Free tensor words and ordered algebra encoding

A list remembers the order of its entries. Tensor coherence lets us change its parentheses while preserving that order. Order-preserving maps then encode multiplication by merging consecutive entries, and encode a unit by inserting an empty block. These are different universal constructions: free tensor words have only componentwise maps, whereas the category of finite ordinals also has merges and insertions.

We use categories and functor categories in fixed suitable universes. A tensor category here may have no unit. A unit, when needed, has the constraints established in Tensor duality and coherent inverses. All algebra morphisms preserve both multiplication and unit.

1. Tensor structures without a unit

A semigroupal category is a category $\mathcal T$, a bifunctor $\otimes$, and a natural isomorphism $a_{X,Y,Z}:(X\otimes Y)\otimes Z \longrightarrow X\otimes(Y\otimes Z)$ satisfying the pentagon. More explicitly, its two maps from $((XY)Z)W$ to $X(Y(ZW))$ obey $\begin{gathered} a_{X,Y,ZW}\,a_{XY,Z,W}\\ {}=(1_X\otimes a_{Y,Z,W})\\ {}\circ a_{X,YZ,W}\\ {}\circ(a_{X,Y,Z}\otimes1_W). \end{gathered}\tag{1.1}$ Juxtaposition denotes tensor product; composition is read from right to left. A monoidal category adds a unit and its coherent constraints. Keeping these notions separate matters for free nonempty words.

A strong tensor functor $F:\mathcal T\to\mathcal U$ in the semigroupal sense has natural isomorphisms $\xi^F_{X,Y}:F(XY)\longrightarrow FX\otimes FY.$ Its associativity requirement is $\begin{gathered} (1_{FX}\otimes\xi^F_{Y,Z})\,\xi^F_{X,YZ}\\ {}\circ F(a_{X,Y,Z})\\ {}=a_{FX,FY,FZ}\, (\xi^F_{X,Y}\otimes1_{FZ})\\ {}\circ\xi^F_{XY,Z}. \end{gathered}\tag{1.2}$ A tensor transformation $\theta:F\to G$ is a natural transformation satisfying $\xi^G_{X,Y}\,\theta_{XY} =(\theta_X\otimes\theta_Y)\,\xi^F_{X,Y}. \tag{1.3}$ It need not be invertible. There is no unit condition in (1.3). When source and target have units, a unital functor has its coherent unit identification $s_F:F(I)\to I$; a unital tensor transformation additionally satisfies $s_G\,\theta_I=s_F.\tag{1.4}$ The standard monoidal definitions in Etingof–Gelaki–Nikshych–Ostrik, §2.4 use the inverse comparison $J=\xi^{-1}$. Their unit requirements cannot be inserted into the semigroupal definition. We will also distinguish (1.3) from (1.4) when classifying algebras.

For a functor between categories with units, the source convention calls it unital when $F(I)$, with multiplication $\rho_F=F(\rho)J_{I,I}$, is a unit. This gives the same coherent functor notion. Indeed put $d_X=F(l_X)J_{I,X}$. Functor associativity and the source left-unit equation give $(1_{FI}\otimes d_X)\,a_{FI,FI,FX} =\rho_F\otimes1_{FX}.$ The canonical left constraint for the unit $FI$ satisfies this equation too. Since tensoring on the left with $FI$ is faithful, it uniquely determines $d_X$. Thus $d_X$ is that constraint. Likewise $F(r_X)J_{X,I}$ is the right constraint, by the source right-unit equation and faithfulness of tensoring on the right with $FI$. The unique multiplication-compatible identification $s_F:FI\to I$, proved in the preceding lesson, transports these constraints to those of the target unit. Hence requiring coherent unit comparisons adds no object-level restriction to that source convention. It is the separate morphism condition (1.4) that excludes zero transformations.

There are two ways to reverse a tensor structure. On $\mathcal T^{\mathrm{op}}$, keep $XY$ as the object product and use the opposite arrow corresponding to $a_{X,Y,Z}^{-1}$ as associator. Reversing all arrows in the original pentagon gives its pentagon. On $\mathcal T$ itself put $X\otimes_rY=YX$ and $\begin{gathered} a^r_{X,Y,Z}=a_{Z,Y,X}^{-1},\\ Z(YX)\longrightarrow(ZY)X. \end{gathered} \tag{1.5}$ Its pentagon is the original pentagon on $W,Z,Y,X$, with both routes inverted. Thus neither construction requires a braiding.

2. Free nonempty words and the precise uniqueness statement

For a category $\mathcal I$, define $\mathsf W^+(\mathcal I)$ to have objects the finite nonempty lists of objects of $\mathcal I$. Morphisms between two lists of length $n$ are $n$-tuples of component morphisms; there are no morphisms between different lengths. Composition and identities are componentwise. Concatenation on objects and morphisms is strictly associative, so $\mathsf W^+(\mathcal I) =\coprod_{n\geq1}\mathcal I^n$ is semigroupal. The inclusion $i:\mathcal I\to\mathsf W^+(\mathcal I)$ sends an object to its one-letter word.

No unit exists. If a unit word had length $r\geq1$, tensoring it with a one-letter word could not be isomorphic to that word, since the lengths would be $r+1$ and $1$. If $\mathcal I$ is empty there are no words and hence no possible unit.

We use the pinned open coherence proof explained in Tensor actions and absolute algebra presentations, §1; compare EGNO, Theorem 2.9.2. Here is its interface for a category without a unit. Adjoin a new object $e$, with only its identity morphism, no morphisms to or from the old objects, and set $eX=Xe=X$, including on morphisms. An associator with an $e$ input is the identity; otherwise it is the old associator. A pentagon with no $e$ is the old pentagon. With an $e$ input, erase that input: both paths are the same three-object associator, or identities if fewer than three old objects remain. The unit constraints are identities. This produces a monoidal category containing $\mathcal T$ fully faithfully. Canonical coherence therefore applies to every nonempty ordered word in $\mathcal T$, without assuming that $\mathcal T$ already has a unit.

Proposition 2.1. Restriction to letters is an equivalence $\begin{gathered} \operatorname{TensorFun} (\mathsf W^+(\mathcal I),\mathcal T)\\ \simeq\operatorname{Fun}(\mathcal I,\mathcal T), \end{gathered} \tag{2.1}$ where the morphisms on the left are all transformations satisfying (1.3).

Proof. Given $\varphi:\mathcal I\to\mathcal T$, evaluate a word by its left-associated product: $E_\varphi(x_1,\ldots,x_n) =((\varphi x_1\otimes\varphi x_2) \otimes\cdots)\otimes\varphi x_n.$ For $n=1$ this means $\varphi x_1$. Apply the same iterated tensor bifunctor to component morphisms. Bifunctoriality proves preservation of identities and composition. The comparison $E_\varphi(uv)\longrightarrow E_\varphi(u)\otimes E_\varphi(v) \tag{2.2}$ is the canonical reassociation of the same ordered factors. It is the identity when $v$ has length one; for longer $v$, first split off its last letter, apply the preceding comparison tensored with that letter, and then apply the associator. Naturality follows from naturality of the associator. Both routes in (1.2) are canonical reassociations of one ordered word, so coherence proves the tensor-functor equation. Restriction of $E_\varphi$ is exactly $\varphi$.

For any strong tensor functor $F$ on words, put $F_1=Fi$. Repeatedly split off the last letter using $\xi^F$. This gives a natural isomorphism $\begin{gathered} c^F_w:F(w)\longrightarrow E_{F_1}(w),\\ c^F_{(x)}=1_{F(x)}. \end{gathered} \tag{2.3}$ It is a tensor isomorphism. To check this, split the right-hand word $v$ in $uv$ one letter at a time. The length-one case is the definition of $c^F$. The next step is exactly (1.2) for $F$; induction gives compatibility with (2.2) for every $u,v$.

Given a natural transformation $\eta:F_1\to G_1$, its only possible tensor extension is $\theta_w=(c^G_w)^{-1} \bigl(\eta_{x_1}\otimes\cdots\otimes\eta_{x_n}\bigr)c^F_w. \tag{2.4}$ Use left-associated products in the middle. This is natural on every tuple of morphisms, because $\eta$ and both $c$'s are natural. Naturality of reassociation makes the iterated tensor of $\eta$'s commute with (2.2). Conjugating by the tensor isomorphisms $c^F,c^G$ proves (1.3). Conversely, (1.3) forces (2.4), by induction on word length. Restriction is therefore full and faithful, and the construction $E_\varphi$ proves essential surjectivity. $\square$

Thus an extension is unique up to a unique isomorphism compatible with a specified identification on letters. An unspecified extension may have automorphisms. Nor does coherence say that every isomorphism between two parenthesizations is the same: it says that the canonical composites of associators are the same. Exercise 1 makes both qualifications visible.

3. Examples and actions at their exact scope

The standard examples in EGNO, §2.3 and Stacks, monoidal categories provide the following interfaces.

For a commutative ring $k$, all $k$-modules with $\otimes_k$ form a monoidal category with unit $k$. No finite-generation or flatness assumption is needed for the ordinary tensor bifunctor or its associator.
A monoid $M$, regarded as a discrete category, has tensor product its multiplication and unit its identity. A semigroup gives the corresponding example without a chosen unit.
For a unital $k$-algebra $A$, the category of $k$-central $A$-bimodules, equivalently modules over $A\otimes_kA^{\mathrm{op}}$, is monoidal under $\otimes_A$, with unit $A$. The balanced relation identifies the left and right $k$-actions on the tensor product, so it stays in this category. Reassociation sends $(x\otimes y)\otimes z$ to $x\otimes(y\otimes z)$; the balanced relations respect this map and its reverse. Every pentagon route sends a fourfold elementary tensor to the same ordered tensor. Such tensors generate the iterated products, proving the equality. This also checks the module example by taking $A=k$.
For any category $\mathcal C$, $\operatorname{End}(\mathcal C)$ is strict monoidal under composition, with unit $1_{\mathcal C}$. The tensor of natural transformations is horizontal composition; its functoriality is the interchange law.
A category with finite products is monoidal under any chosen products, with its terminal object as unit. Each associator is the unique map preserving the three projections. Each pentagon route preserves all four projections, so product uniqueness proves coherence. The unit identities have the same proof with the projections involving the terminal object.
A category with finite coproducts is monoidal under chosen coproducts, with its initial object as unit. Reverse the preceding argument: each route agrees after all summand inclusions, so coproduct uniqueness proves the pentagon and unit identities.
For any group $G$ and field $k$, all $k$-linear $G$-representations are monoidal under the diagonal action. On elementary tensors $g(v\otimes w)=gv\otimes gw$; the group law and tensor relations make this an action. Reassociation is $G$-equivariant on elementary tensors, and its pentagon is the module pentagon. The trivial representation $k$ is the unit. Infinite-dimensional representations are included.
The category $\mathsf W^+(\mathcal I)$ is the strict semigroupal example just constructed, including its absence of a unit.
Bounded cohomological complexes of $k$-modules use ordinary total tensor product. Its degree $n$ term is $\bigoplus_{p+q=n}A^p\otimes_kB^q$, with differential $d(a\otimes b)=d(a)\otimes b+(-1)^p a\otimes d(b)$. We retain the ordinary chain associator and its three differential signs from the published chain-level tensor-coherence interface. Specialize its sheaf calculation to a one-point space with coefficient ring $k$. If the inputs occupy degrees $[a,b]$ and $[c,d]$, their ordinary tensor occupies $[a+c,b+d]$, so this restricts to bounded complexes. The unit is $k$ in degree zero. This example uses the ordinary complex category and ordinary tensor product.

There are also two basic actions. The evaluation of endofunctors makes $\operatorname{End}(\mathcal C)$ act on $\mathcal C$: $(F\circ G)(Z)=F(G(Z))$, so its comparison is the identity. Every semigroupal $\mathcal T$ acts on itself by left tensoring. Send $X$ to $X\otimes-$ and a morphism $f$ to $f\otimes1$. The comparison at $Z$ is $a_{X,Y,Z}$; its action equation is the pentagon (1.1). If $\mathcal T$ has a unit, $l_Z:IZ\to Z$ makes this action unital. This checks the two examples against the action convention of Tensor actions and absolute algebra presentations.

4. Ordinal sum has no braiding

Let $\mathsf O$ be the category of finite totally ordered sets, including the empty set, and order-preserving maps. Work first in its skeleton: write $\langle n\rangle=\{0,\ldots,n-1\}$ for $n\geq0$. The sum is $\langle p\rangle+\langle q\rangle=\langle p+q\rangle$. For maps $f:\langle p\rangle\to\langle p'\rangle$ and $g:\langle q\rangle\to\langle q'\rangle$, set $\begin{gathered} (f+g)(i)=f(i),\quad0\leq i<p,\\ (f+g)(p+j)=p'+g(j),\\ 0\leq j<q. \end{gathered}\tag{4.1}$ This preserves order. The formula proves both bifunctoriality and strict associativity by evaluating on each block. The empty ordinal is a strict unit. On all finite ordered sets use tagged disjoint union, with every element of the first block before the second. Reassociation preserves the ordered positions of all entries; its pentagon and units follow by the same position check. The unique increasing enumeration gives a strong monoidal equivalence with the skeleton.

Although $\sigma+\tau$ and $\tau+\sigma$ have the same cardinality and a unique order isomorphism, those isomorphisms are not natural in $\sigma,\tau$. Indeed take $f:\langle1\rangle\to\langle2\rangle,\quad f(0)=0, \qquad g=1_{\langle1\rangle}.$ Every order isomorphism between two skeletal ordinals of the same size is the identity. Naturality of a proposed exchange would therefore require $f+g=g+f$ as maps $\langle2\rangle\to\langle3\rangle$. But $\begin{aligned} (f+g)(1)&=2,\\ (g+f)(1)&=1. \end{aligned} \tag{4.2}$ Thus ordinal sum admits no braiding at all. A braiding's components would necessarily be precisely these unique order isomorphisms, already ruled out by (4.2). This distinction agrees with Weber, §4.1: finite ordinals with monotone maps classify arbitrary monoids; finite sets with all maps, a different category, classify commutative monoids.

5. The canonical algebra classifier and its morphisms

An algebra object here means $A\in\mathcal T$, with $m:A\otimes A\to A$ associative and $u:I\to A$ a two-sided unit. There is no requirement of additivity, a braiding, or commutative multiplication. Algebra morphisms $h:A\to B$ satisfy $hm_A=m_B(h\otimes h)$ and $hu_A=u_B$.

The full fibre construction below proves the canonical classifier discussed in Weber, §4.1 and Proposition 4.1.4: $\operatorname{Alg}(\mathcal T) \simeq \operatorname{MonFun}(\mathsf O,\mathcal T). \tag{5.1}$ Here $\mathcal T$ is any monoidal category and the right side has strong unital functors and unital tensor transformations. The empty ordinal corresponds to the unit, and the one-element ordinal to the algebra. The following full fibre calculation checks the interface, including empty fibres and morphisms.

Define $A^0=I$, $A^1=A$, and use left-associated products for higher powers. Put $m_0=u$, $m_1=1_A$, and recursively $m_{r+1}=m(m_r\otimes1_A)$ for $r\geq1$. If a block has consecutive subblocks of lengths $r_1,\ldots,r_s$, then $\begin{gathered} m_s(m_{r_1}\otimes\cdots\otimes m_{r_s})\\ {}=m_{r_1+\cdots+r_s}. \end{gathered} \tag{5.2}$ The displayed sources use canonical reassociation and unit maps. For positive $r_i$, induction on the block lengths applies associativity to move each successive multiplication into the left-associated product. A zero $r_i$ inserts $u$; a unit law removes it, leaving the same assertion with that subblock deleted. If all $r_i$ vanish, successive applications of $m(u\otimes u)=u$, with the unit identification $I\otimes I\simeq I$, give (5.2). For $s=0$, both sides mean $u$. Thus every case, including an empty block of empty subblocks, is covered.

For a monotone map $f:\langle p\rangle\to\langle q\rangle$, the fibres are consecutive, in output order. Write $r_j=|f^{-1}(j)|$. Define $\begin{aligned} F_A(\langle p\rangle)&=A^p,\\ F_A(f)&=\bigotimes_{j=0}^{q-1}m_{r_j}. \end{aligned} \tag{5.3}$ The source of the latter map is identified with the ordered product of its fibre blocks. If $q=0$, such a map exists only when $p=0$, and it is sent to $1_I$. Identity maps have singleton fibres, so give identity morphisms. For $g:\langle q\rangle\to\langle r\rangle$, each fibre of $gf$ is the consecutive union of the fibres of $f$ indexed by a fibre of $g$. Apply (5.2) in every output block: it says exactly $F_A(g)F_A(f)=F_A(gf)$. Bifunctoriality of tensor combines the equalities between blocks. This proves functoriality.

The canonical comparison $A^{p+q}\to A^p\otimes A^q$ makes $F_A$ strong unital. Both associativity routes are reassociations of the same ordered factors. Its naturality follows from the disjoint fibre blocks of $f+g$; empty blocks use the unit constraints. The unit comparison is $F_A(0)=I$.

Conversely, take a strong unital $F$, with $s_F:F(0)\to I$. Let $c_2:\langle2\rangle\to\langle1\rangle$ be the unique map and $c_0:\langle0\rangle\to\langle1\rangle$ the empty map. Set $\begin{aligned} A&=F(1),\\ m&=F(c_2)(\xi^F_{1,1})^{-1},\\ u&=F(c_0)s_F^{-1}. \end{aligned} \tag{5.4}$ The two consecutive ways of collapsing a three-element ordinal to one element are equal monotone maps. Applying $F$ and (1.2) gives associativity of $m$. The maps $c_0+1$ and $1+c_0$, followed by $c_2$, are the identity of the one-element ordinal. Applying $F$, its coherent unit comparison and $\xi$ gives the two unit laws.

Iterate $\xi^F$, using $s_F$ at $0$, to obtain natural tensor identifications $F(p)\simeq A^p$. Any monotone map is the ordinal sum of its fibre-collapse maps $c_{r_j}:\langle r_j\rangle\to\langle1\rangle$. For $r_j>1$, successive binary collapses express $c_{r_j}$; $c_1$ is the identity and $c_0$ supplies the unit. Functoriality and the strong comparisons therefore identify $F(f)$ with (5.3). The exceptional map to $0$ is its identity. This proves that (5.4) and (5.3) recover each other up to a unital tensor isomorphism.

For an algebra map $h:A\to B$, its powers, with $h^0=1_I$, give a natural transformation $F_A\to F_B$. The identities $hm_r=m_r(h^{\otimes r})$ follow by induction from the multiplication and unit conditions, proving naturality on every fibre. The powers also satisfy tensor compatibility and (1.4).

Conversely, a unital tensor transformation is determined on positive ordinals by its component $h$ at $1$, using (1.3), and on $0$ by (1.4). Naturality for $c_2$ gives the multiplication condition; naturality for $c_0$ gives the unit condition. The transformation is exactly the powers of $h$ under the preceding identifications. This proves full faithfulness and completes the checked equivalence (5.1).

The adjective “unital” on transformations is essential. In $\mathcal T=\operatorname{Mod}(k)$ for a nonzero commutative ring $k$, the transformation whose every component is zero is natural and satisfies (1.3) between any two strong unital functors. It fails (1.4). Thus evaluation at $1$ with those unrestricted transformations does not give the algebra-morphism classification. There is even no abstract equivalence between those two categories: every such functor has distinct zero and identity endomorphisms, because its value at $0$ is isomorphic to $k\neq0$. The zero algebra is a valid unital algebra object and has only one endomorphism. An equivalence would preserve that endomorphism set. This supplies a categorical counterexample, beyond the failure of the proposed evaluation functor.

6. Graded exercises with complete solutions

Exercise 1 (easy). Let $\mathcal I$ have one object and only its identity, and let its image be the line $k$ over a field with an element $\lambda\in k^\times\setminus\{1\}$. Describe the free nonempty words and exhibit a nontrivial tensor automorphism of their evaluation functor. Explain why it does not contradict Proposition 2.1.

Solution. There is one word of each length $n\geq1$, and only identity morphisms. Concatenation adds lengths. Its evaluation is $k^{\otimes n}$. Multiply this vector space by $\lambda^n$. These maps are invertible and natural; $\lambda^{p+q}=\lambda^p\lambda^q$ gives (1.3), since reassociation commutes with scalar multiplication. At length one the map is multiplication by $\lambda$, so it is nontrivial. Proposition 2.1 permits exactly the automorphisms of the one-letter image. If that component is required to be the identity, every power is the identity, recovering the qualified uniqueness.

Exercise 2 (moderate). Let $A$ be a unital, possibly noncommutative $k$-algebra. Take $f:\langle4\rangle\to\langle3\rangle$ with values $0,0,2,2$, and $g:\langle3\rangle\to\langle2\rangle$ with values $0,1,1$. Compute $F_A(f)$, $F_A(g)$ and $F_A(gf)$ on elementary tensors, and verify composition.

Solution. The fibres of $f$ have lengths $2,0,2$, so $F_A(f)(a\otimes b\otimes c\otimes d) =ab\otimes1_A\otimes cd.$ The fibres of $g$ have lengths $1,2$, so $F_A(g)(x\otimes y\otimes z)=x\otimes yz$. Its value on the preceding tensor is $ab\otimes(1_Acd)=ab\otimes cd$. The composite $gf$ has values $0,0,1,1$, giving precisely $F_A(gf)(a\otimes b\otimes c\otimes d)=ab\otimes cd$. Elementary tensors generate, so equality holds as a linear map. No factors were exchanged and no commutativity was used. The empty fibre was necessary: omitting its unit would give the wrong codomain for $F_A(f)$.

Exercise 3 (hard). For an algebra $A$ over a field $k$, classify all tensor endomorphisms of $F_A$ satisfying only (1.3), and identify those satisfying (1.4).

Solution. On the empty ordinal such an endomorphism is a scalar $t:k\to k$. Tensor compatibility at $0+0$ gives $t^2=t$. At $1$ it is a linear map $h:A\to A$; tensor compatibility at $0+1$ gives $h=th$, and all positive components must be $h^{\otimes n}$. Naturality for $c_2$ and $c_0$ gives $hm=m(h\otimes h),\qquad hu=t\,u.$ Over a field an idempotent scalar is $0$ or $1$. If $t=0$, then $h=0$ and every component is zero. If $t=1$, the displayed conditions say exactly that $h$ is a unital algebra endomorphism, and its powers give the transformation by §5. Those transformations and the all-zero transformation exhaust the possibilities: the fibre description verifies naturality for every map. Condition (1.4) requires $t=1$, so retains precisely the unital algebra endomorphisms. For the zero algebra the positive components of its identity also vanish, but its empty component is $1_k$, distinguishing it from the unrestricted all-zero transformation.

Exercise 4 (expert). Prove that comonoids in any monoidal category $\mathcal T$ are classified by strong unital functors $\mathsf O^{\mathrm{op}}\to\mathcal T$ and unital tensor transformations. State the functor on a reversed monotone map and prove its composition rule, including empty fibres.

Solution. A comonoid has $\delta:C\to C\otimes C$ coassociative and $e:C\to I$ a counit. The opposite-category construction in §1 makes $\mathcal T^{\mathrm{op}}$ monoidal with the same object product. A comonoid in $\mathcal T$ is exactly an algebra in $\mathcal T^{\mathrm{op}}$. Apply (5.1) there. A functor $\mathsf O\to\mathcal T^{\mathrm{op}}$, on taking opposites, is a functor $\mathsf O^{\mathrm{op}}\to\mathcal T$; its comparison reverses direction but remains invertible. Natural transformations reverse direction too. Consequently one must take the opposite of both classified categories; this restores the direction of comonoid morphisms.

Explicitly, put $C^0=I$, $\delta_0=e$, $\delta_1=1_C$, and let $\delta_r:C\to C^r$ be iterated comultiplication. For $f:\langle p\rangle\to\langle q\rangle$ with fibre lengths $r_j$, send its opposite arrow to $C^q\longrightarrow C^p,\qquad \bigotimes_{j=0}^{q-1}\delta_{r_j}.$ The outputs are reassociated in the original input order. A zero fibre applies the counit. For composable $f,g$, each block of $gf$ is the consecutive union of the corresponding blocks of $f$. The reverse of (5.2), proved by coassociativity and the two counit laws, says that expanding first by a fibre of $g$ and then by its fibres of $f$ equals the single expansion by their union. It includes all-zero fibres by the counit law, and the empty target has only its identity. Tensoring these identities proves functoriality on the reversed composite. A comonoid map $h$ has the powers $h^{\otimes n}$ and identity component at $0$; conversely naturality at the reversed $c_2,c_0$ forces preservation of $\delta,e$. This proves the asserted equivalence with the required morphism direction and unit condition.

Original exposition and solutions are dedicated to CC0 1.0. References provide the canonical interfaces; their text is not reproduced. Self-checked; no independent review has occurred.