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The primes $2,3,5$ do not divide the elliptic-curve discriminant $-91$, so all three reductions are nonsingular. Count affine solutions and add the point at infinity. At $p=2$ the right side takes values $0,1$ at $x=0,1$, while $y^2+y=0$ for both values of $y$. This gives two affine points, namely $(0,0),(0,1)$.

At $p=3$, the right-side values at $x=0,1,2$ are $0,0,2$. The corresponding numbers of solutions to $y^2+y$ are $2,2,1$, giving five affine points. At $p=5$, the right-side values at $x=0,1,2,3,4$ are $0,3,0,2,0$, with respective solution counts $2,0,2,2,2$, giving eight affine points. Therefore

At $P_1=(0,0)$ the tangent slope is $-2$, so its equation is $y=-2x$. Substitution leaves $x^3=0$; this tangent meets the curve three times at $P_1$. The chord-and-tangent group law therefore givesIn particular $P_1$ has exact order three. The chord through $P_1$ and $P_2=(-2,-4)$ is $y=2x$. Substitution gives $x(x+2)(x-2)=0$, so the third intersection is $(2,4)$. Negating it gives

The projective completion has identity $O=[0:1:0]$. For this Weierstrass equation of an elliptic curve, negation isFor distinct affine points $P,Q$, draw their chord; for $P=Q$, use the tangent with its intersection multiplicity. If the third intersection is $R$, define $P\oplus Q=-R$. A vertical chord pairs $P$ with $-P$ and gives their sum $O$; also $P\oplus O=P$. A vertical tangent at a point satisfying $2y+1=0$ gives $2P=O$.

For a nonvertical line $y=mx+b$, substitution into the cubic givesThus if $P=(x_1,y_1)$ and $Q=(x_2,y_2)$, the sum has coordinatesHere $m=(y_2-y_1)/(x_2-x_1)$ for a chord, and implicit differentiation givesfor a nonvertical tangent. These formulas include the repeated-root case. Associativity can be seen without a lengthy coordinate calculation: a line cuts the cubic in a divisor equivalent to $3(O)$, and $P\mapsto [(P)-(O)]$ identifies the curve with its degree-zero divisor class group. The chord-and-tangent operation becomes addition of divisor classes, hence is associative.

The given relation impliesTake isogeny degrees and cancel the positive integer $\mathrm{deg}\pi$. Multiplicativity of isogeny degree givesAs above, $\mathrm{deg}(1-{\varphi }_i)=\#E_i({\mathbb{F}}_q)$ and $\mathrm{deg}{\varphi }_i=q$. The degree parallelogram law on $\mathrm{End}(E_2)$ now yieldsSubstitution provesEquivalently, their Frobenius traces are negatives of one another. The isogeny degree argument works over the algebraic closure, so it does not wrongly assume that $\pi$ is defined over ${\mathbb{F}}_q$; the stated quadratic-extension hypothesis is compatible with the anticommutation relation.

Write $N=\#E({\mathbb{F}}_q)$. The Hasse bound isWe prove it using isogeny degrees of endomorphisms, with no assumption about a pre-existing Frobenius eigenvalue estimate. The standard elliptic isogeny facts we use are these: isogeny degree is multiplicative under composition; $\mathrm{deg}[m]=m^2$; assigning isogeny degree zero to the zero homomorphism makes isogeny degree a quadratic form, satisfying the degree parallelogram law; and the isogeny degree of a separable isogeny equals the number of its geometric kernel of an isogeny points.

Let $\varphi$ be the $q$-power Frobenius isogeny. It has isogeny degree $q$ and zero differential. Therefore $1-\varphi$ has differential the identity and is a separable isogeny. Its kernel of an isogeny consists exactly of the points fixed by Frobenius, including the identity, soPut $t=q+1-N$. Polarization of the isogeny degree quadratic form gives, for all integers $m,n$,The cross coefficient is fixed by the case $m=n=1$, where the isogeny degree is $N=1-t+q$; the coefficients of $m^2,n^2$ are $\mathrm{deg}1=1$ and $\mathrm{deg}\varphi =q$.

Every isogeny degree is nonnegative. For $n\ne 0$, divide by $n^2$ to obtain $u^2-tu+q\ge 0$ for every rational $u=m/n$. Continuity and density of the rationals imply the same inequality for real $u$. Its minimum, attained at $u=t/2$, is $q-t^2/4$, so $t^2\le 4q$. This is precisely the stated bound. Equality is allowed; a zero value of the quadratic form can occur when the two endomorphisms involved are rationally dependent.

Choose compatible primitive roots of unity ${\zeta }_{p^{n+1}}$, with ${\zeta }_{p^{n+2}}^p={\zeta }_{p^{n+1}}$. Use the slightly faster-converging series permitted by “or otherwise”:The terms have field absolute value $p^{-2n}$, so the partial sums are a Cauchy sequence in ${\overline{\mathbb{Q}}}_p$. We show that an algebraic limit would contradict the growth of cyclotomic extension degrees. The faster weights make the strict conjugate-distance estimate work uniformly also at $p=2$.

Put $L_N={\mathbb{Q}}_p({\zeta }_{p^{N+1}})$. Its degree is $p^N(p-1)$: the translated cyclotomic polynomial ${\Phi }_{p^{N+1}}(1+T)$ has constant term $p$ and reduces modulo $p$ to $T^{p^N(p-1)}$, so is Eisenstein. To verify the reduction, write it as ${\sum }_{j=0}^{p-1}(1+T{)}^{j{p}^N}$ and use characteristic $p$ to reduce this to ${\sum }_{j=0}^{p-1}(1+T^{p^N}{)}^j=T^{p^N(p-1)}$. All roots lie in $L_N$, and its Galois automorphisms are ${\sigma }_a(\zeta )={\zeta }^a$ for units $a$ modulo $p^{N+1}$.

For a nonidentity automorphism, put $r=v_p(a-1)$, choosing a representative with $0\le r\le N$. It fixes the terms with $n<r$. For $n\ge r$, the root ${\zeta }_{p^{n+1}}^{a-1}$ has order $p^{n+1-r}$, so the root-of-unity valuation formula, with $v_p(p)=1$, givesThis is strictly increasing as $n$ increases: successive differences are $2-p^{-(n-r+1)}>0$. Thus the first changed term is the unique term of smallest valuation and cannot cancel. ConsequentlyIn particular no nonidentity automorphism fixes $A_N$, so ${\mathbb{Q}}_p(A_N)=L_N$. Also the distance from $A_N$ to every distinct conjugate is at least $p^{-2N-1/(p-1)}$.

Suppose the Cauchy sequence converged to $A\in {\overline{\mathbb{Q}}}_p$. The tail estimate givesApply Krasner's lemma with $\alpha =A_N$, $\beta =A$. It forces $L_N={\mathbb{Q}}_p(A_N)\subseteq {\mathbb{Q}}_p(A)$ for every $N$. The right-hand side has finite degree, while $[L_N:{\mathbb{Q}}_p]=p^N(p-1)$ is unbounded, a contradiction. ThereforeThis proves algebraic closure of the p-adic numbers is not complete by an explicit algebraic Cauchy sequence and a fully quantified conjugate-separation argument.

Krasner's lemma states: let $K$ be complete for a Non-Archimedean absolute value, let $\alpha$ be a separable algebraic element over $K$, and let $\beta$ be algebraic over $K$, all in a fixed algebraic closure with the extended field absolute value. Ifthen $K(\alpha )\subseteq K(\beta )$. If $\alpha \in K$, the conclusion is automatic and there is no conjugate-distance condition to check.

To prove it, put $F=K(\beta )$ and take a finite splitting field $N/F$ for the minimal polynomial of $\alpha$ over $F$. That is a separable polynomial, so $N/F$ is Galois. The finite extension $F$ is complete, and uniqueness of extensions of an field absolute value from a complete non-Archimedean field makes every $\sigma \in \mathrm{Gal}(N/F)$ an isometry. It fixes $\beta$, soIf $\sigma (\alpha )\ne \alpha$, it is another $K$-conjugate and this inequality contradicts the strict hypothesis. Every such automorphism therefore fixes $\alpha$. The Fundamental theorem of Galois theory gives $\alpha \in F$, provingOnly $\alpha$ needs to be a separable algebraic element: the proof works even when $\beta$ is inseparable, because the splitting field is taken over $K(\beta )$.

The valuation ring is $\mathcal{O}=\{x\in K:|x|\le 1\}$, and $\mathfrak{m}=\{x\in K:|x|<1\}$ is an ideal by the ultrametric inequality. Its complement inside $\mathcal{O}$ consists exactly of elements of field absolute value one, whose inverses also lie in $\mathcal{O}$. Every proper ideal must avoid units and hence lie in $\mathfrak{m}$. Therefore $\mathfrak{m}$ is the unique maximal ideal: $\mathcal{O}$ is a local ring. Also its fraction field is $K$, since every nonzero element outside the ring has its inverse inside it.

If $z\in K$ is integral over $\mathcal{O}$, it satisfies a monic equation $z^n+a_{n-1}{z}^{n-1}+\cdots +a_0=0$ with $|a_i|\le 1$. Were $|z|>1$, the leading term would have strictly larger field absolute value than every other term, so the ultrametric inequality would prevent cancellation to zero. Thus $z\in \mathcal{O}$, proving the valuation ring is integrally closed.

For the ideal criterion, assume the valuation is nontrivial and write $v=-\log |\cdot |$. If its value group is discrete, normalize it to $\mathbb{Z}$ and choose a uniformizer $\pi$ of valuation one. In any nonzero ideal $I$, the nonnegative integer valuations have a least value $m$. Choose $a\in I$ of that value. For every $b\in I$, $v(b/a)\ge 0$, so $b\in a\mathcal{O}$. Hence $I=(a)=({\pi }^m)$ and $\mathcal{O}$ is a principal ideal domain.

Conversely, if $\mathcal{O}$ is a principal ideal domain, its nonzero maximal ideal is $(\pi )$. For every element $x$ of positive valuation, $x\in \mathfrak{m}$ implies $x=\pi y$ with $y\in \mathcal{O}$, so $v(x)\ge v(\pi )>0$. Thus $\gamma =v(\pi )$ is the least positive value. For any value $t$, subtract $\lfloor t/\gamma \rfloor \gamma$. The remainder belongs to the value group and lies in $[0,\gamma )$, so is zero. The value group is therefore $\gamma \mathbb{Z}$, and the valuation is discrete. We have proved the principal ideal criterion for a nontrivial rank-one valuation ring.

The nontriviality qualification matters under the standard definition of discrete valuation, which requires value group isomorphic to $\mathbb{Z}$. With the trivial field absolute value, $\mathcal{O}=K$ is still a principal ideal domain, but the value group is zero and there is no uniformizer. Thus the literal unrestricted equivalence has a field exception. If “discrete” instead includes the trivial value group as a discrete subgroup of $\mathbb{R}$, that exceptional case satisfies the equivalence too.

Take an arbitrary $f\in C({\mathbb{Z}}_p,{\mathbb{Q}}_p)$. Compactness of ${\mathbb{Z}}_p$ bounds its field absolute value, so some $p^{m}f$ takes values in ${\mathbb{Z}}_p$. Apply part (ii) to obtain a continuous $h$ with $h(x+1)-h(x)=p^{m}f(x)$. Then $g=p^{-m}h$ belongs to the same continuous-function space and satisfies $f=g(\cdot +1)-g$.

Translation invariance by one alone now gives $L(f)=L(g(\cdot +1))-L(g)=0$. Since $f$ was arbitrary,This proves translation-invariant linear forms on p-adic continuous functions vanish without assuming that the linear form itself is continuous. A proof using only density of polynomials would not suffice for an arbitrary, potentially discontinuous linear form; surjectivity of the forward difference operator on the entire function space is what is needed.

Let $c_n$ be the Mahler coefficients of $f$. Define its discrete antiderivative byBecause $c_n\to 0$ and the binomial polynomials have field absolute value at most one on ${\mathbb{Z}}_p$, this series converges uniformly to a continuous ${\mathbb{Z}}_p$-valued function. Every summand vanishes at zero. Pascal's identity gives $\left(\genfrac{}{}{0ex}{}{x+1}{n+1}\right)-\left(\genfrac{}{}{0ex}{}{x}{n+1}\right)=\left(\genfrac{}{}{0ex}{}{x}{n}\right)$, so taking differences through the uniformly convergent series proves $g(x+1)-g(x)=f(x)$. For nonnegative integers, induction from $g(0)=0$ gives precisely $g(m)={\sum }_{j=0}^{m-1}f(j)$. Thus this is the required continuous extension, unique because the nonnegative integers are dense in ${\mathbb{Z}}_p$.

The Mahler expansion of $g$ has coefficient zero in degree zero and coefficient $c_{m-1}$ in degree $m\ge 1$. This discrete antidifferentiation on the p-adic integers is the mechanism behind the next part.

The Mahler theorem states that a function $f:{\mathbb{Z}}_p\to {\mathbb{Q}}_p$ is continuous if and only if it has a unique expansionand the expansion converges uniformly. Moreover $\Vert f{\Vert }_{\infty }={\sup }_n|c_n{|}_p$; in particular $f$ takes values in ${\mathbb{Z}}_p$ exactly when all $c_n$ belong to ${\mathbb{Z}}_p$. Here $\left(\genfrac{}{}{0ex}{}{x}{0}\right)=1$ and $\left(\genfrac{}{}{0ex}{}{x}{n}\right)=x(x-1)\cdots (x-n+1)/n!$. These binomial polynomials are continuous and integer-valued on ${\mathbb{Z}}_p$, since they are integer-valued on the dense nonnegative integers. The coefficient condition $c_n\to 0$ therefore makes the series uniformly convergent.

Evaluate at a nonnegative integer $m$. Terms with $n>m$ vanish, giving the finite triangular relation $f(m)={\sum }_{n=0}^m\left(\genfrac{}{}{0ex}{}{m}{n}\right)c_n$. Binomial inversion gives the Mahler coefficientsTo obtain the generating function, multiply the last finite identity by $T^n/n!$ and compare coefficients. Setting $n=j+k$ yieldsThe Mahler coefficient exponential generating function is an identity in ${\mathbb{Q}}_p[[T]]$: each coefficient is a finite sum. It does not assert convergence for every $p$-adic substitution for $T$.

For a finite extension of fields complete for a Non-Archimedean absolute value, use the unique extended field absolute value. The extension is unramified when its residue extension is a separable field extension andEquivalently, it has ramification index one, a separable field extension of residue fields, and no defect. For general nondiscrete valued fields, merely saying that the ramification index is one is insufficient; the displayed degree equality is part of the unramified condition.

Put $n=[L:K]=[k_L:k_K]$. Since $\bar{x}$ generates the residue extension, $1,\bar{x},\dots ,{\bar{x}}^{n-1}$ are linearly independent over $k_K$. Their lifts $1,x,\dots ,x^{n-1}$ are linearly independent over $K$: a nonzero relation could be divided by a coefficient of largest field absolute value, giving an integral relation whose reduction has at least one nonzero coefficient, contrary to residue independence. They therefore form a $K$-basis of $L$.

For any $y={\sum }_{j=0}^{n-1}{a}_j{x}^j$, choose $a_r$ of largest field absolute value. All coefficients of $y/a_r$ belong to ${\mathcal{O}}_K$, and at least one is a unit. Its reduction is a nonzero linear combination of the residue basis, so $y/a_r$ is a unit of ${\mathcal{O}}_L$. This proves the residue-basis norm formula for an unramified extensionHence $y\in {\mathcal{O}}_L$ exactly when every $a_j\in {\mathcal{O}}_K$. We concludeThe last equality also uses the fact that every polynomial in the integral element $x$ is integral. The proof requires neither a uniformizer nor discrete valuation, and applies to every residue generator specified in the question.

Every nonzero P-adic number has a unique decomposition $z=p^{m}u$ with $m\in \mathbb{Z}$ and $u\in {\mathbb{Z}}_p^{\times }$. The group homomorphism kernel of reduction from the unit group onto ${\mathbb{F}}_p^{\times }$ is $1+p{\mathbb{Z}}_p$, the group of principal units. Each nonzero residue class has a unique lift $\omega$ satisfying ${\omega }^{p-1}=1$, by the simple-root Hensel lemma proved in question 3. Uniqueness makes these lifts multiplicative. They form the cyclic group ${\mu }_{p-1}$, and every unit has a unique factorizationThus ${\mathbb{Q}}_p^{\times }=p^{\mathbb{Z}}\times {\mu }_{p-1}\times (1+p{\mathbb{Z}}_p)$.

For odd $p$, the p-adic logarithm and p-adic exponential function give inverse group isomorphisms $1+p{\mathbb{Z}}_p\leftrightarrow p{\mathbb{Z}}_p$. To verify the convergence domain, for $t\in p{\mathbb{Z}}_p$ the p-adic logarithm terms have valuations $n{v}_p(t)-v_p(n)\to \infty$, and the exponential terms have valuations $n{v}_p(t)-v_p(n!)\to \infty$, since $v_p(n!)\le (n-1)/(p-1)$ and $p>2$. In both series every term after the linear one has strictly larger valuation than the linear term. Consequently $\log (1+t)\in p{\mathbb{Z}}_p$ and $\exp (t)\in 1+p{\mathbb{Z}}_p$. The formal identities $\log (vw)=\log v+\log w$ and $\exp (\log v)=v$, $\log (\exp t)=t$ hold on these convergent domains, as follows by multiplying the convergent series or passing to their formal identities termwise.

Choose a generator $\zeta$ of ${\mu }_{p-1}$. An explicit isomorphism isThe inverse reads off the valuation, the root-of-unity unit factor, and $p^{-1}\log (v)$. Both directions are continuous with the standard product topology, so this is also a topological group isomorphism. The odd-prime condition is essential for using all of $1+p{\mathbb{Z}}_p$ as the p-adic logarithm-isomorphism domain.

An inner model of ZF is a transitive class $N$ containing all ordinals and satisfying ZF with the inherited membership relation. The “all ordinals” requirement is important: an arbitrary transitive set model or a rank segment is not an inner model in this sense.

Every such inner model contains the ambient constructible universe $L$. Here is the reason. Inductively $L_{\alpha }^N=L_{\alpha }$. The zero stage is the same empty set. If the stages at $\alpha$ agree, satisfaction for the common set structure $(L_{\alpha },\in )$ is absolute: atomic first-order formulas agree, Boolean operations agree, and quantified variables range over the identical set. The codes of finite first-order formulas are also the same because a transitive model of ZF has the actual natural numbers. Consequently definability with parameters from $L_{\alpha }$ produces exactly the same subsets in both models, so their successor constructible stages agree. At a limit ordinal both constructions take the union of the same earlier stages. This proves agreement at every stage and hence $L\subseteq N$.

The constructible universe theorem provides models of ZF satisfying $V=L$ and the axiom of choice, whenever ZF is consistent. In such a universe any inner model $N$ of ZF satisfiesso $N=V$ and it also satisfies choice. ThereforePassing to $L$ proves the consistency of choice, but the same downward method cannot prove its negation from an arbitrary starting model: it has no possible output when the starting universe is constructible. This is the precise obstruction intended in an inner-model independence argument. It is not a claim that every inner model of every choice universe satisfies choice; symmetric submodels of suitable forcing extensions can fail choice, and that method first changes the ambient universe.

Use the bounded successor characterization of finite ordinals. Let $\mathrm{Ord}(x)$ assert that $x$ is transitive, each of its members is transitive, and any two members are equal or comparable by membership. This is a bounded first-order formula; foundation makes its linear membership order well-founded, so it is precisely the Von Neumann ordinal predicate. Defineand putFor an ordinal, having a greatest member $y$ says exactly $x=y\cup \{y\}$. The displayed definition makes no reference to an infinite inductive set or to $\omega$; all quantifiers are bounded within the candidate and its members. Occurrences of $x=\varnothing$ can also be expressed as $(\forall u\in x)u\ne u$, so no unbounded quantifier is hidden in empty-set notation.

Every usual natural number satisfies this predicate, by induction: it is a finite ordinal, and every nonzero initial segment is a successor. Conversely, suppose an ordinal $x$ is not a usual natural number. Ordinal comparability gives $x\ge \omega$. If $x=\omega$, it has no greatest member, contradicting ${\mathrm{Succ}}_0(x)$. If $x>\omega$, then $\omega \in x$ has no greatest member, contradicting the corresponding bounded requirement on members of $x$. Hence $\mathrm{Nat}(x)$ holds exactly for the usual natural numbers. The axiom of infinity and separation therefore collect this class as the usual set $\omega$. This is a bounded definition of membership in the natural-number class; it does not claim that the infinite set $\omega$ itself can be produced without the infinity axiom.

We prove the Ehrenfeucht-Mostowski theorem by ultraproducts, with no application of Ramsey theorem. Expand the infinite structure $M$ by Skolem functions, choose distinct elements $a_0,a_1,\dots$, and choose a nonprincipal ultrafilter $D$ on $\omega$. Let $I$ be the desired total index order and put $J={\omega }^I$.

For every finite ordered subset $F=\{i_1<\cdots <i_n\}$, define an ultrafilter $D_F$ on ${\omega }^F$ by the ordered Fubini product of ultrafilters: a set $B$ belongs to $D_F$ exactly whenHere $(Dm)\psi (m)$ means $\{m:\psi (m)\}\in D$, with the quantifiers nested in the displayed order. The resulting collection is an ultrafilter, by induction using the ultrafilter laws for negation and conjunction. For $F\subseteq H$, the inverse image of $B\subseteq {\omega }^F$ under coordinate projection belongs to $D_H$ exactly when $B\in D_F$: quantifiers on unused coordinates leave a truth value unchanged.

Consequently there is a well-defined ultrafilter on the Boolean algebra of finite-coordinate cylinders in $J$, declaring a cylinder large by this test. Coherence proves finite intersection closure and ensures that the empty cylinder is not large. Extend its generated filter to an ultrafilter $U$ on all subsets of $J$. Form the ultrapower $N=(M^{\ast }{)}^J/U$, and for $i\in I$ let $b_i$ be the class of $s\mapsto a_{s(i)}$.

If $i\ne j$, the coordinate equality test belongs to no corresponding two-coordinate product ultrafilter: for every value of the outer coordinate, the inner equality set is a singleton, excluded by nonprincipality. Hence the $b_i$ are distinct. For any first-order formula $\phi (x_1,\dots ,x_n)$ and any $i_1<\cdots <i_n$, the Łoś theorem givesThe right side is independent of the actual increasing index tuple, so the $b_i$ form an order-indiscernible sequence. Constant functions embed $M^{\ast }$ elementarily into $N$. Take the Skolem hull of the generators. The Tarski-Vaught test gives an elementary substructure, and transporting terms along an index-order automorphism gives a well-defined automorphism of the hull, with inverse obtained by transporting along the inverse index automorphism. Equality of term representations is preserved by indiscernibility. This is the indiscernible and automorphism conclusion of Ehrenfeucht–Mostowski, obtained entirely from ultrafilters and Łoś's theorem. The ordered Fubini products need not be symmetric; order indiscernibility is exactly what their coherence supplies.

For a Boolean algebra $B$, let $U(B)$ be its set of ultrafilters, equivalently the Boolean homomorphisms $B\to 2$. Regard $U(B)$ as an object of ${\mathbf{Set}}^{\mathrm{op}}$. A homomorphism $k:B\to B^{\prime }$ acts by inverse image of ultrafilters, a set map $U(B^{\prime })\to U(B)$, hence an arrow $U(B)\to U(B^{\prime })$ in the opposite category.

A Boolean homomorphism $h:B\to \mathcal{P}(A)$ gives the functionConversely, a function $v:A\to U(B)$ gives $h(b)=\{a:b\in v(a)\}$. The ultrafilter laws are exactly what makes this map preserve the Boolean operations, including complements and $0,1$. These constructions are inverse and natural, proving the Boolean ultrafilter-power-set adjunctionIts unit is ${\eta }_B(b)=\{u\in U(B):b\in u\}$.

Take $B=\mathcal{P}(\mathbb{N})$, an object in the image of the right adjoint. There is a nonprincipal ultrafilter on $\mathbb{N}$, by extending the cofinite filter in the usual choice setting. The subset $P\subset U(B)$ consisting of all principal ultrafilters is not ${\eta }_B(b)$ for any $b\subseteq \mathbb{N}$: containing the principal ultrafilter at every integer forces $b=\mathbb{N}$, but ${\eta }_B(\mathbb{N})=U(B)$ also contains the nonprincipal ultrafilters. Thus ${\eta }_B$ is not surjective and is not an isomorphism. By the necessary condition proved in the root solution, this adjunction is not idempotent.

The left adjoint is discretization $D$: give the underlying set of $X$ the discrete topology. Write $J$ for indiscretization. Every function $DX\to Y$ is continuous, and every function $X\to JY$ is continuous. The common set of underlying functions supplies the natural bijectionThe discrete-indiscrete adjunction has induced monad $JD$, which assigns the indiscrete topology to the original set. Applying it twice gives the same space, and its multiplication is the identity on that set. Thus the adjunction is idempotent. Its dual comonad $DJ$, discretization, is idempotent as well.

The left adjoint to the morphism-set functor is the free category on independent arrows:Each copy has two objects, their identities, and one nonidentity arrow from the first object to the second. A functor $L(S)\to \mathcal{C}$ is specified by one arbitrary arrow of $\mathcal{C}$ for each $s$, since that arrow specifies the images of both objects as well. This proves the natural adjunction bijection with functions $S\to \mathrm{Mor}(\mathcal{C})$.

The induced monad on sets has $T(S)=S\times \{\text{source identity},\text{generator},\text{target identity}\}$. For a singleton $S$, its multiplication is a function from the nine-element set $T^{2}S$ to the three-element set $TS$, so it cannot be invertible. This adjunction is not idempotent.

Let $J:\mathbf{CompHaus}\hookrightarrow \mathbf{Top}$ be the inclusion. Both categories are locally small categories. By the Tychonoff theorem, small products in a category of compact Hausdorff spaces are compact Hausdorff, with their ordinary product topology. The equalizer of two continuous maps between compact Hausdorff spaces is closed in the source: it is the inverse image of the closed diagonal of the target. Hence it is again compact Hausdorff. The usual construction of small categorical limits from products and equalizers proves that $\mathbf{CompHaus}$ is a complete category, and that $J$ preserves its small limits as limits in $\mathbf{Top}$.

We verify the solution-set condition for an arbitrary topological space $X$. Put $\kappa =|X|$ and $\theta =2^{2^{\kappa }}$. Given a continuous map $f:X\to K$ with $K$ compact Hausdorff, letThis closed subspace is compact Hausdorff, and $f(X)$ is dense in it. The allowed cardinal estimate therefore gives $|Y|\le \theta$. The map $f$ factors through the continuous map $X\to Y$ and the subspace inclusion $Y\hookrightarrow K$.

Choose a fixed set $S$ of cardinality $\theta$. Consider all compact Hausdorff topologies on subsets of $S$, together with all continuous maps from $X$ into the resulting spaces. There are only a set of such topologies and functions. Every $Y$ above is homeomorphic to one of these spaces after relabelling its underlying set. Consequently every $X\to JK$ factors through a member of this set of maps. This proves the solution-set condition, including $X=\varnothing$.

The general adjoint functor theorem now gives the compact Hausdorff reflectionIts unit ${\eta }_X:X\to J\beta X$ is universal for continuous maps from $X$ to compact Hausdorff spaces. For arbitrary $X$, this is a reflection, and the unit is not required to be a topological embedding.