Monoidal Categories and Monoidal Functors

In $\mathcal{Z}$, every integer is invertible for composition, so any chosen integers can serve as the components of $\alpha ,\lambda ,\rho$. The tensor prescription is a bifunctor because $((m^{\mathrm{\prime }}\otimes n^{\mathrm{\prime }})\circ (m\otimes n))=(m^{\mathrm{\prime }}+n^{\mathrm{\prime }})+(m+n)=(m^{\mathrm{\prime }}+m)+(n^{\mathrm{\prime }}+n)=((m^{\mathrm{\prime }}\circ m)\otimes (n^{\mathrm{\prime }}\circ n))$. Thus only the pentagon and triangle equations need to be checked.

First choose $\alpha =0$, $\lambda =1$, and $\rho =0$. Then the pentagon reads $0+0=0+0+0$, so it holds, while the triangle becomes $0+1=0$, namely $1=0$, so it fails. Hence the pentagon does not imply the triangle.

Now choose $\alpha =1$, $\lambda =0$, and $\rho =1$. Then the triangle reads $1+0=1$, so it holds, while the pentagon becomes $1+1=1+1+1$, namely $2=3$, so it fails. Hence the triangle does not imply the pentagon.

The two witnesses prove that neither axiom is a consequence of the other.

Let $X\otimes Y:=X\times Y$ and let the unit object be $1$. On morphisms, define $(f,g)\mapsto f\times g$ by the universal property of the binary product. Because product pairings are unique, identities and compositions are preserved componentwise, so $\times :\mathcal{C}\times \mathcal{C}\to \mathcal{C}$ is a bifunctor.

Let ${\lambda }_X:(1\times X)\to X$ and ${\rho }_X:(X\times 1)\to X$ be the canonical product projections. If ${!}_X:X\to 1$ is the unique map from [L2], then the inverse of ${\lambda }_X$ is the pairing $⟨{!}_X,1_X⟩:X\to 1\times X$, and the inverse of ${\rho }_X$ is $⟨1_X,{!}_X⟩:X\to X\times 1$. Hence $\lambda$ and $\rho$ are natural isomorphisms.

For objects $X,Y,Z$, let ${\alpha }_{X,Y,Z}:((X\times Y)\times Z)\to X\times (Y\times Z)$ be the unique arrow whose three composite projections are the obvious first, second, and third projections. Define ${\alpha }_{X,Y,Z}^{-1}$ similarly. By the same uniqueness argument, these arrows are inverse and natural.

Both sides of the pentagon are arrows from $(((W\times X)\times Y)\times Z)$ to $W\times (X\times (Y\times Z))$. They have the same four composites with the terminally iterated product projections, so [L1] makes them equal. The same argument on the two projections from $(1\times (X\times Y))$ to $X$ and $Y$ shows the triangle equation.

Therefore $(\mathcal{C},\times ,1,\alpha ,\lambda ,\rho )$ is a monoidal category. The coproduct statement is proved by the same construction with pairings replaced by copairings and terminal replaced by initial, using the dual clauses already present in Products and coproducts as limits and colimits of discrete diagrams, including their existence-and-uniqueness equations and Initial object, terminal object, and zero object.

The cartesian product of sets and the singleton set give binary products and a terminal object in $\mathbf{S}\mathbf{e}\mathbf{t}$, so [L1] yields a monoidal structure on $\mathbf{S}\mathbf{e}\mathbf{t}$.

In $\mathbf{C}\mathbf{a}\mathbf{t}$, the product category is the categorical binary product and the one-object category is terminal, so [L1] yields a monoidal structure on $\mathbf{C}\mathbf{a}\mathbf{t}$.

If $\mathcal{C}$ is complete, then [L3] gives the required terminal object and binary products, and [L1] makes $\mathcal{C}$ cartesian monoidal.

Therefore $\mathbf{S}\mathbf{e}\mathbf{t}$, $\mathbf{C}\mathbf{a}\mathbf{t}$, and every complete category are cartesian monoidal.

On ${\mathrm{End}}_{\mathcal{C}}(\mathbf{1})$ let $f\circ g$ be ordinary composition and define $f\ast g:={\lambda }_{\mathbf{1}}\circ (f\otimes g)\circ {\lambda }_{\mathbf{1}}^{-1}$. Composition is associative with unit $1_{\mathbf{1}}$. Naturality of $\lambda$ at $f$ gives ${\lambda }_{\mathbf{1}}\circ (1_{\mathbf{1}}\otimes f)=f\circ {\lambda }_{\mathbf{1}}$, so $1_{\mathbf{1}}\ast f=f$. Naturality of $\rho$ at $f$ gives ${\rho }_{\mathbf{1}}\circ (f\otimes 1_{\mathbf{1}})=f\circ {\rho }_{\mathbf{1}}$, and [L2] identifies ${\rho }_{\mathbf{1}}$ with ${\lambda }_{\mathbf{1}}$. Hence $f\ast 1_{\mathbf{1}}=f$ as well, so $\ast$ has the same unit $1_{\mathbf{1}}$.

For $f,g,h,k\in {\mathrm{End}}_{\mathcal{C}}(\mathbf{1})$, one has $((f\circ g)\ast (h\circ k))={\lambda }_{\mathbf{1}}\circ ((f\circ g)\otimes (h\circ k))\circ {\lambda }_{\mathbf{1}}^{-1}={\lambda }_{\mathbf{1}}\circ ((f\otimes h)\circ (g\otimes k))\circ {\lambda }_{\mathbf{1}}^{-1}=(f\ast h)\circ (g\ast k)$, so the interchange law holds.

By [L4], the operations $\circ$ and $\ast$ coincide and their common value is commutative. Since composition is already associative and unital, [L3] shows that composition makes ${\mathrm{End}}_{\mathcal{C}}(\mathbf{1})$ into a commutative monoid.

If $\mathcal{C}$ is strict and has one object, then tensor and composition on endomorphisms are the same literal operation. Thus its endomorphism monoid is commutative; [L5] supplies the converse comparison with an ordinary one-object category.

Therefore the endomorphisms of the tensor unit form a commutative monoid.

By [L1] and [L2], $[\mathcal{C},\mathcal{C}]$ is a legitimate category whose objects are endofunctors of $\mathcal{C}$ and whose morphisms are natural transformations.

Define the tensor product on objects by $F\otimes G:=F\circ G$, and on morphisms by whiskered horizontal composition of natural transformations. The unit object is the identity functor $1_{\mathcal{C}}$.

Composition of functors is literally associative and unital, so $((F\otimes G)\otimes H)=F\otimes (G\otimes H)$ and $1_{\mathcal{C}}\otimes F=F=F\otimes 1_{\mathcal{C}}$ on the nose. The associator and unitors are therefore identity transformations.

Whiskering respects identities and compositions, so the tensor on morphisms is a bifunctor.

Hence $[\mathcal{C},\mathcal{C}]$ is strict monoidal under composition.

By [L2] and [L3], the endofunctors of $\mathcal{C}$ form a strict monoidal category whose tensor is composition and whose unit is $1_{\mathcal{C}}$. Thus a monoid-object structure on $T$ is exactly data $\eta :1_{\mathcal{C}}\Rightarrow T$ and $\mu :T\circ T\Rightarrow T$ with the usual associative and unital diagrams, because the associator and unitors are identities.

The equations described in step 1.1 are precisely $\mu \circ T\mu =\mu \circ \mu T$ and $\mu \circ T\eta =1_T=\mu \circ \eta T$, which are exactly the monad laws in [L1]. Hence every monoid object in $[\mathcal{C},\mathcal{C}]$ is a monad.

Conversely, if $(T,\eta ,\mu )$ is a monad, then [L1] gives exactly the same two diagrams, so the same data make $T$ a monoid object in $[\mathcal{C},\mathcal{C}]$.

Therefore the two notions are equivalent for small $\mathcal{C}$.

For every module $(X,a)$, the identity $1_X$ is a module morphism because $1_X\circ a=a=a\circ (1_M\otimes 1_X)$. Thus identities exist.

If $f:(X,a)\to (Y,b)$ and $g:(Y,b)\to (Z,c)$ are module morphisms, then $(g\circ f)\circ a=g\circ (f\circ a)=g\circ b\circ (1_M\otimes f)=c\circ (1_M\otimes g)\circ (1_M\otimes f)=c\circ (1_M\otimes (g\circ f))$, so $g\circ f$ is again a module morphism.

Because composition in $\mathcal{C}$ is associative by [L1], the module morphisms with these identities and composites form a category.

By [L1], the tensor is $M\times M$ and the unit is a terminal object $1$. So [L2] says exactly that a monoid object on $M$ is data $\mu :M\times M\to M$ and $e:1\to M$ satisfying the usual associative and left and right unit diagrams, with the only extra notation being the canonical rebracketing isomorphism $((M\times M)\times M)\cong M\times (M\times M)$.

In $\mathbf{S}\mathbf{e}\mathbf{t}$, a morphism $1\to M$ is the choice of an element $e\in M$, and a morphism $M\times M\to M$ is a binary operation on the underlying set. The three diagrams from step 1.1 then say exactly $\mu (\mu (x,y),z)=\mu (x,\mu (y,z))$ and $\mu (e,x)=x=\mu (x,e)$ for all $x,y,z\in M$.

Therefore monoid objects in a cartesian monoidal category are associative unital multiplications internal to that cartesian structure, and in $\mathbf{S}\mathbf{e}\mathbf{t}$ they are exactly ordinary monoids.

By [L1], every abelian group is a $\mathbb{Z}$-module and every group homomorphism is $\mathbb{Z}$-linear. Thus the tensor product over $\mathbb{Z}$ of two abelian groups is again an abelian group, and the tensor-product maps are morphisms in $\mathbf{A}\mathbf{b}$.

The associativity isomorphism $((A{\otimes }_{\mathbb{Z}}B){\otimes }_{\mathbb{Z}}C)\cong A{\otimes }_{\mathbb{Z}}(B{\otimes }_{\mathbb{Z}}C)$ and the unit isomorphisms $\mathbb{Z}{\otimes }_{\mathbb{Z}}A\cong A$ and $A{\otimes }_{\mathbb{Z}}\mathbb{Z}\cong A$ are exactly the maps supplied by [L3] and [L4] for the commutative ring $\mathbb{Z}$.

By [L5], homomorphisms $A{\otimes }_{\mathbb{Z}}B\to C$ are the same as balanced maps $A\times B\to C$. By step 1.1 and [L1], balanced over $\mathbb{Z}$ is exactly bilinear for abelian groups, so the same universal property identifies homomorphisms $A{\otimes }_{\mathbb{Z}}B\to C$ with bilinear maps $A\times B\to C$. Together with step 1.2 this is the monoidal structure on $\mathbf{A}\mathbf{b}$.

Therefore $\mathbf{A}\mathbf{b}$ is monoidal under ${\otimes }_{\mathbb{Z}}$ with unit object $\mathbb{Z}$.

By [L1] and [L2], a monoid object structure on $A$ is a homomorphism $\mu :A{\otimes }_{\mathbb{Z}}A\to A$ and a homomorphism $\eta :\mathbb{Z}\to A$. The map $\mu$ corresponds to a bilinear operation $(x,y)\mapsto x\cdot y$ on $A$, and $\eta$ is determined by the element $1:=\eta (1)\in A$.

The monoid-object associativity and unit diagrams from [L1] translate under the correspondence in [L2] to $(x\cdot y)\cdot z=x\cdot (y\cdot z)$ and $1\cdot x=x=x\cdot 1$. Because $\mu$ is a group homomorphism in each variable, the operation is bilinear, hence distributes over the abelian-group addition on both sides.

Thus the additive abelian-group structure on $A$, together with multiplication $\cdot$ and unit $1$, satisfies exactly the axioms in [L3], so $A$ is a ring. Conversely, any ring yields such a bilinear multiplication and unit homomorphism, hence a monoid object in $\mathbf{A}\mathbf{b}$.

Therefore monoid objects in abelian groups are exactly rings.

By [L1], the modules under discussion already form a category. Because $R$ is commutative, every left $R$-module is canonically an $(R,R)$-bimodule, so $M{\otimes }_{R}N$ is again an $R$-module.

The associator $((L{\otimes }_{R}M){\otimes }_{R}N)\to L{\otimes }_R(M{\otimes }_{R}N)$ and the unitors $R{\otimes }_{R}M\to M$ and $M{\otimes }_{R}R\to M$ are exactly the isomorphisms supplied by [L2] and [L3].

By [L4], tensoring homomorphisms is functorial in each variable, so ${\otimes }_R$ is a bifunctor on $R\text{-}\mathbf{M}\mathbf{o}\mathbf{d}$. Steps 1.1 and 1.2 therefore provide the data of a monoidal category.

Hence modules over a commutative ring form a monoidal category under tensor product.

By [L2], $P$ becomes a category whose objects are the elements of $P$ and whose morphisms are the order relations.

Define $x\otimes y:=x\wedge y$ on objects. Because meet is monotone in each variable, the unique arrows $x\to x^{\mathrm{\prime }}$ and $y\to y^{\mathrm{\prime }}$ induce the unique arrow $x\wedge y\to x^{\mathrm{\prime }}\wedge y^{\mathrm{\prime }}$, so $\otimes$ is a bifunctor.

The universal property of meet gives $(x\wedge y)\wedge z=x\wedge (y\wedge z)$ and $\mathrm{\top }\wedge x=x=x\wedge \mathrm{\top }$ as equalities of objects in the poset. Since each hom-collection has at most one arrow, the associator and unitors are automatically identity morphisms.

Therefore the associated category is strict monoidal.

The hypotheses of [F1] are exactly those in [L1], so [F1] applies to the present monoidal category.

Therefore the displayed identity holds for every pair $X,Y$.

The hypotheses of [F1] are exactly those in [L1], so [F1] applies to the present monoidal category.

Therefore the displayed identity holds for every pair $X,Y$.

The monoidal-category hypotheses required by [F1] are exactly the ones under which [L1] and [L2] were proved, so [F1] applies to the present category.

Therefore ${\lambda }_{\mathbf{1}}={\rho }_{\mathbf{1}}$.

Therefore the two unitors agree on the tensor unit.

Define the composite structure maps by $(GF{)}_0:=G(F_0)\circ G_0$ and $(GF{)}_{2;X,Y}:=G(F_{2;X,Y})\circ G_{2;F(X),F(Y)}$. These are the only typed composites from $G(F(X))\otimes G(F(Y))$ to $GF(X\otimes Y)$ and from the unit of $\mathcal{E}$ to $GF(\mathbf{1})$.

Paste the associativity square for $G$ with the image under $G$ of the associativity square for $F$. The outside rectangle is exactly the associativity axiom for $(GF{)}_2$. Pasting the two unit squares gives the left and right unit axioms for $(GF{)}_0$. Hence $GF$ is lax monoidal.

If both $F$ and $G$ are strong, then every map used in step 1.1 is an isomorphism, so $(GF{)}_0$ and $(GF{)}_2$ are isomorphisms. If both are strict, those maps are identities, so the composite structure maps are identities too.

Therefore composition preserves laxness, strength, and strictness.

Define ${\mu }^{\mathrm{\prime }}:=F(\mu )\circ F_{2;M,M}$ and ${\eta }^{\mathrm{\prime }}:=F(\eta )\circ F_0$. These are the only typed maps with the displayed source and target.

For associativity, paste the lax associativity square for $F$ with the image under $F$ of the monoid-object associativity square for $\mu$. Both composites from $(F(M)⊠F(M))⊠F(M)$ to $F(M)$ equal $F(\mu \circ (1_M\otimes \mu )\circ {\alpha }_{M,M,M})$ with the same inserted $F_2$-maps. Hence ${\mu }^{\mathrm{\prime }}$ is associative.

The left and right unit diagrams for ${\eta }^{\mathrm{\prime }}$ are obtained in the same way by pasting the lax unit squares with the image under $F$ of the two unit diagrams for $(M,\mu ,\eta )$. Thus ${\eta }^{\mathrm{\prime }}$ is a two-sided unit for ${\mu }^{\mathrm{\prime }}$.

Therefore $(F(M),{\mu }^{\mathrm{\prime }},{\eta }^{\mathrm{\prime }})$ is a monoid object in $\mathcal{D}$.

For $n=1$ there is only the word $x_1$, so $a_1=1$.

If $n\ge 2$, every word has a unique outermost decomposition $(u\otimes v)$, where $u$ uses the first $k$ letters and $v$ uses the remaining $n-k$ letters for a unique $k$ with $1\le k\le n-1$. Conversely, every such pair $(u,v)$ produces a word on $n$ letters.

Therefore the words on $n$ letters are partitioned by the value of $k$, and for fixed $k$ there are $a_k{a}_{n-k}$ choices. Summing over $k$ gives $a_n={\sum }_{k=1}^{n-1}{a}_k{a}_{n-k}$.

The recurrence in step 2.1 is exactly the Catalan recurrence after the index shift $a_n=C_{n-1}$, and step 1.1 matches the initial value $C_0=1$. Hence $a_n=C_{n-1}$ for all $n\ge 1$.