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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.

The tensor differential. Use homological grading, so each differential lowers degree by one. The tensor product of chain complexes hasfor $x\in C_p$. This Koszul sign rule givesThus the graded tensor product is a chain complex.

The Hom differential. Write $M_j={\prod }_p{\mathrm{Hom}}_R(C_p,C_{p+j}^{\prime })$. For a degree-$j$ element of this graded Hom complex of chain complexes, use the prescribed differentialThe next application uses degree $j-1$, soIn degree zero, $d_{M}f=f{d}_C-d_{C^{\prime }}f$, so its kernel consists exactly of chain maps. A degree-one element $h$ has $d_{M}h=h{d}_C+d_{C^{\prime }}h$, which is exactly the change between two chain maps related by a chain homotopy. ConsequentlyThis is natural: precomposition and postcomposition by chain maps preserve both degree-zero cycles and degree-zero boundaries.

The dual complex and the sign adjustment. Define the reversed dual chain complexHere $d_X\varphi$ is a functional on $C_{1-i}$, so it belongs to $X_{i-1}$, and $d_X^2=0$. The absence of an additional sign in $d_X$ is intentional.

Interpret finite generation of the free chain complexes as finiteness of their total graded free modules. Then only finitely many degrees occur, and the finite-free evaluation isomorphisms assemble into a graded isomorphismwhere $\varphi \in X_i$, $y\in C_j^{\prime }$, and this component is zero outside $C_{-i}$. The dual module construction and evaluation make $\psi$ natural. Finite rank is needed for evaluation to be an isomorphism; finite total support also makes the sums on the tensor side agree with the products on the Hom functor side.

Under $\psi$, the Hom functor differential has the formThe usual tensor product of chain complexes instead has second coefficient $(-1{)}^i$. SetThis is integer-valued for every $j\in \mathbb{Z}$, including negative $j$, and satisfies $\rho (j)-\rho (j-1)=j-1$. Conjugating the usual tensor differential by $A$ leaves the first term unchanged and changes its second coefficient to $(-1{)}^{i+j-1}$. Hence $d_M=\psi AD{A}^{-1}{\psi }^{-1}$, giving the sign conjugation for the tensor-Hom identificationThis is an isomorphism of chain complexes, not merely of their homology. If one instead assumes only degreewise finite rank with unbounded grading, the ordinary tensor need not identify with the product defining $M$; the finite-total convention is essential to this last conclusion.

We use the James reduced product theorem and the Bott–Samelson theorem: if $X$ is a connected based CW complex with free integral homology, then $JX\to \Omega \Sigma X$ is a weak equivalence, and its induced Pontryagin ring isHere $\Sigma$ is the reduced suspension, $T$ is the tensor algebra, and the multiplication is induced by concatenating James words, hence by concatenating loops. Taking $X=S^{2n}$ supplies a single generator $x$ of degree $2n$, sois free, with one generator in every degree $2nk$ and zero in the other degrees.

The diagonal makes this a homology coalgebra. The generator $x$ is a primitive homology class:There are no nontrivial lower positive degrees in which its reduced diagonal could land. Compatibility of the diagonal with loop multiplication givesThe degree of $x$ is even, so the two tensor factors commute without a Koszul sign rule.

The universal coefficient theorem for cohomology has no Ext terms here because homology is free. Let $a_k$ be the cohomology class dual to $x^k$, with $a_0=1$ and $|a_k|=2nk$. The cup product is dual to the diagonal, henceTherefore the integral cohomology ring is the divided power algebraExplicitly $H^{2nk}=\mathbb{Z}{a}_k$ for $k\ge 0$, and all other cohomology groups vanish. In particular $a_1^k=k!a_k$: over the integers this is not a polynomial ring on $a_1$. This is the integral cohomology of an odd-sphere loop space.

For the requested homology multiplication, use the homology cross product followed by concatenation:The constant loop gives the degree-zero unit. Loop concatenation is associative up to homotopy, which suffices for associativity on homology; Moore loops can make the space-level operation strictly associative. The Künneth theorem identifies the tensor-product homology because its groups are free abelian groups. The Bott–Samelson theorem identifies this product with word multiplication, so $x^i\ast x^j=x^{i+j}$. The Pontryagin ring of an odd-sphere loop space has presentationWith only one generator the tensor algebra has the indicated polynomial presentation. The homology product and the divided-power cohomology product are different operations.

A super vector space is a vector space graded by the two-element group. Degree-preserving maps and the usual graded tensor product form a monoidal category. Over a field of characteristic different from two, this category has two different symmetries: ordinary interchange of tensor factors and interchange multiplied by $(-1{)}^{pq}$ on homogeneous factors of degrees $p,q$. The latter is the Koszul sign rule.