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Use ${\beta }_{\mathrm{th}}=1/(k_{B}T)$ for inverse temperature to distinguish it from the order-parameter exponent. The decaying two-point correlation function needed for the response identity is connected:Away from the critical point in a short-range Ising model pure phase, the correlation length controls the exponential long-distance falloff, $G(r)\sim A(r)e^{-|r|/\xi }$ with a possible algebraic prefactor. The full two-spin expectation below the transition instead tends to $M^2$. The spin-mixture contribution to zero-field susceptibility explains why a symmetric mixture is unsuitable for defining a finite connected correlation length in the ordered phase.

Let $F=-{\beta }_{\mathrm{th}}^{-1}\log Z$ be the total Helmholtz free energy. Take magnetization and susceptibility per site, soDifferentiating the Boltzmann weights at fixed temperature givesTranslation invariance then yields the correlation-function susceptibility sum ruleTotal magnetization and total susceptibility multiply these expressions by $N$; per physical volume they have an additional factor $a^{-D}$ relative to the per-site convention. These normalizations do not change the critical powers.

For the subsequent scaling requests, let $F_s$ denote singular free energy per site; the fixed extensive normalization is suppressed. The labels $+$ and $-$ denote $t>0$ and $t<0$, with distinct scaling functions and amplitudes on the two sides. Below the transition a one-sided derivative at $h=0$ selects a spontaneous-magnetization branch. Writing $x=h/|t{|}^{\Delta }$, differentiation of the scaling hypothesis for critical phenomena givesThe field-independent $I_{\pm }$ makes no contribution to these field derivatives, although it can change heat-capacity amplitudes. Assuming nonzero leading amplitudes and the indicated pure-phase limits, comparison with the exponent definitions givesThis is the Rushbrooke scaling relation. A temperature-independent critical isotherm requires the large-$|x|$ magnetization scaling function to grow as $\mathrm{sgn}(x)|x{|}^{\beta /\Delta }$: otherwise $M_s=|t{|}^{\beta }{\mathcal{M}}_{\pm }(x)$ would retain a temperature power as $t\to 0$ at fixed small $h$. Therefore $1/\delta =\beta /\Delta$. Since the preceding two equalities also give $\Delta =\beta +\gamma$, the Widom scaling relation follows:For the hyperscaling relation, additionally assume that one correlation volume supplies an order-one singular free-energy contribution and that there is a single isotropic correlation length, with no dangerously irrelevant coupling spoiling this argument. Then $F_s\sim {\xi }^{-D}\sim |t{|}^{D\nu }$. Two temperature derivatives give the singular heat capacity, soHyperscaling is an additional physical assumption, not a consequence of the two-variable equation alone, and generally fails above the upper critical dimension.

Finally the correlation scaling function $f_G(x)$ falls exponentially, up to algebraic factors, for $x\to \infty$ in a massive short-range pure phase. At $x\to 0$ it approaches a finite nonzero value in the scaling regime. Approximating the lattice sum at distances much larger than $a$, and denoting the area of the unit sphere by $S_{D-1}$, givesIf $\eta <2$ and the integral has a finite nonzero limiting value, microscopic distances contribute only a regular background and the leading divergence is ${\xi }^{2-\eta }$. Thus the Fisher scaling relation isThese statements concern leading singular powers; logarithmic corrections and cancellations of amplitudes require separate treatment.

For the requested ordinary critical point take $u>0$, $r=r_{0}t$, and the quartic Landau free energy $U(M)=r{M}^2/2+u{M}^4/4-hM$. At $h=0,t<0$, minimization gives $M^2=-r/u$. Choosing the positive pure thermodynamic phase,For the magnetic susceptibility, differentiate the equation of state at fixed temperature:Both sides diverge as $|t{|}^{-1}$, with different amplitudes, soThe one-sided field limit below the transition selects a pure phase; differentiating the symmetric finite-volume mixture would instead include its macroscopic switching response.

Put $U_0=G$ and $U_{j+1}=U_j^p$, where powers mean closed generated subgroups. The finite-quotient arguments of Question 3 pass to a powerful pro-p group: in particular $U_j$ is powerfully embedded, the power map is onto $U_{j+1}$, andFor the root assertion, apply the finite result to every finite quotient and use the finite intersection property on the closed sets of possible roots in the compact group. Each $U_j$ is finitely generated by the appropriate powers of a finite generating tuple, and $U_j/U_{j+1}$ is finite and elementary abelian. Thus the $U_j$ are open. They form an identity-neighbourhood basis: in any finite continuous quotient their images form a terminating lower p-series, so some $U_j$ is contained in the kernel. In particular ${\bigcap }_j{U}_j=1$. Collection also givesOne can see the extra index directly by induction from $[U_a,G]\subseteq U_{a+1}$: collect $[v,w^p]$ for $w\in U_b$. Its leading factor $[v,w{]}^p$ advances one layer, and the remaining iterated commutators, with their binomial coefficients, lie at least as deep. Products and limits preserve the inclusion. This is the usual strongly central property of the lower p-series.

A uniform pro-p group is a finitely generated powerful pro-p group for which all the indices $|U_j:U_{j+1}|$ equal $|G:G^p|$. The preliminary congruence gives surjective linear mapsIf $G$ is uniform, these maps are bijective. If $x\ne y$, write $y=xu$ and choose the unique layer $j$ with $u\in U_j\setminus U_{j+1}$. Bijectivity of ${\theta }_j$ and the power congruence imply $y^p(x^p{)}^{-1}\in U_{j+1}\setminus U_{j+2}$, so $x^p\ne y^p$. Thus the power map is injective. A finite-order element of a pro-p group has $p$-power order; repeated injectivity of the power map excludes every nonidentity such element. Hence a uniform group is a torsion-free group.

Conversely, suppose some ${\theta }_j$ has a nonzero kernel. Choose $x\in U_j\setminus U_{j+1}$ with $x^p\in U_{j+2}$. At step $r\ge j+1$, if the current $x_r$ satisfies $x_r^p\in U_{r+1}$, surjectivity of ${\theta }_r$ supplies $u_r\in U_r$ withSet $x_{r+1}=x_r{u}_r$. The power congruence gives $x_{r+1}^p\in U_{r+2}$. Since the corrections tend to $1$ in the complete profinite group, the $x_r$ converge to $z$, with $z^p=1$ and $z\equiv x(\mathrm{mod}{U}_{j+1})$. In particular $z\ne 1$. A torsion-free group cannot have such an element, so every ${\theta }_j$ is injective and all layer indices are equal. This proves the root correction criterion for uniformity andFor a uniform group the power maps are continuous bijections $U_j\to U_{j+1}$, hence homeomorphisms. Their inverses define the unique roots used below. Comparison at the first differing layer also proves the power congruences in a uniform pro-p group:

The intrinsic Lie algebra of a uniform pro-p group has the same underlying set as $G$, zero equal to the group identity, and operationsHere $x^{\lambda }$ is the limit of integral powers approaching $\lambda$. The first root is taken in $U_n$ and the second in $U_{2n}$; the commutator actually lies in $U_{2n+1}$. The following faithful algebra realization verifies convergence and the Lie algebra axioms, rather than assuming them as part of the definition.

Choose a minimal tuple $a_1,\dots ,a_d$ of topological generators. The layer isomorphisms ${\theta }_j$ show that the images of $a_i^{p^j}$ are bases of $U_j/U_{j+1}$. The ordered-coordinate argument of Question 3 therefore gives a continuous bijectionExistence follows by successive digits and compactness; uniqueness follows by the first digit at which the coordinates differ, because its images are linearly independent in the corresponding power layer. Thus this map is a homeomorphism.

Here are details of the norm needed on the group algebra. In ${\mathbb{Z}}_p[[G]]$ put $b_i=a_i-1$. Its elements have unique ordered expansions ${\sum }_{\alpha }{c}_{\alpha }{b}_1^{{\alpha }_1}\cdots b_d^{{\alpha }_d}$ with $c_{\alpha }\in {\mathbb{Z}}_p$. To verify this description, identify the ordered coordinates of $G$ with ${\mathbb{Z}}_p^d$. Elements of the completed group algebra are integral measures on this space; their values on the binomial polynomials ${\prod }_i\left(\genfrac{}{}{0ex}{}{{\lambda }_i}{{\alpha }_i}\right)$ are precisely the coefficients $c_{\alpha }$. The finite-difference identity pairs the ordered monomial $b^{\beta }$ with this binomial polynomial as ${\delta }_{\alpha \beta }$. The Mahler expansion of a continuous function therefore proves both existence and uniqueness of the measure with the specified coefficients. This also proves that the ordinary rational group algebra embeds in these expansions.

Choose $1/(p-1)<s<1$ and setThis defines the weighted group algebra of a uniform pro-p group. Its multiplication is submultiplicative in the valuation convention, $v(AB)\ge v(A)+v(B)$. Indeed, $a_i^p-1=p{b}_i+\left(\genfrac{}{}{0ex}{}{p}{2}\right)b_i^2+\cdots +b_i^p$ has valuation $1+s$; collection of ordered coordinates gives the same lower bound for $g-1$ when $g\in U_1$. Interchanging two adjacent $b_i$ in a product introduces a correction coming from $[a_i,a_j]-1$, of valuation at least $1+s>2s$. Iterating the interchange rule never lowers the total weight; the corrections have strictly increasing weight, so infinite collection converges. The same argument with $p^j$-th powers showsThe equality follows from the first nonzero coordinate digit: its leading terms are $p^j{\lambda }_i^{\prime }{b}_i$, at least one with unit ${\lambda }_i^{\prime }$, and cannot cancel in the ordered expansion. Completing the rational group algebra in this norm produces a complete normed algebra $A$ in which $G$ embeds faithfully and with its original topology.

Since $v(g-1)\ge s>1/(p-1)$, the seriesconverge on the relevant balls in $A$ and are inverse there. Their convergence follows from $v_p(r!)\le (r-1)/(p-1)$ and $v_p(r)\le (r-1)/(p-1)$; formal one-variable identities prove that they are inverse. Put $X=\log x$ and $Y=\log y$. The Baker--Campbell--Hausdorff formula in $A$ now givesbecause every term of degree at least two tends to zero after this rescaling. Similarly expansion of $\exp (-p^{n}X)\exp (-p^{n}Y)\exp (p^{n}X)\exp (p^{n}Y)$ givesThe leading term is the indicated associative-algebra commutator and all terms of degree at least three tend to zero. These expressions are the logarithms of the unique roots in the intrinsic formulas. The image $\log G$ is compact and hence closed in $A$, so both limits belong to $\log G$, and the inverse logarithm takes the limits back to $G$. Also $\log (x^{\lambda })=\lambda \log x$. Consequently $\log$ identifies the intrinsic operations with addition, scalar multiplication and commutation in $A$. The Jacobi identity follows by expanding the associative products; thus $L_G$ is indeed a Lie algebra over ${\mathbb{Z}}_p$.

The commutator roots in the intrinsic formula belong to $U_1$, so their limit gives $[L_G,L_G]\subseteq \log U_1=p{L}_G$. Iterating the power congruence gives $x^{p^n}{y}^{p^n}\equiv (xy{)}^{p^n}(\mathrm{mod}{U}_{n+1})$. Taking the unique roots shows that intrinsic addition modulo $U_1$ is the original group product modulo $U_1$. Consequently $L_G/p{L}_G$ is the original elementary abelian quotient $G/U_1$, and the images of the $a_i$ form a basis. Successively removing digits expresses every element uniquely as an additive sum ${\sum }_i{\lambda }_i{a}_i$: existence uses completeness, while a nonzero relation would have a first nonzero digit contradicting the basis in $U_j/U_{j+1}$. Hence $L_G$ is a free module of rank $d$ over the p-adic integers and is a powerful Lie lattice.

Conversely a powerful Lie algebra over p-adic integers, for odd $p$, is a finite free ${\mathbb{Z}}_p$-module $L$ with a bilinear alternating Lie bracket satisfying the Jacobi identity and $[L,L]\subseteq pL$. DefineFor completeness, integrality and convergence can be verified without guessing which individual denominators are units. In ${\mathbb{Q}}_p\otimes L$ set $M=p^{-1}L$. Then $[M,M]\subseteq M$, so its integral universal enveloping algebra has an ordered basis by the Poincaré-Birkhoff-Witt theorem. Its coefficientwise $p$-adic norm is submultiplicative, and $L=pM$ consists of elements of valuation at least one. The homogeneous degree-$r$ terms in $\log (\exp X\exp Y)$ have coefficients of valuation at least $-\lfloor (r-1)/(p-1)\rfloor$: expand the logarithm, use $v_p(k!)\le (k-1)/(p-1)$ on each nonconstant exponential factor, and $v_p(\ell )\le (\ell -1)/(p-1)$ on a product of $\ell$ such factors. Therefore evaluating degree $r$ on $X,Y\in pM$ gives valuation at leastwhich tends to infinity and is at least two when $r\ge 2$. Each homogeneous term is a rational Lie polynomial by the formal Baker--Campbell--Hausdorff formula. The ordered enveloping-algebra basis shows that its intersection with ${\mathbb{Q}}_{p}M$ is exactly $M$, so the estimate places the sum in $L$, with the non-linear part in $pL$.

Associativity follows from the identity $\exp (X\ast Y)=\exp X\exp Y$ in the completed universal enveloping algebra. Its identity is $0$, its inverse is $-X$, and $X^{\ast m}=mX$, since the bracket of $X$ with itself is zero. The same coefficient estimate for the group commutator puts it in $pL$. Conjugation in the BCH group is $\exp (\mathrm{ad}X)$; each term $\mathrm{ad}(X{)}^r/r!$ preserves $p^{j}L$, because each bracket adds a factor of $p$ and $v_p(r!)<r$. Thus the submodules $p^{j}L$ are normal subgroups for $\ast$, their intersection is zero, and their finite quotients have $p$-power order. Thus $(L,\ast )$ is a powerful pro-p group, withA module basis topologically generates this group: it generates the quotient $L/pL$, and successive power digits lift that generation through all $p^{j}L$. The BCH construction therefore gives a uniform pro-p group.

Throughout this question $p$ is odd, $A^p$ means the subgroup generated by the $p$-th powers of elements of $A$, and the commutator of group elements convention is $[x,y]=x^{-1}{y}^{-1}xy$. A subgroup $N$ is powerfully embedded in $G$ when $[N,G]\subseteq N^p$. In particular $N$ is normal, since $n^g=n[n,g]\in N$. A finite p-group $G$ is a powerful p-group when $[G,G]\subseteq G^p$.

First put $M=N^p$. It is normal in $G$. By the reduction permitted in the question, it suffices to prove $[M,G]\subseteq M^p$ after factoring out $[M,G,G]$. In that quotient, $[M,G]$ is central. For $n\in N$ and $g\in G$, put $c=[n,g]\in M$. Then $[c,n]$ is central, and the elementary identity $[ab,g]=[a,g{]}^b[b,g]$ gives, by induction,Both $c$ and $[c,n]$ belong to $M$, and $p$ divides $p(p-1)/2$ because $p$ is odd. Hence $[n^p,g]\in M^p$. The subgroup $M^p$ is normal in $G$, so the commutator identity extends this inclusion from the generating elements $n^p$ to every element of $M$. Applying the allowed reduction proves $N^p$ is powerfully embedded in $G$.

For $H=⟨N,x⟩$, normality of $N$ gives $H=N⟨x⟩$. Its quotient by $N$ is cyclic. Modulo $[N,H]$, the image of $N$ is central and the remaining generator is $x$, so that quotient is abelian. ThereforeThus $⟨N,x⟩$ is powerful.

The lower p-series is $G_1=G$, $G_{i+1}=G_i^p[G_i,G]$. For the later calculations write $P_j=G_{j+1}$, so $P_0=G$. Part (i) will show $P_{j+1}=P_j^p$ and $[P_j,G]\subseteq P_{j+1}$. A useful power lifting in a powerful p-group congruence isIndeed, modulo $P_{j+2}$ the commutators $[u,x]$ are central, lie in $P_{j+1}$ and have exponent dividing $p$. Collecting the $p$ copies of $xu$ therefore gives only the correction $[u,x{]}^{\left(\genfrac{}{}{0ex}{}{p}{2}\right)}$, which is trivial. This argument also covers $j=0$. The congruence is the reason roots and coordinates can be corrected one power layer at a time.

A topological group is a group with a topology for which $(x,y)\mapsto xy$ and $x\mapsto x^{-1}$ are continuous; we use the Hausdorff convention. Translations and inversion are then homeomorphisms.

For the final generator-extraction assertion, use the profinite Frattini subgroupFor a finitely generated pro-p group, the Burnside basis theorem says that $G/\Phi (G)$ is a finite-dimensional vector space over ${\mathbb{F}}_p$ and a subset topologically generates $G$ precisely when its image spans that vector space. In particular its dimension is the minimal number $d(G)$ of topological generators. A given $d$-element generating set shows $d(G)\le d$. The image of any other generating set spans the same finite-dimensional vector space, so it contains a basis of $d(G)$ elements. Choose their preimages in that generating set and apply the Burnside basis theorem again. Every generating set contains a generating subset of size $d(G)\le d$. This argument also applies to an infinite generating set.

A censoring observation means an event time is not observed exactly: the information supplies a bound or interval instead. Under right censoring, one observes $X=\min (T,C)$ and the event indicator $\delta =1_{\{T\le C\}}$; if $\delta =0$, one knows only $T>C$. Left censoring gives an upper bound, while interval censoring locates the event between two observations.

Uninformative censoring means that, conditional on the covariates used in the analysis, the censoring mechanism supplies no additional information about the event time. Independence $T\perp C$ within those covariate strata is a sufficient formulation: subjects remaining in the observed risk set have the same future hazard as the corresponding event-free population. A fixed study cutoff unrelated to prognosis is an example of administrative censoring.

Informative censoring occurs when that condition fails. For example, students at high risk of leaving may also become harder to contact before their departure is formally recorded; censoring at loss of contact selectively removes high-risk students. Ordinary Kaplan–Meier estimator and Nelson–Aalen estimator calculations then use an unrepresentative risk set. The distinction determines whether ordinary survival estimates can be interpreted as population survival. Merely knowing that an observation is censored does not establish independence.

The total processing time ${\sum }_{i=1}^{40}{t}_i$ is constant for a single machine, so only the initial setup and changeover sum depends on the order. Use the dummy-job reduction for sequence-dependent setup times. Add vertex $0$, and setLet the binary variable $x_{ij}$ indicate that $j$ follows $i$ in the augmented tour. A Hamiltonian cycle through $0$ specifies exactly one schedule, read starting after $0$. An exact integer programming formulation isThe degree equations alone describe a cycle cover of a directed graph, possibly several separate graph cycles. The subtour elimination constraints exclude every proper directed graph cycle; the cover is therefore a single tour through all jobs. Conversely every schedule supplies a feasible tour with exactly its completion time as objective. Exponentially many written inequalities are permitted in this exact formulation; an implementation may add them only when violated.

For three parallel machines, the relevant objective is the parallel-machine makespan with sequence-dependent setups, not simply total setup cost. Partition jobs into three ordered lists ${\pi }_r$. For a nonempty list definean empty machine has $C_r=0$. Minimize $C_{\max }$ subject to $C_r\le C_{\max }$ for each machine and each job assigned once. This can be written as a mixed-integer extension with assignment variables, one dummy tour per machine and conditional subtour cuts.

A practical strategy starts with a balanced assignment, sequences each machine using the single-machine methods, and improves by relocating or swapping jobs between machines together with within-machine 2-opt moves. Score each move by its change in ${\max }_r{C}_r$. Branch and bound can combine machine-assignment decisions with within-machine assignment bounds. For nonnegative times, the total processing time divided by three and the largest individual processing time are valid lower bounds, supplemented by setup-dependent route bounds. Balancing processing and changeovers jointly is essential; independently minimizing the three tour costs does not by itself minimize completion time.

For the linear program in maximization form, a nonnegative multiplier $y$ of the resource inequalities giveswhenever $A^{T}y\ge c$. Equivalently, the Lagrangian is $b^{T}y+(c-A^{T}y{)}^{T}x$; its supremum over $x\ge 0$ is finite exactly when $A^{T}y\ge c$. Minimizing this bound gives the dual linear programThe displayed inequality proves weak duality directly.

For the final simplex tableau, the basic-variable order is $(x_2,x_1)$. With $z_1,z_2$ interpreted as the ordinary unpriced slack variables, its constraint equations giveThe bottom row is read in the reduced-cost convention $r_j=c_j-c_B^T{B}^{-1}{A}_j$, with right side minus the basic objective value. It says the reduced costs of $z_1,z_2$ are $0,-1/2$. Thus optimality requires $z_2=0$, while $z_1$ can increase without changing the objective until $x_2$ reaches zero. The complete primal optimum set is

Reconstructing a linear program from a final simplex tableau exposes a source inconsistency. The slack columns areBecause the first basic column is that of $x_2$, the original constraint data areThe slack reduced costs give $y^T=(0,1/2)$ andConsequently the recovered homogeneous linear program isEvery point in the optimum segment has value $15/2$. Since the segment contains a point with both variables positive, complementary slackness forces both dual column inequalities to be equalities. They give the unique dual optimum

The printed bottom-right entry is $-3$, but it must be $-15/2$ for this linear objective and these constraint rows. Thus no original problem in the stated homogeneous standard form has exactly the entire printed tableau. Changing that one entry to $-15/2$ gives the reconstructed problem above. Alternatively, if an affine objective is permitted, $5x_1-\frac{5}{2}{x}_2-\frac{9}{2}$ has the displayed optimum value $3$ and unchanged reduced costs; the corresponding affine dual objective is $b^{T}y-\frac{9}{2}$. Both interpretations have exactly the same primal and dual optimizer sets. The discrepancy is in the objective constant, not in the optimal segment.

For a cochain complex $E$, define the shift of a cochain complex byFor a cochain map $a:E\to F$, put $(\Sigma a{)}^i=a^{i+1}$. This is again a cochain map, so the construction is a functor. The sign ensures the standard cone identities; the shifted differential squares to zero. Iteration gives $E[r{]}^i=E^{i+r}$ and differential $(-1{)}^r{d}_E$, and $H^i(E[r])=H^{i+r}(E)$.

For $\varphi :E\to F$, its algebraic mapping cone isApplying this differential twice gives first component $d_F^{2}f+d_F\varphi e-\varphi d_{E}e=0$, since $\varphi$ is a cochain map, and second component $d_E^{2}e=0$. The mapsare cochain maps: $d_{C}u(f)=(d_{F}f,0)=u(d_{F}f)$ and $v(d_C(f,e))=-d_{E}e=d_{\Sigma E}v(f,e)$. They are maps of complexes of sheaves, because every displayed operation is a sheaf morphism in each degree.

The bounded below derived category of sheaves $D(X)=D^{+}(X)$ can be constructed by first quotienting the category of bounded below cochain complexes by cochain homotopy and then formally inverting all quasi-isomorphisms. A morphism may be represented by a roof $E\stackrel{\sim }{\leftarrow }{E}^{\prime }\to F$. Alternatively, take injective resolutions of sheaves and use homotopy classes of maps between the resulting injective complexes. The cohomology sheaf functors descend to the localization, because they already identify homotopic maps and turn quasi-isomorphisms into isomorphisms. If $E\cong 0$ in $D(X)$, applying these functors gives ${\mathcal{H}}^i(E)=0$ for every $i$. Conversely, if all these sheaves vanish, the unique map $E\to 0$ is a quasi-isomorphism, which becomes invertible in $D(X)$. HenceAn exact triangle in a derived category is a triangle isomorphic to a cone triangleIn particular the last arrow is part of the structure; a mere sequence of three objects with successive composite zero does not suffice.

Suppose $0\to E\stackrel{\varphi }{\to }F\stackrel{\psi }{\to }G\to 0$ is degreewise exact. DefineThen $r(d_C(f,e))=\psi (d_{F}f)+\psi \varphi (e)=d_G\psi (f)=d_{G}r(f,e)$, so $r$ is a cochain map. It is natural under maps of short exact sequences. It is degreewise surjective as a sheaf map, and its kernel can be identified with $E^i\oplus E^{i+1}$ using $(a,e)\mapsto (\varphi (a),e)$. In these coordinates the kernel differential isThere is a contracting homotopy $h^i(a,e)=(0,a)$ in degree $i-1$:Thus the kernel has zero cohomology sheaves. The cohomology-sheaf exact sequence is the sequence of the short exact sequence of complexes $0\to \ker r\to C(\varphi )\to G\to 0$, or equivalently its ordinary complex-cohomology sequence at every stalk. It follows that $r$ induces isomorphisms of all cohomology sheaves, provingUnder this isomorphism the cone triangle becomes the exact trianglewhere $\delta =v\circ r^{-1}$ in the derived category.

For the final construction write $U=S^2\setminus S^1$ and set $A=i_{!}{\mathbb{Q}}_U$, $B={\mathbb{Q}}_{S^2}$, $C=j_{\ast }{\mathbb{Q}}_{S^1}$, each initially placed in degree zero. The stated sequence $0\to A\to B\to C\to 0$ is exact on stalks: on $U$ its first map is an isomorphism and its last sheaf is zero; on the equator its first sheaf is zero and the second map is an isomorphism. Take its connecting morphismThe source has only ${\mathcal{H}}^0(C)=C$, whereas the target has only ${\mathcal{H}}^{-1}(A[1])=A$. Therefore every induced map ${\mathcal{H}}^i(\mu )$ is zero: in any degree at least one of its source and target is zero. This is a ghost morphism in a derived category.

To prove the derived morphism itself is nonzero, apply derived global sections, giving the hypercohomology sequence. The extension by zero identifies $H^{\ast }(S^2;A)$ with the compactly supported cohomology of $U$, and closed direct image identifies $H^{\ast }(S^2;C)$ with $H^{\ast }(S^1;\mathbb{Q})$. The complement $U$ is two open disks, each homeomorphic to ${\mathbb{R}}^2$, soThe degree-one part of the resulting exact sequence isThus ${\mu }_{\ast }$ is injective, in particular nonzero. With the hemispheres oriented from $S^2$, the last map is $(a,b)\mapsto a+b$, so an appropriately oriented equator generator maps to $(1,-1)$. A zero morphism in the derived category of sheaves would induce a zero map after derived global sections and taking cohomology. Consequently $\mu \ne 0$, although ${\mathcal{H}}^i(\mu )=0$ for every $i$. This establishes explicitly that the collection of cohomology-sheaf functors detects zero objects but is not faithful on morphisms.

The Lie group–Lie algebra correspondence separates local differential structure from global topology. For a Lie group $G$, define $LG=T_{e}G$ as a real vector space. Extend each $X\in T_{e}G$ to the left-invariant vector fieldThe commutator of two such vector fields is again left-invariant. SetThe vector-field commutator is bilinear, antisymmetric and satisfies the Jacobi identity, so this makes $T_{e}G$ a Lie algebra. For a matrix group this bracket is $XY-YX$.

For a smooth Lie group homomorphism $\varphi :G_1\to G_2$, its differential at the identity is the linear mapThe identity $\varphi \circ L_g=L_{\varphi (g)}\circ \varphi$ shows that the corresponding left-invariant fields are $\varphi$-related. Brackets of related vector fields are related, henceThe chain rule also gives $L(\psi \circ \varphi )=L\psi \circ L\varphi$ and $L(\mathrm{id})=\mathrm{id}$. Thus differentiation is a functor from Lie groups and smooth homomorphisms to real Lie algebras and bracket-preserving linear maps.

To see what information the differential retains, apply $\varphi$ to a one-parameter subgroup. Its initial tangent is $L\varphi (X)$, so uniqueness yieldsIf $G_1$ is connected, a small exponential neighbourhood generates it. Every $g\in G_1$ is a finite product $\exp X_1\cdots \exp X_k$, and thereforeFor an already existing homomorphism this expression is independent of the chosen product because it equals $\varphi (g)$. Two homomorphisms with the same differential agree on that neighbourhood and hence everywhere. This proves that a Lie group homomorphism determined by its differential requires a connected source.

On a disconnected source only the restriction to $G_1^{\circ }$ is determined. For instance a nontrivial finite discrete group has zero Lie algebra, so its identity and trivial endomorphisms have the same zero differential but are different maps. The discussion of all components must therefore include additional global data. The remaining parts explain the corresponding existence and topology conditions.

For an oriented real vector bundle $\pi :E\to X$ of rank $d$, the Thom isomorphism theorem gives a unique Thom class $u_E\in H^d(D(E),S(E);\mathbb{Z})$ restricting to the chosen generator in each fibre, and isomorphismsEquivalently use the pair $(E,E\setminus X)$, identifying $X$ with the zero section. After forgetting relative supports, pull the Thom class back along the zero section $s$ to define the Euler class:In this expression the relative class is first mapped to absolute cohomology.

A compact embedded submanifold equipped with a coorientation $Y\subset X$ has an oriented normal bundle ${\nu }_Y$ of rank $d$. A tubular neighborhood and excision identify its Thom class with a class $u_Y\in H^d(X,X\setminus Y)$. Its image in $H^d(X)$ is the cohomology class of a cooriented submanifold ${\epsilon }_Y$. In particular it restricts to zero on $X\setminus Y$. If the ambient manifold is oriented, orient $Y$ by the normal-first convention. The class is its Poincare dual, characterized byfor the appropriate complementary degree. This is the stated integration identity in cohomological notation.

First prove the Lefschetz fixed-point theorem for an oriented closed n-manifold $X$. Orient its diagonal by the identification $\Delta \cong X$, and choose the normal orientation so this agrees with the normal-first convention in the product. This produces the integral class ${\epsilon }_{\Delta }$. To compute the part that contributes to the trace, tensor with $\mathbb{Q}$: the trace on a finitely generated integral cohomology group means its trace on the free quotient, equivalently on rational cohomology; torsion contributes nothing.

Choose a basis $e_{p,j}$ of $H^p(X;\mathbb{Q})$ and its Poincare duality pairing dual basis $e_{p,j}^{\vee }\in H^{n-p}(X;\mathbb{Q})$, normalized by $⟨e_{p,i}⌣e_{p,j}^{\vee },[X]⟩={\delta }_{ij}$. The rational image of the cohomology class of the diagonal isIndeed, pairing this expression against $\alpha \times \beta$ with degrees $p,n-p$ picks out the p-summands. The cross-product multiplication contributes $(-1{)}^{p^2}$ and moving $\alpha$ past $e_{p,j}^{\vee }$ contributes $(-1{)}^{p(n-p)}$. Together with $(-1{)}^{np}$ their product is $+1$. The result is $⟨\alpha ⌣\beta ,[X]⟩$, exactly the diagonal's defining identity. The rational Künneth theorem and nondegeneracy of the duality pairing determine the class. An integral torsion component, if present, is not determined by this rational expression and does not affect any evaluation below.

For the graph map ${\Gamma }_f=(1,f):X\to X\times X$, write $f^{\ast }{e}_{p,j}={\sum }_i{A}_{ij}{e}_{p,i}$. Pulling back the expression and evaluating gives the graph-diagonal formula for the Lefschetz number:If $f$ has no fixed point, its graph lies in $(X\times X)\setminus \Delta$. The relative Thom class restricts to zero there, so ${\Gamma }_f^{\ast }{\epsilon }_{\Delta }=0$ and $L(f)=0$. The contrapositive proves the required fixed-point assertion, without requiring smoothness of $f$ or transversality of its graph.

The printed hypothesis does not require $X$ to be orientable. For a nonorientable $X$, the diagonal's normal bundle is $TX$, so an ordinary integral coorientation cannot be assumed. We extend the proved theorem as follows, rather than apply the untwisted diagonal computation incorrectly. We first construct a smooth embedding in Euclidean space using finitely many charts ${\chi }_i:U_i\to {\mathbb{R}}^n$ and smooth functions ${\rho }_i$ supported in those charts, with at least one ${\rho }_i(x)>0$ at every point. The map $\Psi (x)=({\rho }_i(x),{\rho }_i(x){\chi }_i(x){)}_i$, with zero extensions, is injective: a positive first coordinate identifies a chart in which its remaining coordinates recover $x$. Its differential is injective as well, since $d{\rho }_i(v)=0$ and $d({\rho }_i{\chi }_i)(v)=0$ force $d{\chi }_i(v)=0$ in a chart with ${\rho }_i>0$. Compactness makes this injective immersion a homeomorphism onto its image. This finite-chart Euclidean embedding of a compact smooth manifold, enlarged by zero coordinates if necessary, has positive codimension. A compact tubular disc neighbourhood $W$ is an oriented N-manifold with boundary, inheriting its orientation from the ambient Euclidean space. Its double of a manifold $M$ is closed and oriented. Projection of each copy of $W$ to $X$ agrees on their common boundary, giving $r:M\to X$; inclusion into the first copy gives $i:X\to M$ with $ri=1_X$.