$ as follows:
Below, we’ll show that if a natural isomorphism $\mathord {\otimes _{\mathsf{Sets}_{*}}}\cong \mathord {\wedge }$ exists, then it must be unique. This will show that the isomorphism constructed above is equal to the isomorphism $\operatorname {\mathrm{id}}^{\otimes }_{\mathsf{Sets}_{*}|X,Y}\colon X\otimes _{\mathsf{Sets}_{*}}Y\to X\wedge Y$ from before.
Constructing an Isomorphism $\operatorname {\mathrm{id}}^{\otimes }_{\mathbb {1}}\colon \mathbb {1}_{\mathsf{Sets}_{*}}\to S^{0}$
We define an isomorphism $\operatorname {\mathrm{id}}^{\otimes }_{\mathbb {1}}\colon \mathbb {1}_{\mathsf{Sets}_{*}}\to S^{0}$ as the composition
\[ \mathbb {1}_{\mathsf{Sets}_{*}}\overset {\rho ^{\mathsf{Sets}_{*},-1}_{\mathbb {1}_{\mathsf{Sets}_{*}}}}{\underset {\scriptstyle \mathord {\sim }}{\dashrightarrow }}\mathbb {1}_{\mathsf{Sets}_{*}}\wedge S^{0}\overset {\operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|\mathbb {1}_{\mathsf{Sets}_{*}}}}{\underset {\scriptstyle \mathord {\sim }}{\dashrightarrow }}\mathbb {1}_{\mathsf{Sets}_{*}}\otimes _{\mathsf{Sets}_{*}} S^{0}\overset {\lambda '_{ S^{0}}}{\underset {\scriptstyle \mathord {\sim }}{\dashrightarrow }}S^{0} \]
in $\mathsf{Sets}_{*}$.
Monoidal Left Unity of the Isomorphism $\mathord {\otimes _{\mathsf{Sets}_{*}}}\cong \mathord {\wedge }$
We have to show that the diagram 
Since all subdiagrams commute, so does the boundary diagram, i.e. the diagram $(\dagger )$ above. As a result, the diagram
it follows that the diagram
Here’s a step-by-step showcase of this argument: 
We then have
\begin{align*} \lambda ^{\prime ,-1}_{X}(x) & = [\lambda ^{\prime ,-1}_{X}\circ [x]](1)\\ & = [(\operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathbb {1}|\mathsf{Sets}_{*}}\wedge \operatorname {\mathrm{id}}_{X})\circ \operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|S^{0},X}\circ \lambda ^{\mathsf{Sets}_{*},-1}_{X}\circ [x]](1)\\ & = [(\operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathbb {1}|\mathsf{Sets}_{*}}\wedge \operatorname {\mathrm{id}}_{X})\circ \operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|S^{0},X}\circ \lambda ^{\mathsf{Sets}_{*},-1}_{X}](x) \end{align*}
for each $x\in X$, and thus we have
\[ \lambda ^{\prime ,-1}_{X}=(\operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathbb {1}|\mathsf{Sets}_{*}}\wedge \operatorname {\mathrm{id}}_{X})\circ \operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|S^{0},X}\circ \lambda ^{\mathsf{Sets}_{*},-1}_{X}. \]
Taking inverses then gives
\[ \lambda ^{\prime }_{X}=\lambda ^{\mathsf{Sets}_{*}}_{X}\circ \operatorname {\mathrm{id}}^{\otimes }_{\mathsf{Sets}_{*}|S^{0},X}\circ (\operatorname {\mathrm{id}}^{\otimes }_{\mathbb {1}|\mathsf{Sets}_{*}}\wedge \operatorname {\mathrm{id}}_{X}), \]
showing that the diagram
Braidedness of the Isomorphism $\mathord {\otimes _{\mathsf{Sets}_{*}}}\cong \mathord {\wedge }$
We have to show that the diagram 
it follows that the boundary diagram, i.e. diagram $(\S )$, also commutes. Next, consider the diagram
it follows that the boundary diagram, i.e. diagram $(\ddagger )$, also commutes. Next, consider the diagram
it follows that the boundary diagram, i.e. diagram $(\dagger )$ also commutes. Taking inverses for the diagram $(\dagger )$, we see that the diagram
it follows that the diagram
\begin{align*} [\operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|X,Y}\circ \sigma ^{\mathsf{Sets}_{*},-1}_{X,Y}](y,x) & = [\operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|X,Y}\circ \sigma ^{\mathsf{Sets}_{*},-1}_{X,Y}\circ ([y]\wedge [x])](1,1)\\ & = [\sigma ^{\prime ,-1}_{X,Y}\circ \operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|Y,X}\circ ([y]\wedge [x])](1,1)\\ & = [\sigma ^{\prime ,-1}_{X,Y}\circ \operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|Y,X}](y,x) \end{align*}
for each $(y,x)\in Y\wedge X$, and thus we have
\[ \operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|X,Y}\circ \sigma ^{\mathsf{Sets}_{*},-1}_{X,Y}=\sigma ^{\prime ,-1}_{X,Y}\circ \operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|Y,X}. \]
Taking inverses then gives
\[ \sigma ^{\mathsf{Sets}_{*}}_{X,Y}\circ \operatorname {\mathrm{id}}^{\otimes }_{\mathsf{Sets}_{*}|X,Y}=\operatorname {\mathrm{id}}^{\otimes }_{\mathsf{Sets}_{*}|Y,X}\circ \sigma ^{\prime }_{X,Y}, \]
showing that the diagram
Monoidal Right Unity of the Isomorphism $\mathord {\otimes _{\mathsf{Sets}_{*}}}\cong \mathord {\wedge }$
We have to show that the diagram 
\[ \sigma '_{S^{0},\mathbb {1}_{\mathsf{Sets}_{*}}}\colon S^{0}\otimes _{\mathsf{Sets}_{*}}\mathbb {1}_{\mathsf{Sets}_{*}}\to \mathbb {1}_{\mathsf{Sets}_{*}}\otimes _{\mathsf{Sets}_{*}}S^{0} \]
as the composition
Next, consider the diagram
. In this diagram:
Since all subdiagrams commute, so does the boundary diagram, i.e. the diagram $(\dagger )$ above. As a result, the diagram
it follows that the diagram
Here’s a step-by-step showcase of this argument: 
We then have
\begin{align*} \rho ^{\prime ,-1}_{X}(a) & = [\rho ^{\prime ,-1}_{X}\circ [x]](1)\\ & = [(\operatorname {\mathrm{id}}_{X}\wedge \operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathbb {1}|\mathsf{Sets}_{*}})\circ \operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|S^{0},X}\circ \rho ^{\mathsf{Sets}_{*},-1}_{X}\circ [x]](1)\\ & = [(\operatorname {\mathrm{id}}_{X}\wedge \operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathbb {1}|\mathsf{Sets}_{*}})\circ \operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|S^{0},X}\circ \rho ^{\mathsf{Sets}_{*},-1}_{X}](a) \end{align*}
for each $a\in X$, and thus we have
\[ \rho ^{\prime ,-1}_{X}=(\operatorname {\mathrm{id}}_{X}\wedge \operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathbb {1}|\mathsf{Sets}_{*}})\circ \operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|S^{0},X}\circ \rho ^{\mathsf{Sets}_{*},-1}_{X}. \]
Taking inverses then gives
\[ \rho ^{\prime }_{X}=\rho ^{\mathsf{Sets}_{*}}_{X}\circ \operatorname {\mathrm{id}}^{\otimes }_{\mathsf{Sets}_{*}|S^{0},X}\circ (\operatorname {\mathrm{id}}_{X}\wedge \operatorname {\mathrm{id}}^{\otimes }_{\mathbb {1}|\mathsf{Sets}_{*}}), \]
showing that the diagram
Monoidality of the Isomorphism $\mathord {\otimes _{\mathsf{Sets}_{*}}}\cong \mathord {\wedge }$
We have to show that the diagram 
it follows that the boundary diagram, i.e. diagram $(\ddagger )$, also commutes. Consider now the diagram
it follows that the boundary diagram, i.e. diagram $(\dagger )$, also commutes. Taking inverses on the diagram $(\dagger )$, we see that the diagram
it follows that the diagram
\begin{align*} \left[(\operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|X,Y}\otimes _{\mathsf{Sets}_{*}}\operatorname {\mathrm{id}}_{Z})\circ \operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|X\wedge Y,Z}\right.\\ \left.{}\circ \alpha ^{\mathsf{Sets}_{*},-1}_{X,Y,Z}\right](x,(y,z)) & = \left[(\operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|X,Y}\otimes _{\mathsf{Sets}_{*}}\operatorname {\mathrm{id}}_{Z})\circ \operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|X\wedge Y,Z}\right.\\ & \phantom{={}} \mkern 4mu\left.{}{}\circ \alpha ^{\mathsf{Sets}_{*},-1}_{X,Y,Z}\circ ([x]\wedge ([y]\wedge [z]))\right](1,(1,1))\\ & = \left[\alpha ^{\prime ,-1}_{X,Y,Z}\circ (\operatorname {\mathrm{id}}_{X}\wedge \operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|Y,Z})\right.\\ & \phantom{={}} \mkern 4mu\left.{}\circ \operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|X,Y\wedge Z}\circ ([x]\wedge ([y]\wedge [z]))\right](1,(1,1))\\ & = [\alpha ^{\prime ,-1}_{X,Y,Z}\circ (\operatorname {\mathrm{id}}_{X}\wedge \operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|Y,Z})\circ \operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|X,Y\wedge Z}](x,(y,z)) \end{align*}
\[ (\operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|X,Y}\otimes _{\mathsf{Sets}_{*}}\operatorname {\mathrm{id}}_{Z})\circ \operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|X\wedge Y,Z}\circ \alpha ^{\mathsf{Sets}_{*},-1}_{X,Y,Z}=\alpha ^{\prime ,-1}_{X,Y,Z}\circ (\operatorname {\mathrm{id}}_{X}\wedge \operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|Y,Z})\circ \operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathsf{Sets}_{*}|X,Y\wedge Z}. \]
\[ \alpha ^{\mathsf{Sets}_{*}}_{X,Y,Z}\circ \operatorname {\mathrm{id}}^{\otimes }_{\mathsf{Sets}_{*}|X\wedge Y,Z}\circ (\operatorname {\mathrm{id}}^{\otimes }_{\mathsf{Sets}_{*}|X,Y}\otimes _{\mathsf{Sets}_{*}}\operatorname {\mathrm{id}}_{Z})=\operatorname {\mathrm{id}}^{\otimes }_{\mathsf{Sets}_{*}|X,Y\wedge Z}\circ (\operatorname {\mathrm{id}}_{X}\wedge \operatorname {\mathrm{id}}^{\otimes }_{\mathsf{Sets}_{*}|Y,Z})\circ \alpha ^{\prime }_{X,Y,Z}, \]
Uniqueness of the Isomorphism $\mathord {\otimes _{\mathsf{Sets}_{*}}}\cong \mathord {\wedge }$
Let $\phi ,\psi \colon -_{1}\otimes _{\mathsf{Sets}_{*}}-_{2}\Rightarrow -_{1}\wedge -_{2}$ be natural isomorphisms. Since these isomorphisms are compatible with the unitors of $\mathsf{Sets}_{*}$ with respect to $\wedge $ and $\otimes $ (as shown above), we have
\begin{align*} \lambda '_{Y} & = \lambda ^{\mathsf{Sets}_{*}}_{Y}\circ \phi _{S^{0},Y}\circ (\operatorname {\mathrm{id}}^{\otimes }_{\mathbb {1}|\mathsf{Sets}}\otimes _{\mathsf{Sets}}\operatorname {\mathrm{id}}_{Y}),\\ \lambda '_{Y} & = \lambda ^{\mathsf{Sets}_{*}}_{Y}\circ \psi _{S^{0},Y}\circ (\operatorname {\mathrm{id}}^{\otimes }_{\mathbb {1}|\mathsf{Sets}}\otimes _{\mathsf{Sets}}\operatorname {\mathrm{id}}_{Y}). \end{align*}
Postcomposing both sides with $\lambda ^{\mathsf{Sets}_{*},-1}_{Y}$ and then precomposing both sides with $\operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathbb {1}|\mathsf{Sets}}\otimes _{\mathsf{Sets}}\operatorname {\mathrm{id}}_{Y}$ gives
\begin{align*} \lambda ^{\mathsf{Sets}_{*},-1}_{Y}\circ \lambda '_{Y}\circ (\operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathbb {1}|\mathsf{Sets}}\otimes _{\mathsf{Sets}}\operatorname {\mathrm{id}}_{Y}) & = \phi _{S^{0},Y},\\ \lambda ^{\mathsf{Sets}_{*},-1}_{Y}\circ \lambda '_{Y}\circ (\operatorname {\mathrm{id}}^{\otimes ,-1}_{\mathbb {1}|\mathsf{Sets}}\otimes _{\mathsf{Sets}}\operatorname {\mathrm{id}}_{Y}) & = \psi _{S^{0},Y}, \end{align*}
and thus we have
\[ \phi _{S^{0},Y}=\psi _{S^{0},Y} \]
for each $Y\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$. Now, let $x\in X$ and consider the naturality diagrams
\begin{align*} \phi _{X,Y}(x,y) & = [\phi _{X,Y}\circ ([x]\wedge \operatorname {\mathrm{id}}_{Y})](1,y)\\ & = [([x]\otimes _{\mathsf{Sets}_{*}}\operatorname {\mathrm{id}}_{Y})\circ \phi _{S^{0},Y}](1,y)\\ & = [([x]\otimes _{\mathsf{Sets}_{*}}\operatorname {\mathrm{id}}_{Y})\circ \psi _{S^{0},Y}](1,y)\\ & = [\psi _{X,Y}\circ ([x]\wedge \operatorname {\mathrm{id}}_{Y})](1,y)\\ & = \psi _{X,Y}(x,y) \end{align*}
for each $(x,y)\in X\wedge Y$. Therefore we have
\[ \phi _{X,Y}=\psi _{X,Y} \]
for each $X,Y\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$ and thus $\phi =\psi $, showing the isomorphism $\mathord {\otimes _{\mathsf{Sets}_{*}}}\cong \mathord {\times }$ to be unique.
The symmetric monoidal structure on the category $\mathsf{Sets}_{*}$ of Proposition 7.5.9.1.1 is uniquely determined by the following requirements:
More precisely, the full subcategory of the category $\mathcal{M}_{\mathbb {E}_{\infty }}(\mathsf{Sets}_{*})$ of spanned by the symmetric monoidal categories $\left(\phantom{\mathrlap {\lambda ^{\mathsf{Sets}_{*}}}}\mathsf{Sets}_{*}\right.$, $\otimes _{\mathsf{Sets}_{*}}$, $\mathbb {1}_{\mathsf{Sets}_{*}}$, $\lambda ^{\mathsf{Sets}_{*}}$, $\rho ^{\mathsf{Sets}_{*}}$, $\left.\sigma ^{\mathsf{Sets}_{*}}\right)$ satisfying Item 1 and Item 2 is contractible.
Proof of Corollary 7.5.10.1.2.
Since $\mathsf{Sets}_{*}$ is locally presentable (), it follows from that Corollary 7.5.10.1.2 is equivalent to the existence of an internal Hom as in Item 1 of Theorem 7.5.10.1.1. The result then follows from Theorem 7.5.10.1.1.
The symmetric monoidal structure on the category $\mathsf{Sets}_{*}$ is the unique symmetric monoidal structure on $\mathsf{Sets}_{*}$ such that the free pointed set functor
\[ (-)^{+} \colon \mathsf{Sets}\to \mathsf{Sets}_{*} \]
admits a symmetric monoidal structure, i.e. the full subcategory of the category $\mathcal{M}_{\mathbb {E}_{\infty }}(\mathsf{Sets}_{*})$ of spanned by the symmetric monoidal categories $\left(\phantom{\mathrlap {\lambda ^{\mathsf{Sets}_{*}}}}\mathsf{Sets}_{*}\right.$, $\otimes _{\mathsf{Sets}_{*}}$, $\mathbb {1}_{\mathsf{Sets}_{*}}$, $\lambda ^{\mathsf{Sets}_{*}}$, $\rho ^{\mathsf{Sets}_{*}}$, $\left.\sigma ^{\mathsf{Sets}_{*}}\right)$ with respect to which $(-)^{+}$ admits a symmetric monoidal structure is contractible.
Proof of Corollary 7.5.10.1.3.
Let $(\otimes _{\mathsf{Sets}_{*}},\mathbb {1}_{\mathsf{Sets}_{*}},\lambda ^{\mathsf{Sets}_{*}},\rho ^{\mathsf{Sets}_{*}},\sigma ^{\mathsf{Sets}_{*}})$ be a symmetric monoidal structure on $\mathsf{Sets}_{*}$ such that $(-)^{+}$ admits a symmetric monoidal structure with respect to $\otimes _{\mathsf{Sets}_{*}}$ and $\wedge $. We have isomorphisms
\begin{align*} X\otimes _{\mathsf{Sets}_{*}}Y & \cong (X^{-})^{+}\otimes _{\mathsf{Sets}_{*}}(Y^{-})^{+}\\ & \cong (X^{-}\times Y^{-})^{+}\\ & \cong (X^{-})^{+}\wedge (Y^{-})^{+}\\ & \cong X\wedge Y, \end{align*}
all natural in $X$ and $Y$. Now, since $\wedge $ preserves colimits in both variables and $\mathord {\otimes _{\mathsf{Sets}_{*}}}\cong \mathord {\wedge }$, it follows that $\otimes _{\mathsf{Sets}_{*}}$ also preserves colimits in both variables, so the result then follows from Corollary 7.5.10.1.2.
