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For a quasi-compact separated scheme and a global regular function $f$, global sections on $X_f$ are the localization of global sections by $f$. Choose a finite affine open cover. Its intersections are affine, so the sheaf gluing axiom describes global sections as the kernel of the difference map between two finite products of affine coordinate rings. Exactness of localization preserves this kernel, and localization commutes with those finite products. Localizing the same cover gives the claimed isomorphism. This applies before knowing whether $X$ itself is affine.

For a chain map $f:C\to D$, use $M_i=C_{i-1}\oplus D_i$ and $d(x,y)=(d_{C}x,(-1{)}^{i}f(x)+d_{D}y)$. The chain map identity makes $d^2=0$. The short exact sequence of chain complexes $0\to D\to M\to C_{\ast -1}\to 0$ has connecting homomorphism $(-1{)}^i{f}_{\ast }$ in degree $i$. Its long exact sequence in homology shows that the homology of $M$ vanishes precisely when every $f_{\ast }$ is an isomorphism.

For a continuous self-map $f:X\to X$, the mapping torus has the Wang sequence even when $f$ is not a homeomorphism. Cover the time circle by a seam neighborhood and an interval away from the seam. Their inverse images retract to $X$: the seam neighborhood is a mapping cylinder, and the other is a product. Their intersection retracts to two copies of $X$. The Mayer–Vietoris sequence map on homology is, after consistent choices of signs,Invertible changes of coordinates reduce it to the direct sum of $\mathrm{id}-f_{\ast }$ and an isomorphism. Canceling the isomorphism summands gives the Wang sequence. This proof does not assume a fiber bundle structure for a noninvertible $f$.

A cartesian square of unital rings $A=A_1{\times }_{A_0}{A}_2$ is a Milnor square when at least one map $A_i\to A_0$ is surjective. Finitely generated projective modules over $A$ can then be assembled from projective modules over the two upper rings together with an isomorphism of their reductions over $A_0$.

For the nested partial unary operation category tower, each one-step forgetful functor is a monadic adjunction right adjoint: it reflects isomorphisms and creates its split coequalizers. The Beck monadicity theorem identifies the first Eilenberg-Moore category with the next category in the tower. A long composite has that same first monad because free extension of nested partial unary operations leaves no new top fixed points. Repeating the comparison forgets one fewer operation at each step. The remaining forgetful functor is not full until no operation remains to forget, giving monadic length $m-n$.

If $e:X\to Y$ is a monomorphism and has the left lifting property against monomorphisms, apply its property to the square with both vertical arrows $e$ and both horizontal arrows identities. The filler is a two-sided inverse, so $e$ is an isomorphism.

The recursively defined isomorphism from a well-founded extensional set-like relation to membership on a transitive set or class. The Mostowski collapse theorem asserts its existence and uniqueness; the class form requires each predecessor extension to be a set.

For a surjective module endomorphism of a Noetherian module, the chain $\ker f\subseteq \ker f^2\subseteq \cdots$ stabilizes at some $N$. If $f(x)=0$, choose $y$ with $f^N(y)=x$. Then $y\in \ker f^{N+1}=\ker f^N$, giving $x=0$. Thus the endomorphism is also injective and is an isomorphism.

For an integral scheme, take on each affine chart the integral closure of its coordinate ring in its function field. Integral closure commutes with localization, so these rings glue to a normal scheme with an integral birational morphism to the original scheme. This is the normalization. It need not be finite without additional hypotheses; for an algebraic curve of finite type over a field it is finite. It is an isomorphism exactly when the original scheme is normal.

Over $k={\mathbb{F}}_p$, the Nottingham group is $\mathcal{N}(k)=\{f(t)=t+{\sum }_{i\ge 2}{a}_i{t}^i:a_i\in k\}$ with composition and the $t$-adic topology. Equivalently it is the group of continuous $k$-automorphisms of the Laurent series field $k((t))$ that act trivially on $(t)/(t^2)$. With ordinary series composition, the isomorphism sends $f$ to the automorphism $h(t)\mapsto h(f^{-1}(t))$; this inverse avoids reversing the group product. It is a pro-p group, recovered from its finite truncations, and contains elements of arbitrarily large finite $p$-power order.

For an odd map between spheres $f:S^n\to S^m$ and its quotient $\bar{f}$, there is an isomorphism ${\gamma }_n\cong {\bar{f}}^{\ast }{\gamma }_m$ of real tautological line bundles. With $x$ a unit vector, send $tx$ to $tf(x)$; replacing $x$ by $-x$ and $t$ by $-t$ gives the same vector. This proves both linearity on fibers and well-definedness. Therefore ${\bar{f}}^{\ast }{w}_1({\gamma }_m)=w_1({\gamma }_n)$. The mod-two cohomology ring of real projective space then implies $n\le m$, since a vanishing $(m+1)$st power must pull back to a vanishing power.

Choose a connected proper incompressible surface $S$ with nonempty boundary as the first cut in a three-manifold hierarchy of $N$. Homotope $f$ relative to $\partial M$ into transverse position and let $F=f^{-1}(S)$.

We first make $F$ incompressible. If $D$ is a compression disk for $F$, its boundary maps to a loop in $S$ which is null-homotopic in $N$. Since $S$ is fundamental-group-injective, this loop is null-homotopic in $S$. Fill it in $S$ and alter $f$ in a small neighbourhood of $D$ so that its inverse image is obtained by compressing $F$ along $D$. The alteration is a homotopy: the filling and the old disk give a sphere map in $N$, which is null-homotopic by the asphericity proved above. A product collar of $S$ specifies the two sides of the modification. The integer ${\sum }_j\max (0,1-\chi (F_j))$ decreases at every essential compression; sphere components can be removed separately. A resulting sphere component of $F$ bounds a three-ball in $M$ by three-manifold irreducibility; move the map of that three-ball to one side of $S$ to remove it. The required extension and homotopy exist because the higher homotopy groups of $N$ vanish. Continuing leaves an incompressible $F$ with no spherical components.

For each component $F_j$, the map on fundamental groups to $S$ is injective. Indeed, its composition with the injection ${\pi }_1(S)\to {\pi }_1(N)$ equals the composition of ${\pi }_1(F_j)\to {\pi }_1(M)$ with the isomorphism $f_{\ast }$. Both maps in the latter composition are injective. A closed nonspherical orientable $F_j$ cannot occur: its fundamental group would inject into ${\pi }_1(S)$, a free group, whereas every subgroup of a free group is free and a closed positive-genus surface group is not free. Thus every component of $F$ has boundary.

Apply the permitted surface-map homotopy result to $F_j\to S$, fixing its boundary. A fundamental-group-injective map of compact surfaces is homotopic either to a covering or, in the annulus case here, to a map into one target boundary circle. The exceptional annulus case is impossible: its two distinct boundary circles would map onto the same circle of $\partial S$, contradicting the injectivity of the original boundary map. If $S$ is a disk, injectivity makes each $F_j$ a disk, and the unique boundary circle directly gives a single disk homeomorphism. Otherwise each component is homotopic to a cover of $S$.

Every point of $\partial S$ has exactly one preimage on $\partial F$, since $f{|}_{\partial M}$ is a homeomorphism. Every cover component contributes its positive degree to the number of preimages on every target boundary circle. Thus there is exactly one component and its covering degree is one. We have proved that $F\to S$ can be homotoped to a homeomorphism, relative to its boundary.

Extend this surface homotopy to $M$, and use the two-sided collars to make $f$ a product homeomorphism on a neighbourhood of $F$. The transverse normal direction is a degree-one local crossing, so the collars on its two sides match the two sides of $S$; any superfluous folds can be removed in the normal interval. We may now cut $M$ along $F$ and $N$ along $S$, obtaining a map whose restriction to the entire cut boundary is a homeomorphism. The other hypotheses also survive: the cut manifolds are irreducible, have nonempty boundary, and the inclusion of every cut component into its uncut manifold is fundamental-group-injective by the normal form of the van Kampen theorem. A loop killed by the cut map is consequently killed by $f_{\ast }$ and hence is trivial in its source cut component.

First show that the boundary map is onto and has degree $\pm 1$, and that the fundamental-group injection is actually an isomorphism. An injective map between closed surfaces is a homeomorphism onto an open-and-closed subset by invariance of domain. Thus each component of $\partial M$ maps homeomorphically to a distinct component of $\partial N$.

Orient $M$ and $N$, and writeApply the boundary map in the relative homology sequence. The boundary of the right side has coefficient $d$ on every boundary component of $N$. On the left, an image boundary component has coefficient $+1$ or $-1$, and an omitted component has coefficient zero. Since at least one component is present, $d=\pm 1$; then no component can be omitted and all signs agree. Therefore

Let $H=f_{\ast }{\pi }_1(M)$ and lift $f$ to the connected cover $p:N_H\to N$ corresponding to $H$. If the cover were noncompact, then $H_3(N_H,\partial N_H;\mathbb{Z})=0$. The lifted relative fundamental class would have zero boundary, but its boundary is a nonzero sum of distinct boundary fundamental classes, since the original boundary map is injective and the boundary components are compact. This is impossible. The cover therefore has a finite number $r$ of sheets. Degree multiplicativity givesHence $r=1$. The image of $f_{\ast }$ is the whole target fundamental group, and the assumed injection makes $f_{\ast }$ an isomorphism.

A form of Beck's monadicity theorem compatible with equivalence-based monadic adjunctions is this. A functor $U:\mathcal{D}\to \mathcal{C}$ is monadic exactly when it has a left adjoint, is a conservative functor, and $\mathcal{D}$ has coequalizers of all pairs whose images under $U$ admit a split coequalizer, with $U$ preserving these coequalizers. In the strict lifting formulation, one states creation of these coequalizers. Such a pair is a functor-split coequalizer pair; the splitting is required in the base category, not necessarily upstairs.

A split coequalizer fork $A\begin{array}{c}\stackrel{r}{\to }\\ \underset{s}{\to }\end{array}B\stackrel{q}{\to }Q$ has maps $u:B\to A$ and $t:Q\to B$ such thatEvery functor preserves this coequalizer: if $hr=hs$, then $h=hru=hsu=htq$, and $q$ is split epic, so the factor through $q$ is unique. The same equations remain true after applying any functor.

For necessity, let $U^T:{\mathcal{C}}^T\to \mathcal{C}$ be the monad algebra forgetful functor. It has the free-algebra left adjoint. It reflects isomorphisms, because if an algebra map $f$ is invertible in $\mathcal{C}$, the equation $fa=bTf$ rearranges to $f^{-1}b=aT(f^{-1})$.

If two algebra maps $r,s:(A,a)\rightrightarrows (B,b)$ have an underlying split coequalizer $q:B\to Q$, then $Tq$ and $T^{2}q$ are also coequalizers. Since $qb$ equalizes $Tr,Ts$, there is a unique $c:TQ\to Q$ satisfyingAfter composing with the epimorphism $q$, the equation $c{\eta }_Q=1_Q$ reduces to the unit law for $b$. After composing with the epimorphism $T^{2}q$, the equation $cT(c)=c{\mu }_Q$ reduces to $bT(b)=b{\mu }_B$. Thus $(Q,c)$ is an algebra and $q$ an algebra map. If an algebra map $h:(B,b)\to (Z,z)$ equalizes $r,s$, its unique underlying factor $k:Q\to Z$ satisfies $kc=zTk$: this follows by precomposing with the epimorphism $Tq$. Therefore this is a coequalizer upstairs as well. This proves that the monad algebra forgetful functor creates split coequalizers. The same properties transport through the comparison equivalence of categories, with the object-lifting convention distinguished in Question 9.

For sufficiency, suppose the three conditions hold, choose $F⊣U$, and let $T=UF$ with comparison $K$. For each algebra $(X,a)$, consider the pairIts image has coequalizer $a:TX\to X$. It is split: take $r={\mu }_X$, $s=T(a)$, $q=a$, $u={\eta }_{TX}$ and $t={\eta }_X$. The monad and algebra laws giveBy hypothesis the pair has a coequalizer $q:FX\to D_X$ and $Uq$ is also a coequalizer. Identify $U{D}_X$ with $X$ using the unique coequalizer isomorphism, so $Uq=a$. Since $q$ is carried by $K$ to an algebra map from the free algebra, the transported action $b:TX\to X$ satisfiesBut $T(a)$ is a split epimorphism, with right inverse $T({\eta }_X)$, so $b=a$. Hence $(X,a)\cong K(D_X)$: the comparison is essentially surjective.

Next, for every $D\in \mathcal{D}$, put $a_D=U{\epsilon }_D$. The counit of an adjunction coequalizesThis follows from naturality of the counit. The image pair has the split coequalizer $a_D$, as above. If $q:FUD\to Q$ is the hypothesized preserved coequalizer, the induced map $h:Q\to D$ has $Uh$ invertible because both $Uq$ and $a_D$ coequalize the same pair. Conservativity makes $h$ invertible. Therefore ${\epsilon }_D$ itself is a coequalizer.

Let $f:K(D)\to K(E)$ be an algebra map, so $f{a}_D=a_{E}Tf$. Set $g={\epsilon }_{E}F(f):FUD\to E$. Under the adjunction transpose bijection, the two composites of $g$ with the displayed presentation pair correspond respectively to $f{a}_D$ and $a_{E}Tf$. They are equal, so $g$ factors uniquely through ${\epsilon }_D$, giving $h:D\to E$ with $h{\epsilon }_D=g$. Applying $U$ yieldsSince $a_D$ is split epic, $Uh=f$. Thus $K$ is full. If $Uh=Uk$, the transposes of $h{\epsilon }_D$ and $k{\epsilon }_D$ are equal, so those composites agree. The counit is epic, hence $h=k$; thus $K$ is faithful.

The comparison is full, faithful and essentially surjective, so it is an equivalence of categories. This proves monadicity. The argument also explains the role of each hypothesis: split base presentations construct all algebras, and reflection of isomorphisms turns the counit presentations into actual coequalizers upstairs.

An isomorphism $f:A\to B$ gives a natural isomorphism $h_f:h_A\to h_B$ by postcomposition, with inverse $h_{f^{-1}}$.

Conversely suppose $\alpha :h_A\to h_B$ is a natural isomorphism, with inverse $\beta$. Since the Yoneda embedding is a full and faithful functor, there are unique $f:A\to B$ and $g:B\to A$ with $h_f=\alpha$ and $h_g=\beta$. Their composites satisfyFaithfulness gives $gf=1_A$ and $fg=1_B$. ThusThe naturality requirement ensures that all incoming morphisms are respected; unrelated componentwise bijections alone would not justify the conclusion.

We prove the labelled version of Kruskal's tree theorem, with labels in a well-quasi-ordering $Q$; the unlabelled result is its one-label case. A label-monotone tree embedding is an injective map of finite rooted trees preserving lowest common ancestors and increasing labels. Its root may map below the host root, and an edge may map to a path. These conventions distinguish rooted-tree homeomorphic embedding from the stricter adjacency-preserving relation, which is not well-quasi-ordered.

We will use two elementary consequences of well-quasi-ordering. Every infinite sequence in a well-quasi-ordering has an infinite nondecreasing subsequence: color pairs by whether the earlier element precedes the later one, apply Ramsey's theorem, and exclude an all-bad infinite set. Restricting successively in each coordinate proves finite product closure of well-quasi-orderings.

For completeness, here is the required proof of Higman's lemma. If finite words over $Q$ admitted a bad sequence, choose one $w_0,w_1,\dots$ whose next word always has the least length admitting a bad continuation of the fixed prefix. No word is empty. Write $w_i=v_i{a}_i$, separating the last letter. The set of prefixes $\{v_i\}$ is a well-quasi-ordering. Otherwise select a bad sequence of these prefixes with strictly increasing owner indices $n_0<n_1<\cdots$; this is possible because every finite initial set of owners supplies only finitely many prefixes, and a bad sequence cannot repeat a word. Thenis bad. An earlier original word embedding into $v_{n_j}$ would also embed into $w_{n_j}$, contradicting the original badness; comparisons inside the new tail are excluded by its choice. But $v_{n_0}$ is shorter than $w_{n_0}$, contradicting minimality. Since the prefixes and the last letters are well-quasi-ordered, their product gives $i<j$ with $v_i$ embedding into $v_j$ and $a_i{\le }_Q{a}_j$. Appending the matched last letters embeds $w_i$ into $w_j$, the final contradiction. Thus finite words over a WQO are WQO.

Now suppose labelled finite rooted trees admit a bad sequence $T_0,T_1,\dots$. Choose it minimally in vertex count at each successive position. Let $\mathcal{S}$ be all immediate child subtrees of its members. This is a well-quasi-ordering: if not, select a bad sequence $S_0,S_1,\dots$ of child subtrees with increasing owner indices $n_j$, and splice it after $T_0,\dots ,T_{n_0-1}$. Every child subtree embeds into its owner, so a comparison from the prefix to the tail would contradict original badness. The splice is bad and its first replacement is strictly smaller, contradicting the minimal bad sequence choice.

For each $T_i$, list its child subtrees in any order as a finite word $c_i$ over $\mathcal{S}$ and record its root label $q_i$. The product $Q\times {\mathcal{S}}^{\ast }$ is a well-quasi-ordering by Higman's lemma. Hence some $i<j$ has $q_i{\le }_Q{q}_j$ and a subsequence embedding of $c_i$ into $c_j$. Match the roots and use the chosen subtree embeddings in distinct target branches. Their union preserves lowest common ancestors and labels, and stretches only edge paths. Thus $T_i$ embeds into $T_j$, a contradiction. We have proved

To deduce Friedman's finite form of Kruskal's theorem, fix $k\in \mathbb{N}$. At position $i=0,1,\dots$ permit unlabelled trees of at most $k+i$ vertices. Up to isomorphism only finitely many trees are possible at each position. Form the finite bad-sequence tree whose nodes are finite bad sequences satisfying these bounds, ordered by extension. It is finitely branching. If there were arbitrarily long admissible bad sequences, König infinity lemma would give an infinite branch, contradicting Kruskal's tree theorem. ThereforeOne can instead use a fixed finite set of labels, giving a correspondingly parameterized finite statement by exactly the same proof. The crucial feature is a uniform bound $N$ for each initial size allowance, not merely termination of one chosen sequence.

Put $x=I+E_{12}$, $y=I+E_{23}$ and $z=I+E_{13}$. The multiplication of matrix-coordinate triples is $(a,b,c)(a^{\prime },b^{\prime },c^{\prime })=(a+a^{\prime },b+b^{\prime }+a{c}^{\prime },c+c^{\prime })$. In particular $z$ is central, $xy=zyx$, and every element has the unique normal form $y^c{z}^b{x}^a$. ConsequentlyThe distinct normal-form monomials are the group basis, so this is an isomorphism, not merely a surjection from a presented algebra. This is the integral Heisenberg group algebra as a skew Laurent ring.

Here is the needed skew Hilbert basis theorem, including its proof. If $A$ is a left Noetherian ring and $\sigma$ an automorphism, let $J$ be a left ideal of $A[t;\sigma ]$, where $ta=\sigma (a)t$. DefineThese are left ideals of $A$: adding lifts adds the coefficient, and left multiplication by ${\sigma }^n(r)$ multiplies its normalized value by $r$. Multiplication of a lift by $t$ proves $I_n\subseteq I_{n+1}$. The chain stabilizes at some $N$. Choose finite generators $u_{nj}$ for each $I_n$, $0\le n\le N$, and lifts $f_{nj}\in J$ with $t^n$ coefficient ${\sigma }^n(u_{nj})$.

For $f\in J$ of degree $m\ge N$, write its normalized leading coefficient as ${\sigma }^{-m}(\mathrm{lc}f)={\sum }_j{r}_j{u}_{Nj}$. Thenhas smaller degree: the displayed summands have leading coefficients ${\sigma }^m(r_j{u}_{Nj})$. For $m<N$ use the generators for $I_m$ without the shift. Induction on degree expresses every element of $J$ in terms of the finitely many chosen lifts. Thus $A[t;\sigma ]$ is a left Noetherian ring. Passing to the opposite ring replaces $\sigma$ by ${\sigma }^{-1}$ and proves the corresponding right-sided result.

This proof includes the ordinary Hilbert basis theorem when $\sigma$ is the identity. Starting with the field $k$, it makes $k[y,z]$ Noetherian; localization at powers of $y,z$ makes $B$ Noetherian as well. Finally, for a left ideal $L$ in $B[x,x^{-1};\sigma ]$, its intersection $J$ with $B[x;\sigma ]$ has finitely many generators. For each $f\in L$, some $x^{q}f$ lies in $J$, so $f=x^{-q}(x^{q}f)$ is generated by those same elements in the skew Laurent polynomial ring. Right multiplication clears powers for a right ideal. Therefore $kG$ is a left Noetherian ring and a right Noetherian ring.

Let $X$ be a connected compact Riemann surface of genus $g$. Choose an oriented symplectic basis $a_1,\dots ,a_g,b_1,\dots ,b_g$ of $H_1(X,\mathbb{Z})$, with $a_i\cdot b_j={\delta }_{ij}$. For a closed complex differential form $\alpha$, write $A_i(\alpha )={\int }_{a_i}\alpha$ and $B_i(\alpha )={\int }_{b_i}\alpha$. The Riemann bilinear relations begin with the identityTo prove it, cut $X$ along the chosen cycles to obtain a fundamental polygon $P$. On this simply connected polygon, a closed differential form has a primitive $F$, with $dF=\alpha$. Stokes theorem gives ${\int }_P\alpha \wedge \beta ={\int }_{\partial P}F\beta$. The two copies of each edge have opposite orientations, and their values of $F$ differ by the period acquired in passing between them. Pairing the copies of the $a_i$ and $b_i$ edges gives, respectively, $-B_i(\alpha )A_i(\beta )$ and $A_i(\alpha )B_i(\beta )$. Summing proves the identity; the convention $a_i\cdot b_i=1$ fixes its sign.

For holomorphic one-forms $\omega ,\eta$, their wedge product is zero because locally both are multiples of $dz$. Thus the first holomorphic Riemann bilinear relations isFor a nonzero holomorphic one-form $\omega =f(z)dz$, the second relation follows by applying the bilinear identity to $\omega ,\overline{\omega }$:In particular, a holomorphic one-form with all $a$-periods zero must vanish. Since $\dim H^0(X,K_X)=g$, the map taking a holomorphic one-form to its $a$-periods is an isomorphism. We can therefore choose normalized holomorphic one-forms ${\omega }_1,\dots ,{\omega }_g$ with ${\int }_{a_i}{\omega }_j={\delta }_{ij}$. Put $B_{ij}={\int }_{b_i}{\omega }_j$. The first relation applied to ${\omega }_j,{\omega }_k$ gives $B_{jk}=B_{kj}$, and the second applied to ${\sum }_j{c}_j{\omega }_j$ gives $2{\overline{c}}^{,t}(\mathrm{Im}B)c>0$ for $c\ne 0$. ConsequentlyThese conclusions make the construction of the Jacobian variety work. Integration identifies the image of $H_1(X,\mathbb{Z})$ in $H^0(X,K_X{)}^{\ast }$ with the period lattice $\Lambda ={\mathbb{Z}}^g+B{\mathbb{Z}}^g$. The real-linear map $(m,n)\mapsto m+Bn$ is invertible: its imaginary part is $Yn$, and $Y$ is invertible. Hence $\Lambda$ is a discrete lattice of full real rank $2g$, andis a compact complex torus, rather than a non-Hausdorff quotient. Changing the symplectic basis changes the displayed coordinates but not this intrinsic quotient.

The Riemann bilinear relations also supply its polarization of a complex torus. Use the Hermitian form $H(z,w)=z^t{Y}^{-1}\overline{w}$, linear in the first argument, and set $E=-\mathrm{Im}H$. Symmetry of $B$ givesThus $E$ is an integral positive Riemann form on a complex torus. Its matrix on the displayed lattice basis is the unimodular symplectic matrix. The corresponding positive holomorphic line bundle, equivalently the one constructed from theta automorphy in Solution 4, gives a principal polarization of a complex torus. By the Kodaira embedding theorem this makes the complex torus projective, hence an abelian variety. The normalization $E=-\mathrm{Im}H$ here is important because $H$ was chosen linear in its first argument. If $g=0$, the spaces of holomorphic one-forms and periods are zero and the Jacobian variety is a point.

Identify ${\mathrm{Pic}}^{g-1}(X)$ with the Jacobian variety using the base point and the translation of Solution 4. A line bundle $L$ of degree $g-1$ corresponds to the point $z=u(D)-\kappa$ for any divisor representation $L\simeq {\mathcal{O}}_X(D)$. The Riemann singularity theorem isIn particular, the theta divisor is singular exactly at the degree-$g-1$ line bundles with at least two independent sections. We prove the equality, including the upper bound on multiplicity.

Choose a fixed effective divisor $A$ of degree $N\ge g$, and a local holomorphic family $L_{\eta }$ representing nearby points of ${\mathrm{Pic}}^{g-1}(X)$. Such a family can be constructed locally from the holomorphic exponential sequence, by varying line-bundle transition functions through exponentials of Čech cohomology representatives. By Serre duality, $H^1(L_{\eta }(A))$ is dual to $H^0(K_X\otimes L_{\eta }^{-1}(-A))$, whose degree is $g-1-N<0$, so it vanishes. The Riemann-Roch theorem gives $h^0(L_{\eta }(A))=N$. By cohomology and base change for line bundles on a curve, these spaces form a holomorphic vector bundle $E$ of rank $N$. The spaces $H^0(L_{\eta }(A){|}_A)$ form another holomorphic vector bundle $F$ of rank $N$. Evaluation gives a holomorphic square matrix $T(\eta )$ locally, with exact sequenceConsequently $\det T(\eta )=0$ exactly when $L_{\eta }$ has a nonzero section, namely on the support of the theta divisor.

At the point $L$, put $r=h^0(L)=h^1(L)$, the equality being Riemann-Roch theorem. An invertible minor of size $N-r$ persists near this point. Holomorphic row and column operations, or the Schur complement, givewhere $M$ is an $r\times r$ holomorphic matrix. This is the Schur complement presentation of a theta singularity. Its linear term in a tangent direction $\xi \in H^1({\mathcal{O}}_X)$ is the cup product mapTo see this concretely, vary transition functions $g_{ij}$ of $L$ to $g_{ij}(1+ϵ{\xi }_{ij})$. A section $s_i$ lifts to $s_i+ϵs_i^{\prime }$ precisely when its first-order compatibility equation can be solved; the obstruction is the cocycle ${\xi }_{ij}{s}_j$. This is $\xi ⌣s$. Passing to the kernel and cokernel of $T(0)$ gives exactly the derivative of the reduced matrix $M$, proving the assertion without assuming a multiplicity formula.

Before extracting the order of $\det M$, we verify that $\det T$ is a reduced equation of the theta divisor. Generic points of $W_{g-1}$ have $h^0(L)=1$ by Solution 3. At such a point, both $H^0(L)$ and $H^0(K_X\otimes L^{-1})$ are one-dimensional. Their nonzero sections $s,t$ have nonzero product $st\in H^0(K_X)$. Choose $\xi \in H^1({\mathcal{O}}_X)$ with $⟨\xi ,st⟩\ne 0$ under Serre duality. The scalar $d{M}_0(\xi )$ is nonzero, so $\det T$ has a simple zero generically on the irreducible theta divisor. Since its support is exactly that irreducible hypersurface, it defines the reduced divisor. Solution 4 independently proved the same reduced structure for the Riemann theta function. On the smooth ambient Jacobian variety, local equations of the same reduced divisor differ by a nonvanishing holomorphic unit. Hence the orders of the local theta equation and $\det M$ agree everywhere.

Every entry of $M$ vanishes at the point, so each term in its determinant has order at least $r$. This proves ${\mathrm{mult}}_z\Theta \ge r$. To obtain equality, we exhibit an invertible cup-product direction for a line bundle on a curve. Set $V=H^0(L)$ and $W=H^0(K_X\otimes L^{-1})$, both of dimension $r$. There are $r$ distinct points $p_1,\dots ,p_r$ at which evaluation is an isomorphism for $V$: choose them inductively, since a surviving nonzero section has only finitely many zeros. The same holds for $W$. The nonvanishing of each evaluation determinant defines a nonempty open subset of the irreducible variety $X^r$. Their intersection, also avoiding the diagonals, is therefore nonempty. Choose points in it and local coordinates and compatible trivializations of $L$ and $K_X\otimes L^{-1}$ at each point.

Let $\xi$ be the connecting homomorphism class of the meromorphic principal parts ${\lambda }_i/t_i$ at these points, with all ${\lambda }_i\ne 0$. The residue form of Serre duality givesIf $S$ and $U$ are the two invertible evaluation matrices, the matrix of this pairing is $S^t\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_r)U$, and is invertible. Thus the cup product map $\xi ⌣-:H^0(L)\to H^1(L)$ is an isomorphism. Along the corresponding analytic parameter line,with nonzero leading coefficient. The order is at most $r$, completing the proof of the Riemann singularity theorem.

This argument also identifies the tangent cone to a theta divisor: its degree-$r$ equation is $\det (\xi ⌣-)$, which is not the zero polynomial. For $r=1$, the tangent hyperplane is $⟨\xi ,st⟩=0$, in agreement with the annihilator description of the derivative of the Abelian sum map. For $r\ge 2$ all first derivatives vanish, proving singularity. In genus $1$ the theta divisor is a single smooth point of multiplicity $1$. In genus $0$ the effective degree-$-1$ locus is empty, so there is no singularity statement at a point of it.