Braided and Symmetric Monoidal Categories

Because [L1] says each $c_{X,Y}$ is an isomorphism, the family $c_{X,Y}^{\mathrm{\prime }}=c_{Y,X}^{-1}$ is well defined. Inverting the naturality square for $c$ and swapping the variable names shows that $c^{\mathrm{\prime }}$ is natural in both variables.

Substitute $c_{X,Y}^{\mathrm{\prime }}=c_{Y,X}^{-1}$ into the first hexagon for $c^{\mathrm{\prime }}$. After reversing the arrows, this equation is exactly the second hexagon for $c$ from [L1] with the object names permuted. Hence the first hexagon holds for $c^{\mathrm{\prime }}$.

The same calculation with the roles of the two hexagons reversed shows that the second hexagon for $c^{\mathrm{\prime }}$ is the first hexagon for $c$ written backwards. Therefore $c^{\mathrm{\prime }}$ satisfies both hexagon identities and is a braiding.

The present hypotheses are exactly those of [F1], so the two displayed unit-compatibility identities hold for every object $X$.

By [L1], the morphisms $c_{\mathbf{1},X}$ and $c_{X,\mathbf{1}}$ are isomorphisms. Step 1.1 therefore gives $c_{\mathbf{1},X}={\rho }_X^{-1}{\lambda }_X,c_{X,\mathbf{1}}={\lambda }_X^{-1}{\rho }_X\mathrm{.}$ These two formulas are inverse to each other, so $c_{\mathbf{1},X}=c_{X,\mathbf{1}}^{-1}$.

By [L2], both hexagon identities from [L1] lose all associators. They become $c_{X,Y\otimes Z}=(1_Y\otimes c_{X,Z})\circ (c_{X,Y}\otimes 1_Z)$ and $c_{X\otimes Y,Z}=(c_{X,Z}\otimes 1_Y)\circ (1_X\otimes c_{Y,Z})$.

Naturality of $c_{X,-}$ with respect to the morphism $c_{Y,Z}:Y\otimes Z\to Z\otimes Y$ gives $(c_{Y,Z}\otimes 1_X)\circ c_{X,Y\otimes Z}=c_{X,Z\otimes Y}\circ (1_X\otimes c_{Y,Z})\mathrm{.}$

Expand the left-hand occurrence of $c_{X,Y\otimes Z}$ in step 2.1 by the first formula of step 1.1, and expand $c_{X,Z\otimes Y}$ by the same formula with $Y$ and $Z$ interchanged. This yields $(c_{Y,Z}\otimes 1_X)\circ (1_Y\otimes c_{X,Z})\circ (c_{X,Y}\otimes 1_Z)=(1_Z\otimes c_{X,Y})\circ (c_{X,Z}\otimes 1_Y)\circ (1_X\otimes c_{Y,Z}),$ which is the stated Yang-Baxter relation. The special case $X=Y=Z$ is the displayed braid equation.

Assume first that the given hexagon is ${\alpha }_{Y,Z,X}\circ c_{X,Y\otimes Z}\circ {\alpha }_{X,Y,Z}=(1_Y\otimes c_{X,Z})\circ {\alpha }_{Y,X,Z}\circ (c_{X,Y}\otimes 1_Z)\mathrm{.}$ Inverting this equality gives ${\alpha }_{X,Y,Z}^{-1}\circ c_{X,Y\otimes Z}^{-1}\circ {\alpha }_{Y,Z,X}^{-1}=(c_{X,Y}^{-1}\otimes 1_Z)\circ {\alpha }_{Y,X,Z}^{-1}\circ (1_Y\otimes c_{X,Z}^{-1})\mathrm{.}$

Replace $(X,Y,Z)$ in step 1.1 by $(Z,X,Y)$. By [L1], the symmetry equation implies $c_{U,V}^{-1}=c_{V,U}$ for all objects $U,V$. After that substitution, step 1.1 becomes ${\alpha }_{Z,X,Y}^{-1}\circ c_{X\otimes Y,Z}\circ {\alpha }_{X,Y,Z}^{-1}=(c_{X,Z}\otimes 1_Y)\circ {\alpha }_{X,Z,Y}^{-1}\circ (1_X\otimes c_{Y,Z}),$ which is exactly the second braiding hexagon.

If instead the second hexagon is given, the same argument with the roles of the two hexagons reversed proves the first. Hence under the symmetry axiom either hexagon implies the other.

For each pair $X,Y$, the two coordinate projections from $X\times Y$ define a unique morphism ${\tau }_{X,Y}:X\times Y\to Y\times X$ with first projection ${\pi }_2$ and second projection ${\pi }_1$. The same universal property defines its inverse ${\tau }_{Y,X}$, so the family is a natural isomorphism.

To check the first hexagon, compare both composites from $X\times (Y\times Z)$ to $(Y\times Z)\times X$ after composing with the three product projections. Each route sends $(x,y,z)$ to $(y,z,x)$, so the two maps are equal. The second hexagon is the same coordinate permutation written on $((X\times Y)\times Z)$ and is checked in the same way.

Swapping twice returns every pair to itself, so ${\tau }_{Y,X}\circ {\tau }_{X,Y}=1_{X\times Y}$. By [L2], the cartesian swap braiding is therefore a symmetry.

The tensor unit belongs to ${\mathcal{Z}}_2(\mathcal{C})$: for every $Y$, [L3] gives $c_{Y,\mathbf{1}}^{-1}=c_{\mathbf{1},Y}$, hence $c_{Y,\mathbf{1}}\circ c_{\mathbf{1},Y}=1_{\mathbf{1}\otimes Y}$.

Suppose $X$ and $Y$ lie in ${\mathcal{Z}}_2(\mathcal{C})$, and let $Z$ be arbitrary. By [L4], the canonical associators in the two hexagons may be suppressed after transporting both sides between the same parenthesised source and target. In that coherent notation, the hexagons give $c_{Z,X\otimes Y}\circ c_{X\otimes Y,Z}=(1_X\otimes c_{Z,Y})(c_{Z,X}\otimes 1_Y)(c_{X,Z}\otimes 1_Y)(1_X\otimes c_{Y,Z})\mathrm{.}$

Transparency of $X$ gives $(c_{Z,X}\otimes 1_Y)(c_{X,Z}\otimes 1_Y)=1_{X\otimes Z}\otimes 1_Y$. Substituting this into step 1.2 leaves $(1_X\otimes c_{Z,Y})(1_X\otimes c_{Y,Z})=1_X\otimes (c_{Z,Y}\circ c_{Y,Z}),$ which is the identity by transparency of $Y$. Hence $X\otimes Y$ lies in ${\mathcal{Z}}_2(\mathcal{C})$.

Because the subcategory is full and is closed under tensor product by step 2.1 and under the unit by step 1.1, tensor products of its morphisms and the ambient associator and unitors all remain in it. The ambient braiding also restricts to it. For central objects $X,Y$, their defining condition says $c_{Y,X}\circ c_{X,Y}=1_{X\otimes Y}$. By the definition in [L2], the restricted braiding is therefore a symmetry. Thus ${\mathcal{Z}}_2(\mathcal{C})$ is symmetric monoidal.

Apply [L1] and [L2] to obtain a monoidal equivalence $F:\mathcal{C}\to \mathcal{D}$ with $\mathcal{D}$ strict. By [F1], the braiding $c$ transports across the full strong monoidal equivalence to a braiding $c^{\mathrm{s}\mathrm{t}\mathrm{r}}$ on $\mathcal{D}$.

The transported braiding is characterized by the compatibility square $J_{Y,X}\circ c_{F(X),F(Y)}^{\mathrm{s}\mathrm{t}\mathrm{r}}=F(c_{X,Y})\circ J_{X,Y}\mathrm{.}$ Thus [L3] makes $F$ braided monoidal, and $\mathcal{C}$ is braided-monoidally equivalent to the strict braided monoidal category $(\mathcal{D},c^{\mathrm{s}\mathrm{t}\mathrm{r}})$.

By [L1], the presentation of $B_2$ has one generator and no defining relator, so every element of $B_2$ is a power of ${\sigma }_1$. Thus $B_2$ is cyclic.

With no relators present, the presented group is the free group on the singleton alphabet $\{{\sigma }_1\}$. For $m>0$ the power ${\sigma }_1^m$ is represented by a word of $m$ copies of ${\sigma }_1$; for $m<0$ it is represented by $\mathrm{\mid }m\mathrm{\mid }$ copies of the formal inverse. Each is a nonempty reduced word, distinct from the empty identity word by [L2]. Hence ${\sigma }_1^m\mathrm{\ne }1$ for every nonzero integer $m$.

Therefore the cyclic group $B_2=⟨{\sigma }_1⟩$ is infinite, so it is infinite cyclic.

Fix $n\ge 2$ and let $G_n$ be the group given by the displayed generators and relations. The adjacent transpositions ${\tau }_i=(ii+1)$ satisfy ${\tau }_i^2=1$. Transpositions with disjoint supports commute, and the two length-three words in ${\tau }_i,{\tau }_{i+1}$ both exchange $i$ with $i+2$ and fix $i+1$. Thus every relator holds in $S_n$. By [F2] there is a homomorphism ${\pi }_n:G_n\to S_n$ sending $s_i$ to ${\tau }_i$, and [F1] makes it surjective.

Put $H=⟨s_1,\dots ,s_{n-2}⟩\le G_n$, with $H=\{1\}$ when $n=2$. For $1\le j<n$ set $r_j=s_{n-1}{s}_{n-2}\cdots s_j$, and set $r_n=1$. The defining relations give the following right-multiplication rules for $1\le i<n$: if $i\le j-2$, then $r_j{s}_i=s_i{r}_j$ by commuting $s_i$ past every factor of $r_j$; if $i=j-1$, then $r_j{s}_i=r_{j-1}$; if $i=j$, then $r_j{s}_i=r_{j+1}$ by $s_j^2=1$; and if $i>j$, then $r_j{s}_i=s_{i-1}{r}_j$. For the last rule commute the final $s_i$ left past $s_j,\dots ,s_{i-2}$, replace $s_i{s}_{i-1}{s}_i$ by $s_{i-1}{s}_i{s}_{i-1}$, and commute the leading $s_{i-1}$ left past $s_{i+1},\dots ,s_{n-1}$. For $j=n$, the rules are $r_n{s}_i=s_i{r}_n$ when $i\le n-2$ and $r_n{s}_{n-1}=r_{n-1}$. Every coefficient on the left of a resulting $r_k$ lies in $H$.

The union $T={\bigcup }_{j=1}^{n}H{r}_j$ contains $1=r_n$ and, by step 1.2, is stable under right multiplication by every generator $s_i$. Since each $s_i$ is its own inverse, every word in the generators lies in $T$, so $G_n=T$ and $\mathrm{\mid }{G}_n\mathrm{\mid }\le n\mathrm{\mid }H\mathrm{\mid }$. The defining relations of the presented group $G_{n-1}$ hold among $s_1,\dots ,s_{n-2}$, hence [F2] gives a surjection $G_{n-1}\twoheadrightarrow H$; this includes $G_1=\{1\}$ when $n=2$. Induction yields $\mathrm{\mid }{G}_n\mathrm{\mid }\le n!$. Independently, a bijection of an $n$-element set has $n$ choices for the first image, then $n-1$ for the second, and so on, hence $\mathrm{\mid }{S}_n\mathrm{\mid }=n!$ by [F3]. The surjection ${\pi }_n$ of step 1.1 is therefore a bijection and an isomorphism. Transport it through [L1] to obtain exactly the stated presentation on the library's underlying set.

If $n=0$ or $1$, there are no adjacent transpositions and the only bijection of the underlying set is the identity. Its group is presented by the empty set of generators and relators.

By [L3], the adjacent transpositions in $S_n$ satisfy the braid relations and the distant-commutativity relations from [L1]. Therefore [L4] extends the assignment ${\sigma }_i\mapsto (ii+1)$ to a homomorphism ${\pi }_n:B_n\to S_n$.

The image of ${\pi }_n$ contains every adjacent transposition in the one-based model, so the transported generation clause of [L2] makes ${\pi }_n$ surjective. Conjugating its values by ${\kappa }_n^{-1}$ preserves multiplication and surjectivity and sends $(ii+1)$ to $(i-1i)$, giving the stated zero-based version.

By [L1] and [L3], choose a braided monoidal equivalence from $\mathcal{C}$ to a strict braided monoidal category $\mathcal{D}$. Because the braiding of $\mathcal{C}$ is involutive, the transported braiding on $\mathcal{D}$ is also involutive, so $\mathcal{D}$ is strict symmetric.

In the strict symmetric category $\mathcal{D}$, every canonical map between two tensor words with the same ordered letters is built from adjacent swaps of neighboring factors. By [F1] and [L4], the resulting composite depends only on the permutation carrying the source word to the target word, not on the chosen decomposition of that permutation into adjacent transpositions.

Transport this canonical map back across the equivalence chosen in step 1.1. Faithfulness of an equivalence preserves uniqueness, so the resulting canonical map in $\mathcal{C}$ depends only on the same permutation. Hence any two canonical symmetric composites with the same source, target, and underlying permutation are equal.

Any two written forms of the same labelled tensor expression differ only by a choice of brackets and a permutation of the listed occurrences.

By [L1], those two parenthesised reorderings are canonically and uniquely isomorphic. Therefore suppressing both the brackets and the order does not change the resulting labelled tensor expression except by that unique canonical identification.

Take $X:=1$ in the braid category. By [L1], the canonical braiding on $X\otimes X$ is ${\sigma }_1$, so its even powers are the endomorphisms ${\sigma }_1^{2m}$ of the object $2$.

By [L2], the elements $1,{\sigma }_1^2,{\sigma }_1^4,\dots$ are pairwise distinct in $B_2$. Therefore the canonical endomorphisms $1,c_{X,X}^2,c_{X,X}^4,\dots$ are pairwise distinct in this braided monoidal category.

If every formal diagram built from associators and braidings commuted in every braided monoidal category, then the morphisms in step 2.1 would all agree. They do not, so the symmetric-form coherence slogan fails in the braided setting.

Define $F_X(n):=X^{\otimes n}$ on objects, with $F_X(0):=\mathbf{1}$. For each generator ${\sigma }_i\in B_n$, define $F_X({\sigma }_i)$ to be the morphism $1_X^{\otimes (i-1)}\otimes c_{X,X}\otimes 1_X^{\otimes (n-i-1)}:X^{\otimes n}\to X^{\otimes n},$ where the braiding acts only on the $i$th and $(i+1)$st tensor factors.

By [L3], the neighboring generators from step 1.1 satisfy the braid relation. Local braidings on disjoint tensor factors commute because in a strict monoidal category they act on separate coordinates. Therefore the Artin relations of $B_n$ hold, and [L4] extends the assignment of step 1.1 uniquely to a homomorphism $B_n\to \mathrm{Aut}(X^{\otimes n})$ for every $n$.

The family from step 2.1 defines a functor $\mathcal{B}\to \mathcal{C}$ because morphisms exist only between equal objects in $\mathcal{B}$, and group multiplication in each $B_n$ is respected by the constructed homomorphism. By construction $F_X(m+n)=F_X(m)\otimes F_X(n)$ on objects, the tensor of braids is sent to juxtaposition of local braidings, and the standard braiding ${\beta }_{m,n}$ of $\mathcal{B}$ is sent to the corresponding block braiding in $\mathcal{C}$. Hence $F_X$ is strict braided monoidal.

Any strict braided monoidal functor $\mathcal{B}\to \mathcal{C}$ sending $1$ to $X$ must send $n=1^{\otimes n}$ to $X^{\otimes n}$ and each generator ${\sigma }_i$ to the local braiding on the $i$th and $(i+1)$st factors. Since the ${\sigma }_i$ generate every $B_n$, such a functor must equal $F_X$. Therefore $F_X$ is unique.

By [L1], replace the given braided monoidal category by a braided-monoidally equivalent strict one. Equality may be checked there because an equivalence is faithful on morphisms.

In the strict model, every formal canonical endomorphism of $X^{\otimes n}$ is built only from the local braidings of adjacent copies of $X$, together with identities, tensoring, and composition. Reading those local crossings before interpretation gives a braid word and hence an element of $B_n$. By [L2], interpreting the formal composite is exactly applying the unique strict braided monoidal functor from the braid category that sends the generating object to $X$.

If the two formal canonical composites have the same underlying braid, then step 2.1 identifies their interpretations as the images of the same morphism in the free braid category under the same braided functor. Hence they are equal in the strict model, and therefore equal in the original braided category by step 1.1 and [L3].

If the two underlying braids are equal, then [L1] says the two canonical composites agree in every braided monoidal category.

Conversely, suppose the two canonical composites agree in every braided monoidal category. Apply this to the braid category from [L2]. There the canonical composites are literally the corresponding braids, so equality of the composites forces equality of the underlying braids.

Therefore the two universal statements are equivalent.

Using [L3], define on $A\otimes B$ the multiplication and unit displayed in the statement. The formula is typed because $1_A\otimes c_{B,A}\otimes 1_B$ rewrites $A\otimes B\otimes A\otimes B$ as $A\otimes A\otimes B\otimes B$, after which ${\mu }_A\otimes {\mu }_B$ lands in $A\otimes B$.

The associativity and unit axioms for this multiplication are exactly the braided interchange identities proved in [F1]: one repeatedly moves the middle $B$-tensorand past the middle $A$-tensorand by the braiding, and the two possible threefold rearrangements agree because the hexagons express the Yang-Baxter compatibility needed for those moves. Hence $A\otimes B$ is again a monoid object.

If $f:A\to A^{\mathrm{\prime }}$ and $g:B\to B^{\mathrm{\prime }}$ are monoid morphisms, then $f\otimes g$ preserves the units and multiplications because tensoring respects composition and the braiding is natural. Thus tensoring extends to morphisms. The ambient associator and unit object of $\mathcal{C}$ are monoid morphisms by the same coherence computation recorded in [F1], so they supply the associator and unit data for $\mathrm{Mon}(\mathcal{C})$. Therefore $\mathrm{Mon}(\mathcal{C})$ is monoidal.

By [L1] and [L2], $\mathrm{Mon}(\mathcal{C})$ is already a monoidal category.

The ambient symmetry $c_{A,B}:A\otimes B\to B\otimes A$ is a monoid morphism between the tensor-product monoids because the symmetry is involutive and natural, so it commutes with the braided interchange formula defining multiplication. Its square is the identity because it already is in $\mathcal{C}$. Therefore these maps provide a symmetric braiding on $\mathrm{Mon}(\mathcal{C})$.

By [L1], $\mathcal{B}$ is a braided monoidal category. Its braiding on the generating object $1\otimes 1=2$ is the crossing braid ${\sigma }_1\in B_2$.

By [L2], ${\sigma }_1$ has infinite order, so in particular ${\sigma }_1^2\mathrm{\ne }{1}_2$. Therefore the square of the braiding on $1\otimes 1$ is not the identity.

A symmetric braiding must square to the identity on every tensor product. Step 2.1 shows that this fails in $\mathcal{B}$, so $\mathcal{B}$ is braided but not symmetric.

If the displayed claim were true, then any two canonical endomorphisms of the same tensor word built from associators and braidings would agree.

The morphisms $1,c_{X,X}^2,c_{X,X}^4,\dots$ from [L1] are all such canonical endomorphisms of $X\otimes X$, yet [L1] says they are pairwise distinct in a specific braided monoidal category. This contradicts step 1.1.

Therefore the claim is false.

Apply the claim to the braid category from [L1]. Then there would exist a braided monoidal equivalence from $\mathcal{B}$ to a strict braided category $\mathcal{D}$ whose braiding is the identity.

A strict braided category with identity braiding is symmetric, because its braiding certainly squares to the identity. Transporting that symmetric structure back across the supposed braided equivalence would make $\mathcal{B}$ symmetric as well.

Step 2.1 contradicts [L1], which says the braid category is not symmetric. Therefore the claim is false.