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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.

Interpret the connectivity hypothesis in its usual based sense: $X$ is $(n-2)$-connected, with $n\ge 2$ for the nontrivial suspension range. Form the suspension as two cones $C_{-}X$ and $C_{+}X$ meeting in $X$. Both cones are contractible. Their relative homotopy sequences give ${\pi }_j(C_{\pm }X,X)\cong {\pi }_{j-1}(X)$ in the relevant degrees, so both cone pairs are $(n-1)$-connected.

The homotopy excision theorem says that if two CW pairs meeting in a connected subcomplex are $p$- and $q$-connected, $p,q\ge 1$, the map from the first relative pair to the union relative to the second part is an isomorphism in degrees $i<p+q$ and a surjection in degree $p+q$. Here $p=q=n-1$, givingas an isomorphism for $i<2n-2$ and a surjection for $i=2n-2$. Contractibility identifies the domain with ${\pi }_{i-1}(X)$ and the target with ${\pi }_i(\Sigma X)$. The comparison under these identifications is the suspension map: a sphere representative is coned in each half and the two cones give its suspended sphere representative. Hence the Freudenthal suspension from two cones provesFor connected CW complexes, reduced and unreduced suspension have the same based homotopy type, so either standard suspension convention gives this result. In the low-dimensional range, pointed sets are used where necessary; the displayed positive-degree homomorphism assertion concerns the usual groups.

For $S^r$ with $r\ge 2$, cellular approximation makes every sphere map of degree below $r$ null-homotopic: it can be homotoped into the point skeleton of the CW structure having cells only in degrees zero and $r$. The Hurewicz theorem therefore identifies its first possible nonzero homotopy group with homology. Its cellular chain complex gives $H_r(S^r;\mathbb{Z})=\mathbb{Z}$. For $r=1$, lifting a based loop to the universal cover $\mathbb{R}\to S^1$ classifies it by an integer endpoint displacement, so ${\pi }_1(S^1)=\mathbb{Z}$; the degree-one Hurewicz map is abelianization and is an isomorphism here. ThusAlternatively, after the $r=2$ Hurewicz base case, the suspension isomorphisms above propagate the generator to every higher sphere.

Use finitely generated right projective modules throughout. The group K0 of a ring is the Grothendieck group of their direct-sum monoid, with $[P\oplus Q]=[P]+[Q]$. The group K1 of a ring is ${\mathrm{GL}}_{\infty }(A)/E_{\infty }(A)$, equivalently the abelianization of the stable general linear group. Stabilization is $g\mapsto \mathrm{diag}(g,1)$; $E_{\infty }$ is generated by elementary transvections. The stable elementary subgroup is the commutator subgroup: elementary transvections are commutators using a third index, and stable block interchange and elementary row identities make the quotient abelian.

Write $A_0$ for the lower-right ring, printed as $A^{\prime }$ in the source. A Milnor square has unital commuting maps, is a pullbackand has at least one of $j_1,j_2$ surjective. Interchange the two sides if necessary so that $j_1$ is surjective.

We first establish the patching fact used in the proof. For projective $P_i$ and an isomorphism $\varphi :P_1\otimes A_0\to P_2\otimes A_0$, setFor free modules denote the result by $M_g$. The block matrix identityexpresses $\mathrm{diag}(g,g^{-1})$ as elementary matrices. Their entries lift through $j_1$, and their elementary lifts remain invertible. A change of basis on the first side then trivializes this gluing. Thus $M_g\oplus M_{g^{-1}}\cong A^{2r}$, proving that $M_g$ is finitely generated projective. Notice that we lifted elementary matrices, not arbitrary invertible matrices.

For general $P_1,P_2$, choose $P_1\oplus Q_1\cong A_1^r$ and $P_2\oplus Q_2\cong A_2^s$. The reductions of $Q_1\oplus A_1^s$ and $Q_2\oplus A_2^r$ are isomorphic: insert the common reduced $P_i$ and reorder the two complements. Patch them by such an isomorphism. The sum with $M(P_1,P_2,\varphi )$ now patches two free modules of equal size, so is projective by the free case, making each summand projective. The same free change of basis followed by these direct-summand projections shows that extension to $A_i$ recovers $P_i$. Conversely, write an $A$-projective as the image of an idempotent over a finite free module. Since matrices over the pullback are compatible pairs of matrices, applying the compatible idempotents to the free pullback reconstructs this projective from its two reductions. Compatible module isomorphisms patch as well. This proves Milnor patching of projective modules in the form needed below.

The diagonal maps in the sequence are extension of scalars; the two maps to $A_0$ are differences $j_{1\ast }-j_{2\ast }$. Define the connecting map byIt is unchanged by stabilization or elementary multiplication: an elementary factor lifts through $j_1$ and changes a free basis in the gluing. The identity $\mathrm{diag}(g,h)=\mathrm{diag}(gh,I)\mathrm{diag}(h^{-1},h)$ and the preceding elementary factorization show $M_g\oplus M_h\cong M_{gh}\oplus A^r$ after matching sizes. Thus $\partial$ is an additive homomorphism on $K_1(A_0)$.

Exactness at $K_1(A_1)\oplus K_1(A_2)$. If the two reduced classes agree, choose stabilized representatives $g_1,g_2$ of the same size. Their reductions differ by an elementary matrix $e$ over $A_0$. Lift $e$ to an elementary matrix over $A_1$ and adjust $g_1$ without changing its $K_1$ class. Its reduction now equals that of $g_2$. The compatible pair and its compatible inverse give an invertible matrix over $A$, so the original pair of classes is in the diagonal image. The reverse containment follows from commutativity of the square.

Exactness at $K_1(A_0)$. A gluing obtained from a change of basis over either upper ring patches a free module, so the difference-map image is killed by $\partial$. Conversely, $\partial [g]=0$ means that $M_g$ becomes free after adding a common projective summand. Add a complement of that summand to obtain an actual isomorphism $M_g\oplus A^s\cong A^{r+s}$. Its two local bases express $g\oplus I_s=j_2(v_2)j_1(v_1{)}^{-1}$. Therefore $[g]=j_{2\ast }[v_2]-j_{1\ast }[v_1]$, which lies in the image of the difference map. This uses stable equality of projective classes, not cancellation of arbitrary projectives.

Exactness at $K_0(A)$. Connecting classes reduce to zero on both sides. Conversely, write a class in the kernel as $[P]-[A^r]$ by adding a complement to its negative projective term. Its reductions have the same classes as $A_i^r$. Equality in $K_0(A_i)$ becomes an isomorphism after a finite common free summand, by taking a complement to the auxiliary projective. Use one sufficiently large free summand on both sides. The stabilized $P$ is then locally free with gluing $g$, so the original class is $[M_g]-[A^{r+s}]=\partial [g]$.

Exactness at $K_0(A_1)\oplus K_0(A_2)$. Write a kernel pair as $([P_1]-[A_1^r],[P_2]-[A_2^r])$, using complements and padding to make the free sizes equal. Their reduced positive classes agree. Adding a common finite free summand makes those reductions isomorphic. Patch the resulting projectives, and subtract the matching free $A$-module. Its restrictions are the desired pair. The reverse containment is immediate because both reductions to $A_0$ agree.

These arguments prove the Mayer-Vietoris sequence in algebraic K-theoryNo zero is inserted at either end; this fragment does not assert injectivity at the first term or surjectivity at the last.

For the group ring of the cyclic group, take the cyclotomic Milnor squareThe maps send the cyclic generator to $1$ and to $\zeta$; the lower map sets $\zeta =1$, and the right map reduces modulo $p$. Since ${\Phi }_p(1)=p$, $\mathbb{Z}[\zeta ]/(\zeta -1)={\mathbb{F}}_p$. To verify the pullback, represent an element by a polynomial of degree at most $p-1$. If both evaluations vanish, it is a constant multiple of the minimal polynomial of an algebraic element ${\Phi }_p$, and evaluation at one forces that constant to be zero. Conversely, represent $b\in \mathbb{Z}[\zeta ]$ by $f$ of degree below $p-1$. For a compatible integer $a$, $a-f(1)$ is divisible by $p$; replacing $f$ by $f+[(a-f(1))/p]{\Phi }_p$ yields the pair $(a,b)$. The right map is surjective, so this is a Milnor square.

The hinted cyclotomic units are $u_k=1+\zeta +\cdots +{\zeta }^{k-1}$. If $kl\equiv 1(\mathrm{mod}p)$ with $1\le l<p$, the geometric sum ${\sum }_{j=0}^{l-1}{\zeta }^{kj}$ is their inverse, because ${\zeta }^{kl}=\zeta$. Their residues are $k\in {\mathbb{F}}_p^{\times }$, covering every unit of the common field. Gaussian elimination identifies $K_1({\mathbb{F}}_p)$ with ${\mathbb{F}}_p^{\times }$ by determinant. Thus $K_1(\mathbb{Z}[\zeta ])\to K_1({\mathbb{F}}_p)$ is surjective, and the connecting map vanishes.

The remaining exact sequence injects $K_0(\mathbb{Z}[C_p])$ into $\mathbb{Z}\oplus K_0(\mathbb{Z}[\zeta ])$, with image $\{(r,x):r=\mathrm{rank}x\}$. Here $K_0(\mathbb{Z})=K_0({\mathbb{F}}_p)=\mathbb{Z}$, and reduction of a projective over the integral domain $\mathbb{Z}[\zeta ]$ has dimension its rank: localize at $(\zeta -1)$, where the projective is free, and compare its residue-field dimension with its fraction-field dimension. Projection onto the second factor is therefore a bijection on this kernel. Hence extension along the cyclotomic evaluation givesThe essential step is reduction of cyclotomic units at the ramified prime; surjectivity of a ring map alone would not imply surjectivity on $K_1$.

Choose for each $j\in \mathcal{J}$ a categorical limit $L_j$ of $F(-,j)$, with projections ${\lambda }_{i,j}:L_j\to F(i,j)$. For $v:j\to k$, the morphisms $F(1_i,v){\lambda }_{i,j}$ form a categorical cone over $F(-,k)$, since the two actions of the product functor $F$ commute. Define $L(v):L_j\to L_k$ by the equationsExistence and uniqueness follow from the universal property. The identity morphism of $L_j$ satisfies the equations for $v=1_j$, and $L(w)L(v)$ satisfies those for $wv$, so uniqueness gives $L(1_j)=1_{L_j}$ and $L(wv)=L(w)L(v)$. Hence $L:\mathcal{J}\to \mathcal{D}$ is a functor.

If $L_j^{\prime }$ with projections ${\lambda }_{i,j}^{\prime }$ are other choices, there is a unique isomorphism $u_j:L_j\to L_j^{\prime }$ carrying every projection to the corresponding projection. The same equations show $u_{k}L(v)=L^{\prime }(v)u_j$, because their composites with all ${\lambda }_{i,k}^{\prime }$ agree. Thus the choices give a unique natural isomorphism compatible with the chosen categorical limit categorical cones. Without the compatibility condition, uniqueness of an arbitrary natural isomorphism is not asserted.

For $D:\mathcal{I}\to \mathcal{C}$, define the functor $\mathrm{Cone}(-,D):{\mathcal{C}}^{\mathrm{op}}\to \mathbf{Set}$ bywhere $\Delta X$ is the constant diagram functor; a morphism into $X$ acts by precomposition on every categorical cone leg. A categorical limit is a representation of a functor of the formThe image of $1_L$ is the universal categorical cone $\lambda :\Delta L\to D$. By naturality of the representing isomorphism, the image of $h:X\to L$ is $\lambda \circ \Delta h$. Thus this definition is exactly existence and uniqueness of factorization of every categorical cone through $\lambda$. Conversely that universal property gives the representing natural bijection.

A precise commutation of iterated categorical limits statement is the following. For small categories $\mathcal{I},\mathcal{J}$ and $F:\mathcal{I}\times \mathcal{J}\to \mathcal{D}$, if every column categorical limit and the resulting $\mathcal{J}$-limit exist, thenIf every row categorical limit and the resulting $\mathcal{I}$-limit also exist, thenEach isomorphism is the unique one identifying all projections to $F(i,j)$. Indeed, the preceding part proves that each iterated categorical limit has the universal property of the whole product diagram in a category. Its uniqueness gives the canonical isomorphism. Equivalently, categorical cones with vertex $X$ over the product diagram in a category are exactly compatible families, indexed by $j$, of categorical cones over $F(-,j)$; regrouping the same compatibility equations cannot change their representing object. In a complete category all the stated existence hypotheses hold for small $\mathcal{I},\mathcal{J}$. The theorem is about categorical limits with categorical limits, and makes no unconditional claim about commuting a categorical limit with a colimit.

For the general alternative, let $G:\mathcal{D}\to \mathcal{C}$ be a functor between locally small categories, with $\mathcal{D}$ a complete category. The Freyd general adjoint functor theorem says that $G$ has a left adjoint if and only if it preserves small categorical limits and satisfies the solution-set condition. Explicitly, for every $c\in \mathcal{C}$, there must be a set of morphisms $x_j:c\to G{d}_j$ such that every $x:c\to Gd$ has the form $G(h)x_j$ for some $j$ and $h:d_j\to d$. Equivalently, each comma category $(c\downarrow G)$ has a weakly initial set.

We prove the initial-object lemma for complete categories with a weakly initial set. Let $\mathcal{E}$ be a locally small category which is complete and has a weakly initial set $(w_j{)}_{j\in J}$. Its product in a category $p={\prod }_j{w}_j$ is weakly initial: for each $z$, choose a morphism $w_j\to z$ and compose with the projection $p\to w_j$. The hom-set $\mathcal{E}(p,p)$ is a set. Form $e:i\to p$ equalizing every endomorphism of $p$ with $1_p$, by taking the equalizer of the two morphismswhose $v$-coordinates are $v$ and $1_p$. Then $ve=e$ for every endomorphism $v$, and $e$ is monic. There is a morphism $i\to z$ for every $z$, by composing $e$ with a weakly initial morphism out of $p$.

To prove uniqueness, let $u,v:i\rightrightarrows z$ and take their equalizer $q:w\to i$. Weak initiality of $p$ gives $a:p\to w$. The composite $eqa:p\to p$ is an endomorphism, so $eqae=e$. Cancel the monomorphism $e$ to obtain $qae=1_i$. Thus $q$ is both monic and split epic, hence an isomorphism: if $qk=1$, cancellation from $qkq=q$ gives $kq=1$. Since $uq=vq$, it follows that $u=v$. Therefore $i$ is an initial object, proving the lemma fully.

Now assume categorical limit preservation and the solution-set condition. The comma category $(c\downarrow G)$ is locally small. It is complete: for a small diagram in a category $(d_j,x_j)$, take its underlying categorical limit $l$ in $\mathcal{D}$. Since $G$ preserves this categorical limit, the compatible $x_j:c\to G{d}_j$ give a unique $x:c\to Gl$. The same universal property shows that $(l,x)$ is the categorical limit in the comma category. Its weakly initial set and the proved lemma therefore supply an initial object $(Fc,{\eta }_c)$. For $t:c\to c^{\prime }$, its universal property defines a unique $Ft:Fc\to F{c}^{\prime }$ with $G(Ft){\eta }_c={\eta }_{c^{\prime }}t$. Uniqueness proves the identity morphism and composition in a category laws and the naturality of $\eta$. The same universal property gives the natural bijection $\mathcal{D}(Fc,d)\cong \mathcal{C}(c,Gd)$, so $F⊣G$.

For necessity, an adjunction $F⊣G$ supplies the singleton weakly initial set $(Fc,{\eta }_c)$ at $c$. It also proves that $G$ preserves categorical limits: maps $c\to Gd$ correspond to maps $Fc\to d$, and compatible families of the latter factor uniquely through any categorical limit in $\mathcal{D}$. Hence the image categorical cone under $G$ has the required universal property. This proves both directions of the general alternative.

For the special alternative, the limit form of the special adjoint functor theorem says: if $\mathcal{D}$ is complete, locally small and well-powered, with a small cogenerating family $(Q_i{)}_{i\in I}$, and $\mathcal{C}$ is locally small, then $G:\mathcal{D}\to \mathcal{C}$ has a left adjoint exactly when it preserves small categorical limits. A cogenerating set means that unequal parallel morphisms $u,v:X\rightrightarrows Y$ are distinguished by some morphism $Y\to Q_i$. We prove the needed solution-set condition, so that the general theorem applies.

Fix $x:c\to Gd$. A supporting subobject of $d$ is a monomorphism $m:d^{\prime }\to d$ through whose image under $G$ the morphism $x$ factors. There is at least one, namely $1_d$. Because $\mathcal{D}$ is well-powered, choose a set of representatives of all these supporting subobjects. A limit-preserving functor preserves monomorphisms: a map is monic exactly when its diagonal into its self-pullback in a category is an isomorphism, and both the pullback in a category and this isomorphism are preserved. Consequently each factorization of $x$ through $Gm$ is unique. Form the intersection of subobjects of the supporting representatives, as a wide pullback in a category over $d$. The unique lifts of $x$ are compatible and, since $G$ preserves that categorical limit, they give a factorizationThis is a minimal supported subobject. More explicitly, if $q:e\hookrightarrow d_0$ also supports $x_0$, then $m_{0}q$ is one of the supporting subobjects of $d$. The intersection therefore factors through $m_{0}q$, giving $r:d_0\to e$ with $m_{0}qr=m_0$. Monicity of $m_0$ gives $qr=1_{d_0}$; monicity of $q$ then makes $q$ invertible.

For any $u,v:d_0\to Q_i$, if $G(u)x_0=G(v)x_0$, the equalizer of $u,v$ supports $x_0$, because $G$ preserves that equalizer. Minimality makes it invertible, so $u=v$. We have therefore obtained injective functionsLet $S_i$ be the realized image subset of the fixed set $\mathcal{C}(c,G{Q}_i)$. Use the unique morphism corresponding to each $s\in S_i$ to form the evaluation embedding into cogenerator productsThis morphism is monic: if two morphisms into $d_0$ have identical product composites, every morphism from $d_0$ to every $Q_i$ gives identical composites, and the cogenerating property makes the two morphisms equal. There is only a set of possible families of subsets $(S_i)$, and hence a set of possible products $P_S$. Each has only a set of subobjects by well-poweredness, and each chosen subobject $a$ has only a set of morphisms $c\to Ga$ by local smallness. Take all these pairs $(a,z:c\to Ga)$ as a weakly initial set. Every original $(d,x)$ receives a morphism from a pair representing its $(d_0,x_0)$, through $m_0$. Thus the set is weakly initial in $(c\downarrow G)$.

The general theorem now supplies a left adjoint. Necessity again follows from preservation of categorical limits by a right adjoint. This proves the special alternative, including its size argument. Using only the realized subsets $S_i$ avoids assuming that an arbitrary object admits a morphism into every cogenerator. Both adjoint functor theorem alternatives are proved.

We use rooted-tree homeomorphic embedding: an injective vertex map preserves ancestors and lowest common ancestors, while an edge may map to a longer path. The root may map below the host root. Write $T⪯U$ for this relation. The assertion of Kruskal's tree theorem is that finite rooted trees form a well-quasi-ordering: every infinite sequence has $i<j$ with $T_i⪯T_j$.

We first supply the word lemma needed in the proof. In a well-quasi-order $Q$, every sequence has an infinite nondecreasing subsequence. Indeed, if no term had infinitely many later terms above it, successive choices beyond the finite sets of later dominators would give a bad sequence. Some term therefore has infinitely many later dominators; repeat the argument inside that infinite subsequence, which is still well-quasi-ordered, to build a chain.

Now order finite words over $Q$ by subsequence embedding with coordinatewise increase. Suppose a bad word sequence exists and choose one $w_0,w_1,\dots$ minimally: at each position use a word of least length allowing a bad continuation of the fixed prefix. No word is empty. Write $w_i=v_i{a}_i$, removing its last letter. Choose indices $i_0<i_1<\cdots$ with $a_{i_0}\le a_{i_1}\le \cdots$. The sequencewould also be bad. An old word embedding into $v_{i_j}$ embeds into $w_{i_j}$, contrary to the original badness; and $v_{i_j}⪯v_{i_l}$ would extend, using the last-letter comparison, to $w_{i_j}⪯w_{i_l}$. But $v_{i_0}$ is shorter than $w_{i_0}$, contradicting minimality. This proves Higman lemma, including its empty-word case.

Suppose next that there is a bad sequence of finite rooted trees, and choose a minimal bad sequence $T_0,T_1,\dots$ by vertex count. Let $\mathcal{S}$ consist of all proper rooted subtrees hanging below vertices of the $T_i$. This collection is well-quasi-ordered. Otherwise choose a bad sequence $S_0,S_1,\dots$ in $\mathcal{S}$. Its containing-tree indices can be selected strictly increasing, say $S_j$ is a proper subtree of $T_{i_j}$: a finite initial collection of containing trees has only finitely many proper subtrees, and a bad sequence cannot repeat an isomorphism class. Thenis bad. No earlier $T_l$ can embed into $S_j$, because $S_j⪯T_{i_j}$ and this would contradict the original sequence. The new tail is bad by construction. Since $|S_0|<|T_{i_0}|$, this contradicts minimality.

List each tree's immediate root subtrees in any chosen order. These are finite words in the well-quasi-order $\mathcal{S}$, so the word lemma gives $i<j$ for which the child word of $T_i$ embeds into that of $T_j$. Send the root of $T_i$ to the root of $T_j$, and use the chosen embeddings inside the selected, distinct target branches. The connecting paths from the target root to the images of the source children have disjoint interiors; lowest common ancestors in different branches are preserved. This constructs $T_i⪯T_j$, the final contradiction. Thus finite rooted trees are well-quasi-ordered by homeomorphic embedding. If labels lie in an arbitrary well-quasi-order, the same proof uses label-monotone embeddings; compare root labels and child words in their product well-quasi-order. The product property follows by taking a nondecreasing subsequence in the first coordinate and a good pair in the second.

For Friedman's finite form of Kruskal's theorem, fix $k$. Consider all finite bad prefixes $(T_1,\dots ,T_l)$ satisfying $|T_i|\le k+i$, up to rooted-tree isomorphism. They form a prefix tree. At each node there are finitely many possible next trees, because there are only finitely many rooted-tree isomorphism types on at most $k+l+1$ vertices. If prefixes of arbitrary length existed, the König infinity lemma would give an infinite branch, hence an infinite bad sequence, contrary to what we proved. Its height is therefore bounded. ConsequentlyThe same finite-prefix argument works with a fixed finite label alphabet, and more generally with any prescribed size bound $b(i)$ having only finitely many possibilities at each position. These are finite existence statements; no numerical estimate on $N$ is needed for the conclusion.