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Since the hom-sets in these categories are either empty or singleton, an adjunction $f⊣g$ is precisely the conditionThe hom-set bijections are then the unique possible ones, and naturality is automatic. This is a Galois connection between the two posets.

Equivalently, the monotone functions satisfy the unit and counit inequalities $x\le gf(x)$ and $fg(y)\le y$. The equivalence implies these by taking $y=f(x)$ and $x=g(y)$. Conversely, if the inequalities hold, $f(x)\le y$ implies $x\le gf(x)\le g(y)$, while $x\le g(y)$ implies $f(x)\le fg(y)\le y$.

For a categorical presheaf $P$ on the open subsets, define the functors on objects byTo prove $L⊣\Delta$, a natural transformation $t:P\to \Delta A$ must satisfyfor every open $U$, since the restrictions of $\Delta A$ are identities. Thus it is determined by the function $P(\varnothing )\to A$. Conversely such a function defines all $t_U$ by that equation, and the categorical presheaf restriction laws make the components compatible. HenceFor the other adjunction, a transformation $s:\Delta A\to P$ is determined by its component $s_S:A\to P(S)$, sinceEvery function $A\to P(S)$ supplies compatible components by this formula. Therefore $\mathrm{Nat}(\Delta A,P)\cong \mathrm{Set}(A,P(S))$, giving the adjoints to constant presheaves on open sets:These are adjoints of the stated constant-presheaf functor, with its value at every open set including the empty one; no sheafification is involved.

Enumerate $A$ as $a_1,\dots ,a_a$. A function is specified by choosing the image of each of these elements independently, with $b$ possibilities each. The multiplication principle gives $\boxed{b^a\text{ functions}}$.

For an injective function, successive images must be distinct. If $a\le b$, there are $b$ choices for the first, $b-1$ for the second, and so on. If $a>b$, the pigeonhole principle makes injectivity impossible. Thus the number of injective functions isThe product in the first case is a falling factorial.

Let $\Omega$ be the set of all functions from the $n$-element domain to the $k$-element codomain, and let $C_j$ consist of the functions which omit codomain element $j$. A function is surjective precisely when it lies outside ${\bigcup }_j{C}_j$.

For $J\subseteq [k]$ of size $i$, the intersection ${\bigcap }_{j\in J}{C}_j$ consists of maps taking all their values in the remaining $k-i$ elements, so it has $(k-i{)}^n$ members. There are $\left(\genfrac{}{}{0ex}{}{k}{i}\right)$ choices of $J$. Applying the complement form of the inclusion-exclusion principle givesWhen $k=n$, every fibre of a surjective function has at least one member, and the $n$ fibres partition an $n$-element domain. Thus every fibre has exactly one member and the function is a bijection. The count of these bijections is $n!$ by the successive distinct-image choices from part (a). Hence

Put ${\mathcal{C}}_0=\mathbf{Set}$. The initial free construction $L_0:\mathbf{Set}\to {\mathcal{C}}_1$ sends $A$ to $A\times \mathbb{N}$ with ${\alpha }_1(a,k)=(a,k+1)$. A function $f:A\to UB$ extends uniquely by $(a,k)\mapsto {\beta }_1^k(f(a))$. Its induced monad is $T_{0}A=A\times \mathbb{N}$, with unit $a\mapsto (a,0)$ and multiplication $(a,k,l)\mapsto (a,k+l)$. An algebra for a monad is determined by the action of $1\in \mathbb{N}$, so its Eilenberg-Moore category is ${\mathcal{C}}_1$. Thus this first free–forgetful adjunction is a monadic adjunction without needing the given assumption for $n=0$.

The monadic length counts successive Eilenberg-Moore comparison functor steps until the comparison becomes an equivalence. Fix $0\le m<n$. For $m\ge 1$, the free construction in the preceding part adds the $(m+1)$st operation with no fixed points. For $m=0$, the free first operation also has no fixed points. Consequently the left adjoint $F_{n,m}$ to the composite forgetful functor $U_{n,m}:{\mathcal{C}}_n\to {\mathcal{C}}_m$ is obtained by the one-step free construction, followed by empty higher operations. In particular, its underlying object of ${\mathcal{C}}_m$, unit and multiplication are exactly those of the one-step adjunction:as monads, not merely as object functions. The multiplication agrees because both counits evaluate the same first new operation and the same chains; no later operation has a value on a free object.

By the assumed one-step monadic adjunction property, or the explicit argument above when $m=0$, the Eilenberg-Moore category ${\mathcal{C}}_m^{T_m}$ is equivalent to ${\mathcal{C}}_{m+1}$. Under this equivalence the comparison ${\mathcal{C}}_n\to {\mathcal{C}}_m^{T_m}$ is precisely $U_{n,m+1}$. Indeed, its algebra action evaluates each newly adjoined chain using ${\alpha }_1$, and evaluates the first new value using ${\alpha }_{m+1}$; it retains that operation and ignores all higher ones. The comparison has a left adjoint by the same explicit free extension. Thus the whole monadic tower for nested partial unary operations iswith the remaining comparison at stage $m$ equal to $U_{n,m}$, and stage $n$ the identity.

It does not terminate earlier. If $m<n$, take the same underlying two-element set in two objects of ${\mathcal{C}}_n$, making every operation up to $m$ the identity. In the first object make all remaining operations identities; in the second make ${\alpha }_{m+1}$ exchange the two elements, and leave all subsequent operations undefined. Both satisfy the nesting rule. The identity function is a morphism after forgetting to ${\mathcal{C}}_m$, but is not a morphism between the original objects. Thus $U_{n,m}$ is not a full functor, and cannot be an equivalence. Therefore

Use a Fourier transform parallel to the boundary. Splitting its traces into right- and left-supported parts gives Half-range Fourier transforms $F_{+}$ and $F_{-}$, analytic respectively above and below a common inversion contour. A small positive imaginary part of the wavenumber implements the limiting absorption principle and separates the incident poles and outgoing branch points. Solve the transformed differential equation normal to the boundary using its outgoing or decaying solution. The boundary conditions then yield a Wiener-Hopf equation of the formFactor its Wiener-Hopf kernel as $K=K_{+}{K}_{-}$, with each factor analytic and nonzero in its designated half-plane. Divide by $K_{-}$ and split $H/K_{-}=H_{+}+H_{-}$, allocating poles according to the incident-wave prescription. ThenThe two sides continue to a common entire function. Their growth at infinity, inherited from the field's edge behavior, determines that function, usually a constant or a polynomial; in a decaying case it is zero. Solve for the unknown transforms and apply Fourier inversion, preserving the outgoing contour prescription.

The principal analytic results used are as follows. The Paley–Wiener theorem and its half-line Laplace-transform version relate support on a half-line to holomorphy in a half-plane, with growth or square-integrability bounds. The Cauchy integral formula and Sokhotski–Plemelj theorem provide additive splitting: a Cauchy integral of a sufficiently decaying jump has upper and lower boundary values with that prescribed difference. Applying the same construction to a continuous logarithm of a nonvanishing scalar Wiener-Hopf kernel gives Wiener-Hopf factorization when its winding number is zero; a nonzero index requires an explicit index factor. The identity theorem gives analytic continuation across the common strip. Finally Liouville's theorem makes a bounded entire function constant; its polynomial-growth extension bounds the possible degree. The physical edge and radiation conditions select the remaining constants. Zeros, branch cuts, nonzero index and insufficient decay must be accounted for rather than discarded during splitting.

Interpret the longitudinal delta covariance as a white-noise limit. It is a generalized random field, not a pointwise unit-variance function. Let $B_x(\mathbf{r})$ have independent longitudinal increments withThe smooth-medium limit gives the Stratonovich equation $dE=DEdx+iaE\circ dB$. Conversion to Itô form contributes the local variance drift:The Itô stochastic integral has zero mean, so coherent attenuation in a white-noise random medium satisfiesHere $U_x=e^{ix{\Delta }_{\perp }/(2k)}$ is the Fresnel propagator. Equivalently,For a constant entrance envelope this is simply its constant value times $e^{-\gamma x}$.

Put $B_n(r)={\mu }^{2}A(r)$, the transverse covariance coefficient of the refractive-index fluctuations. The spectrum required here is the transverse spectrum of that longitudinal covariance coefficient, or equivalently the longitudinal-zero-frequency slice of a prelimit stationary medium. Radial symmetry refers to the transverse plane; exact white noise in one direction is not full three-dimensional isotropy. With the Cartesian Fourier convention,angular integration gives $S_F(\nu )=2\pi {\int }_0^{\infty }{B}_n(r)J_0(\nu r)rdr$. Its inverse givesThe printed transform hint mixes two different normalizations. Its radial transform and inverse constitute the self-reciprocal Hankel pair, whose spectrum is $S_H=S_F/(2\pi )$, rather than the unnormalized Cartesian transform also displayed there. Using that Hankel convention consistently yields insteadThese are the same physical solution after converting spectra. The distinction is Fourier-Hankel normalization for an isotropic spectrum; $J_0(0)=1$ fixes the zero-separation covariance in either convention.

Away from zero, the product and composition of the displayed smooth functions are continuous and differentiable. At zero,so $f$ remains continuous. Hence the continuity set is $\boxed{\mathbb{R}}$.

The difference quotient at zero is $f(h)/h=\sin (1/h)$. For $h_n=(\pi /2+2\pi n{)}^{-1}$ it equals one, and for $k_n=(3\pi /2+2\pi n{)}^{-1}$ it equals minus one. Both sequences tend to zero, so no derivative exists there. For $x\ne 0$, the product rule and chain rule giveThus the differentiability set is $\boxed{\mathbb{R}\setminus \{0\}}$.

At zero, $|f(x)-f(0)|=|x|\to 0$, so the function is continuous. At a nonzero $x_0$, approach along rational and irrational sequences; density of the rational numbers and density of the irrational numbers give limiting values $x_0$ and $-x_0$, which are different. Thus the continuity set is $\boxed{\{0\}}$.

At zero, the difference quotient is one for rational nonzero increments and minus one for irrational increments, so the derivative does not exist. Elsewhere differentiability implies continuity, and continuity already fails. Thus the differentiability set is $\boxed{\varnothing }$.

The fundamental theorem of calculus connects the Riemann integral with differentiation. Its first part says that for continuous $h:[a,b]\to \mathbb{R}$, the accumulation function $H(x)={\int }_a^{x}h(t)dt$ is continuous on $[a,b]$ and differentiable inside it, withTo prove this, boundedness of $h$ gives $|H(y)-H(x)|\le \Vert h{\Vert }_{\infty }|y-x|$, proving continuity. For an interior point and a sufficiently small nonzero increment $u$,Its absolute value is at most ${\sup }_{|t-x|\le |u|}|h(t)-h(x)|$, which tends to zero by continuity of $h$ at $x$. This proves the derivative formula. The same estimate gives the appropriate one-sided endpoint derivatives.

The second part says that if $Q$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $Q^{\prime }=h$ with $h$ continuous, thenIndeed $(Q-H{)}^{\prime }=0$, so the mean value theorem makes $Q-H$ constant, and evaluating at the endpoints proves the formula. A useful stronger version only requires $Q^{\prime }$ to be Riemann integrable. For a partition $a=x_0<\cdots <x_N=b$, the mean value theorem on each subinterval provides ${\xi }_j\in (x_{j-1},x_j)$ such thatAs the mesh tends to zero, these tagged Riemann sums converge to ${\int }_a^b{Q}^{\prime }$, proving that version too. Its integrability hypothesis must not be omitted.

An integrable integrand need not give an everywhere differentiable accumulation function. For a counterexample, let $f(t)=0$ for $t<1/2$ and $f(t)=1$ for $t\ge 1/2$. This bounded step function is Riemann integrable, butIts left and right derivatives at $1/2$ are zero and one. Thus $F$ need not be differentiable at every interior point. The first part of the fundamental theorem of calculus does apply at every point where the integrand is continuous.

A derivative need not have a Riemann integral. Define on all of $\mathbb{R}$At zero $f(h)/h=h\sin (1/h^2)\to 0$, so $f^{\prime }(0)=0$; elsewhere the product rule and chain rule giveAt $x_n=(2\pi n{)}^{-1/2}$, this equals $-2/x_n\to -\infty$. Hence $f^{\prime }$ is unbounded on $[0,1]$, whereas a properly Riemann integrable function on a compact interval must be bounded. The requested Riemann integral therefore need not exist, even though $f$ is differentiable everywhere. This example is an unbounded derivative obstruction to Riemann integrability; it does not contradict the theorem's version that explicitly assumes integrability of the derivative.

The function $1/x$ is decreasing and positive. Therefore on $[k,k+1]$ it is at most $1/k$, and the Riemann integral givesSumming from $k=1$ to $n-1$ yieldsFor $n=1$ both sides are zero; for $n>1$ the inequality is actually strict. The same integral test for convergence gives ${\sum }_{k=N}^{n-1}1/k\ge \log (n/N)$.

The Dirichlet boundary condition gives $B=-a^2/6$, so the unique radial solution isIts radial Laplacian is one, it is smooth at the origin, and its boundary value is zero. The uniqueness result in part (i) shows that this is also the only solution among all classical functions on the ball, not merely among spherically symmetric ones.

For a spherically symmetric function in three dimensions, the radial Laplacian is ${\nabla }^{2}u=r^{-2}(r^2{u}^{\prime }{)}^{\prime }$. On $0<r\le a$, integrate $(r^2{u}^{\prime }{)}^{\prime }=r^2$ twice to obtainThe region includes the origin. A classical solution must be regular there, so $A=0$: the $1/r$ term is singular and is not a solution of the stated full-ball problem. Hence the complete family on the ball isThe boundary cases below use this regular family. Allowing punctured-ball solutions would change the problem and the compatibility calculation.

A differential one-form $Pdx+Qdy$ is an exact differential if there is a differentiable function $F$ with $P=F_x$ and $Q=F_y$, so the form equals $dF$. For continuously differentiable $P,Q$, equality of mixed partial derivatives gives the necessary conditionA local differential condition need not by itself give a global potential on a domain with holes; no such issue arises for the explicit primitives here.

For the first form, $P_y=2y=Q_x$ and direct differentiation givesFor the second, $P_y=2y$ but $Q_x=y^2$. These are not equal on any nonempty open subset of the plane, so that form is not an exact differential. Nevertheless,Thus one suitable pair is $f(x,y)=x{e}^y$, $g(y)=y^2{e}^{-y}$. This factorization is smooth even at $y=0$. On regions where $y\ne 0$, its reciprocal factor $e^y/y^2$ is an integrating factor which turns the original form into $df$.