Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 24 (Category Theory)

We use the general adjoint functor theorem in its solution-set form. Let $G:\mathcal{D}\to \mathcal{C}$ be a functor between locally small categories, with $\mathcal{D}$ a complete category. Then $G$ has a left adjoint exactly when it preserves all small categorical limits and, for each $C\in \mathcal{C}$, there is a set-indexed familysuch that every $u:C\to GD$ factors as $Gf{u}_i$ for some $i$ and some $f:D_i\to D$. This is the solution-set condition. The indexing family is a set, whereas the entire comma category $(C\downarrow G)$ can be large.

We first prove the initial-object lemma for complete categories with a weakly initial set. Let $\mathcal{E}$ be locally small and complete, with a weakly initial set $(E_i{)}_{i\in I}$. Its product in a category $W={\prod }_i{E}_i$ is weakly initial: follow a projection by a map from the corresponding $E_i$ to any desired object. The hom-set $\mathcal{E}(W,W)$ is a set, so we can form the simultaneous equalizerThis is a small limit; equivalently, equalize the family of endomorphisms with the family of identity maps into a product of copies of $W$. Every object receives a map from $E$, since it receives one from $W$.

To prove uniqueness, let $a,b:E\rightrightarrows B$, and take their equalizer $j:V\to E$. Weak initiality of $W$ gives $t:W\to V$. Since $ejt$ is an endomorphism of $W$, the defining property of $e$ givesThe monomorphism $e$ cancels, yielding $jte=1_E$. Thus $j$ is both a monomorphism and a split epimorphism, hence an isomorphism. Its equalizer property now implies $a=b$. Therefore $E$ is initial.

Fix $C$. Small categorical limits in $(C\downarrow G)$ are obtained from the limits in $\mathcal{D}$: preservation by $G$ supplies the unique map from $C$ to the image of the limiting object. The comma category is locally small, and the solution family is a weakly initial set in it. The lemma supplies an initial object $(LC,{\eta }_C:C\to GLC)$.

For $k:C\to C^{\prime }$, initiality gives a unique arrow $Lk:LC\to L{C}^{\prime }$ satisfying $G(Lk){\eta }_C={\eta }_{C^{\prime }}k$. Uniqueness proves preservation of identities and composition, so $L$ is a functor. Initiality also gives natural bijectionswhich prove $L⊣G$.

Conversely, if $L⊣G$, the singleton ${\eta }_C:C\to GLC$ is a solution set by the adjunction bijection. To see that $G$ preserves a small limit $D={\lim }_j{D}_j$, use the adjunction and the limit property to obtain, naturally in $C$,This is precisely the universal property of the image limit cone. It proves the necessity of both hypotheses.

Take the category of ordinals in reverse order ${\mathbf{Ord}}^{\mathrm{op}}$, with a unique arrow $\alpha \to \beta$ exactly when $\alpha \ge \beta$, and the unique functorBoth categories are complete categories. A small limit in the ordinal category is the supremum of the ordinals occurring in the diagram; the empty supremum is $0$, its terminal object. The functor $G$ preserves all these categorical limits, and both categories are locally small categories.

A left adjoint would select an initial object of ${\mathbf{Ord}}^{\mathrm{op}}$, meaning an ordinal at least as large as every ordinal. No such object exists. More specifically, any set of candidate ordinals has an upper bound $\gamma$, and none of its members maps to $\gamma +1$. Thus the solution-set condition fails. This example isolates the genuine smallness requirement in the general adjoint functor theorem.

Let $J:\mathbf{CompHaus}\hookrightarrow \mathbf{Top}$ be the inclusion. Both categories are locally small categories. By the Tychonoff theorem, small products in a category of compact Hausdorff spaces are compact Hausdorff, with their ordinary product topology. The equalizer of two continuous maps between compact Hausdorff spaces is closed in the source: it is the inverse image of the closed diagonal of the target. Hence it is again compact Hausdorff. The usual construction of small categorical limits from products and equalizers proves that $\mathbf{CompHaus}$ is a complete category, and that $J$ preserves its small limits as limits in $\mathbf{Top}$.

We verify the solution-set condition for an arbitrary topological space $X$. Put $\kappa =|X|$ and $\theta =2^{2^{\kappa }}$. Given a continuous map $f:X\to K$ with $K$ compact Hausdorff, letThis closed subspace is compact Hausdorff, and $f(X)$ is dense in it. The allowed cardinal estimate therefore gives $|Y|\le \theta$. The map $f$ factors through the continuous map $X\to Y$ and the subspace inclusion $Y\hookrightarrow K$.

Choose a fixed set $S$ of cardinality $\theta$. Consider all compact Hausdorff topologies on subsets of $S$, together with all continuous maps from $X$ into the resulting spaces. There are only a set of such topologies and functions. Every $Y$ above is homeomorphic to one of these spaces after relabelling its underlying set. Consequently every $X\to JK$ factors through a member of this set of maps. This proves the solution-set condition, including $X=\varnothing$.

The general adjoint functor theorem now gives the compact Hausdorff reflectionIts unit ${\eta }_X:X\to J\beta X$ is universal for continuous maps from $X$ to compact Hausdorff spaces. For arbitrary $X$, this is a reflection, and the unit is not required to be a topological embedding.

I would defend the assertion as a description of what good abstraction achieves, provided it is not taken to mean that finding the abstraction is easy. Category theory makes recurring arguments consequences of a common universal property, instead of requiring a new calculation in every setting.

The Yoneda lemma is an instructive example. A natural transformation from $\mathcal{C}(c,-)$ to a set-valued functor $F$ is determined by its value on $1_c$; naturality forces its value at $u:c\to d$ to be $F(u)$ of that element. Once the right question is asked, the proof is short. Yet the statement explains why representable functors encode objects faithfully, and why a functor can be reconstructed from its elements. The simplicity belongs to the organized argument, not to a claim that these consequences were obvious beforehand.

Likewise, the fact that a right adjoint preserves categorical limits does not need separate proofs for products, equalizers and pullbacks in every example. The adjunction converts a map into the proposed limit into a compatible family of maps, and the natural bijection converts that family back. This one argument explains preservation of many different constructions. It also makes the canonical comparison maps clear, which matters more than merely guessing an isomorphism of the resulting objects.

The monoidal coherence theorem shows a different benefit. Complicated calculations with bracketed tensor products can become transparent because every structural rebracketing is the unique map between its formal source and target. However, the proof that the pentagon and triangle suffice is real work. Suppressing brackets is justified by that proof; it cannot serve as a substitute for it. Abstraction is valuable precisely because it separates an essential compatibility argument from subsequent routine bookkeeping.

There is also a necessary caution. The general adjoint functor theorem requires a solution-set condition; small-limit preservation by itself does not produce a left adjoint. The reverse-ordinal example makes the missing condition concrete. Similarly, the monad of an adjunction need not capture all the structure in the original category: the tower of nested partial unary operation categories has arbitrary finite monadic length, despite its long composites sharing their first induced monad. These examples prevent formal language from hiding a size problem or an unjustified reconstruction claim.

The persuasive interpretation is that abstraction makes the reason for a result visible and reusable. It often turns the final step into a direct consequence of definitions, while leaving the discovery of those definitions and the verification of their hypotheses as substantial mathematics.

For a locally small category $\mathcal{C}$, the covariant Yoneda lemma gives a bijectionnatural in $c$ and the functor $H:\mathcal{C}\to \mathbf{Set}$. The inverse sends $x\in H(c)$ to the natural transformation whose component at $d$ maps $u:c\to d$ to $H(u)x$. Naturality of $\tau$ at $u$ forces this formula, proving both inverseness and uniqueness.

Write $\mathcal{E}=\int F$ for the category of elements of $F$. Its objects are $(c,x)$ with $x\in F(c)$, and an arrow $(c,x)\to (d,y)$ is an arrow $u:c\to d$ satisfying $F(u)x=y$.

Given an object $\alpha :P\to F$ of the slice category $[\mathcal{C},\mathbf{Set}]/F$, defineAn arrow $u:(c,x)\to (d,y)$ acts by the restriction of $P(u)$; naturality of $\alpha$ makes this restriction well-defined. A slice morphism $v:P\to Q$ restricts to a map between the corresponding fibers and thus gives a natural transformation on $\mathcal{E}$.

Conversely, given $H:\mathcal{E}\to \mathbf{Set}$, setand let ${\alpha }_H:P_H\to F$ be projection to $x$. The functor laws and naturality follow from those for $F$ and $H$. On morphisms use the maps on each summand. Taking a fiber of this disjoint union recovers $H(c,x)$ canonically. Taking the disjoint union of the fibers of $\alpha$ recovers $P(c)$ by $(x,p)\mapsto p$, naturally in $c$ and in the slice object. These are inverse natural isomorphisms, provingThis is the slice of a set-valued functor category construction.

Now assume $\mathcal{C}$ is a small category, so $\mathcal{E}$ is small. Define a diagram in a categoryFor $u:(c,x)\to (d,y)$ in $\mathcal{E}$, its reverse arrow acts by precomposition $\mathcal{C}(d,-)\to \mathcal{C}(c,-)$. There is a cocone to $F$ with component $g\mapsto F(g)x$. To verify its colimit property directly, use pointwise colimits of sets. At $b$, a colimit element has a representative $(c,x,g:c\to b)$, and the induced map to $F(b)$ sends it to $F(g)x$. This map is surjective, since $z\in F(b)$ is represented by $(b,z,1_b)$. Moreover, the element-category arrow $g:(c,x)\to (b,F(g)x)$ gives precisely the colimit relationThus all representatives with the same image are equal in the colimit, proving injectivity. The resulting bijections are natural in $b$. We have proved the canonical colimit presentation of a covariant set-valued functor:

A Cartesian closed category has finite products in a category and an exponential object $H^G$ for each pair, representing maps from a product with $G$. In $[\mathcal{C},\mathbf{Set}]$, the terminal functor and binary products are computed pointwise. Define the exponential of covariant set-valued functors bySmallness makes this a set. If $u:c\to d$, precomposition with $\mathcal{C}(d,-)\times G\to \mathcal{C}(c,-)\times G$ defines $(H^G)(u)$. Functoriality is immediate from associativity of composition. Its evaluation map of an exponential object isFor $u:c\to d$, naturality of $\tau$ shows $H(u){\tau }_c(1_c,z)={\tau }_d(u,G(u)z)$, which is exactly the naturality equation for evaluation.

Given $\theta :P\times G\to H$, define its currying at $p\in P(c)$ byNaturality of $\theta$ proves that this is a natural transformation in $b$, and its compatibility with arrows $c\to d$ proves that $\hat{\theta }:P\to H^G$ is natural. Evaluation recovers $\theta$ by setting $f=1_c$. Conversely, for $\sigma :P\to H^G$, its naturality impliesso currying the evaluation composite recovers $\sigma$. These formulas give the required natural bijection, hence $[\mathcal{C},\mathbf{Set}]$ is Cartesian closed.

Finally, for every $F$ the small category $\mathcal{E}=\int F$ has a Cartesian closed functor category by the same construction. The slice equivalence above transports its terminal object, products and exponentials to $[\mathcal{C},\mathbf{Set}]/F$. The original functor category also has pointwise finite limits. Therefore it is locally Cartesian closed. Every construction here is explicit; no adjoint functor theorem is used.

For an adjunction $L:\mathcal{C}\rightleftarrows \mathcal{D}:R$, with unit $\eta$ and counit $\epsilon$, call the adjunction idempotent when the multiplicationof its induced monad is invertible. We show that this idempotent adjunction condition is equivalent to invertibility of $\delta =L\eta R:LR\to (LR{)}^2$, the comultiplication of the induced comonad.

For an idempotent monad $T$, the unit laws implyIf $a:TA\to A$ is a monad algebra, naturality of the unit and its algebra unit law giveThus ${\eta }_A$ is an isomorphism. Apply this to the canonical algebra $R{\epsilon }_B:T(RB)\to RB$: its unit law is a triangle identity for an adjunction. It follows that ${\eta }_{RB}$, and therefore $L{\eta }_{RB}$, is invertible for every $B$. Hence $\delta$ is invertible.

Apply the same argument to the dual adjunction $R^{\mathrm{op}}⊣L^{\mathrm{op}}$. It gives the converse. Idempotence is self-dual: the induced monad is idempotent exactly when the induced comonad is idempotent. In particular, idempotence requires the unit to be invertible on every object in the image of the right adjoint.

The left adjoint to the morphism-set functor is the free category on independent arrows:Each copy has two objects, their identities, and one nonidentity arrow from the first object to the second. A functor $L(S)\to \mathcal{C}$ is specified by one arbitrary arrow of $\mathcal{C}$ for each $s$, since that arrow specifies the images of both objects as well. This proves the natural adjunction bijection with functions $S\to \mathrm{Mor}(\mathcal{C})$.

The induced monad on sets has $T(S)=S\times \{\text{source identity},\text{generator},\text{target identity}\}$. For a singleton $S$, its multiplication is a function from the nine-element set $T^{2}S$ to the three-element set $TS$, so it cannot be invertible. This adjunction is not idempotent.

The left adjoint is discretization $D$: give the underlying set of $X$ the discrete topology. Write $J$ for indiscretization. Every function $DX\to Y$ is continuous, and every function $X\to JY$ is continuous. The common set of underlying functions supplies the natural bijectionThe discrete-indiscrete adjunction has induced monad $JD$, which assigns the indiscrete topology to the original set. Applying it twice gives the same space, and its multiplication is the identity on that set. Thus the adjunction is idempotent. Its dual comonad $DJ$, discretization, is idempotent as well.

For a Boolean algebra $B$, let $U(B)$ be its set of ultrafilters, equivalently the Boolean homomorphisms $B\to 2$. Regard $U(B)$ as an object of ${\mathbf{Set}}^{\mathrm{op}}$. A homomorphism $k:B\to B^{\prime }$ acts by inverse image of ultrafilters, a set map $U(B^{\prime })\to U(B)$, hence an arrow $U(B)\to U(B^{\prime })$ in the opposite category.

A Boolean homomorphism $h:B\to \mathcal{P}(A)$ gives the functionConversely, a function $v:A\to U(B)$ gives $h(b)=\{a:b\in v(a)\}$. The ultrafilter laws are exactly what makes this map preserve the Boolean operations, including complements and $0,1$. These constructions are inverse and natural, proving the Boolean ultrafilter-power-set adjunctionIts unit is ${\eta }_B(b)=\{u\in U(B):b\in u\}$.

Take $B=\mathcal{P}(\mathbb{N})$, an object in the image of the right adjoint. There is a nonprincipal ultrafilter on $\mathbb{N}$, by extending the cofinite filter in the usual choice setting. The subset $P\subset U(B)$ consisting of all principal ultrafilters is not ${\eta }_B(b)$ for any $b\subseteq \mathbb{N}$: containing the principal ultrafilter at every integer forces $b=\mathbb{N}$, but ${\eta }_B(\mathbb{N})=U(B)$ also contains the nonprincipal ultrafilters. Thus ${\eta }_B$ is not surjective and is not an isomorphism. By the necessary condition proved in the root solution, this adjunction is not idempotent.

For $L⊣U:\mathcal{D}\to \mathcal{C}$, let $T=UL$ be the induced monad. The Eilenberg-Moore comparison functor isThe adjunction is a monadic adjunction if this functor is an equivalence of categories. To define monadic length, start with $K_0=U$ and form the comparison $K_1$ into the Eilenberg-Moore category of its induced monad. When $K_1$ has a left adjoint, repeat this construction, and continue with the subsequent comparison adjunctions. The length is the least number of comparison steps after which the comparison is an equivalence; an equivalence has length zero and a monadic non-equivalence has length one. The successive left adjoints will all be explicit in this example.

The precise monadicity theorem states: a functor $U$ with a left adjoint is monadic if and only if it reflects isomorphisms and every parallel pair in $\mathcal{D}$ whose image admits a split coequalizer in $\mathcal{C}$ has a coequalizer in $\mathcal{D}$ preserved by $U$. Here a split coequalizer of $f,g:B\rightrightarrows C$ consists of $q:C\to Q$, $s:Q\to C$ and $t:C\to B$ satisfyingThese identities imply the coequalizer property and survive every functor. We shall identify the relevant algebra categories directly, which proves monadicity without having to check quotient constructions separately.

Write $U_m^{m+1}:{\mathcal{C}}_{m+1}\to {\mathcal{C}}_m$ for one step in the nested partial unary operation category tower. For $m=0$, its left adjoint isA function $f:A\to UB$ extends uniquely by $(a,k)\mapsto {\omega }_1^{k}f(a)$, proving the adjunction.

For $m\ge 1$, take $A\in {\mathcal{C}}_m$ and letThe free extension of nested partial unary operations has underlying setKeep the old operations on $A$. On every new chain set ${\omega }_1(b_{a,k})=b_{a,k+1}$, and leave the other old operations undefined on new points. Define the new operation by ${\omega }_{m+1}(a)=b_{a,0}$ for $a\in S_m(A)$, and nowhere else. This is a valid object: the new chains have no fixed points under ${\omega }_1$, so all later operations are correctly undefined on them; the new top operation has exactly the required domain.

If $f:A\to U_m^{m+1}B$ preserves the old operations, it carries $S_m(A)$ into $S_m(B)$. Its unique extension isThe formula preserves the first operation along each chain, the new operation at its old fixed-point source, and all the other defined operations. Conversely these requirements force the formula. This proves the one-step adjunction in every case.

Let $T_m=U_m^{m+1}{F}_m^{m+1}$. For $m=0$, this is the usual free unary-operation monad: $T_{0}A=A\times \mathbb{N}$, with multiplication $(a,j,k)\mapsto (a,j+k)$. An algebra action $h:A\times \mathbb{N}\to A$ obeys $h(a,0)=a$ and $h(h(a,j),k)=h(a,j+k)$. It is uniquely determined by the total unary operation $v(a)=h(a,1)$, since $h(a,k)=v^k(a)$. Algebra morphisms are exactly functions commuting with $v$. Thus ${\mathcal{C}}_0^{T_0}\cong {\mathcal{C}}_1$.

For $m\ge 1$, no new chain point belongs to $S_m(T_{m}A)$, so $S_m(T_{m}A)=S_m(A)$. The set $T_m^{2}A$ consists of $A$ and two copies of the free chain at every $a\in S_m(A)$. Multiplication is the identity on $A$ and folds both copies onto the corresponding chain in $T_{m}A$.

An algebra action $h:T_{m}A\to A$ must be the identity on the old copy of $A$, by its unit law. Preservation of ${\omega }_1$ forcesThe choices $v(a)\in A$, for $a\in S_m(A)$, are otherwise arbitrary. There are no higher old operations to preserve on new points. Associativity is automatic: on an inner chain both $h\mu$ and $h{T}_{m}h$ apply the displayed formula, and on an outer chain $T_{m}h$ sends $b_{a,k}$ to the inner chain point with the same label because $h(a)=a$; multiplication does the same. Therefore $h\mu =h{T}_{m}h$ on every point.

The data of the algebra are consequently exactly a partial operation $v$ whose domain is $S_m(A)$, that is, an object of ${\mathcal{C}}_{m+1}$. For a map $f:A\to B$ in ${\mathcal{C}}_m$, the algebra-morphism equation $fh=h^{\prime }{T}_{m}f$ at $b_{a,0}$ says precisely $f(v(a))=v^{\prime }(f(a))$; equality on the rest of the chain then follows from preservation of ${\omega }_1$. Thus we have identified objects and morphisms, not just their underlying sets:Under this isomorphism the comparison functor for $F_m^{m+1}⊣U_m^{m+1}$ is the identity identification. Hence every one-step forgetful adjunction is monadic.

Now take $n>m$. In the free extension just constructed, the newly added operation has no fixed points: it sends an old point to a distinct new point, or, for $m=0$, shifts along a free chain. Therefore all further operations can only be undefined. Adding them freely adds no more points. The same extension formula gives a left adjoint $F_m^n$ to $U_m^n:{\mathcal{C}}_n\to {\mathcal{C}}_m$.

After forgetting to ${\mathcal{C}}_m$, its induced monad is exactly $T_m$, independent of $n$. This includes the unit and multiplication: the unit is the old-point inclusion, and the counit formula gives the same addition of chain indices for $m=0$ or the same folding of the two chain copies for $m\ge 1$. Its comparison functor, after the identification ${\mathcal{C}}_m^{T_m}\cong {\mathcal{C}}_{m+1}$, is exactlyIndeed, the comparison action on the first point of each free chain recovers ${\omega }_{m+1}$, while the higher operations are forgotten. Thus each comparison step recovers exactly one additional operation. After $n$ steps from sets, the comparison is the identity of ${\mathcal{C}}_n$.

No earlier step is an equivalence. For $0\le m<n$, put all of the first $m$ operations equal to the identity on a three-element set. In one ${\mathcal{C}}_n$ structure choose ${\omega }_{m+1}$ to be the cycle $(012)$, and in another choose its reverse $(021)$. Both have no fixed points at this stage, so every higher operation is undefined. The identity function between their underlying ${\mathcal{C}}_m$ objects is a morphism there, but does not preserve ${\omega }_{m+1}$. Hence $U_m^n$ is not full, and cannot be an equivalence. This also applies for $m=0$, when there are no old operations. We conclude

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.