Monoidal Categories and Monoidal Functors - Examples

By [L1], tensor in $\mathbf{S}\mathbf{e}\mathbf{t}$ is cartesian product, so the displayed set for $X\otimes Y$ is correct.

The map ${\alpha }_{X,Y,Z}$ sends each element of $((X\times Y)\times Z)$ to the same triple with the other bracketing, and its inverse is $(x,(y,z))\mapsto ((x,y),z)$.

The maps ${\lambda }_X(\ast ,x)=x$ and ${\rho }_X(x,\ast )=x$ are bijections with inverses $x\mapsto (\ast ,x)$ and $x\mapsto (x,\ast )$. So this concrete example realizes the cartesian monoidal structure exactly as claimed.

Start with an element $(((w,x),y),z)$. Along the left side of the pentagon, the first associator gives $((w,x),(y,z))$ and the second gives $(w,(x,(y,z)))$.

Along the right side, the first associator gives $((w,(x,y)),z)$, the second gives $(w,((x,y),z))$, and the third gives $(w,(x,(y,z)))$.

Both composites send $(((w,x),y),z)$ to the same element $(w,(x,(y,z)))$, so the pentagon commutes in this concrete cartesian example.

By [L1], ${\mathcal{C}}_M$ is a category with identity $e$ and composition given by $\cdot$.

Because $M$ is commutative, $((m^{\mathrm{\prime }}\otimes n^{\mathrm{\prime }})\circ (m\otimes n))=(m^{\mathrm{\prime }}\cdot n^{\mathrm{\prime }})\cdot (m\cdot n)=(m^{\mathrm{\prime }}\cdot m)\cdot (n^{\mathrm{\prime }}\cdot n)=((m^{\mathrm{\prime }}\circ m)\otimes (n^{\mathrm{\prime }}\circ n))$, so tensor is a bifunctor. Associativity and unit are literal equalities because there is only one object and the tensor on morphisms is exactly the monoid product.

Thus ${\mathcal{C}}_M$ is a one-object strict monoidal category. The theorem in [L2] explains why commutativity is the right hypothesis for this strict example.

Each displayed word uses the letters $A,B,C,D$ exactly once, contains no unit symbol, and differs only by the placement of parentheses, so each is a valid parenthesised tensor word of the kind counted in [L2].

The list has five elements, matching [L2]. Therefore it is complete.

The pentagon starts at $(((A\otimes B)\otimes C)\otimes D)$ and ends at $A\otimes (B\otimes (C\otimes D))$, passing through the other three displayed bracketings exactly as the coherence diagram predicts.

By [L1], the free-monoid construction on $\mathbf{S}\mathbf{e}\mathbf{t}$ comes with unit $\eta$ and multiplication $\mu$ satisfying the monad equations. The formulas in the Example conjugate the functor action and structure maps by the bijections $b_X$, so functoriality, naturality, and the monad equations are preserved. Hence $(T,{\eta }^{\mathrm{\prime }},{\mu }^{\mathrm{\prime }})$ is a monad on $\mathcal{S}$.

Because $\mathcal{S}$ has only three objects, it is small, so [L2] applies to the monad from step 1.1. Therefore the data $(T,{\eta }^{\mathrm{\prime }},{\mu }^{\mathrm{\prime }})$ are exactly the structure maps of a monoid object in the endofunctor category $[\mathcal{S},\mathcal{S}]$.

Therefore this transported free-monoid monad is a concrete example of a monoid object in a small endofunctor category.

Direct images preserve identities and composition, so the displayed action on functions makes $\mathcal{P}$ a functor. By [L1] and [L3], the map ${\mathcal{P}}_2$ is well typed: if $A\subseteq X$ and $B\subseteq Y$, then $A\times B\subseteq X\times Y$. For functions $f:X\to X^{\mathrm{\prime }}$ and $g:Y\to Y^{\mathrm{\prime }}$, one has $(f\times g)(A\times B)=f(A)\times g(B)$, so ${\mathcal{P}}_2$ is natural. The unit map ${\mathcal{P}}_0$ is also well typed because a singleton set has exactly one subset equal to itself.

For subsets $A\subseteq X$, $B\subseteq Y$, and $C\subseteq Z$, both sides of the lax associativity axiom send $((A,B),C)$ to the subset of $X\times (Y\times Z)$ consisting of all $(x,(y,z))$ with $x\in A$, $y\in B$, and $z\in C$. The left and right unit axioms likewise send $(\ast ,A)$ and $(A,\ast )$ back to $A$. Hence $\mathcal{P}$ is lax monoidal.

Take $X=Y=\{0,1\}$. The diagonal subset $\mathrm{\Delta }=\{(0,0),(1,1)\}\subseteq X\times Y$ is not of the form $A\times B$ for subsets $A\subseteq X$ and $B\subseteq Y$. So ${\mathcal{P}}_2$ is not surjective, hence not an isomorphism.

Therefore the power-set functor is lax monoidal but not strong.

Assume one could make the monoidal structure on $\mathcal{S}$ strict merely by identifying isomorphic objects. Then the associator at $D$ would be an identity in the sense excluded by [L1].

By [L1], that would force all endomorphisms $D\to D$ to be equal. But $1_D$ and any constant endomorphism are different maps, since $D$ has at least two elements.

Hence $\mathcal{S}$ is a counterexample to the claim.

The category $\mathcal{C}$ is small, so [L1] applies. Its endofunctors are determined by their action on the two objects because every morphism is an identity.

The composites are easy to compute: $S\circ S=1_{\mathcal{C}}$, $T\circ S=T$, and $S\circ T$ is the constant functor at $1$. Hence $((S\circ T)\circ S)=S\circ (T\circ S)$ as literal equalities of functors, and $1_{\mathcal{C}}\circ S=S=S\circ 1_{\mathcal{C}}$.

This concrete pair therefore realizes composition as a strict tensor product.