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In a symmetric monoidal category, a convolution-invertible scalar pairing $\gamma :H\otimes H\to I$ on a bimonoid inducing the comodule braiding $u\otimes v\mapsto \sum v_{(0)}\otimes u_{(0)}\gamma (u_{(1)},v_{(1)})$. Its multiplicativity, normalization and commutation axioms express the hexagon, unit and colinearity conditions. The notation abbreviates morphisms and ambient symmetries, not a requirement for elements.

A preframe has finite meets and joins of nonempty directed sets, with finite meets distributing over these joins. A preframe homomorphism preserves both operations. Its preframe tensor product $P\#Q$ has generators $p\#q$ and relations making $(p,q)\mapsto p\#q$ a homomorphism separately in each variable. Equivalently,In particular $1\#q=p\#1=1$. The tensor unit is the free preframe on one generator: classically it is $2$, and constructively it is the frame $\Omega$ of truth values, with generator $0$.

Regard a frame $L$ as a commutative monoid object in preframes, with multiplication $x\ast y=x\vee y$ and unit $0$. Binary join preserves finite meets and directed joins separately. Conversely take a commutative monoid object $(M,\ast ,e)$ in preframes and putThis set contains $e$ and $1$, since preservation of the empty meet gives $1\ast x=1$. If $x,y\in R(M)$, then $x\ast y\ge x,y$, andThus finite meets are inherited. If $D\subseteq R(M)$ is nonempty and directed, separate preservation of directed joins givesbecause a common upper bound $t\in D$ gives $d\ast d^{\prime }\le t\ast t=t$. Directed joins are inherited too. Associativity and commutativity show that $x\ast y$ is idempotent; it is the least idempotent upper bound of $x,y$. Therefore $e$ is the bottom and $\ast$ is binary join on $R(M)$. Arbitrary joins are directed joins of finite products, including the empty product $e$. Finallyso binary distributivity, together with directed distributivity, proves that $R(M)$ is a frame.

A homomorphism from a frame join monoid to $M$ sends every element to an idempotent above $e$ and therefore factors uniquely through $R(M)$. The factor preserves finite joins and directed joins, hence arbitrary joins. Conversely the inclusion $R(M)\to M$ preserves the preframe and monoid operations. Also a preframe monoid homomorphism between frames is automatically a frame homomorphism. This proves the full embedding and its right adjoint: $\mathbf{Frm}$ is a coreflective subcategory of $\mathbf{CMon}(\mathbf{PFrm})$, with coreflector $R$.

The coproduct of commutative monoid objects in a symmetric monoidal category is their tensor: the maps are $x\mapsto x\#e_N$, $y\mapsto e_M\#y$, and a pair $f,g$ induces $x\#y\mapsto f(x)\ast g(y)$. For two frames $L,K$, the induced multiplication on pure generators isEvery pure generator is idempotent and above $e$. Those properties are preserved by finite meets and directed joins, as just proved, so hold throughout the generated preframe. It is therefore already a frame, rather than requiring a further coreflection. ConsequentlyRectangles in the localic product are instead $i_L(a)\wedge i_K(b)=(a\#0)\wedge (0\#b)$.

For a filtered colimit in a category of frames, form the colimit $P$ in preframes, with stage maps $i_j$. It can be presented by stage elements, all their finite-meet and directed-join relations, and the transition relations. Since tensoring is a left adjoint, it preserves colimits. Filteredness makes the diagonal diagram cofinal in the product index diagram, so the stage multiplications induce $P\#P\to P$. The units agree. Each stage element is idempotent above the common unit; generation again implies $P=R(P)$. The resulting frame is the filtered colimit in a category in $\mathbf{Frm}$; its binary join is induced by computing two representatives at a common later stage. Thus binary coproducts and filtered colimits of frames are computed in preframes, with the indicated induced join operation.

We now prove compactness. For a compact locale, its frame $L$ has a top test $q_L:L\to \Omega$, $q_L(a)=[a=1]$, preserving finite meets and directed joins. For compact $L,K$, tensor the tests and use the unit isomorphism $\Omega \#\Omega \cong \Omega$ to obtain a preframe map $q:L\#K\to \Omega$ withIt reflects top on every pure generator. The property $q(x)=1⟹x=1$ is closed under finite meets and directed joins: for a directed join, preservation of the join says that its truth value is the existential disjunction of the individual truth values, so some term has value $1$. Since pure generators generate the preframe, $q$ reflects top everywhere. Applying $q$ to a directed cover of $1$ therefore shows that a term of that cover is $1$. This proves compactness of the binary coproduct frame, hence of the binary product locale.

For a filtered diagram of compact frames, define on stage $j$These truth-valued maps preserve finite meets: witnesses can be taken to a common later stage, using filteredness. They preserve directed joins: compactness at the witnessing later stage selects a member of the directed family. They are compatible with transitions, again by filteredness, so induce a preframe map $q:P\to \Omega$. On stage generators, $q(i_j(a))=1$ implies $i_j(a)=1$. The same generation argument proves that $q$ reflects top throughout $P$. Thus the filtered colimit frame is compact.

Finally the arbitrary coproduct of the frames $\mathcal{O}(X_i)$ is the filtered colimit of their finite coproducts, indexed by finite subsets of the index set; the empty finite coproduct is $\Omega$, itself compact. Finite coproducts are compact by the binary argument, and their filtered colimit is compact by the preceding paragraph. Opposing the categories givesThis proves the localic Tychonoff theorem through preframes, without choosing points or invoking the axiom of choice.

In the stable homotopy category, a ring spectrum is a unital associative monoid object for the smash product of spectra. Thus it consists of a topological spectrum $E$ with multiplication and unitwhere $\mathbb{S}$ is the sphere spectrum, and the identities areThe canonical associativity and unit identifications of the symmetric monoidal category are understood in these formulas. All identities are identities of maps in the stable homotopy category.

A commutative ring spectrum also satisfieswhere $\tau :E\wedge E\to E\wedge E$ interchanges the two factors. This is the ring-object convention; a coherently structured ring spectrum has additional point-set or higher-homotopy data. In particular homotopy-category commutativity alone is not an assertion of such a coherent enhancement.

A monoidal category consists of a category $\mathcal{C}$, a bifunctor $\otimes :\mathcal{C}\times \mathcal{C}\to \mathcal{C}$, a monoidal unit object $I$, and natural isomorphismsThe associator and unitors satisfy the pentagon identity for a monoidal categoryand the triangle identity for a monoidal categoryComposition is written from right to left. Naturality of these isomorphisms is part of the definition.

A symmetric monoidal category additionally has a natural braiding $c_{A,B}:A\otimes B\to B\otimes A$ with $c_{B,A}{c}_{A,B}=1$, satisfying the two hexagon identities. Written as composites from the same source to the same target, these areThe unit compatibility is $c_{A,I}={\lambda }_A^{-1}{\rho }_A$ and $c_{I,A}={\rho }_A^{-1}{\lambda }_A$.

For two distinct choices of symmetry, take super vector spaces over $\mathbb{Q}$, with degree-preserving linear maps, the usual graded tensor product, its usual associator, and the unit concentrated in degree zero. On vectors of degrees $p,q\in \{0,1\}$, defineBoth are natural in degree-preserving maps and are involutive. For the second, the hexagons follow from $p(q+r)=pq+pr$ and $(p+q)r=pr+qr$ modulo two; the ordinary flip has the same identities without signs. Degree zero of the unit gives unit compatibility. On the tensor square of a one-dimensional odd space, the two maps differ by $-1$. Thus the same monoidal structure can carry two different symmetries; the second uses the Koszul sign rule.

We prove the monoidal coherence theorem by a terminating normalization argument. Its precise formal assertion is that any two composites of associators, unitors, their inverses, and their tensor products with identities, between the same formal tensor expressions, are equal. The expressions must have the same ordered list of object letters after unit symbols are removed. This does not assert that different symmetries are equal, nor does it treat an accidental equality of actual objects as a new structural identification.

Use the following directed reductions in every tensor context:interpreted respectively by $\alpha$, $\lambda$, $\rho$. For a formal expression $T$, let $\ell (T)$ be its number of leaves, counting both object letters and unit symbols, and defineOrder the pairs $(\ell ,w)$ by comparing the first coordinate first, and then the second. A unit reduction decreases the first coordinate. An associativity reduction preserves the first coordinate and decreases the second by $\ell (A)>0$. The same calculation holds in any context, since associativity preserves the size of the rewritten subtree. Hence there is no infinite directed reduction path.

An irreducible expression has no unit next to another tensor factor and no composite left tensor factor. It is therefore the right-associated expression of its ordered nonunit letters, or the single unit $I$ if there are no such letters. In particular the normal form is uniquely determined by the ordered word.

We next show that every pair of first reductions has a commuting completion. Reductions in disjoint subexpressions commute by bifunctoriality of the monoidal tensor product. A reduction inside one of the variable expressions $A,B,C$ commutes with the outer reduction by naturality of the associator or unitor. The genuine overlaps are exactly the following cases.

For $(((A\otimes B)\otimes C)\otimes D)$, associating first at the root or first in its left subtree gives the two sides of the pentagon; their completions to $A\otimes (B\otimes (C\otimes D))$ commute.

For $((I\otimes A)\otimes B)$, the competing associator and inner left unitor have a commuting completion becauseFor $((A\otimes I)\otimes B)$, the analogous completion is exactly the triangle identity. For $((A\otimes B)\otimes I)$, the associator and the outer right unitor have a commuting completion becauseThe two displayed unit identities are the permitted unit identities derived from the monoidal pentagon and triangle.

The final overlap is $I\otimes I$, where both unitors apply. They agree as a consequence of the same identities. Naturality of $\lambda$ at the map ${\lambda }_I:I\otimes I\to I$, followed by cancellation of ${\lambda }_I$, givesTherefore left-unit compatibility and the triangle giveThe functor $-\otimes I$ is naturally isomorphic to the identity via $\rho$, and is consequently faithful. Thus ${\lambda }_I={\rho }_I$, so this overlap commutes too. There are no other nonvariable overlaps among the three reduction patterns.

Termination and these commuting completions imply equality of all normalization arrows, not merely uniqueness of the normal-form object. Here is the induction. Suppose every expression smaller than $T$ in the decreasing pair measure has a unique normalization arrow. Two normalization paths from $T$ start with arrows $r:T\to T_1$ and $s:T\to T_2$. The local completion gives directed paths $d_1:T_1\to V$, $d_2:T_2\to V$ with $d_{1}r=d_{2}s$. Append any normalization ${\nu }_V$ of $V$. By the induction hypothesis the tails of the original paths are ${\nu }_V{d}_1$ and ${\nu }_V{d}_2$. Their composites from $T$ are equal. If $T$ is already normal, its only directed normalization is the identity. This proves the induction.

Write ${\nu }_T:T\to N(T)$ for the resulting unique normalization. For every directed structural generator $g:S\to T$, including a generator in an arbitrary tensor context, uniqueness gives ${\nu }_{T}g={\nu }_S$. Since all structural generators are invertible, the corresponding equation holds for their inverses as well. Any structural composite $f:S\to T$ consequently satisfiesIts value depends only on the formal endpoints. Thus every structural diagram with the specified ordered word commutes, proving the monoidal coherence theorem.

We use right comodules and write their coactions as $u\mapsto \sum u_{(0)}\otimes u_{(1)}$. A coquasitriangular structure gives the braiding on these comodulesIn an arbitrary symmetric monoidal category this notation abbreviates a composite of the two coactions, the ambient symmetry, and $\gamma$; it does not assume that the objects have elements. The coquasitriangular axioms ensure that this composite is a comodule morphism, is invertible using the convolution inverse of $\gamma$, and satisfies the two hexagon laws. More explicitly, in scalar notation those laws come fromwhile the comodule-morphism condition isUnit normalizations give the unit constraints. These descriptions are identities of morphisms in the ambient symmetric category, with the displayed reordering carried by its symmetry.

Apply the preceding self-braiding result to the regular right comodule $(H,\Delta )$. Its Yang–Baxter operator isCompose its braid equation with the three counits. Expanding $R_{23}{R}_{12}{R}_{23}$ and cancelling the leading counit factors givesExpanding $R_{12}{R}_{23}{R}_{12}$ instead givesThey are equal by the Yang–Baxter operator equation. This is the required scalar identity:Indeed, after ${\Delta }^{\otimes 3}$ the ordered factors are $a_{(1)},a_{(2)},b_{(1)},b_{(2)},c_{(1)},c_{(2)}$. The two ambient symmetries on the left produce the three pairings $(a_{(1)},b_{(1)})$, $(a_{(2)},c_{(1)})$, $(b_{(2)},c_{(2)})$. On the right, the central symmetry produces $a_{(1)},a_{(2)},b_{(2)},b_{(1)},c_{(1)},c_{(2)}$, yielding exactly the other three pairings. Hence the calculation identifies the actual morphisms requested, also when the category is not a category of vector spaces.

Let $(\mathcal{V},\otimes ,I)$ be a symmetric monoidal category. A $\mathcal{V}$-enriched category has objects, hom-objects $\mathcal{C}(A,B)\in \mathcal{V}$, composition morphismsand unit morphisms $I\to \mathcal{C}(A,A)$ satisfying the associative and unit diagrams. Its underlying ordinary category has hom-sets

If $\mathcal{V}$ is closed, take its internal hom $[A,B]$ as hom-object. Composition is the transpose of evaluationand the unit is the transpose of $I\otimes A\cong A$. This is the self-enrichment of a closed symmetric monoidal category.