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Construct the Euclidean lattices in an orthogonal coordinate system whose squared axis lengths areThus an integer coordinate vector $z$ has squared norm ${\sum }_i{a}_i{z}_i^2$. These four numbers are linearly independent over $\mathbb{Q}$: the four sign changes of $\sqrt{2},\sqrt{3}$ in their field extension isolate the four coefficients of a rational relation. The independence will distinguish the two flat tori; equality of their length spectra will hold for every positive choice of axis weights.

LetHere and below these integer coordinates are interpreted in the weighted orthogonal axes. Direct multiplication gives $S^T=-S$, $S^2=-3I$, andIn particular both $L_{\pm }$ are full-rank Euclidean lattices. Denote their four generator columns by $v_i^{\pm }$. Modulo $3$ the two matrices have the same columns, and the nonzero words of the given ternary tetracode $T$ are exactlyThese are the reductions of $\pm v_i^{\pm }$, respectively. The first and fourth displayed words are the specified generators, and the middle two arise from their sum and difference, up to sign. Thus reduction modulo $3$ maps either Euclidean lattice onto precisely $T$.

To construct a length-preserving correspondence, find the kernels $M_{\pm }=L_{\pm }\cap 3{\mathbb{Z}}^4$ of this reduction. The inverse matrix is $A_{\pm }^{-1}=A_{\pm }^T/12$. Hence $3w\in L_{\pm }$ exactly when $A_{\pm }^{T}w\equiv 0(\mathrm{mod}4)$. These congruences are equivalent toFor clarity, the first $+$ congruence is $-(w_1+w_2+w_3+w_4)=0(\mathrm{mod}4)$; subtracting the others gives evenness of each required pair sum. For the $-$ matrix the first congruence is $2w_1-(w_1+w_2+w_3+w_4)=0(\mathrm{mod}4)$; the remaining congruences again give equality of the four parities. Substitution proves the converse in each case.

Let $R_i$ be the orthogonal reflection changing only coordinate $i$. If the coordinates of $w$ are even, this changes their sum by a multiple of $4$; if all are odd, it changes the sum by $2$ modulo $4$. ThereforeEach Euclidean lattice is the disjoint union of its kernel and the eight cosets $\pm v_i^{\pm }+M_{\pm }$, since $T$ has nine words. Define a map $\Phi :L_{+}\to L_{-}$ by using $R_1$ on $M_{+}$ and $R_i$ on each of $v_i^{+}+M_{+}$ and $-v_i^{+}+M_{+}$. Each piece maps bijectively onto the corresponding $-$ coset. Thus $\Phi$ is a bijection satisfyingIt also commutes with $z\mapsto -z$. The closed geodesics of a flat torus have lengths $\Vert z\Vert$ for nonzero translation vectors $z$ in its Euclidean lattice. Consequently the two length spectra agree, including the multiplicities of translation classes and with either consistent orientation convention. If only primitive closed geodesics are counted, the same conclusion follows by removing iterates in increasing order of length: the total multiplicity at $\ell$ is the sum of primitive multiplicities at $\ell /k$, and there are only finitely many such contributions below any fixed length. The correspondence need not itself preserve primitiveness to give this conclusion.

It remains to prove the two flat tori are not isometric. An isometry of flat tori lifts to an affine Euclidean isometry; its linear part of an affine map must carry $L_{+}$ onto $L_{-}$. In integer coordinates write this part as $B$. For every $z\in L_{+}$, both $z$ and $Bz$ have integer coordinates and have equal squared norms. Rational independence of the $a_i$ impliesApply this to $z=A_{+}n$ for all $n\in {\mathbb{Z}}^4$. Two squared linear forms agreeing on all integer points agree as polynomials; their difference factors, so one form is the other or its negative. Since $A_{+}$ is invertible, $B$ is consequently a diagonal sign change $D=\mathrm{diag}({\epsilon }_1,\dots ,{\epsilon }_4)$.

Modulo $3$, such a $D$ must preserve $T$. Each nonzero word of $T$ has exactly three nonzero coordinates, and for each choice of zero coordinate there are just two such words, differing by an overall sign. Preserving these words forces the three signs on each support to agree. The four overlapping supports then force ${\epsilon }_1={\epsilon }_2={\epsilon }_3={\epsilon }_4$. Thus only $D=I$ and $D=-I$ are possible. Neither changes $L_{+}$.

Finally $v_1^{+}\notin L_{-}$: it has the same ternary word as $v_1^{-}$, but their difference is $6e_1=3(2,0,0,0)$, which fails the congruence defining $M_{-}$. Hence $L_{+}\ne L_{-}$ and no such linear isometry exists. This completes the tetracode length-isospectral lattice construction.

Use the covariant Yoneda embedding $Y(A)=\mathcal{C}(A,-)$. An arrow $f:B\to A$, viewed as an arrow $A\to B$ in the opposite category, induces precomposition $h\mapsto hf$. Identities and composition are preserved because composition in $\mathcal{C}$ is associative.

The covariant Yoneda lemma is the bijectionIts inverse sends $x\in F(A)$ to ${\alpha }_B(h)=F(h)(x)$. This is a natural transformation: for $k:B\to C$, functoriality gives $F(k){\alpha }_B(h)=F(kh)(x)={\alpha }_C(kh)$. Conversely, naturality of an arbitrary $\alpha$ at $h:A\to B$ gives ${\alpha }_B(h)=F(h){\alpha }_A(1_A)$. Evaluation at the identity morphism therefore makes the two constructions inverse. Their formulas also prove naturality in $F$ and, contravariantly, in $A$.

Taking $F=Y(B)$ identifies natural transformations $Y(A)\to Y(B)$ with $\mathcal{C}(B,A)$. Thus $Y$ is full and faithful, and reflects isomorphisms; two image objects are isomorphic exactly when the original objects are isomorphic. It also carries existing colimits in $\mathcal{C}$ to categorical limits of covariant representable functors: maps out of a colimit are exactly compatible families of maps out of its diagram. Equivalently, this embedding preserves existing categorical limits in ${\mathcal{C}}^{\mathrm{op}}$. The direction matters: the paper uses covariant representables, rather than the more usual contravariant Yoneda embedding.

A representation of a functor $F$ is $(A,x)$ with $x\in F(A)$ such that $h\mapsto F(h)(x)$ is a bijection $\mathcal{C}(A,B)\to F(B)$ for every $B$. These bijections are natural by functoriality. The element $x$ is its universal element.

If $(B,y)$ is another representation, universality gives unique arrows $f:A\to B$ and $g:B\to A$ satisfying $F(f)x=y$ and $F(g)y=x$. Consequently $F(gf)x=x=F(1_A)x$, so $gf=1_A$ by injectivity of the representing bijection. Similarly $fg=1_B$. Therefore $f$ is the unique compatible isomorphism. This proves uniqueness of functor representations; uniqueness refers to an isomorphism carrying the specified universal elements to one another, rather than to every isomorphism between the underlying objects.

For a diagram $D:\mathcal{J}\to \mathcal{C}$, a limit-preserving functor carries every limiting cone over a diagram to a limiting categorical cone. A limit-reflecting functor has the converse property: a categorical cone is limiting whenever its image is limiting. A limit-creating functor uniquely lifts every specified limiting categorical cone over the image diagram to a categorical cone over $D$, and the lift is limiting. These definitions concern diagrams of the stipulated shape; creation includes the lifting requirement, not merely reflection.

Let $(L,(p_j))$ be a categorical limit of $D$. The mapis a bijection: the right side consists exactly of compatible families of arrows from $A$, and the categorical limit's universal property gives their unique factorization through $L$. The bijection is induced by the categorical limit projections, so it proves that covariant representables preserve limits, including the empty diagram and its terminal object.

An adjunction $F⊣G$ is a family of bijectionsnatural in both variables. Its adjunction unit and adjunction counit are ${\eta }_A={\Phi }_{A,FA}(1_{FA})$ and ${\epsilon }_B={\Phi }_{GB,B}^{-1}(1_{GB})$. Naturality of the correspondence givesFor $f:A\to A^{\prime }$, naturality in both variables evaluates $\Phi (Ff)$ in two ways, giving $GFf{\eta }_A={\eta }_{A^{\prime }}f$. Thus $\eta$ is a natural transformation; the dual calculation gives naturality of $\epsilon$.

Applying the inverse correspondence to ${\eta }_A$ and the correspondence to ${\epsilon }_B$ yields the triangle identities for an adjunction:They express that transposing an identity morphism and transposing back returns that identity.

Define the correspondence from the given natural transformations by $u\mapsto Gu{\eta }_A$, with candidate inverse $v\mapsto {\epsilon }_{B}Fv$. For $u:FA\to B$, naturality of $\epsilon$ and a triangle identity for an adjunction giveFor $v:A\to GB$, naturality of $\eta$ and the other triangle identity giveThese are inverse bijections. Naturality of $\eta$, $\epsilon$, $F$ and $G$ makes the bijections natural in $A$ and $B$, so they define an adjunction with the required unit and counit. Uniqueness follows from the formulas in the previous part: any adjunction with that unit and counit must have exactly these transposition maps.

The currying adjunction for small categories uses the bijectionIt sends $H$ to the functor $d\mapsto H(d,-)$, with an arrow $u:d\to d^{\prime }$ inducing the natural transformation whose $c$-component is $H(u,1_c)$. Conversely, for $K:\mathcal{D}\to [\mathcal{C},\mathcal{E}]$, define its uncurried functor by $(d,c)\mapsto K(d)(c)$ andNaturality of $K(u)$ allows the two factors to be interchanged in the appropriate order, giving functoriality. The constructions are inverse and natural in $\mathcal{D}$ and $\mathcal{E}$. Hence $-\times \mathcal{C}$ is a left adjoint to $[\mathcal{C},-]$ on the category of small categories.

The unit sends $d$ to the functor $c\mapsto (d,c)$ and $u$ to the transformation $(u,1_c)$. The counit is evaluation $[\mathcal{C},\mathcal{E}]\times \mathcal{C}\to \mathcal{E}$: $(H,c)\mapsto H(c)$ and $(\alpha ,v)\mapsto H^{\prime }(v){\alpha }_c={\alpha }_{c^{\prime }}H(v)$.

A representation of a functor $(A,x)$ for $G$ says precisely that for every object $(B,y)$ of $(1\downarrow G)$ there is a unique arrow $f:A\to B$ with $Gf(x)=y$. This is exactly the initial object property for $(A,x)$ in that comma category. Conversely, an initial object supplies these unique arrows, hence the representing bijections $\mathcal{C}(A,B)\cong GB$.

For the monad $(T,\eta ,\mu )$, an algebra for a monad is $(A,a:TA\to A)$ with $a{\eta }_A=1_A$ and $aT(a)=a{\mu }_A$. A morphism of algebras for a monad $f:(A,a)\to (B,b)$ satisfies $fa=bTf$. These objects and arrows form the Eilenberg-Moore category ${\mathcal{C}}^T$.

The free algebra functor sends $A$ to $(TA,{\mu }_A)$ and $f$ to $Tf$. The monad identities verify the algebra laws. For the forgetful functor $U$, the free adjunction iswith inverse $v\mapsto bTv$. The algebra law makes the latter an monad algebra morphism. The identities $h{\eta }_A\mapsto bT(h{\eta }_A)=h{\mu }_{A}T{\eta }_A=h$ and $bTv{\eta }_A=b{\eta }_{B}v=v$ prove the bijection.

In the category of adjunctions inducing a fixed monad, objects are adjunctions $F⊣G$ with their induced monad identified with $T$. A morphism to $F^{\prime }⊣G^{\prime }$ is a functor $H$ between the right-hand categories satisfying $G^{\prime }H=G$, $HF=F^{\prime }$ and compatibility with units and counits. With these strict identifications, defineThe triangle identities give the unit algebra law; naturality of $\epsilon$ at ${\epsilon }_B$ gives the multiplication law. Naturality at $f$ makes $Gf$ an monad algebra morphism. Moreover $UK=G$, $KF=F^T$ because $G{\epsilon }_{FA}={\mu }_A$, and $K{\epsilon }_B$ is the free-adjunction counit at $K(B)$. Hence $K$ is a morphism into the Eilenberg-Moore adjunction.

For any other such $H$, its underlying object at $B$ must be $GB$. Counit compatibility forces its algebra action to be $G{\epsilon }_B$, and the forgetful functor, which is a faithful functor, forces $H(f)=Gf$. Thus $H=K$. This proves terminality of the Eilenberg-Moore adjunction. If adjunctions are specified only up to coherent isomorphisms, the same argument gives uniqueness up to the corresponding compatible natural isomorphism.

The crude monadicity theorem in its reflexive-coequalizer form says: if $F⊣G:\mathcal{D}\to \mathcal{C}$, $G$ is a conservative functor, $\mathcal{D}$ has reflexive coequalizers, and $G$ preserves them, then the Eilenberg-Moore comparison functor $K:\mathcal{D}\to {\mathcal{C}}^T$, for $T=GF$, is an equivalence of categories. Requiring all coequalizers to exist and be preserved is a stronger sufficient form.

For an algebra $(A,a)$, form in $\mathcal{D}$ the coequalizerThe pair is a reflexive pair with common section $F{\eta }_A$: both composites are the identity by the algebra unit law and the triangle identity. An arrow $u:FA\to B$ transposes to $v=Gu{\eta }_A:A\to GB$. The two composites $uFa$ and $u{\epsilon }_{FA}$ transpose to $va$ and $G{\epsilon }_{B}Tv$, respectively. Indeed the latter transpose is $Gu{\mu }_A{\eta }_{TA}=Gu$, while $G{\epsilon }_{B}Tv=Gu{\mu }_{A}T{\eta }_A=Gu$ by counit naturality. Thus $u$ equalizes the pair exactly when $v$ is an monad algebra morphism $(A,a)\to K(B)$. The coequalizer property gives natural bijectionsThese define $L$ on arrows by uniqueness, so $L⊣K$. Its unit has underlying arrow $\beta =Gq{\eta }_A:A\to GL(A,a)$.

By preservation, $Gq$ coequalizes $Ta$ and ${\mu }_A$ in $\mathcal{C}$. The action $a:TA\to A$ is a split coequalizer of that pair: take $s={\eta }_A$ and $t={\eta }_{TA}$, with ${\mu }_{A}t=1_{TA}$ and $Tat={\eta }_{A}a$. Thus there is a unique isomorphism $\alpha :GL(A,a)\to A$ with $\alpha Gq=a$. We have $\alpha \beta =a{\eta }_A=1_A$. AlsoConsequently $\beta \alpha Gq=Gq$, and epimorphic cancellation gives $\beta \alpha =1$. The adjunction unit $\beta$ is an monad algebra morphism; its invertible underlying arrow has an algebra-morphism inverse. Thus the unit of $L⊣K$ is an isomorphism.

For its counit $c_B:LK(B)\to B$, the triangle identity gives $K(c_B){\beta }_{K(B)}=1_{K(B)}$. Hence $K(c_B)$, and therefore $G(c_B)$, is an isomorphism. Since $G$ is a conservative functor, $c_B$ is an isomorphism too. Both unit and counit of $L⊣K$ are invertible, which proves the claimed equivalence. This proof uses only reflexive coequalizers in $\mathcal{D}$ and split coequalizers in $\mathcal{C}$.

Take the graphs $G_{\zeta }=\{(\delta ,S_{\zeta }(\delta )):\delta <\kappa \}$. Each has size $\kappa$, and any two meet in fewer than $\kappa$ points, because their functions are eventually strictly ordered. Fix a bijection $b:\kappa \times \kappa \to \kappa$ and let $A_{\zeta }=b[G_{\zeta }]\in [\kappa {]}^{\kappa }$.

Given $I\in [{\kappa }^{+}{]}^{\kappa }$, enumerate it as $\{{\zeta }_{\eta }:\eta <\kappa \}$ and trim the sets in this order:The deleted part is a union of fewer than $\kappa$ pairwise intersections, each of size less than $\kappa$. Regularity ensures its size is less than $\kappa$. If $\nu <\eta$, then $A_{{\zeta }_{\eta }}^{\ast }$ is disjoint even from the full earlier $A_{{\zeta }_{\nu }}$, hence from its trimmed version. Thus all the retained sets are pairwise disjoint and each lost fewer than $\kappa$ elements. This provesThe argument is the essential disjointness of small subfamilies of a regular-cardinal almost disjoint family.

Write ${◊}_S^{\prime }$ for the countable-family assertion in the PDF; its prime is not a superscript $c$. A single stationary diamond at a regular cardinal sequence gives such families by taking singletons, so ${◊}_S\Rightarrow {◊}_S^{\prime }$.

Conversely enumerate each countable family as $E_{\alpha }(n)$, $n<\omega$, padding finite families and allowing the empty set as a default. Fix a bijection $\pi :\lambda \times \omega \to \lambda$. The previous part applied to $B(\xi )=\{\pi (\xi ,n):n<\omega \}$ gives a club set $D$ on which $\pi [\alpha \times \omega ]\subseteq \alpha$. Define, for each $n$, a single candidate sequenceSuppose no candidate sequence witnesses ${◊}_S$. For each $n$ choose $X_n\subseteq \lambda$ and a club set $C_n$ such that $D_{\alpha }^n\ne X_n\cap \alpha$ for every $\alpha \in S\cap C_n$. Code all the counterexamples intoThe countable-family hypothesis guesses $X$ on a stationary set. Choose a guessing $\alpha$ in the club set $D\cap {\bigcap }_n{C}_n$, and choose $n$ with $E_{\alpha }(n)=X\cap \alpha$. For every $\xi <\alpha$, closure under $\pi$ givesThus $D_{\alpha }^n=X_n\cap \alpha$, contradicting $\alpha \in C_n$. At least one candidate is a diamond sequence, provingThis is the countable-family diamond equivalence, for every regular uncountable $\lambda$ and stationary $S\subseteq \lambda$.

Use $\mathbb{N}=\{1,2,\dots \}$. If a bijection $f:\mathbb{N}\to \mathbb{N}$ maps $S$ onto the even natural numbers, its restriction is a bijection from $S$ onto an infinite set. Moreover, its restriction to the complement of $S$ maps onto the odd natural numbers. Both sets are therefore infinite.

Conversely, suppose that $S$ and its complement are infinite. Applying the increasing enumeration of an infinite subset of natural numbers to each givesFor completeness, successively choose the least unused member. The process never stops because the set is infinite, and it lists every member: an omitted member would have infinitely many selected natural numbers below it, which is impossible. DefineThis is injective on each list, their images are disjoint, and together their images cover $\mathbb{N}$. Thus it is a bijection and maps $S$ onto $2\mathbb{N}$. Both infinitude conditions are necessary and sufficient. If a convention includes zero in $\mathbb{N}$, index each list from zero and use $2j$, $2j+1$ instead.

A set is a countable set if it admits an injective function into the natural numbers; this includes finite sets and countably infinite sets. A countably infinite set can be listed by a bijection from $\mathbb{N}$.

Let $Z$ be the set of infinite binary sequences. Suppose it were countable and list all its members as $s^{(1)},s^{(2)},\dots$, writing $s_n^{(j)}$ for the $n$th entry of row $j$. A finite listing could be repeated to make such a list as well. Define the binary sequence $y$ by $y_n=1-s_n^{(n)}$. It differs from the $n$th row in coordinate $n$, for every $n$. Thus it belongs to $Z$ but is missing from the list, a contradiction. The set $Z$ is uncountable, by Cantor's diagonal argument.