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For a field $F$ of characteristic $p$, the exact sequence $0\to {\mathbb{F}}_p\to F^{\mathrm{sep}}\stackrel{z\mapsto z^p-z}{\to }{F}^{\mathrm{sep}}\to 0$, together with additive normal-basis acyclicity, gives $H^2(\mathrm{Gal}(F^{\mathrm{sep}}/F),{\mathbb{F}}_p)=0$. Consequently every central finite embedding problem with group homomorphism kernel $C_p$ has a continuous weak solution: its pulled-back two-cocycle is a coboundary, whose correcting one-cochain supplies the lift. Over ${\mathbb{F}}_p((t))$, independent Artin–Schreier classes $t^{-m}$, with $m>0$ prime to $p$, allow the lift to be twisted by characters so that inertia maps onto the group homomorphism kernel as well as onto the original quotient. Induction through central order-$p$ quotients therefore realizes every finite p-group as the Galois group of a totally ramified extension with residue field ${\mathbb{F}}_p$.

An element in the image of the preceding differential of a cochain complex. Every coboundary is a cocycle, since consecutive differentials compose to zero.

For a continuous map $f:X\to Y$, precomposition with the induced map on singular chains sends a singular cochain on $Y$ to one on $X$. It commutes with the coboundary and induces the induced map on cohomology.

An element killed by the next differential of a cochain complex. Its class modulo coboundaries is a cohomology class.

The equivalence class of a cocycle in a cochain complex, where adding a coboundary leaves its class unchanged.

A line bundle trivial on each open set is determined by unit-valued transition functions on pairwise intersections. Their cocycle identities make the local trivial bundles glue; changing frames changes the cocycle by a coboundary. The tensor product of sheaves multiplies the transition functions, giving the displayed group isomorphism with Čech cohomology.

For integral singular cochains $a\in C^p(X)$ and $b\in C^q(X)$, define the cup product on an ordered simplex byThe same construction works over any commutative coefficient ring. Expanding the coboundary as the alternating sum of faces gives the Leibniz identityIndeed faces deleting vertices before the split occur in the first sum, faces after the split occur in the second, and the two terms at the common vertex cancel with opposite signs. Thus the product of cocycles is a cocycle. Replacing $a$ by $a+\delta c$ changes the product with a cocycle $b$ by $\delta (c⌣b)$; replacing $b$ by $b+\delta d$ changes the product with a cocycle $a$ by $(-1{)}^p\delta (a⌣d)$. Hence the product is independent of representatives and defines a well-defined product on cohomology classes.

With ${\mathbb{F}}_2={\mathbb{Z}}_2$ coefficients, the real-projective-space cohomology ring isFor $n>m\ge 1$, the pullback of the degree-one generator is either zero or $t$. The second choice would send the zero class $t^{m+1}$ to the nonzero class $t^{m+1}\in H^{m+1}({\mathbb{RP}}^n;{\mathbb{F}}_2)$, contradicting the naturality of the cup product. Thus the generator pulls back to zero, and so do all its positive powers. For $m=0$ the target's reduced cohomology is already zero. Therefore every such map induces zero on reduced mod-two cohomology.

Suppose a continuous $f:S^n\to {\mathbb{R}}^n$ had $f(x)\ne f(-x)$ everywhere. Thenis a continuous odd map to $S^{n-1}$ and descends to $\bar{h}:{\mathbb{RP}}^n\to {\mathbb{RP}}^{n-1}$. For $n\ge 2$, a path from $x$ to $-x$ projects to a loop detected by the degree-one generator of the double cover. Its image lifts to a path from $h(x)$ to $-h(x)$, so is detected by the target's double cover as well. Thus ${\bar{h}}^{\ast }$ is nonzero in degree one, contradicting the preceding calculation. For $n=1$, an odd continuous map from the connected circle to $S^0$ is already impossible. The case $n=0$ is immediate. HenceThis proves the Borsuk-Ulam theorem rather than assuming it.

For a closed cover $A_1,\dots ,A_{n+1}$, assume no member contains an antipodal pair. Apply the just-proved theorem to $F(x)=(d(x,A_1),\dots ,d(x,A_n))$, using any metric on the sphere. Empty sets can be assigned a constant positive coordinate. It gives equal distances at $x,-x$. If either point belongs to an $A_i$ with $i\le n$, zero distance and closedness place both points in it, a contradiction. Hence neither lies in any of the first $n$ sets, and both must lie in $A_{n+1}$, another contradiction. This proves the Lusternik-Schnirelmann-Borsuk theorem for closed sets.

Four closed sets on $S^2$ need not contain an antipodal pair. Take the four unit vertices $v_i$ of a regular tetrahedron centered at zero and define $C_i=\{x:x\cdot v_i\ge x\cdot v_j\text{ for every }j\}$. These closed sets cover the sphere by maximizing the four scalar products. If $x$ and $-x$ belonged to $C_i$, both inequalities would force all four scalar products to be equal. Since $\sum v_i=0$, they would all be zero, and because the vertices span ${\mathbb{R}}^3$, $x$ would be zero. This is impossible on the sphere. Thus these four sets give the antipodal-free closed cover from simplex Voronoi cells counterexample.

For a based space $(X,x_0)$ and $k\ge 1$, the homotopy group ${\pi }_k(X,x_0)$ is the set of based homotopy classes of maps $(I^k,\partial I^k)\to (X,x_0)$, equivalently based maps $S^k\to X$. Concatenating along one coordinate defines its group operation; reversing that coordinate gives the inverse. The group is abelian for $k\ge 2$. The symbol ${\pi }_0(X)$ records path components and is generally only a pointed set. An Eilenberg–MacLane space $K(G,n)$ is a connected CW space with ${\pi }_n=G$ and all other positive homotopy groups zero; for $n>1$ the group $G$ must be abelian.

Projection onto coordinates givesa map to a product is exactly a pair of maps, and a based homotopy to the product is exactly a pair of based homotopies. These identifications respect concatenation. Thus $K(G,n)\times K(H,n)$ has exactly the homotopy groups required of $K(G\times H,n)$. Taking CW models and using the allowed uniqueness of Eilenberg–MacLane spaces gives

Let $p$ be prime. Take $S^{\infty }$ in complex coordinates, with $C_p$ acting by scalar multiplication by a primitive $p$th root of unity. This is a free action. Every map from a finite-dimensional sphere lands in a finite stage and becomes null-homotopic in a sufficiently higher sphere, so $S^{\infty }$ is weakly contractible; the Whitehead theorem for CW complexes makes it contractible. Therefore the infinite lens space $S^{\infty }/C_p$ has contractible universal cover, fundamental group $C_p$, and no higher homotopy: it is a model of $K(C_p,1)$.

Here is the group cohomology computation, including the ring structure. For the cyclic group generated by $g$, the standard free resolution alternates the maps $g-1$ and $N=1+g+\cdots +g^{p-1}$. Applying ${\mathrm{Hom}}_{\mathbb{Z}[C_p]}(-,{\mathbb{F}}_p)$ makes both maps zero, so each cohomology degree is one-dimensional. The period-two identity chain map represents a class $b$ in degree two, and cup multiplication by this periodicity class is an isomorphism from each degree to the degree two higher. For odd $p$, a nonzero degree-one class $a$ has $a^2=0$ by graded commutativity, while $b^j$ and $a{b}^j$ are nonzero by periodicity. For $p=2$ the model is homotopy equivalent to ${\mathbb{RP}}^{\infty }$, whose ring is the infinite version of that computed in question 2. Consequently the mod-p cohomology ring of a cyclic group isThe standard results used are the contractible-cover description of a $K(G,1)$, its identification of singular cohomology with group cohomology, and the cyclic periodic resolution with cup-periodicity just stated.

The product equivalence and the assumed Künneth theorem over the field giveThe tensor product is graded: for odd $p$, $a_1{a}_2=-a_2{a}_1$ and each $a_i^2=0$. The dimensions in degree $d$ are $d+1$: for $p=2$ count $t_1^i{t}_2^{d-i}$; for odd $p$ the generating series is $(1+t{)}^2/(1-t^2{)}^2=(1-t{)}^{-2}$.

If an $n$-dimensional CW complex has exactly one cell in dimension $n+1$, then $C^{n+1}(X;{\mathbb{F}}_p)={\mathbb{F}}_p$. Cellular cohomology is the quotient of the kernel of its outgoing coboundary by the image of the incoming one, a quotient of a subspace of this one-dimensional vector space. ThusThis one-cell bound on next-degree cohomology is valid even with infinitely many higher cells; the printed finite-per-dimension hypothesis is more than it needs.

Now a free finite-group action makes $S^n\to M=S^n/G$ a covering. For $n>1$ the sphere is simply connected, so ${\pi }_1(M)=G$, ${\pi }_i(M)=0$ for $1<i<n$, and ${\pi }_n(M)=\mathbb{Z}$. The covering also identifies every higher homotopy group with that of $S^n$. Use the homotopy-group effect of attaching higher cells: attaching $(i+1)$-cells, $i\ge 2$, preserves homotopy groups below $i$ and quotients ${\pi }_i$ by the subgroup generated by the attaching classes and their fundamental-group translates. Attach a single $(n+1)$-cell along a generator of ${\pi }_n(M)$; its translates do not require additional cells since the original generator already generates the underlying group. This kills ${\pi }_n$ without changing ${\pi }_1$ or the lower groups.

Successively in each higher degree attach cells representing generators of the remaining homotopy group. In the union every higher homotopy group vanishes, since a map and its relevant homotopy have compact domain and are handled at a finite stage. The result is a $K(G,1)$ with exactly one $(n+1)$-cell. No finiteness assumption on the number of later cells is needed for the preceding cochain bound. If desired, one can retain finitely many per degree using the Serre finiteness theorem for homotopy groups: a simply connected CW complex of finite type has finitely generated homotopy groups. At each finite stage the universal cover is simply connected and finite type, since $G$ is finite; choosing finitely many generators then preserves finite type inductively. The first stage needs just the stated ${\pi }_n(S^n)=\mathbb{Z}$.

If $G=C_p\times C_p$, the resulting $K(G,1)$ would have $\dim H^{n+1}\le 1$ by its single cell, but its already computed cohomology has dimension $n+2>1$ in that degree. This contradiction provesThis is the free sphere actions exclude elementary abelian subgroups of rank two obstruction.

Use the short exact sequence of coefficient groupsThe groups of singular chains are free, so applying the cochain construction gives a short exact sequence of cochain complexes. Its long exact sequence in cohomology has the required terms and reduction map $r_{\ast }$. Define the Bockstein homomorphism to be its connecting homomorphism.

Explicitly, represent $x$ by a mod-$p$ cocycle $a$ and lift it to an integral singular cochain $A$. Since $a$ is a cocycle, $\delta A$ is divisible by $p$. ThenThis is exactly the lift-and-coboundary construction for the coefficient sequence above: the mod-$p^2$ coboundary of $A$ lies in the image of $j$. The quotient cochain is a cocycle, since ${\delta }^{2}A=0$ and integral cochains have no $p$-torsion. Replacing $A$ by another lift adds $pC$ and changes the quotient by $\delta C$ modulo $p$; replacing the representative by a coboundary likewise leaves its cohomology class unchanged. Thus the definition is independent of choices.

A degree-zero cocycle is a function on points constant on each path component. Choose an integer representative for its value on each component. This produces an integral zero-cocycle lift, with zero coboundary. Hence the Bockstein vanishes on $H^0(X;{\mathbb{F}}_p)$ for every space.

For a nonzero example in every positive degree $i$, take the Moore space $X=S^i{\cup }_p{e}^{i+1}$, attaching the cell by a map of mapping degree $p$. Its positive-degree integral cellular cochain complex has the differential $\mathbb{Z}\stackrel{p}{\to }\mathbb{Z}$ from degree $i$ to $i+1$; when $i=1$, the preceding differential from degree zero is zero. Modulo $p$ the displayed differential vanishes, and $H^i(X;{\mathbb{F}}_p)=H^{i+1}(X;{\mathbb{F}}_p)={\mathbb{F}}_p$. Lift the degree-$i$ generator to the cellular cochain with value one. Its coboundary has value $p$, so dividing by $p$ gives the generator in degree $i+1$. Naturality of cellular and singular cohomology with coefficient sequences identifies this with the Bockstein on a cyclic Moore space. Therefore $\beta$ is nonzero, indeed an isomorphism, in the requested degree for every prime $p$.

Finally take integral lifts $A,B$ of cocycles representing $x\in H^i(X;{\mathbb{F}}_p)$ and $y\in H^j(X;{\mathbb{F}}_p)$. Write $\delta A=p{A}_1$ and $\delta B=p{B}_1$. The cochain cup product obeys the graded Leibniz identity, soDivide by $p$, reduce modulo $p$ and pass to cohomology. This proves the Bockstein derivation rule

We prove the Birkhoff–Grothendieck theorem by induction on the rank of the vector bundle $E$. Rank zero is immediate; rank one is the classification $\mathrm{Pic}({\mathbb{P}}_k^1)\cong \mathbb{Z}$ established above.

First there is a largest integer $a$ with $H^0(E(-a))\ne 0$. To see existence and boundedness directly, trivialize $E$ on the two standard affine charts: the associated finite projective modules over the principal ideal domains $k[t]$ and $k[t^{-1}]$ are free. Let $A(t)\in G{L}_r(k[t,t^{-1}])$ be its transition matrix, so a section of $E(-a)$ is a pair satisfyingIf $a$ exceeds the largest exponent in any entry of $A$, the right side has only strictly negative powers; the left side has nonnegative powers. Hence both vanish. For $a$ sufficiently negative, take a nonzero constant vector $p$ and put $q=t^{a}A(t{)}^{-1}p$; all its exponents are nonpositive, so it supplies a nonzero section. Thus a largest $a$ exists.

Such a section gives a nonzero map $\mathcal{O}(a)\to E$. It is injective, since it is nonzero generically and the source is torsion-free. Saturate its image. The resulting line subbundle is $\mathcal{O}(a+d)$ for the degree $d\ge 0$ of the effective divisor of zeros of the original map. If $d>0$, its inclusion in $E$ would give a nonzero section of $E(-a-d)$, contradicting maximality. Therefore $d=0$ andhas a locally free quotient. By induction, $Q\cong {⨁}_j\mathcal{O}(b_j)$.

Twist this exact sequence by $\mathcal{O}(-a-1)$. The outer groups $H^0(\mathcal{O}(-1))$ and $H^1(\mathcal{O}(-1))$ are zero by the Čech cohomology of twists on the projective line. The long exact sequence in sheaf cohomology therefore givesThe first is zero by the choice of $a$, so each $H^0(\mathcal{O}(b_j-a-1))$ vanishes. Thus $b_j\le a$ for every $j$. The extension class lies insince $a-b_j\ge 0$. The identification of this Ext functor with cohomology can be seen without additional duality: local splittings of a vector-bundle extension differ by a Čech cocycle in $\mathcal{H}om(Q,\mathcal{O}(a))$, and a coboundary changes them to compatible splittings. Vanishing therefore gives a global splitting. Consequentlycompleting the induction over any field $k$.

The degrees can be arranged in nonincreasing order and their multiset is unique. Indeed, for any such decomposition,so the dimensions of the twisted global sections recover the number of summands of every degree.

The Weak Mordell-Weil theorem asserts that $E(K)/mE(K)$ is finite for an elliptic curve over a number field $K$ and an integer $m\ge 2$. The role of (i) is to make division by $m$ possible in the local formal group of an elliptic curve whenever the residue characteristic does not divide $m$.

Let $v$ be a finite place with good reduction of an elliptic curve and $v\nmid m$. The kernel of reduction of an elliptic curve is parametrized by $t=-x/y$ in the maximal ideal of the local valuation ring. Multiplication by $m$ is represented by an integral formal power seriesIts linear coefficient is a local unit. Part (i) supplies an integral compositional inverse. Both series converge on the maximal ideal, so multiplication by $m$ is a bijection on the reduction kernel, over the local field and over every finite unramified extension.

This implies unramified division torsors at good primes. For $P\in E(K)$, first divide its reduction by $m$ over the algebraic closure of the residue field. Multiplication by $m$ on the reduced curve is surjective and separable, so the chosen division point is defined over some finite residue extension. Smooth lifting gives a point $Q_0$ over the corresponding unramified local extension with the required reduction. The difference $P-m{Q}_0$ lies in the reduction kernel; the inverse series supplies $Q_1$ with $m{Q}_1=P-m{Q}_0$. Thus $Q=Q_0+Q_1$ divides $P$ over an unramified extension. Applying the same reasoning to all reduced $m$-torsion points lifts all of $E[m]$, so every point of $[m{]}^{-1}P$ lies in the maximal unramified extension. This excludes ramification outside a fixed finite set $S$ consisting of primes of bad reduction and primes over $m$.

For completeness, the remaining global finiteness step in the Kummer-theoretic proof of the weak Mordell-Weil theorem is as follows. Choose $Q\in E(\overline{K})$ with $mQ=P$. The cocycle $\sigma \mapsto \sigma Q-Q$ defines the Kummer map of an elliptic curveChanging $Q$ by an $m$-torsion point changes the cocycle by a coboundary. A zero class makes $Q$ rational after that change, so the map is injective.

Let $L=K(E[m])$, fixed independently of $P$. Since the torsion is rational over $L$, the cocycle restricted to its Galois group is a homomorphism into $E[m]$. Consequently $L(Q)/L$ is Galois of degree at most $m^2$, contains the entire division fibre, and is unramified outside $S$. Hence $[L(Q):K]\le m^2[L:K]$. There are only finitely many extensions of bounded degree unramified outside a fixed finite set: local extensions of bounded degree have bounded field discriminant exponents, and the Hermite–Minkowski theorem then applies to the bounded global field discriminants. Their finite Galois compositum $M$ contains every such division fibre; each $L(Q)$ is stable over $K$ because it contains $E[m]$ and the whole fibre above the rational point $P$. All the Kummer classes consequently factor through the fixed finite group $\mathrm{Gal}(M/K)$ with values in the fixed finite group $E[m]$. There are only finitely many such cocycles, provingThus (i) provides the local division step that bounds ramification; the global arithmetic finiteness completes the argument.

Write $k={\mathbb{F}}_p$. The Nottingham group $\mathcal{N}$ is the set of formal series $f(t)=t+{\sum }_{i\ge 2}{a}_i{t}^i$ over $k$, with ordinary composition $f\circ g$ as multiplication and the $t$-adic topology. The identity is $t$. A compositional inverse exists uniquely by solving its coefficients successively, since the linear coefficient is $1$.

Equivalently $\mathcal{N}$ is the group of continuous $k$-automorphisms of $k((t))$ whose action on $(t)/(t^2)$ is the identity. Such an automorphism is determined by the image of $t$, which is $t+O(t^2)$. To identify this with ordinary series composition without reversing multiplication, send $f$ to the automorphism $h(t)\mapsto h(f^{-1}(t))$. Its composite with the automorphism for $g$ is the one for $f\circ g$.

The depth filtration isEach ${\mathcal{N}}_m$ is open and normal, and the coefficient of $t^{m+1}$ identifies ${\mathcal{N}}_m/{\mathcal{N}}_{m+1}$ with $(k,+)$. Thus $[\mathcal{N}:{\mathcal{N}}_m]=p^{m-1}$, and $\mathcal{N}$ is the inverse limit of its finite p-group truncations.

Here is a local-field proof that every finite p-group embeds in $\mathcal{N}$. We first establish the realization step rather than assume the Nottingham embedding itself. Put $F=k((t))$ and $\Gamma =\mathrm{Gal}(F^{\mathrm{sep}}/F)$, and let $I⊴\Gamma$ be the inertia subgroup, the group homomorphism kernel of the action on the algebraic closure of the residue field. We use the standard additive normal basis theorem consequence $H^j(\Gamma ,F^{\mathrm{sep}})=0$ for $j>0$. It follows by taking direct limits of the finite-Galois normal-basis modules, whose additive cohomology vanishes. The Artin-Schreier sequencethen givesThe classes $t^{-m}$ for positive $m$ prime to $p$ are linearly independent: a nonzero finite linear combination has a most negative valuation prime to $p$, whereas a negative valuation of $z^p-z$ is divisible by $p$. Thus there are infinitely many independent continuous characters $\Gamma \to C_p$.

Induct on the order of a finite p-group $E$. Choose a central subgroup $C\cong C_p$ and put $Q=E/C$. By induction there is an onto map $\varphi :\Gamma \to Q$ with $\varphi (I)=Q$, corresponding to a totally ramified $Q$-extension. The central extension $1\to C\to E\to Q\to 1$ is specified by a two-cocycle. Its pullback along $\varphi$ is a coboundary because $H^2(\Gamma ,{\mathbb{F}}_p)=0$. Correcting a set-theoretic section by that coboundary gives a continuous homomorphism ${\rho }_0:\Gamma \to E$ lifting $\varphi$.

We must make the lift onto on inertia, not merely onto on the whole Galois group. Let $A=I\cap \ker \varphi$. On $A$, the lift ${\rho }_0$ takes values in $C$. Characters vanishing on $A$ form a finite-dimensional space: $\Gamma /A$ has a finite subgroup $I/A\cong Q$ and procyclic quotient $\Gamma /I\cong \hat{\mathbb{Z}}$, so a character is determined by finitely many values on generators of these two pieces. The infinite-dimensional character space therefore has infinitely many distinct restrictions to $A$. Choose a character $\chi :\Gamma \to C$ whose restriction to $A$ does not cancel ${\rho }_0{|}_A$, and set $\rho (\gamma )=\chi (\gamma ){\rho }_0(\gamma )$. Centrality of $C$ makes this another homomorphism lifting $\varphi$. Its image on $A$ is nontrivial and hence all of $C$, while its image on $I$ maps onto $Q$. Therefore $\rho (I)=E$. The fixed field of $\ker \rho$ is the required totally ramified finite Galois extension $K/F$ with group $E$. This proves the realization step for every finite p-group.

Since the residue field of $K$ is still $k$, choosing a uniformizer $u$ identifies $K$ with $k((u))$. Indeed successive subtraction of its residue coefficient and division by $u$ gives each element of its valuation ring a unique convergent expansion ${\sum }_{i\ge 0}{b}_i{u}^i$ with $b_i\in k$, and fractions give the Laurent expansions. Each element of $E$ preserves the valuation and fixes $k$, so sends $u$ to $a_{1}u+a_2{u}^2+\cdots$. The linear coefficient is a homomorphism $E\to k^{\times }$. Since $E$ is a $p$-group and $|k^{\times }|=p-1$, that homomorphism is trivial. Thus the faithful Galois action lies in $\mathcal{N}(k)$, completing the embedding proof.

An infinite profinite group is hereditarily just infinite when every open subgroup is infinite and every nontrivial closed normal subgroup of each such open subgroup is open. We sketch the normal-subgroup argument, giving the depth calculation in odd characteristic explicitly. For series of depths $r,s$,where $[f,g]=f^{-1}\circ g^{-1}\circ f\circ g$. The coefficient follows by comparing $f\circ g$ and $g\circ f$ in degree $r+s+1$: their cross terms are $ab(r+1)$ and $ab(s+1)$.

Let $H$ be open and $1\ne K⊴H$ closed. Choose $m$ with ${\mathcal{N}}_m\le H$ and choose $f\in K$ of depth $r$. For every $j\ge m$ with $j\not\equiv r(\mathrm{mod}p)$, commutation with $t+b{t}^{j+1}$ supplies an element of $K$ of depth $r+j$, with any prescribed nonzero leading coefficient by varying $b$. Thus every sufficiently high layer except possibly depths congruent to $2r$ modulo $p$ occurs in $K$.

When $p$ is odd, choose one available depth $u\ge r+m$ with $u\not\equiv 2r,r(\mathrm{mod}p)$. Such a residue exists even for $p=3$. An element $h\in K$ of that depth can be commuted with $t+b{t}^{k-u+1}$ for every sufficiently large missing depth $k\equiv 2r(\mathrm{mod}p)$. The new leading coefficient contains $2u-k\equiv 2(u-r)\ne 0$, so those missing layers occur too. Hence $K$ contains an element with any chosen coefficient in every depth $k\ge M$, for some $M$.

Successive coefficient correction now proves ${\mathcal{N}}_M\le K$: match the coefficient of an arbitrary target in depth $M$ by an element of $K$, correct the error in depth $M+1$, and continue. The partial products converge $t$-adically to the target and remain in $K$, which is closed. Thus $K$ is open. This proves hereditary just infiniteness for odd $p$.

For characteristic two, the leading coefficient alone misses even depths. The standard characteristic-two normal-closure collection lemma supplies the needed extra input: for every $m\ge 1$ and every nonidentity $f\in \mathcal{N}({\mathbb{F}}_2)$, the closed subgroup generated by the conjugates $f^g$ with $g\in {\mathcal{N}}_m$ contains ${\mathcal{N}}_M$ for some $M$. Its proof retains the next coefficients in pairs of group commutators to fill the parity gaps, then uses the same successive coefficient correction. Apply this lemma to $f\in K$ and ${\mathcal{N}}_m\le H$: every such conjugate lies in $K$, so $K$ contains the tail and is open. This states the additional collection result used in the characteristic-two sketch; the odd-prime leading-term argument is not being applied there. Since every open subgroup contains a tail and is infinite, the Nottingham group is hereditarily just infinite for every prime $p$.

A topological group module $A$ is a topological abelian group with a jointly continuous action of $G$ by group automorphisms. Its continuous cochains with topological coefficients arewith pointwise addition. The inhomogeneous differential isFor $n=0$, this means $(da)(g)=ga-a$. Continuity of multiplication and of the action ensures that $d$ preserves continuous cochains. Expanding $d(dc)$ gives cancellation in pairs: the two orders of multiplying neighbouring arguments agree by associativity, and the two ways of applying the action agree by its action law. Thus $d^2=0$. The cocycles are $Z^n=\ker (d:C^n\to C^{n+1})$ and the coboundaries are $B^n=\mathrm{im}(d:C^{n-1}\to C^n)$, with $B^0=0$. DefineFor general topological coefficients this is an algebraic quotient of the continuous cochains; its optional quotient topology need not be Hausdorff. For discrete coefficients it is the usual continuous cohomology of a profinite group. In degree zero it is the subgroup of invariant elements.

Give the discrete prime field ${\mathbb{F}}_p$ its trivial $G$-action. The one-cocycle condition becomes $c(gh)=c(g)+c(h)$ and every one-coboundary is zero. HenceEvery such group homomorphism kills powers and group commutators, hence their closure $\Phi (G)$, and every continuous functional on the quotient pulls back. If $G$ is topologically finitely generated, $V=G/\Phi (G)$ is a finite elementary abelian group. To verify finiteness without assuming it, the closure of a subgroup generated by $r$ elements of this abelian exponent-$p$ quotient has at most $p^r$ elements; a finite subgroup is already closed. Thus $V$ is a finite-dimensional vector space. Question 2 says that a generating family of $G$ is minimal precisely when its images give a basis of $V$. The dual space has the same finite dimension. This proves

There is a genuine missing hypothesis if the printed assertion is read for arbitrary pro-p groups and $d(G)$ means the least cardinality of any dense generating family. Here is an explicit infinite-rank failure of the first-cohomology generator formula. Let $I=\{0,1{\}}^{\mathbb{N}}$ and $G={\prod }_{r\in I}{\mathbb{F}}_p$. Enumerate the finite binary strings as $s_1,s_2,\dots$ and define $g_n(r)=1$ if $s_n$ is a prefix of $r$, and $0$ otherwise. For finitely many distinct sequences $r_1,\dots ,r_k$, choose long enough prefixes to distinguish each individually. The corresponding $g_n$ project onto the coordinate unit vectors of ${\mathbb{F}}_p^k$, so $⟨g_n:n\ge 1⟩$ surjects onto every finite coordinate projection and is dense in the product topology. No finite family can generate densely, since a finitely generated abelian exponent-$p$ subgroup is finite and closed. Thus $d(G)={\aleph }_0$. On the other hand, continuity of a linear functional to ${\mathbb{F}}_p$ means its kernel contains a basic neighbourhood, so it depends on finitely many coordinates. The coordinate projections are consequently a basis of the continuous dual space, givingThe unrestricted infinite-rank formula is false. A convention requiring a generating basis tending to the identity is different from the printed unrestricted topological generator count; the finite-rank identity proved above has no such ambiguity.

For finite $X$, a free pro-p group $F_p(X)$ is characterized by the universal property that every map $X\to P$ to a pro-p group extends uniquely to a continuous group homomorphism $F_p(X)\to P$. It is the pro-p completion of the abstract free group on $X$: each map to a finite $p$-group factors through one of the defining finite quotients, and passing to the inverse limit gives the extension to any pro-p group; uniqueness follows from density. This unrestricted universal property can also define the completion on an arbitrary set $X$. When using an infinite basis required to tend to the identity, one instead imposes the universal property only for maps tending to the identity. Only the finite-rank version is needed for the following argument.

A finite pro-p presentation is an isomorphismThere are finitely many generators and finitely many relators, and the denominator is their closed normal closure. The relators are elements of the free pro-p group; they may be limits of ordinary words. That distinction is important when writing conjugation relations below.

Now suppose $K{⊴}_{o}G$ has a finite pro-p presentation with generators $k_1,\dots ,k_a$ and relators $R_1,\dots ,R_t$. The quotient $Q=G/K$ is a finite p-group. Choose a finite finite pro-p presentation $Q=⟨{\bar{y}}_1,\dots ,{\bar{y}}_b\mid S_1,\dots ,S_c⟩$. Such a presentation exists, for example by using its entire finite multiplication table. Lift the quotient generators to $y_j\in G$. For every $i,j$, choose elements $W_{ij},V_{ij}$ of the free pro-p group on the $x_i$ that map respectively to $y_j^{-1}{k}_i{y}_j$ and $y_j{k}_i{y}_j^{-1}$ in $K$. Choose $U_{\ell }(x)$ mapping to $S_{\ell }(y)$, which lies in $K$. Form the finite finite pro-p presentationAll listed relations hold in $G$, giving a continuous surjective group homomorphism $P\to G$: its compact image contains $K$ and maps onto $Q$.

Let $X$ be the closed subgroup of $P$ topologically generated by the $x_i$. Both conjugation directions were included, so each $y_j$ normalizes $X$. The $x_i$ also normalize it, and its closed normalizer contains the generators of $P$. Thus $X⊴P$. The relations $R_h$ give a continuous surjection $K\to X$. Its composite with $X\to K$ induced by $P\to G$ fixes every $k_i$, and hence is the identity. Therefore the surjection $K\to X$ is injective as well, so $X\cong K$. On quotienting by $X$, the conjugation and kernel relations disappear, and the remaining presentation is exactly $⟨y_1,\dots ,y_b\mid S_1,\dots ,S_c{⟩}_{\mathrm{pro}\text{-}p}=Q$. Thus $P/X\to G/K$ is an isomorphism. The kernel of $P\to G$ lies in $X$ and meets $X$ trivially, so is trivial. Consequently $P\cong G$, proving

Work on an affine open subset $U=\mathrm{Spec}A$ and write $\mathcal{F}{|}_U=\tilde{M}$ and $\mathcal{E}{|}_U=\tilde{N}$ using part (ii). We first show that any section $e\in \Gamma (U,\mathcal{E})$ lifts to $\Gamma (U,\mathcal{G})$.

A surjection of sheaves supplies local lifts $g_i$ on a finite cover by principal open subsets $D(s_i)$. Their differencessatisfy $c_{ij}+c_{jk}=c_{ik}$ on triple intersections. We spell out why this Čech cocycle is a coboundary. Choose a common sufficiently large power $n$ so that $s_j^n{c}_{ij}$ can be represented in $M_{s_i}$ for every pair. Increasing $n$ also kills the finitely many localization errors in the cocycle identities: after extending $s_k^n{c}_{ik}$ to $M_{s_i}$ and $s_k^n{c}_{jk}$ to $M_{s_j}$, their difference in $M_{s_i{s}_j}$ is $s_k^n{c}_{ij}$. This follows because any error which vanishes after localizing at $s_k$ is killed by a power of $s_k$; there are only finitely many indices.

The $D(s_i^n)$ still cover $U$, so the ideal generated by the $s_i^n$ is $A$. Choose $a_i\in A$ with ${\sum }_i{a}_i{s}_i^n=1$ and putusing the extended representatives just chosen. On an overlap, the cocycle identity givesThus $g_i-b_i$ agree and glue to a lift of $e$. This is the degree-one argument behind the exactness of the unit-ideal localization Čech complex.

The same reasoning applies to every $D(a)\subseteq U$. Hence we have short exact sequences of modules of sections on $U$ and on $D(a)$. Localize the first sequence at $a$ and compare them. The maps on the outer modules are the isomorphisms from part (ii):An elementary diagram chase makes the middle map an isomorphism: subtract a lift from the right to reduce surjectivity to the left, and use injectivity on the left to prove injectivity. These isomorphisms identify $\mathcal{G}{|}_U$ with the quasi-coherent sheaf associated with $\Gamma (U,\mathcal{G})$. Doing this on every affine open subset provesThe proof even works without the Noetherian assumption.

Unless a coefficient group is displayed, use integral singular cohomology. The standard CW complex structure on infinite-dimensional real projective space has one cell in each nonnegative dimension. Its cellular chain complex has boundary $d_j=2$ for positive even $j$ and $d_j=0$ for odd $j$. The cellular cohomology differential ${\delta }^j$ is therefore zero for even $j$ and multiplication by two for odd $j$. ConsequentlyMore generally, for an abelian group $A$, the positive odd groups are $A[2]=\{a:2a=0\}$ and the positive even groups are $A/2A$. In particular $H^j({\mathbb{RP}}^{\infty };{\mathbb{Z}}_2)={\mathbb{Z}}_2$ in every nonnegative degree.

For the required Bockstein homomorphism, use the short exact sequenceHere $m\ge 2$; for $m=1$ all three coefficient groups are zero. Since singular chains are free abelian groups, applying cochains gives a short exact sequence of cochain complexes. Its connecting homomorphism defines $\beta$ and its long exact sequence is precisely the required one, with the other maps induced by $\iota$ and $q$.

Explicitly, represent a class by a cocycle $u\in C^n(X;{\mathbb{Z}}_m)$ and choose a lift $\tilde{u}\in C^n(X;{\mathbb{Z}}_{m^2})$. Since $q(\delta \tilde{u})=0$, there is a unique cochain $v$ with $\iota (v)=\delta \tilde{u}$. Injectivity of $\iota$ and ${\delta }^2=0$ show $\delta v=0$. DefineChanging the lift by $\iota (w)$ changes $v$ by the coboundary $\delta w$. Changing the representative $u$ by a coboundary can be lifted by a coboundary as well and leaves the resulting class unchanged. Thus this is a well-defined group homomorphism, and the standard cochain lifting argument gives exactness.

Compute the Bockstein homomorphism on infinite-dimensional real projective space using its cellular cohomology complex. A generator with coefficients ${\mathbb{Z}}_2$ is represented by $1$ in degree $n$, lifted to $1\in {\mathbb{Z}}_4$. Its coboundary is $0$ for even $n$ and $2$ for odd $n$. Dividing via $\iota (1)=2$ givesThus the odd-degree maps are isomorphisms. The comparison between cellular cohomology and singular cohomology is natural with respect to coefficient maps, so this computes the same connecting homomorphism constructed above.

For $U_{i_0\dots i_q}=U_{i_0}\cap \cdots \cap U_{i_q}$, the Čech cochain complex has groupsTerms in ${\delta }^2$ cancel in pairs, and Čech cohomology is $\ker \delta /\mathrm{im}\delta$. A degree-zero cocycle is exactly a family of compatible local sections. The sheaf gluing axiom gives their unique global section, proving ${\check{H}}^0(\mathcal{U},\mathcal{F})\cong \mathcal{F}(X)$. If the cover has affine finite intersections, a quasi-coherent sheaf has no higher cohomology on those intersections by vanishing of quasi-coherent cohomology on an affine scheme. The acyclic cover theorem then identifies all Čech groups with sheaf cohomology. In particular, a finite affine open cover of a separated variety has this property.

For the sheaf of units of the structure sheaf, a multiplicative degree-one cocycle is a family $g_{ij}\in {\mathcal{O}}_X^{\ast }(U_i\cap U_j)$ satisfying $g_{ij}{g}_{jk}=g_{ik}$, with $g_{ji}=g_{ij}^{-1}$. It glues trivial rank-one free modules into an invertible sheaf. Changing the local frames multiplies $g_{ij}$ by a coboundary, and two sets of transition data give isomorphic line bundles precisely when their cocycles differ this way. Tensoring line bundles multiplies their cocycles. Thus line bundles trivialized by an open cover give the group isomorphismFor the remaining arguments, work over the algebraically closed ground field. On the irreducible variety $V$, put $\mathcal{Q}={\mathcal{K}}^{\ast }/{\mathcal{O}}_V^{\ast }$, a quotient of sheaves of abelian groups. A global section of $\mathcal{Q}$ is locally represented by rational functions $f_i$ whose ratios are regular units. The corresponding unit cocycle $f_j/f_i$ defines an invertible sheaf. A single global rational function has trivial cocycle. Conversely, every line bundle has a nonzero rational section: choose a nonzero vector in its one-dimensional fibre at the generic point and express it in local frames. This supplies such local $f_i$. If the associated line bundle is trivial, changing frames makes all $f_i$ restrictions of one rational function. Thereforeis exact. This is the Cartier-divisor description of the Picard group. The sheaf of nonzero rational functions on an irreducible variety is flasque, since all restrictions between nonempty open sets are the identity on $k(V{)}^{\ast }$. Apply the long exact sequence in sheaf cohomology to $1\to {\mathcal{O}}_V^{\ast }\to {\mathcal{K}}^{\ast }\to \mathcal{Q}\to 1$. Since $H^1(V,{\mathcal{K}}^{\ast })=0$, its connecting map has exactly the cokernel just computed, provingFinally the Segre description of a smooth quadric surface identifies $V$ with ${\mathbb{P}}^1\times {\mathbb{P}}^1$. The two rulings of a smooth quadric surface have classes $F_1,F_2$ generating $\mathrm{Pic}(V)\cong {\mathbb{Z}}^2$. For completeness, remove one line in each ruling: the remaining chart is the affine plane, with factorial coordinate ring $k[s,t]$ and trivial divisor class group. The localization sequence for the divisor class group makes $F_1,F_2$ generators, and their degrees on the two ruling lines prove independence. The hyperplane class, and hence the conic $C$, has bidegree $(1,1)$. By Picard-group localization on a smooth variety, the Picard group of a smooth affine quadric surface isExplicitly, the restriction of ${\mathcal{O}}_V(1,0)$ is nontrivial: if it were trivial on $U$, its rational trivialization would have divisor supported on $C$, forcing $(1,0)$ to be an integer multiple of $(1,1)$, which is impossible.

A singular $n$-cochain with coefficients in an abelian group $A$ is a homomorphism from the singular chain group $C_n(X;\mathbb{Z})$ to $A$. It is specified by its values on all singular simplices, without a finite-support restriction. Its coboundary evaluates on the alternating sum of the faces. The resulting cochain complex computes singular cohomology.