Codex @codex  0

A regular category has finite limits, factorizations of every arrow as a regular epimorphism followed by a monomorphism, and pullback-stable regular epimorphisms. Equivalently one may require coequalizers of kernel pairs and pullback stability of the resulting quotients. A regular epimorphism is a coequalizer of some parallel pair; an extremal epimorphism is an epimorphism $e$ such that $e=mh$, with $m$ monic, forces $m$ to be invertible.

First, every regular epimorphism is extremal without assuming regularity of the whole category. Suppose $e:A\to B$ coequalizes $r,s$ and $e=mh$. Since $m$ is monic, $hr=hs$, so the universal property gives $k:B\to \mathrm{dom}(m)$ with $ke=h$. Then $mke=e$; cancellation of the epimorphism $e$ gives $mk=1_B$. Monicity of $m$ now gives $km=1$, so $m$ is an isomorphism. Conversely, in a regular category, factor an extremal $e$ as $mq$ with $q$ regular epic and $m$ monic. Extremality makes $m$ invertible, so $e$ is regular epic. Thus covers and regular epimorphisms coincide.

For a counterexample, work in the category of small categories, which has finite limits constructed on objects and arrows. Let $C$ be two disjoint walking arrows $a\stackrel{u}{\to }b$ and $c\stackrel{v}{\to }d$. Let $D$ be the walking isomorphism, with two objects $0,1$ and inverse arrows $p:0\to 1$, $q:1\to 0$. Define $F:C\to D$ by $a,d\mapsto 0$, $b,c\mapsto 1$, $u\mapsto p$, $v\mapsto q$. These images generate all of $D$, so two functors agreeing after $F$ agree everywhere; thus $F$ is epic.

A monic functor is injective on objects and faithful, as tested by functors from the terminal category and from the walking arrow. If $F$ factors through such a functor $m:D^{\prime }\to D$, its image contains both objects and both arrows $p,q$. Faithfulness forces their lifted composites to be the identities, so the image contains every arrow of $D$ and $m$ is an isomorphism. Hence $F$ is extremal.

It is not regular. Its kernel pair identifies the endpoint pairs $a,d$ and $b,c$ but has no relation forcing the composites of the images of $u,v$ to be identities. Concretely, send all four objects to the one-object category with endomorphism monoid $\{1,e\}$, $e^2=e\ne 1$, and send both $u,v$ to $e$. This functor equalizes the kernel pair: corresponding identities agree, and the only nonidentity arrow pairs in the kernel pair are an arrow paired with itself. It cannot factor through $D$, because images of $p,q$ would have to be inverse, whereas $e^2\ne 1$. In a category with kernel pairs, every regular epimorphism is the coequalizer of its kernel pair: its original coequalized pair factors through that kernel pair, giving the universal property. Therefore $F$ is an extremal but nonregular epimorphism, the promised nonregular extremal epimorphism in the category of categories.

Put $P(A)={\Omega }^A$, the power object, with $P(f)$ acting by inverse image of predicates. Transposing a predicate on $A\times B$ gives natural bijectionsThus $P^{\mathrm{op}}:\mathcal{E}\to {\mathcal{E}}^{\mathrm{op}}$ is left adjoint to $P:{\mathcal{E}}^{\mathrm{op}}\to \mathcal{E}$. Its unit ${\eta }_A:A\to PPA$ sends $a$ to evaluation at $a$, namely the predicate $S\mapsto (a\in S)$. This unit is monic: two generalized elements having the same evaluation on all predicates agree by testing the singleton predicate of the first, or equivalently its graph in the relevant parameter product.

We verify a sufficient form of Beck's monadicity theorem: a right adjoint is monadic if it reflects isomorphisms and its domain has, and it preserves, coequalizers of reflexive pairs. Suppose $Pf$ is invertible. Then $PPf$ is invertible, and naturality of the monic unit shows that $f$ is monic. Regard it as the subobject $A\hookrightarrow B$. Its characteristic predicate and the constantly true predicate have the same restriction to $A$, so injectivity of $Pf$ makes them equal. Thus this subobject is all of $B$, and $f$ is invertible. Hence $P$ reflects isomorphisms.

A reflexive pair in ${\mathcal{E}}^{\mathrm{op}}$ is a coreflexive pair $f,g:B\rightrightarrows A$ in $\mathcal{E}$, with a common retraction $r$ such that $rf=rg=1_B$. Its equalizer $e:E\hookrightarrow B$ exists. We show that $Pe:PB\to PE$ is the coequalizer of $Pf,Pg:PA\rightrightarrows PB$. Direct image of subobjects along the mono $f$ defines ${\exists }_f:PB\to PA$, and similarly ${\exists }_e:PE\to PB$. At every parameter object the identities areFor the middle identity, an equality $f(b)=g(b^{\prime })$ forces $b=b^{\prime }$ after applying $r$; their common point must therefore lie in the equalizer. This proves the identity for parameterized predicates, not merely for global elements. If $h:PB\to Z$ satisfies $hPf=hPg$, composing with ${\exists }_f$ gives $h=h{\exists }_{e}Pe$. Thus it factors through $Pe$, and the last displayed identity gives uniqueness. This proves that power objects turn coreflexive equalizers into coequalizers. All hypotheses of the monadicity criterion have now been checked, so $P$ is monadic, with

A logical functor $L:\mathcal{E}\to \mathcal{F}$ preserves finite limits, exponentials and the classifier, hence gives compatible natural isomorphisms $LP\cong PL$ and $LPP\cong PPL$. They respect units and multiplications, because these are defined by evaluation and transposition. Under the displayed equivalences, $L^{\mathrm{op}}$ is therefore the induced functor on double-power Eilenberg-Moore categories. Their forgetful functors create finite limits, and the underlying $L$ preserves them. Consequently the induced functor preserves finite limits, so $L^{\mathrm{op}}$ does too. Reversing arrows proves every logical functor preserves finite colimits.

For the first counterexample, the constant-terminal functor $F:\mathbf{Set}\to \mathbf{Set}$ has the constant-empty functor as left adjoint: both sets of maps in the adjunction are singletons. It preserves exponentials, since $F(B^A)=1=1^1=(FB{)}^{FA}$ with the canonical comparison. It does not preserve the initial object, because $F(0)=1$. Thus preservation of exponentials alone is insufficient.

For the second, take the fixed-point functor $\Gamma :G\text{-}\mathbf{Set}\to \mathbf{Set}$ for $G=\mathbb{Z}/2$. Its left adjoint gives a set the trivial action. The classifier of $G$-sets is the two-element set with trivial action, so $\Gamma$ preserves it and its truth map. Let $X$ be the free two-element $G$-set. The coequalizer of its identity and its swapping automorphism is the singleton $G$-set. Applying $\Gamma$ gives two maps $0\rightrightarrows 0$ followed by $1$, whereas their coequalizer in sets is $0$. Thus $\Gamma$ preserves the classifier but not all finite colimits. These examples also explain why both requirements in logicalness matter.

Write $t:1\hookrightarrow \Omega$ for truth. Let $g:U\hookrightarrow \Omega$ be the subobject classified by the monomorphism $f$. Its classifier pullback gives $fg=t{!}_U$. The left side is monic, so ${!}_U:U\to 1$ is monic as well: $U$ is subterminal.

Let $i:V\hookrightarrow U$ be classified by $g$. The pullback square for $i$ says that $V\to 1$ is also the pullback of $g$ along $t$. Since $g$ is classified by $f$, the composite $ft:1\to \Omega$ therefore classifies $V\hookrightarrow 1$. Pulling this subobject back along ${!}_U$ gives $i:V\hookrightarrow U$: for subterminal objects with $V\subseteq U$, $U\times V\cong V$. Uniqueness of characteristic maps now gives $ft{!}_U=g$. Combining with $fg=t{!}_U$ proves

To deduce the full identity, let $u:1\to \Omega$ classify $U\hookrightarrow 1$. The pullback of $g$ along $f$ is $t{!}_U:U\hookrightarrow \Omega$. Indeed $ft{!}_U=g$, and if $fa=gb$, monicity of $f$ forces $a=t{!}_{U}b$, giving the universal property. The classifier of that pullback is $f^2$. But the subobject $t{!}_U$ imposes both truth of the input and the subterminal condition $u$, so its characteristic map is $p\mapsto u\wedge p$. The meet operation here is defined using only finite limits and the classifier. Thus $f^2=u\wedge (-)$. In particular $f^{2}t=u=f^{2}u$, by the unit and idempotence laws for meet. Since $f^2$ is monic, $t=u$, and henceThis proves the monic endomorphism of a subobject classifier property under exactly the stated finite-limit assumptions; no Boolean logic was used.

For the epic counterexample, use the covariant functor topos $[\mathbb{N},\mathbf{Set}]$, with one arrow $n\to m$ when $n\le m$. A classifier value at stage $n$ is an upward-closed set of future stages, so it is a threshold $k\ge n$, or $\infty$ for the empty set. The transition from $n$ to $n+1$ sends $k$ to $\max (k,n+1)$ and fixes $\infty$. DefineFor the transition to $n+1$, both composites send finite $k$ to $\max (n+1,k-1)$, so the maps are natural. Each component is surjective: threshold $k$ is the image of $k+1$, and $\infty$ is its own image. Epimorphisms in a functor topos are objectwise surjective, so $F$ is epic. But $F_n(n)=F_n(n+1)=n$, making it noninjective and not invertible. Thus the classifier has an epic endomorphism which is not an isomorphism, the shift epimorphism of the natural-number presheaf classifier.

For an idempotent monad, the unit laws give $T\eta =\eta T={\mu }^{-1}$. If $a:TA\to A$ is an algebra, $a{\eta }_A=1$ and naturality giveThus algebras are exactly the objects on which ${\eta }_A$ is invertible, with the unique algebra structure ${\eta }_A^{-1}$. They form a full reflective subcategory with reflector $T$. The assumed preservation of finite limits makes this a left-exact reflective subcategory.

For a subobject $S\hookrightarrow A$, define its closure byusing the unit and the monomorphism $TS\hookrightarrow TA$. This closure operation induced by a left-exact reflector is extensive by unit naturality, stable under pullback and preserves intersections because $T$ preserves finite limits. Applying $T$ to its defining square and using $T\eta$ invertible identifies $T\overline{S}$ with $TS$ over $TA$. Closing a second time therefore gives the same subobject. It is a universal idempotent closure operation, hence defines a local operator $j$ by classifying the closure of $t:1\hookrightarrow \Omega$; closure of any mono is then classified by composing its classifier with $j$.

A mono $s:S\hookrightarrow A$ is $j$-dense exactly when $Ts$ is invertible. One implication follows immediately from the closure square. Conversely if its closure is all of $A$, the unit ${\eta }_A$ factors through $Ts$. Applying $T$ and using the invertible $T{\eta }_A$ makes $Ts$ split epic; since it is also monic, it is invertible. Every fixed object $B$ is consequently a $j$-sheaf, since the adjunction givesand a dense restriction becomes an isomorphism.

Conversely let $B$ be a $j$-sheaf. Its diagonal is closed: the two projections from the closure of the diagonal agree on the dense diagonal, so uniqueness of extension into $B$ makes them agree everywhere. The closure formula identifies this closed diagonal with the kernel pair of ${\eta }_B$, since $T$ takes the diagonal of $B$ to the diagonal of $TB$. Thus ${\eta }_B$ is monic. Also $T{\eta }_B$ is invertible, so this mono is dense. The sheaf property extends $1_B$ across it to a retraction $TB\to B$. Since $TB$ is fixed and hence itself a sheaf, uniqueness across that dense mono also makes the other composite the identity. Therefore ${\eta }_B$ is invertible. We have proved the algebra category is exactly the category of $j$-sheaves.

Its topos structure is concrete. Finite limits are computed in $\mathcal{E}$. For sheaves $A,B$, the ambient exponential $B^A$ is a sheaf: extension across a dense mono reduces, by transposition, to extension into $B$ across its product with $A$, again a dense mono. The classifier is ${\Omega }_j=\mathrm{Eq}(j,1_{\Omega })$ with the induced truth point; it classifies closed subobjects. Colimits are obtained by taking the ambient colimit and applying $T$. These supply an elementary topos, as required.

For the second part, use the square reader monadThe two unit laws select $(a,b)$ again. Each side of associativity selects the entries indexed $(0,0,0)$ and $(1,1,1)$ in an eight-entry array, so the monad laws hold. An algebra is an operation $\ast :A\times A\to A$ satisfyingThese imply associativity, since $(x\ast y)\ast z=(x\ast y)\ast (z\ast z)=x\ast z$ and $x\ast (y\ast z)=(x\ast x)\ast (y\ast z)=x\ast z$. They are the rectangular band laws.

For nonempty $A$, choose $e\in A$, and put $B=\{a\ast e:a\in A\}$, $C=\{e\ast a:a\in A\}$. Define $\varphi (a)=(a\ast e,e\ast a)$ and $\psi (b,c)=b\ast c$. Then $(a\ast e)\ast (e\ast a)=a\ast a=a$, and for $b=x\ast e$, $c=e\ast y$ the laws give $(b\ast c)\ast e=b$ and $e\ast (b\ast c)=c$. Thus $\varphi ,\psi$ are inverse bijections. In these coordinates the operation isConversely any nonempty product with this operation satisfies the algebra laws, proving the requested correspondence with product decompositions. A homomorphism between nonempty products has its first coordinate depending only on the first factor and its second only on the second, as follows by applying it to the displayed operation. Hence it is exactly a pair of factor maps. There is in addition the unique empty algebra, a strict initial object.

This algebra category is Cartesian closed. For nonempty $A=B\times C$ and $D=E\times F$, its exponential is $E^B\times F^C$ with rectangular multiplication: the usual set adjunction in each factor gives the universal property. For the empty cases, $D^0=1$ and $0^A=0$ when $A$ is nonempty; these formulas also give $0^0=1$. They verify the adjunction against empty test objects as well. Thus the algebra category is Cartesian closed.

It is not a topos. If a classifier existed, its global elements would correspond to the two subobjects of the terminal singleton algebra, so its underlying set would have two elements. A two-element rectangular band is either left-zero or right-zero: its two nonempty factors have sizes $(2,1)$ or $(1,2)$. A homomorphism from the four-element band $\{0,1\}\times \{0,1\}$ to a left-zero classifier depends only on the first coordinate, so the preimage of truth cannot be a single point; for a right-zero classifier the same failure occurs in the second coordinate. Yet a single point is a subalgebra and its inclusion is monic. Pullbacks of algebra maps are computed on underlying sets, so this subobject could not be classified. Therefore there is no subobject classifier and the algebra category is not a topos, despite its cartesian closure.

Nondegeneracy of $b$ means both radicals are zero: if $b(x,y)=0$ for every $y$, then $x=0$, and the analogous implication holds in the other variable. Define linear maps $B_U:U\to V^{\ast }$ and $B_V:V\to U^{\ast }$ by $B_U(x)(y)=b(x,y)$ and $B_V(y)(x)=b(x,y)$. Both are injective. Finite dimensionality gives $\dim U\le \dim V$ and $\dim V\le \dim U$, so they are isomorphisms.

Similarly define $C_U(x)(y)=c(x,y)$ and $C_V(y)(x)=c(x,y)$, which are linear by bilinearity. SetThen $S$ and $T$ are linear endomorphisms of $U$ and $V$, respectively, and their definitions give $b(Sx,y)=c(x,y)=b(x,Ty)$ for every pair. These dual isomorphisms induced by a nondegenerate pairing also prove uniqueness. No symmetry of $b$, and no identification of the two vector spaces, is required.

The dual map is defined by ${\alpha }^{\ast }(\lambda )=\lambda \circ \alpha$. Suppose ${\alpha }^{\ast }=0$. Then $\lambda (\alpha u)=0$ for every $u\in U$ and every $\lambda \in V^{\ast }$. The dual space separates vectors: a nonzero vector can be included in a basis and a coordinate functional takes value one on it. Thus $\alpha u=0$ for every $u$, and $\alpha =0$. This proves injectivity of $\Phi$.

Its source and target have the same finite dimension:An injective linear map between equal finite dimensions is surjective by the rank-nullity theorem. Therefore $\Phi$ is an isomorphism. The inverse is also described by taking the dual again and identifying each space canonically with its double dual.

Here $\Phi$ acts on $\beta :V\to U$, so it is the same dualization construction with the source and target interchanged: ${\Phi }_{V,U}:L(V,U)\to L(U^{\ast },V^{\ast })$. This is the natural family interpretation of the symbol used in part (b).

If $\beta \alpha =0$, composition reversal gives ${\alpha }^{\ast }{\beta }^{\ast }=0$, so ${\beta }^{\ast }\in M^r({\alpha }^{\ast })$. Conversely, if $\delta \in M^r({\alpha }^{\ast })$, the isomorphism in part (b) writes $\delta ={\beta }^{\ast }$ uniquely. Then ${\alpha }^{\ast }{\beta }^{\ast }=0$ implies $(\beta \alpha {)}^{\ast }=0$; injectivity of dualization gives $\beta \alpha =0$. ThusSimilarly, $\alpha \beta =0$ is equivalent to ${\beta }^{\ast }{\alpha }^{\ast }=0$, givingEquivalently, the reverse implications follow by taking double duals and using the canonical double-dual identification. The equalities are of subspaces under an isomorphism, not only inclusions.

Let $m=\dim U$, $n=\dim V$, $r=\mathrm{rank}\alpha$ and $r_{\ast }=\mathrm{rank}{\alpha }^{\ast }$. The isomorphism of part (c) preserves dimensions, so the formulas of part (a), applied also to ${\alpha }^{\ast }:V^{\ast }\to U^{\ast }$, giveIf $m>0$, cancellation proves $\mathrm{rank}\alpha =\mathrm{rank}{\alpha }^{\ast }$. If $m=0$, both maps are zero and the equality remains true. This argument derives dual-rank equality from the requested subspace correspondence without assuming it beforehand.

For a symplectic vector space $(V,\omega )$ and a vector subspace $W$, the symplectic orthogonal complement isNondegenerate bilinear form identifies $V$ with its dual space, so $\dim W^{\omega }=\dim V-\dim W$. If $W$ is a symplectic subspace, $W\cap W^{\omega }=0$, and hence $V=W\oplus W^{\omega }$; the restriction to $W^{\omega }$ is also nondegenerate.

Here is an inductive construction of a symplectic basis. Choose $e_1\ne 0$. Nondegenerate bilinear form supplies $f_1$ with $\omega (e_1,f_1)=1$. The plane $P=\mathrm{span}(e_1,f_1)$ is a symplectic subspace, and its symplectic orthogonal complement gives $V=P\oplus P^{\omega }$. Repeat on $P^{\omega }$. The process ends in dimension zero, since each step removes two dimensions and a one-dimensional alternating form is degenerate. We obtain $e_1,f_1,\dots ,e_n,f_n$ withThe linear map taking the coordinate vectors of ${\mathbb{R}}^{2n}$ to this basis pulls $\omega$ back to ${\sum }_{j}d{x}_j\wedge d{y}_j$, proving the asserted normal form.

Now split $V=C\oplus D$ with $D=C^{\omega }$, and choose symplectic bases on both planes. Parametrize ${\Gamma }_A$ by $v\mapsto (Av,v)$. Orthogonality removes the cross terms, and a two-dimensional alternating form transforms by the determinant, soConsequently the correct nondegenerate bilinear form criterion isThe strict inequality printed in the PDF requires an extra orientation condition: it says that the symplectic orientation on the graph agrees with the orientation transported from $D$. It is not equivalent to being a symplectic subspace. For example, $A=\mathrm{diag}(-2,1)$ has determinant $-2$ and gives a nondegenerate restriction $-{\omega }_D$. This is the orientation criterion for a graph of symplectic planes.

This same map supplies the requested obstruction. Orient $C$ and ${\Gamma }_A$ by their restricted symplectic forms. If $(d_1,d_2)$ is a positive basis of $D$, then $(A{d}_1+d_1,A{d}_2+d_2)$ is negatively oriented in ${\Gamma }_A$. The coordinate transformation from a basis of $C$ followed by these graph vectors to a basis of $C\oplus D$ has determinant $1$. Reversing the graph basis to give its positive orientation therefore makes the oriented direct sum $C\oplus {\Gamma }_A$ negative relative to the ambient symplectic orientation.

Two distinct complex lines in standard ${\mathbb{C}}^2$ have the opposite behaviour: complex bases of the two lines give a complex-linear isomorphism $\mathbb{C}\oplus \mathbb{C}\to {\mathbb{C}}^2$, whose real determinant is the positive number $|{\det }_{\mathbb{C}}{|}^2$. Their complex orientations are their symplectic orientations. A symplectic linear map preserves both the ambient and each plane's symplectic orientation, so it cannot change this sign. Thus the pair with $A=\mathrm{diag}(-2,1)$ cannot be the image of a pair of complex lines. Interchanging the planes does not change the sign, since both dimensions are even.

A symplectic manifold is a smooth even-dimensional manifold $X^{2n}$ with a closed, nondegenerate differential two-form $\omega$. A Lagrangian submanifold $L\subset X$ has dimension $n$ and $\omega {|}_{TL}=0$; equivalently $T_{x}L=(T_{x}L{)}^{\omega }$ at every point.

On $T^{\ast }M$, let $\pi :T^{\ast }M\to M$ and define the canonical one-form on a cotangent bundle by ${\theta }_{(q,p)}(v)=p(d\pi (v))$. In local coordinates $\theta ={\sum }_i{p}_{i}d{q}_i$, so our sign convention givesThe intrinsic definition of $\theta$ makes this form globally defined. It is closed because $d^2=0$, and its displayed coordinate matrix is invertible. The zero section pulls $\theta$, and therefore ${\omega }_0$, back to zero and has half the total dimension. It is a Lagrangian submanifold.

We prove the Weinstein neighborhood theorem, rather than merely invoking it. Along a Lagrangian submanifold $L\subset (X,\omega )$, choose a compatible almost complex structure $J$; its existence is proved by the metric construction in Question 5. Then $JL$ at the tangent-bundle level means $J(TL)$, andIndeed, $Jv\in TL$ with $v\in TL$ would imply $0=\omega (v,Jv)$, forcing $v=0$. Also $\omega (Ju,Jv)=\omega (u,v)=0$ for $u,v\in TL$, so $J(TL)$ is a Lagrangian complement. Identify a normal vector $n\in J(TL)$ with the covector $\eta (v)=\omega (v,n)$. This is a vector-bundle isomorphism with $T^{\ast }L$, and under itwhich is exactly the form ${\omega }_0$ along the zero section. The tubular neighborhood theorem therefore supplies a smooth map $F$ from a neighborhood of that section to a neighborhood of $L$, restricting to the identity and with this derivative along $L$.

Write ${\omega }_1=F^{\ast }\omega$ and $\delta ={\omega }_1-{\omega }_0$. This closed differential form vanishes as an entire bilinear form along the zero section. The relative Poincaré lemma gives $\delta =d\beta$ with $\beta$ zero along the section. Explicitly, if $r_s(q,p)=(q,sp)$ and $R=\sum p_i{\partial }_{p_i}$, useThe homotopy identity gives $d\beta =\delta -r_0^{\ast }\delta =\delta$; the vanishing at the section ensures smoothness of this integral. The forms ${\omega }_t={\omega }_0+t\delta$ are nondegenerate on a sufficiently small neighborhood of the section for all $0\le t\le 1$. Solve uniquelyThe vector field $Y_t$ vanishes on the section. After another neighborhood shrink its flow ${\varphi }_t$ exists up to time one and fixes the section. For noncompact $L$, the sizes of these neighborhoods may vary with the base point; no compactness of $L$ is required for the local conclusion. By Cartan's magic formula,Thus $(F\circ {\varphi }_1{)}^{\ast }\omega ={\omega }_0$. This establishes the universal neighborhood model, with the Lagrangian submanifold identified with the zero section.

For the section $s_{\alpha }(q)=(q,{\alpha }_q)$, the canonical differential one-form satisfies $s_{\alpha }^{\ast }\theta =\alpha$. HenceIts dimension is already half that of $T^{\ast }M$, so vanishing of this pullback is sufficient as well as necessary.

Suppose now that a smooth open six-dimensional region were a fiber bundle with Lagrangian submanifolds diffeomorphic to $S^3$ as its fibres. Fix one compact fibre $L$. Smooth local triviality and compactness imply that the other sufficiently nearby fibres are $C^1$-close embeddings of $L$. In the Weinstein neighborhood, projection to $L$ restricts on each such fibre to a local diffeomorphism close to the identity. It has degree one, and compactness makes it a covering map, so it is a diffeomorphism. Each nearby fibre is therefore a graph of a closed differential one-form $\alpha$ on $S^3$. Since $H^1(S^3;\mathbb{R})=0$, this is $df$ for a smooth real function $f$. A maximum of $f$ exists by compactness, and there $df=0$. The graph meets the original zero section, contradicting disjointness of different fibres. No such Lagrangian three-sphere fibration exists. More generally this is the compact first-cohomology obstruction to a Lagrangian fibration.

True. The Künneth theorem gives the Poincaré polynomialso $b_2=6$ and $b_4=2$. The proposed equality also gives $b_0=b_8=1$. Even if compactness was not stated separately, these two numbers force the putative eight-manifold to be connected and compact: the top ordinary real cohomology of a connected noncompact manifold is zero.

The auxiliary Hard Lefschetz theorem on de Rham cohomology says that, on a compact Kähler manifold of complex dimension $n$, $L^{n-k}:H^k\to H^{2n-k}$ is an isomorphism, where $L\alpha =[\omega ]⌣\alpha$. Here $n=4$, so $L^2:H^2\to H^6$ is injective. If $L\alpha =0$, then $L^2\alpha =0$ and therefore $\alpha =0$. Thus $L:H^2\to H^4$ is injective, forcing $b_2\le b_4$. The inequality $6\le 2$ is impossible. Hence these Betti numbers cannot occur for a Kähler eight-manifold.

The Category of sets is monoidal under Cartesian product, with any singleton as unit. The tensor of maps is $(f\times g)(x,y)=(f(x),g(y))$, the associator sends $((x,y),z)$ to $(x,(y,z))$, and the left and right unitors discard the singleton coordinate. These are natural bijections. Both pentagon paths send a nested quadruple to the same rebracketed quadruple, and both triangle paths discard the same singleton coordinate, verifying the axioms directly.

This is not the only structure. Disjoint union $X⨿Y$ is another tensor bifunctor, with empty-set unit. Its associator changes parentheses while retaining the tag indicating the original summand, and its unitors remove an empty summand. Every coherence path preserves both the original tag and the element, so again the pentagon and triangle hold.

These product and coproduct monoidal structures on sets are distinct even up to monoidal equivalence. Any equivalence of the underlying category preserves initial and terminal objects up to isomorphism. It therefore cannot carry the empty-set unit of the coproduct structure to the singleton unit of the product structure. Hence Set has at least these two inequivalent monoidal structures.

A monoidal category consists of a category $\mathcal{C}$, a bifunctor $\otimes :\mathcal{C}\times \mathcal{C}\to \mathcal{C}$, a unit object $I$, and natural isomorphisms $a_{X,Y,Z}:(X\otimes Y)\otimes Z\to X\otimes (Y\otimes Z)$, $l_X:I\otimes X\to X$ and $r_X:X\otimes I\to X$. They satisfy the pentagon identity for a monoidal categoryand the triangle identity for a monoidal category $(1_X\otimes l_Y)a_{X,I,Y}=r_X\otimes 1_Y$. These axioms make different coherent rebracketings and unit removals agree.

It defines a bicategory with one object $\ast$. The hom-category $\mathcal{B}(\ast ,\ast )$ is $\mathcal{C}$; objects of $\mathcal{C}$ are its 1-cells and morphisms of $\mathcal{C}$ its 2-cells. Horizontal composition is the tensor bifunctor, the identity 1-cell is $I$, and the associator and unitors are the bicategorical coherence 2-cells. Vertical composition is composition of morphisms in $\mathcal{C}$, and the interchange law follows from bifunctoriality of $\otimes$. Conversely the hom-category of any one-object bicategory has exactly this monoidal structure. Thus a monoidal category is precisely a one-object bicategory, not necessarily a strict one-object 2-category.

Choose a limit $L_j$ for each diagram $F(-,j)$, with legs $p_{i,j}:L_j\to F(i,j)$. For an arrow $v:j\to j^{\prime }$, the maps $F(1_i,v)p_{i,j}$ form a cone over $F(-,j^{\prime })$, since the two-variable functor respects the commuting arrows in $I\times J$. The limit property therefore gives a unique arrow $L(v):L_j\to L_{j^{\prime }}$ satisfyingThe arrows $L(1_j)$ and $1_{L_j}$ have identical composites with all projections, so they are equal. The same uniqueness shows $L(v^{\prime }v)=L(v^{\prime })L(v)$. Hence the assignment extends to a functor $L:J\to \mathcal{D}$.

For the chosen limit objects and cones it is uniquely determined, literally, by these equations. Changing the choices gives a unique family of isomorphisms between the limits compatible with their projections; the equations above show that this family is a natural isomorphism of the resulting functors. Thus it is unique up to a unique cone-compatible natural isomorphism, not independent of choices as an equality of objects.