Let $V\subseteq U$ be open subsets, and let $s\in {\mathcal{F}}_3(V)$. The assumed flasque-kernel section-lifting lemma lifts $s$ to $\tilde{s}\in {\mathcal{F}}_2(V)$. Since ${\mathcal{F}}_2$ is a flabby sheaf, extend $\tilde{s}$ to a section on $U$, then project to ${\mathcal{F}}_3(U)$. Its restriction is $s$. Thus every restriction map of ${\mathcal{F}}_3$ is surjective, proving that the quotient of two flabby sheaves in a short exact sequence is flabby.

For the construction of sheaf cohomology, embed a sheaf of abelian groups $\mathcal{F}$ into the sheafby sending a section to all its germs. This is injective because a section with every germ zero is zero. Arbitrary families of stalk elements glue point by point, and restriction maps are projections; hence $C(\mathcal{F})$ is a flabby sheaf. Set ${\mathcal{Q}}^0=\mathcal{F}$, ${\mathcal{I}}^j=C({\mathcal{Q}}^j)$ and ${\mathcal{Q}}^{j+1}={\mathcal{I}}^j/{\mathcal{Q}}^j$. Compose the quotient map with the next embedding to obtain the flasque resolutionThe sheaf cohomology groups are the cohomology groups of the resulting cochain complex of global sections:The usual comparison of resolutions identifies this construction with the right derived functors of global sections; the construction above is the Godement resolution.

If $\mathcal{F}$ is itself a flabby sheaf, the quotient argument shows inductively that every ${\mathcal{Q}}^j$ is flabby. The assumed lifting result therefore makesan exact sequence for every $j$. The global-section complex is consequently exact in every positive degree. In particular,This establishes the vanishing from the section-lifting property rather than assuming it in advance.

For the smooth algebraic curve $X$, a divisor on an algebraic curve is a finite formal sum $D={\sum }_P{n}_{P}P$ of closed points with integer coefficients. At each closed point, the local ring of a smooth algebraic curve is a discrete valuation ring; write its normalized discrete valuation as $v_P$. For $f\in k(X{)}^{\ast }$ the principal divisor is $\mathrm{div}(f)={\sum }_P{v}_P(f)P$. Only finitely many coefficients are nonzero: zeros and poles are proper closed subsets in the Noetherian Zariski topology of a curve. The divisor class group is the abelian groupEquivalently, its elements are divisors on an algebraic curve modulo linear equivalence of divisors.

Let $\pi :E\to B$ be an oriented real vector bundle of rank $r$ over a paracompact base. A Thom class is a class $U\in H^r(E,E\setminus B;\mathbb{Z})$ restricting on every oriented fiber pair $({\mathbb{R}}^r,{\mathbb{R}}^r\setminus 0)$ to its positive generator. The Thom isomorphism theorem states that this class exists uniquely with those fiber restrictions and thatHere $B$ is identified with the zero section. Equivalently use the disk/sphere bundle pair. The compact-support version is $H_c^q(B;\mathbb{Z})\cong H_c^{q+r}(E;\mathbb{Z})$, with fiberwise compact support in the Thom construction. If the bundle is not oriented, the integral assertion requires the corresponding orientation local system.

For a closed co-oriented embedded submanifold $Y$ of codimension $r$, co-orientation means an orientation of its normal bundle. A tubular neighborhood identifies a neighborhood of $Y$ with that bundle. Its Thom class, transferred by excision, is a class in $H^r(M,M\setminus Y;\mathbb{Z})$. For compact $Y$ this maps into the direct limit defining compactly supported cohomology, giving $\boxed{{\epsilon }_Y\in H_c^r(M;\mathbb{Z})}$. The construction is independent of the tubular neighborhood and uses the chosen co-orientation. In an oriented ambient manifold it is the Poincare dual of the oriented submanifold. Here “closed submanifold” has the usual compact-without-boundary meaning; if it meant merely a closed subset and $Y$ were noncompact, the supported relative class would still exist but need not have compact support.

We next prove the fixed-point implication without imposing an unprinted orientability assumption on $M$. SetUse a finite triangulation of the closed smooth manifold. The elementary Hopf trace identity says that the alternating trace of a chain endomorphism equals its alternating trace on homology. To verify it, split each finite-dimensional chain space as boundaries, a complement representing homology, and a complement mapping isomorphically onto the preceding boundaries. The traces on these two boundary summands cancel in adjacent degrees, leaving exactly the homology traces. The cohomology traces equal the homology traces because the induced maps are dual over $\mathbb{Q}$.

If $f$ has no fixed points, compactness gives $\delta ={\min }_{x}d(x,f(x))>0$. Choose a triangulation $K$ of mesh less than $\delta /4$ and a sufficiently fine subdivision $K^{\prime }$ admitting a simplicial approximation $F:K^{\prime }\to K$ uniformly within $\delta /4$ of $f$. The subdivision chain map $s:C_{\ast }(K)\to C_{\ast }(K^{\prime })$ sends each simplex to simplices inside it. For a vertex $v$ of such a small simplex, $d(v,F(v))>3\delta /4$, so $F(v)$ cannot lie in the original coarse simplex. Consequently $F_{\#}s$ has zero diagonal coefficient on every coarse simplex. Its trace vanishes in every degree, while it induces $f_{\ast }$ on homology. The trace identity yields $L(f)=0$. Taking the contrapositive provesThis is the Lefschetz fixed-point theorem, established here rather than invoked. For oriented $M$ the same identity is the graph-diagonal formula for the Lefschetz number: $L(f)=⟨(1,f{)}^{\ast }{\epsilon }_{\Delta },[M]⟩$, which vanishes when the graph misses the diagonal. Finally, homotopic maps induce the same cohomology maps, so their Lefschetz numbers agree.

There are genuine normalization inconsistencies in this PDF. In the standard linearized chord diagram model $G_n^{(1)}$, time $n$ gives $n$ edges and total vertex degree $2n$, with self-loops counted twice. The paper calls $2n$ the number of edges. Its right-endpoint hint also uses $\sqrt{i/(2n)}$ rather than $\sqrt{i/n}$. These cannot both describe the standard model; part (c)'s exponent is consequently inconsistent as well. We give the definition, prove the alternative construction and its limits, and explicitly identify the corrected conclusion rather than infer an invalid constant from the printed hint.

Start from the empty graph at time zero, or from one vertex with one self-loop at time one. To obtain $G_t$ from $G_{t-1}$, add vertex $t$ and an edge directed from $t$ to $I_t$, whereA self-choice creates a self-loop. The old vertex degrees sum to $2(t-1)$, so these probabilities sum to one. This is exact preferential attachment, including the outward half of the new edge as weight one for the new vertex; it is not a rule normalized by $2t$ or a process adding two edges at each step. The degree of a vertex here is its total incoming plus outgoing vertex degree. The construction in part (a) has exactly one chord per vertex and makes the edge-count discrepancy explicit.

Identify the discrete cube $Q_n$ with the subsets of $[n]$, with hypercube graph adjacency given by changing one element. Write $N[A]$ for the closed graph neighbourhood and $\partial A=N[A]\setminus A$ for the external vertex boundary. The simplicial order on the discrete cube first orders by set size and then by lexicographic order, with the smallest differing coordinate belonging to the earlier set.

Harper theorem states that, for every set family $A\subseteq Q_n$,Here is the simplicial section compression proof. Induct on $n$, the one-dimensional case being immediate. Split $A$ by a coordinate into sections $A_0,A_1\subseteq Q_{n-1}$. Its neighbourhood sections areReplace $A_0,A_1$ by equally large simplicial initial segments $I_0,I_1$. By induction $|N[I_j]|\le |N[A_j]|$. Also the neighbourhood of a simplicial initial segment is itself a simplicial initial segment: at its last occupied level the extra vertices form the upper shadow of a lexicographic initial segment, again lexicographically initial. Consequently the compressed unions have sizes $\max \{|N[I_0]|,|I_1|\}$ and $\max \{|N[I_1]|,|I_0|\}$, no larger than the original unions. The compression preserves $|A|$ and cannot increase $|N[A]|$.

Repeat in every coordinate until all sections are initial. Every nontrivial compression strictly decreases the sum of the global simplicial positions, so the process terminates. To describe the terminal families for simplicial section compression, suppose an earlier absent vertex $S$ precedes a later present vertex $T$. They cannot share a coordinate value: otherwise they lie in the same compressed section, contradicting its initiality. Thus $T=[n]\setminus S$. No vertex can lie strictly between them in the order, since it would make another absent/present pair that is not complementary. Hence the terminal family is either $C(|A|)$ or that segment with its last vertex $S$ exchanged for its immediate complementary successor $T$.

For odd $n=2m+1$, that exceptional exchange crosses the middle levels: $S=\{m+2,\dots ,2m+1\}$ and $T=\{1,\dots ,m+1\}$. For $m\ge 1$, all sets of size at most $m+1$ remain in the neighbourhood after the exchange, since an $(m+1)$-set has at least two lower neighbours and only one was removed. The new $T$ can only add vertices of size $m+2$.

For even $n=2m$, the only consecutive complementary pair in one level is $S=\{1,m+2,\dots ,2m\}$ and $T=\{2,\dots ,m+1\}$. The original segment contains all sets below size $m$ and all $m$-sets containing 1. For $m\ge 2$, its neighbourhood contains all sets of size at most $m$ and all $(m+1)$-sets containing 1; every such upper set still has an included lower neighbour after the single removal. The new $T$ can only add further neighbours. The cases $n=1,2$ follow directly, with the exceptional families related to the initial ones by a graph automorphism. Thus neither exception improves the neighbourhood, completing the induction.

For the half-cube vertex-boundary extrema, if $n=2m+1$, the initial half cube is all sets of size at most $m$; its boundary is the next level, of size $\left(\genfrac{}{}{0ex}{}{2m+1}{m+1}\right)$. If $n=2m$, it contains all levels below $m$ and the $m$-sets containing 1. The boundary consists of the $m$-sets avoiding 1 and the $(m+1)$-sets containing 1, giving $2\left(\genfrac{}{}{0ex}{}{2m-1}{m}\right)=\left(\genfrac{}{}{0ex}{}{2m}{m}\right)$.

No boundary can exceed the $2^{n-1}$ vertices outside a half cube. Taking all vertices of even Hamming weight attains that bound, because every odd-weight vertex is adjacent to one of them. Therefore, for $n\ge 1$,

A one-dimensional canonical transformation is a locally invertible change of phase-space variables preserving the symplectic two-form: $dQ\wedge dP=dq\wedge dp$. Equivalently it preserves Hamilton equations for the transformed Hamiltonian. Expanding the two-form yieldssoConversely this Jacobian condition makes the transformation locally invertible and symplectic in two dimensions.

An odd composite integer $n>1$ coprime to $b$ is a Fermat pseudoprime to base $b$ if $b^{n-1}\equiv 1(\mathrm{mod}n)$. A Carmichael number is a composite integer satisfying this congruence for every integer $b$ coprime to $n$. The Korselt criterion says that a composite $n$ is a Carmichael number exactly when it is squarefree and $p-1$ divides $n-1$ for every prime divisor $p$ of $n$.

Differentiate under the contour integral, with one consistent branch and convergent endpoint behavior. ThenThe stated zero boundary term therefore makes the integral a solution. On a closed contour this means returning to the same branch; on an open contour it requires the endpoint values to cancel or vanish.

The inductive definition of ordinal addition isfor nonzero limit ordinals $\lambda$. The synthetic definition is the order type of a disjoint copy of $\alpha$ followed by a disjoint copy of $\beta$, retaining the order within each block and declaring every element of the first block smaller than every element of the second.

These are equivalent by transfinite induction in the second argument. An empty second block leaves $\alpha$; a successor block appends one final point; and a limit block is the increasing union of its proper initial blocks, whose order types have the indicated supremum. Thus the synthetic construction satisfies exactly the inductive recursion, which determines the operation uniquely. It is not commutative: $1+\omega =\omega$, whereas $\omega +1>\omega$.

Let the three orbit points be $a<b<c$. Up to reflection of the interval, their images are $F(a)=b,F(b)=c,F(c)=a$. Set $I=[a,b]$ and $J=[b,c]$. Continuity and the intermediate value theorem imply the covering relationsWe first prove the needed interval-covering principle using the permitted pullback fact. If a cyclic list of closed intervals $I_0,\dots ,I_{n-1},I_n=I_0$ obeys $F(I_j)\supseteq I_{j+1}$, pull back successively from the end to obtain a closed interval $K\subseteq I_0$ with $F^n(K)=I_0$ and $F^j(K)\subseteq I_j$. Since $F^n(K)\supseteq K$, the intermediate value theorem gives a fixed point of $F^n$ in $K$: choose preimages of the two endpoints of $K$, which give opposite weak signs of $F^n(x)-x$, irrespective of their order. This point realizes the cyclic itinerary.

For $n\ge 2$, use the word $I{J}^{n-1}$. It obeys the covering relations and therefore yields a point with period dividing $n$. A point away from the shared endpoint $b$ has an unambiguous itinerary, and this word has exactly one $I$ per cycle; it cannot be a repetition of a shorter word. Its least period is consequently $n$. If an orbit touches $b$, it is the original three-cycle. That realizes the case $n=3$ itself. It cannot realize any of the other words: for $n$ not divisible by three it is not fixed by $F^n$, and for a multiple of three larger than three it visits $a\in I\setminus J$ more than once, contradicting the itinerary. Thus the construction works for every $n\ge 2$. For $n=1$, $F(J)\supseteq J$ and the same fixed-point argument gives a fixed point. This proves period three implies all periods: there is an orbit of every positive least period.

For the two five-cycles forced by a three-cycle, use the two cyclic words $IJJJJ$ and $IJIJJ$. Both are admissible. Their points cannot be endpoints from the three-cycle, and because five is prime and the words use both intervals, each has least period five. Their itineraries have respectively one and two visits to the interior of $I$; cyclically shifting the starting point cannot change this count. Hence they give at least two distinct period-five orbits.

For steady unidirectional pressure-driven channel flow, write $G=-dp/dz>0$. The axial momentum equation and no-slip boundary condition areNear an ideal wedge, the constant forcing has an $r^2$ particular solution by scaling. Substitution gives $f^{\prime \prime }+4f=-1$, with $f(\pm \alpha )=0$. For $\cos 2\alpha \ne 0$, the symmetric particular solution isThis corner expansion of pressure-driven sector flow is a particular solution, not necessarily the dominant corner term; a homogeneous corner mode can be larger, as the requested cases below show.

Separation of the symmetric harmonic correction with zero values on the straight walls gives modes $r^{{\lambda }_n}\cos ({\lambda }_n\theta )$, whereRegularity at the vertex excludes negative powers. Orthogonality on $[-\alpha ,\alpha ]$ has norm $\alpha$. To cancel the particular solution on $r=a$, use its cosine coefficientsThe integral follows by product-to-sum and $\sin ({\lambda }_n\alpha )=(-1{)}^n$. ThusThe arc data are satisfied by the cosine expansion, and the Poisson equation by construction. At the omitted thresholds $2\alpha =\pi /2,3\pi /2$, a mode has ${\lambda }_n=2$, so the pure $r^{2}f$ ansatz is resonant. The corresponding finite-limit term in the full sector solution is $(G/\mu )(-1{)}^n{r}^2\log (a/r)\cos 2\theta /(4\alpha )$; the other modes remain regular. This explains rather than hides the division by $\cos 2\alpha$.

For a closed piecewise continuously differentiable path avoiding $a$, the winding number isBreak the parameter interval at the finitely many derivative discontinuities when evaluating this integral. Translating the curve gives $n(\gamma ,a)=n(\gamma -a,0)$, which will also be used below.

A function is a differentiable function at $a_0=(a,b)$ if there is a linear map $L:{\mathbb{R}}^2\to \mathbb{R}$ withThe map $L=D{f}_{a_0}$ is its derivative; equivalently $L(h_1,h_2)=A{h}_1+B{h}_2$ for fixed constants. Since every such linear map is bounded, $|f(a_0+h)-f(a_0)|\le \Vert L\Vert \Vert h\Vert +|r(h)|\to 0$. This proves differentiability implies continuity, not merely continuity along each coordinate axis.

For a two-dimensional incompressible flow $\mathbf{u}=(u,v)$, the continuity equation is $u_x+v_y=0$. Locally, or globally on a simply connected fluid region, this permits a streamfunction withIndeed, the one-form $-vdx+udy$ is closed by incompressibility and hence is locally the differential of $\psi$. Conversely these expressions automatically give zero divergence by equality of mixed derivatives.

A streamline at a fixed time is tangent to the instantaneous velocity field. Along it,Thus streamlines are connected curves on which the streamfunction is constant; wherever $\nabla \psi \ne 0$, the level curves have exactly the required tangent direction. At a stagnation point, the tangent is degenerate and distinct branches of a level set need not form a single streamline. For unsteady flow this argument fixes time: it does not identify streamlines with particle pathlines.

For the unheaded continuation, set $w=\log z$ in the food equation. The two square roots are $w=\pm i\pi /4$, so exponentiating givesBoth values have $|z-1{|}^2=2-\sqrt{2}<1$ and hence lie strictly inside the disk. Both also have the indicated principal logarithms. Allowing the multivalued complex logarithm produces no additional points: the only permitted values of $w$ are still the two square roots already used.

For drink, the PDF squares only the denominator. Put $z=x+iy$. Its numerator is $(3x+iy)/2$ and its unsquared denominator is $(x+3iy)/2$, which is nonzero unless $z=0$. Multiplying the equation by this denominator squared and equating the real part and imaginary part yieldsthat is $x^2-9y^2=2x$ and $y(1-9x)=0$. If $y=0$, the real equation gives $x=0$ or $x=2$, and zero is excluded. If $y\ne 0$, then $x=1/9$, but the real equation would require $9y^2=-17/81$, impossible for real $y$. ThusIt lies on the allowed disk boundary, and direct substitution gives $3/1^2=3$. The denominator-only square matters: squaring the entire quotient would define a different locus.

For the unheaded continuation, write the vector triple product asThe Levi-Civita symbol contraction is ${ϵ}_{ijk}{ϵ}_{klm}={\delta }_{il}{\delta }_{jm}-{\delta }_{im}{\delta }_{jl}$. One way to check its sign is to fix distinct $i,j$: the only nonzero term has $k$ the remaining index, and the pairs $(l,m)=(i,j),(j,i)$ contribute $+1,-1$ respectively. When $i=j$, both sides vanish. Contracting with the vectors now gives both requested notations:If $\mathbf{a}$ and $\mathbf{b}$ are orthogonal, then $\mathbf{a}\cdot \mathbf{b}=0$. With the specified $\mathbf{c}$, the perpendicularity of the cross product gives $\mathbf{a}\cdot \mathbf{c}=|\mathbf{a}{|}^2$. ThereforeThe same answer covers the degenerate case in which one of these two vectors is zero.

Let $R=\overline{u^{\prime }{v}^{\prime }}$ be the eddy momentum flux, $F=\overline{{\rho }^{\prime }{v}^{\prime }}$ the eddy density flux, and $S=d{\rho }_s/dz<0$. The PDF's hydrostatic equation is $g\overline{\rho }=-{\overline{p}}_z$; the converted TeX mistakenly replaces the pressure derivative by a density derivative. With $N_0^2=-gS/{\rho }_0$, defineThe residual mean circulation is nondivergent because $v_y^{\ast }+w_z^{\ast }={\overline{v}}_{ay}+{\overline{w}}_{az}$. Substituting in the original mean momentum and density equations gives the transformed Eulerian mean equationswith the unchanged geostrophic and hydrostatic balances. Here the Eliassen–Palm flux components areThe eddy buoyancy flux has been absorbed into the residual transport, while both types of wave forcing enter the momentum equation through the Eliassen–Palm flux divergence.

For geostrophic disturbances, the Taylor identity for quasi-geostrophic flux givesFor example, zonal averaging removes $\overline{{\psi }_x^{\prime }{\psi }_{xx}^{\prime }}$ and converts the remaining horizontal and vertical terms in $\overline{{\psi }_x^{\prime }{q}^{\prime }}$ into derivatives of $\overline{{\psi }_x^{\prime }{\psi }_y^{\prime }}$ and $(f_0^2/N_0^2)\overline{{\psi }_x^{\prime }{\psi }_z^{\prime }}$, giving precisely the displayed fluxes.

Locally, for a basic flow with nonzero meridional PV gradient $Q_y$, the quasi-geostrophic wave pseudomomentum is $\mathcal{A}=\overline{q^{\prime 2}}/(2Q_y)$. Multiplying the linear perturbation PV equation by $q^{\prime }/Q_y$ giveswhere $\mathcal{D}$ is a PV source or sink. In a locally uniform conservative Rossby wave packet, $\mathsf{F}=\mathcal{A}{\mathbf{c}}_g$. Thus the flux tracks group velocity rather than the generally different phase velocity, with a signed wave activity if $Q_y<0$.

The non-acceleration theorem for quasi-geostrophic waves states that statistically steady conservative waves, with no wave-activity sources, sinks or independent boundary forcing, produce no balanced mean acceleration. Their flux divergence vanishes. In the present stable, impermeable mean-flow problem the homogeneous residual-circulation equation then has the zero solution, so $v^{\ast }=w^{\ast }=0$ and ${\overline{u}}_t=0$. The Eulerian mean flow can still circulate because its eddy-induced correction need not vanish. Dissipation, absorption at a critical layer, time-dependent wave activity or boundary forcing produces a nonzero flux divergence and invalidates the non-acceleration conclusion.