Let $\mathcal{C}$ be a locally small category, let $h_A=\mathcal{C}(-,A)$ be a representable presheaf, and let $X:{\mathcal{C}}^{\mathrm{op}}\to \mathbf{Set}$. The Yoneda lemma gives a bijection, natural in $A$ and $X$,For $x\in X(A)$, define a natural transformation ${\alpha }^x:h_A\to X$ byFor $v:C\to B$, the functor composition law gives $X(v){\alpha }_B^x(f)=X(fv)(x)={\alpha }_C^x(fv)$, proving naturality. Conversely, naturality of any $\alpha :h_A\to X$ along $f:B\to A$ givesThus $\alpha$ is determined by its displayed element, and ${\alpha }_A^x(1_A)=x$ proves that the constructions are inverse.

A natural transformation $\theta :X\to Y$ sends the element to ${\theta }_A(x)$, which corresponds to $\theta {\alpha }^x$. A morphism $u:A^{\prime }\to A$ induces $h_u:h_{A^{\prime }}\to h_A$; precomposition by $h_u$ sends the element to $X(u)(x)$. This proves naturality in both variables and completes the Yoneda lemma proof.

For the example, take a once-punctured torus $P$, an essential nonseparating simple closed curve $c\subset P$, andThe inclusion of the annulus $A$ induces the injection $⟨c⟩\hookrightarrow {\pi }_1(P)$, so $A$ is an incompressible surface. Cutting gives $(P\text{ cut along }c)\times I$, a pair of pants times an interval. A compact connected surface with boundary times an interval is a handlebody of genus $1-\chi (P)$: cut the surface to a disk by proper arcs and reverse those cuts as one-handle attachments. Both the punctured torus and the pair of pants have Euler characteristic $-1$. ThereforeThis self-reproducing incompressible annulus cut also shows why arbitrary incompressible cutting choices need not terminate.

A three-manifold hierarchy is a finite sequence of cuts along properly embedded two-sided incompressible surfaces, with no spherical components, whose final manifold is a disjoint union of three-balls. Disks are permitted. In an orientable ambient manifold two-sided surfaces are orientable; for an ambient manifold which is not orientable we permit two-sided surfaces without assuming their orientability as well.

For $S^1\times F$, first cut along $\{p\}\times F$, obtaining $I\times F$. This surface is two-sided and its fundamental group injects as a direct factor. Next choose essential two-sided curves in $F$ until the cut surface has boundary, and cut along their products with $I$. For orientable $F$ one nonseparating curve suffices. For $F$ which is not orientable and has genus at least two, the boundary of an embedded Möbius band is a two-sided essential curve; neither side is a disk, so its fundamental group injects. Cutting this curve leaves pieces with boundary, including the Möbius band. For each resulting compact surface choose a finite system of proper arcs cutting it into disks. The corresponding vertical disks give the remaining three-manifold hierarchy cuts. Essential annuli have injective fundamental group, and disks have trivial fundamental group, so all cuts are incompressible. The terminal products are $D^2\times I$, hence three-balls. This construction covers the possibilities without orientability in the stated range as well as the orientable ones.

Here is a terminating construction for a compact connected orientable irreducible three-manifold $M$ with nonempty boundary. We use the permitted general results in the following precise forms: a boundary-incompressible non-ball component has a connected two-sided nonseparating incompressible surface with nonempty boundary; and a sufficiently large collection of disjoint closed incompressible surfaces has two parallel components with no intervening member. These are respectively the existence and finiteness results for incompressible surfaces.

First perform essential boundary disk compressions as long as possible. Their number is finite: each nonseparating compression decreases boundary genus, while a separating compression divides a boundary component into two of smaller genus. More explicitly, the sum of $\max (0,2g-1)$ over the closed boundary components decreases at every essential compression. Each cut preserves three-manifold irreducibility. To see this for any incompressible cutter, a sphere in the cut manifold bounds a three-ball in the uncut manifold. A closed nonspherical incompressible component of the cutter cannot lie in this three-ball, since its fundamental group could not inject into the trivial group. A component with boundary cannot lie in an interior three-ball either. Thus the three-ball misses the cutter and survives in the cut manifold. Fundamental-group injectivity of two-sided incompressible surfaces follows from the Loop theorem, applied to their two boundary copies after cutting.

Push the boundary of each non-ball residual component slightly into the original $M$ to make a closed frontier $F_1$, omitting spherical components. This frontier is incompressible on its residual side, since all boundary compressions there have been exhausted. It is also incompressible on the side already cut off. Indeed, a hypothetical compression disk on that side can be made transverse to the earlier cutting surfaces. An innermost intersection circle bounds a disk on its cutting surface by incompressibility; replacing the corresponding subdisk removes the circle, without introducing intersections with earlier cutters. Repeat in the order of the cuts. The disk would finally lie in a product collar of the frontier, where its boundary is fundamental-group-injective, a contradiction. Thus $F_1$ is incompressible in $M$.

If a residual component is not a three-ball, choose the permitted nonseparating incompressible surface with boundary, cut it, and compress the new boundary fully again. Push that boundary in to obtain a new closed frontier $F_2$ disjoint from $F_1$. Continue. Three-manifold irreducibility persists, and the same innermost-circle proof makes every nonspherical frontier incompressible in the original $M$.

We explain why the parallel frontiers forced by the finiteness theorem cannot occur in this construction. The required product fact is: a connected nonseparating incompressible surface in $G\times I$ cannot be proper and disjoint from $G\times \{1\}$. To prove it, intersect the surface with vertical annuli $c\times I$ for essential curves $c$ in a closed $G$. Inessential intersection circles can be removed using incompressibility and the three-balls supplied by three-manifold irreducibility. An outermost intersection arc has both endpoints on the lower boundary; boundary compression along its annular bigon reduces the intersections. At least one resulting component stays nonseparating, since the mod-two relative class of the original surface is the sum of the resulting classes. Essential intersection circles can likewise be removed by surgery along the annulus running to its lower boundary. Retain a nonseparating incompressible component after each surgery. These operations reduce the intersection number and cut the base surface to one with boundary. Now intersect with the vertical disks of a proper arc system cutting that base to a disk. The same innermost-circle and outermost-arc operations make the surface disjoint from these disks. It then lies in a three-ball $D^2\times I$, where any proper two-sided surface separates. This is the contradiction proving the product fact.

If two frontiers in the construction were parallel with no other frontier between, their product region would contain one of the intervening nonseparating cutters, disjoint from the later frontier. The product fact forbids this. Consequently the closed-frontier construction has only finitely many stages. Once a residual irreducible component has only spherical boundary, push one boundary sphere inward. The three-ball it bounds cannot be on the collar side, which contains a boundary component of the manifold. The three-ball and the collar therefore exhaust that component, making it a three-ball. We have constructed a finite three-manifold hierarchy, and can choose all its cutting surfaces to have nonempty boundary.

We will use one further consequence in the map argument: these manifolds are aspherical. Work backwards through the three-manifold hierarchy. A three-ball is contractible. At one gluing stage, the van Kampen theorem normal form identifies the lifts of the cut pieces and cutting surface in the universal cover with a tree of spaces. The lifted pieces are contractible by induction, and the lifted cutting surfaces are contractible because the surfaces have boundary and are fundamental-group-injective. A finite subtree is contractible by removing terminal vertices; every sphere map meets only finitely many pieces. All higher homotopy groups therefore vanish at this stage, and induction proves asphericity from a three-manifold hierarchy. Apply this tree argument separately at each gluing stage.

The following three entries complete a single proof of the relative-boundary homeomorphism assertion.

The Hermitian metric induces pointwise Hermitian inner products on the exterior powers of the complex cotangent bundle, with different bidegrees orthogonal, and a positive volume form $d{V}_g$. Thus the global $L^2$ inner product on smooth differential forms of type $(p,q)$ isLet ${\bar{\partial }}^{\ast }$ be the formal adjoint of the Dolbeault operator, defined by integration by parts, and let ${\Delta }_{\bar{\partial }}=\bar{\partial }{\bar{\partial }}^{\ast }+{\bar{\partial }}^{\ast }\bar{\partial }$. This Dolbeault Laplacian preserves bidegree, is elliptic and nonnegative, and hasOn a compact Hermitian manifold without boundary, the Dolbeault Hodge decomposition on a compact Hermitian manifold says that the smooth harmonic space ${\mathcal{H}}_{\bar{\partial }}^{p,q}=\ker {\Delta }_{\bar{\partial }}$ is finite-dimensional andMore precisely, there are the orthogonal harmonic projection $H$ and a Dolbeault Green operator $G$ taking smooth forms to smooth forms, with $G=0$ on harmonics and ${\Delta }_{\bar{\partial }}G=G{\Delta }_{\bar{\partial }}=I-H$. Elliptic regularity ensures these are statements about smooth forms, not only $L^2$ completions. The Green operator commutes with $\bar{\partial }$ and ${\bar{\partial }}^{\ast }$, since the Laplacian does and its inverse is unique on the orthogonal complement of the kernel.

Expanding $\alpha =H\alpha +{\Delta }_{\bar{\partial }}G\alpha$ givesThe three summands are orthogonal: harmonic forms are killed by both adjoints, and $⟨\bar{\partial }u,{\bar{\partial }}^{\ast }v⟩=⟨{\bar{\partial }}^{2}u,v⟩=0$. If $\alpha$ is $\bar{\partial }$-closed, its coexact component is orthogonal to every closed form; taking its inner product with $\alpha$ shows that component has norm zero. Conversely the first two components are closed. Thus $\boxed{\ker \bar{\partial }={\mathcal{H}}_{\bar{\partial }}^{p,q}\oplus \bar{\partial }{A}^{p,q-1}}$, with out-of-range form spaces taken to be zero.

Write $\chi$ for the red-blue finite colouring of unordered pairs. The two alternatives concern one arithmetic progression or a pair of disjoint arithmetic progressions; they need not be mutually exclusive. For $m=1$, a singleton already has no non-blue edge, so the first alternative holds. Assume henceforth $m\ge 2$.

If the first alternative fails, every $m$-term arithmetic progression contains a red pair. Label each $(a,d)\in {\mathbb{N}}^2$ by one such pair's positions $0\le r<s<m$, choosing the first in a fixed ordering when there are several. This is a finite colouring with at most $\left(\genfrac{}{}{0ex}{}{m}{2}\right)$ colours. The following two branches establish the arithmetic-progression bipartite Ramsey dichotomy.

Use separation of variables in Laplace's equation: writing $\varphi =R(r)\Theta (\theta )$ givesSingle-valuedness requires $2\pi$-periodicity, so the nonconstant angular Fourier modes have integer $n\ge 1$ and are $\cos n\theta ,\sin n\theta$. Their radial solutions are $r^n,r^{-n}$. The zero angular mode has radial solutions $1,\log r$. These give the harmonic Fourier expansions in planar concentric domains used below.

For the annulus with $0<a<b$, constant boundary data suggest a radial harmonic function $A+B\log r$. Enforcing both boundary values givesThis has the required boundary values and satisfies $(r{\varphi }_r{)}_r=0$. It is unique: the difference of two continuous solutions is harmonic in the annulus and zero on both boundary circles, so the maximum principle for harmonic functions, applied to the difference and its negative, makes it zero.

A norm on a real vector space $V$ is a map $p:V\to [0,\infty )$ satisfying $p(x)=0$ exactly when $x=0$, $p(tx)=|t|p(x)$ for real $t$, and $p(x+y)\le p(x)+p(y)$. Its induced metric is $d_p(x,y)=p(x-y)$. The three axioms respectively give definiteness, compatibility with scaling, and the triangle inequality.

In the upper half-plane model, hyperbolic lines are vertical Euclidean rays and semicircles centred on the real axis, equivalently circle arcs meeting the real boundary orthogonally. A semicircle with endpoints $r<s$ is mapped to the imaginary axis byIts determinant is one, and its real endpoints map to $0,\infty$. A vertical line $x=a$ is mapped to the imaginary axis by $z\mapsto z-a$. Thus every hyperbolic line can be carried to a fixed one; composing one such map with the inverse of another proves transitivity of $\mathrm{PSL}(2,\mathbb{R})$ on the lines.

Reflection in $x=a$ is $R_a(z)=2a-\overline{z}$. Reflection in the unit semicircle is $R_{\circ }(z)=1/\overline{z}$. The latter fixes the semicircle pointwise, is an involution, and preserves the hyperbolic metric because $|dw|=|dz|/|z{|}^2$ and $\mathrm{Im}w=\mathrm{Im}z/|z{|}^2$. The vertical reflection similarly preserves $|dz|/\mathrm{Im}z$. Their composition in the specified order is

The Liouville theorem states that every bounded entire function is constant. Let $|f(z)|\le M$ and choose distinct $a,b$. For $R>\max (|a|,|b|)$, partial fractions and the Cauchy integral formula givewith the circle traversed anticlockwise. On that circle,The displayed difference is independent of $R$, so $f(a)=f(b)$. Since $a,b$ were arbitrary, $f$ is constant. This proves the theorem with the requested two-pole contour rather than assuming it.

A function $f:I\to \mathbb{R}$ is uniformly continuous ifThe crucial point is that $\delta$ works throughout the interval, not just near one chosen point.

Yes, it is bounded on the closed unit interval. Choose $\delta$ for $\epsilon =1$, and choose finitely many grid points $t_j$ such that every $x\in [0,1]$ is within $\delta$ of some $t_j$. Then $|f(x)|\le 1+{\max }_j|f(t_j)|$, a finite bound. Equivalently, uniform continuity implies continuity, and a continuous real function on a compact set is bounded.

Yes, a bounded derivative gives uniform continuity on the half-line. If $|f^{\prime }|\le K$, the mean value theorem gives $|f(x)-f(y)|\le K|x-y|$. This is a Lipschitz condition, so for $K>0$ use $\delta =\epsilon /K$. If $K=0$, the function is constant. As usual for a derivative on this closed half-line, continuity at its endpoint is included.

For a bounded function on a nondegenerate compact interval $[a,b]$, a partition $P$ has Darboux sumsThe Riemann integrability criterion isEquivalently the supremum of lower sums and infimum of upper sums agree; their common value is the Riemann integral. A bounded nonintegrable example is the Dirichlet function $\boxed{f=1_{\mathbb{Q}}}$. Every positive-length interval contains rational and irrational points, so every lower sum is zero and every upper sum is $b-a>0$.

To prove the intermediate value theorem in this setting, let $A=\{t\in [a,b]:f(t)<v\}$ and $c=\sup A$. The endpoint inequalities and continuity give an interval immediately to the right of $a$ contained in $A$ and an interval immediately to the left of $b$ disjoint from $A$. Thus $a<c<b$. Choose points of $A$ tending to $c$; continuity gives $f(c)\le v$. If $f(c)<v$, continuity would place some point larger than $c$ in $A$, contradicting the supremum. Hence $\boxed{f(c)=v}$ at an interior point.

For a continuous self-map of $[a,b]$, define $g(t)=f(t)-t$. The range condition implies $g(a)\ge 0$ and $g(b)\le 0$. If either is zero, its endpoint is a fixed point. Otherwise apply the just-proved intermediate value theorem to $-g$ with target zero to obtain $d\in (a,b)$ and $g(d)=0$. ThereforeThis proves the fixed-point property of a closed interval without requiring $f$ itself to be monotone.

Take a right-handed local Cartesian frame with $x$ towards the equator, $y$ azimuthal and $z$ radially outwards in the northern hemisphere. A local radial shear $U_y=Sz$ and a poloidal magnetic field represented by $A\hat{\mathbf{y}}$ give the alpha-Omega dynamo equations $A_t=\alpha B+{\eta }_T{A}_{xx}$ and $B_t=S{A}_x+{\eta }_T{B}_{xx}$. Substitution of a Fourier mode proportional to $e^{pt+ikx}$ gives the displayed dispersion relation. The growing branch has $\mathrm{Re}p=-{\eta }_T{k}^2+\sqrt{|\alpha Sk|/2}$ and $\mathrm{Im}p=\mathrm{sgn}(\alpha Sk)\sqrt{|\alpha Sk|/2}$. Its phase velocity is $-\mathrm{Im}p/k$: it propagates equatorwards when $\alpha S<0$, polewards when $\alpha S>0$. The result assumes local constant coefficients and no meridional circulation in a star; advection and spatially separated field regeneration can alter migration.