An integral counterexample to Gersten’s conjecture

Proof. The cofiber sequence for M, together with \pi_1\mathbb S=\pi_2\mathbb S=\mathbb Z/2, gives \pi_1M=\pi_2M=0. Applying maps into M to this cofiber sequence shows that restriction along the unit \eta:\mathbb S\to M is injective on

Thus 5\operatorname{id}_M=0. The cofiber sequence

therefore permits an extension of \operatorname{id}_M to a pairing \mu. Its restriction along 1\wedge\eta is also the identity: these endomorphisms of M agree after restriction along \eta. Moreover, [\Sigma M,M]=0, by the same cofiber sequence and the two vanishing homotopy groups. The extension is consequently unique. Smashing 5\operatorname{id}_M=0 with any spectrum proves the assertion about coefficients. The products are induced by the usual tensor pairings of K-theory spectra and this fixed pairing on M.□

Proof. Tensoring with a finitely generated projective R-module is exact on coherent modules and preserves every specified support. Tensor product therefore gives compatible pairings by K(R) on the supported, whole, and localized categories. These pairings respect the null composite in the localization diagram and hence its cofiber sequence. On a dévissage category, tensoring is tensoring with the restriction to the residue field. The projection formula for a finite map and the usual flat base-change identity are identities of these tensor functors. For each term B of this localization triangle, use the pairing

where the last arrow uses the integral tensor action and the fixed pairing on M. These pairings commute with all arrows of the triangle. Precomposing with the representative \beta:\mathbb S^2\to K(R)\wedge M gives the asserted diagrams for b; no integral lift of \beta is required. Its degree is even, so moving it across a boundary contributes no sign. Passage to a generic point follows by filtered localization. These assertions concern localization boundaries; the coefficient connecting map \partial_{\mathrm{coef}} is a different map.□

Proof of Lemma 2.3. We use the corrected, strongly convergent motivic spectral sequence with its finite-coefficient multiplicative structure:

This form of the spectral sequence is recorded in [27, VI.4.2, Addendum VI.4.2.1, and VI.4.4]. For the corrected construction, see [22, Introduction and Theorem 6.1] (the locators use the Trudy pagination). The product is the usual K-theory product on the abutment and the motivic cup product on E_2 [4, Proposition 16.1 and Theorem 16.2]. The Moore-space comultiplication used there corresponds by stable duality to a unital pairing on M. Its coefficient product agrees with ours by the uniqueness in Lemma 2.1.

Write an E_2-term as H^s(H,\mathbb Z/5(j)), of total K-degree 2j-s. Terms of negative weight vanish. For j\geq0, the defining motivic complex C_*\mathbb Z_{\mathrm{tr}}(\mathbb G_m^{\wedge j})[-j] is concentrated in cohomological degrees at most j. Global sections on the Nisnevich site of a field are exact, since every covering of the field point has a splitting, and derived reduction modulo five does not increase this upper degree. Thus the field motivic group vanishes when s>j. For s\leq j, the norm-residue and Beilinson–Lichtenbaum comparison theorems identify it with

[25, Theorems 6.16–6.17]. These are the Rost–Voevodsky norm-residue theorem and its motivic comparison consequence, with the completion of the norm-residue argument in [26]; the earlier motivic formulation is [23]. In particular the terms of negative cohomological degree vanish. Under the dimension bounds below, only finitely many terms contribute to each total degree, so the resulting filtrations are finite. In these coordinates a differential d_\rho, for \rho\geq2, goes from (s,j) to (s+2\rho-1,j+\rho-1).

We first derive the algebraically closed case from this same spectral sequence. If \overline H is an algebraic closure of H, only the terms with s=0 survive. In each degree 2j\geq0 there is exactly one term, H^0(\overline H,\mu_5^{\otimes j})\cong\F_5, and in odd degree there is none. There are no possible differentials. Thus the edge map is an isomorphism in every even degree, and

The coefficient sequence surjects K_2(\overline H;5) onto \overline H^\times[5]=\mu_5(\overline H). Both groups have dimension one, so this boundary is an isomorphism. Consequently the restriction of \beta_H is nonzero, and so is its edge image. Multiplicativity now shows that each b^j(1) over \overline H, computed by iterated left multiplication, has nonzero edge image and generates the corresponding even group. This proves the last assertion of the lemma without a separate rigidity input.

By naturality, the edge image \alpha\in H^0(H,\mu_5) of \beta_H is nonzero: restriction on this H^0 group is injective, and its geometric image is the nonzero class just computed. The edge image of b^r(1) is \alpha^{\otimes r}. Since H contains \mu_5, this generates the one-dimensional group H^0(H,\mu_5^{\otimes r}). Hence this whole line consists of permanent cycles for every r\geq1. Only \alpha\ne0 is needed; no choice of an identification of \alpha with the specified primitive root is required.

Suppose first that \operatorname{cd}_5H\leq3. In total degree 2r, the only possible terms are

There is no incoming differential to either term. The only possible outgoing differential from the first is d_2 to an H^3-term, and it vanishes by permanence of the Bott power. Every outgoing differential from the second has cohomological degree at least 5 and is zero. We obtain a two-step filtration, expressed as

Multiplication by \beta_H preserves these filtrations and acts on both graded pieces by tensoring with \alpha. Each such map is an isomorphism, proving (1).

If \operatorname{cd}_5H\leq2, the sole term in total degree 2r+1 is H^1(H,\mu_5^{\otimes(r+1)}), with no possible nonzero differential. Multiplication by \beta_H again acts by tensoring with \alpha. This proves (2). For a finite field, the same calculation applies in every positive degree since its 5-cohomological dimension is one. In degree zero, the class 1 generates K_0(H;5) and b(1)=\beta_H generates K_2(H;5)=H^0(H,\mu_5). This proves (3).□

Proof. Matsumoto’s symbol theorem [10, §5, Theorem 5.10 and Corollary 5.11] gives the classical Steinberg-symbol presentation; see [27, III.6.1] for its modern formulation. The classical-to-Quillen comparison [27, IV.1.10.1 and IV.7.2] identifies this group with K_2(N). The degree-two norm-residue theorem [25, Theorem 6.16] then gives

Use \zeta to identify one copy of \mu_5 with the constant module \Z/5. The Kummer sequence identifies the resulting group H^2(N,\mu_5) with \Br(N)[5]. Write a for the Brauer class corresponding to the class in the lemma.

For a closed point v of \Spec S, the residue characteristic is not five. The Quillen boundary on K_2(N) is the tame symbol. Under norm residue and the chosen twist identification, its reduction modulo five and the Brauer residue agree up to an overall sign depending on boundary conventions; in particular, their vanishing is equivalent. Indeed, write each field element as a power of a uniformizer times a unit. Both residues kill unit–unit symbols, and on a symbol consisting of a unit \epsilon and a uniformizer they give the class of \bar\epsilon, up to the corresponding sign. Bilinearity and the Steinberg relations give the general case [27, V.6.1.2 and V.(6.6)] [9, Proposition 7.10]. Thus a has zero Brauer residue at every finite place v\nmid5.

The completion N_v has finite residue field k_v. For its henselian valuation ring, the kernel of the Brauer residue is \Br(k_v)=0 [9, Proposition 6.7 and Example 6.8(a)]. Consequently a|_{N_v}=0 at every v\nmid5. The infinite completions are complex and have zero Brauer group. Only the single place above five remains. In the global Brauer invariant sequence, the sum of all local invariants is zero, so the invariant at this last place is zero. The local invariant is injective there, and the global restriction map is injective. These are the local and global Brauer statements in [12, IV, Example 2.14(c)–(e), Proposition 4.3 and Remark 4.4(a)]; see also [27, III.6 and VI.8.1.1]. They imply a=0, hence the original class is zero. No tame-residue calculation at the place above five is used.□

Proof. Because S is a Dedekind ring with \Pic(S)=0, every divisor on \Spec S is principal. Equivalently, the normalized valuation map

is surjective. The sums here and below have finite support. Quillen localization and dévissage for S [18, §7, Proposition 3.2 and paragraph 3.4] identify the boundary from K_1(N;5) with this valuation map modulo five. We use the positive normalized valuation convention [27, Example V.6.1.2 and (V.6.6)]. Thus

is surjective.

The number field N has 5-cohomological dimension two [27, VI.§7 and VI.8.2]. The odd-degree clause of Lemma 2.3 therefore makes b^2:K_1(N;5)\to K_5(N;5) an isomorphism. For every v, the finite field k_v contains the primitive image of \zeta, so the finite-field clause makes b^2:K_0(k_v;5)\to K_4(k_v;5) an isomorphism as well. The same restricted class \beta acts on every term. Its compatibility with localization in Lemma 2.2 carries the last surjection to a surjection

In the exact localization segment

the middle arrow is consequently zero. We obtain an injection

We next determine where this injected group can lie. Take x\in K_4(S;5) and write x_N for its image. The even-degree clause of Lemma 2.3 gives b:K_2(N;5)\xrightarrow{\sim}K_4(N;5), so x_N=b(z) for some z\in K_2(N;5). The localization residues of x_N in K_3(k_v;5) vanish because x extends over S. Write \partial_v for the localization boundary at v. Compatibility with b gives

Here \partial_v z\in K_1(k_v;5), and b:K_1(k_v;5)\to K_3(k_v;5) is injective by the finite-field clause. Thus every localization residue of z is zero.

On the other hand, \partial_{\mathrm{coef}}z\in K_1(N)[5]=N^\times[5]=\langle\zeta\rangle. Choose m\in\F_5 such that this boundary equals m[\zeta], and consider the corrected class z-m\beta_N, where \beta_N is the restriction of \beta. Equation (3.1) makes its coefficient boundary zero. Its localization residues are still zero, since \beta extends over S. Exactness of the coefficient sequence identifies this corrected class with an element of K_2(N)/5. Naturality of reduction with localization identifies its coefficient residues with the tame residues of that element. Lemma 3.1 therefore makes the corrected class zero. We have proved

Let \overline N be an algebraic closure of N. By the closed-field clause of Lemma 2.3, the restriction of b(\beta_N) is the nonzero square of a geometric Bott generator in K_4(\overline N;5). Restriction is therefore injective on the \F_5-line spanned by b(\beta_N). Together with (3.2), this proves

It remains to pass from coefficient detection to integral torsion. The same closed-field clause gives K_5(\overline N;5)=0. The coefficient exact sequence then gives K_4(\overline N)[5]=0. In fact K_4(\overline N) has no nonzero element of any finite five-power order: a suitable multiple of such an element would have order five.

Set G=K_4(S), and write \rho for coefficient reduction at each ring. Naturality gives the commutative square

The right vertical arrow is injective by (3.3). If g\in G has finite five-power order, its integral image in K_4(\overline N) is zero. The square therefore gives \rho(g)=0. Since G/5 injects into K_4(S;5), this says g\in5G. This conclusion holds for every element of five-power order, not just for elements killed by five.

Finite generation now rules out a nonzero divisible primary subgroup. Quillen’s theorem makes K_4(\mathcal O_N) finitely generated [17, §1, Theorem 1 and Remark (2)]; see also [27, IV.6.9]. Integral localization at the single place above five has the segment

The two finite-field values are Quillen’s integral computation [16]; see [27, IV.1.13]. Thus G is finitely generated. Write G\simeq\Z^r\oplus P\oplus T, where P is its finite five-primary torsion subgroup and T is finite of order prime to five. We proved P\subseteq5G. If p=5g\in P, the free component of g is zero, and its T-component is zero because multiplication by five is an automorphism of T. Hence g\in P, and P\subseteq5P. The reverse inclusion is automatic. Iterating P=5P to the exponent of the finite group P gives P=0. In particular K_4(S)[5]=0.□

Proof. Lemma 3.2 and the coefficient exact sequence give

Consequently K_5(S)\to K_5(S;5) is surjective, and b^2[u]_5 has an integral lift c. Applying \delta after reduction, then using naturality and bilinearity of the parenthesized product, gives

This proves (3.5); functoriality gives its restriction along \iota. The construction made no equivariant choice of the integral lift c.□

Proof. The sum in (4.1) is zero when m=0, so its nonvanishing implies m>0. Suppose, for a contradiction, that the sum in (4.2) is zero. We first turn this assumed relation into prescribed ordinary tame symbols on the generic curve.

Localization and Bott untwisting. Put J=L(C) and A_U=\Gamma(U,\mathcal O_U). The ring A_U is a localization of the coordinate ring of a smooth affine curve. It is therefore a Noetherian regular domain of dimension at most one, with fraction field J. Since one of the p_i is retained, its dimension is one. Every finitely generated torsion A_U-module is killed by a nonzero element; the corresponding quotient of A_U has dimension at most zero and is Noetherian, hence Artinian. Such a module has finite length, with simple factors the residue fields of the retained closed points. This argument does not require U to be semilocal or the multiplicative set to be finite.

Quillen localization for the torsion category, followed by dévissage and the coefficient convention in Section 2, gives the exact sequence

The second arrow is the sum of coherent pushforwards from the points; the first is the vector of localization residues. These are ordinary localization and dévissage for a multiplicative set [18, §4, Theorem 3 and Corollary 2; §5, Theorems 4–5; §7, Proposition 3.2 and paragraph 3.4]. The index set of this direct sum can be infinite, but each vector has finite support. Every q\in U^{(1)} is a closed point of C, so \kappa(q) is a finite extension of L.

Exactness now supplies \xi\in K_6(J;5) whose residue at p_i is e_i b^2[a_i]_5 and whose other residues in U are zero. The field J has \operatorname{cd}_5J\leq3, and each \kappa(q) has \operatorname{cd}_5\kappa(q)=2. All these fields are perfect and contain the same primitive root by restriction from L. Thus Lemma 2.3 applies to the actual multiplication operations used here: it makes

Choose \xi_0\in K_2(J;5) with \xi=b^2\xi_0. Write \partial_q^{(5)} for a coefficient localization residue. The compatibility in Lemma 2.2 gives \partial_q^{(5)}(b^2\xi_0)=b^2\partial_q^{(5)}\xi_0, with no sign because b has even degree. Injectivity on the residue fields then yields

There is still a coefficient obstruction to representing \xi_0 integrally. Its coefficient boundary belongs to

The last equality holds because L already contains all fifth roots of unity. Hence there is \lambda\in\F_5 with \partial_{\mathrm{coef}}\xi_0=\lambda[\zeta]. Let \beta_U and \beta_J be the restrictions of \beta_L. Naturality gives \partial_{\mathrm{coef}}\beta_J=[\zeta], while all localization residues of \beta_J on U vanish because it extends as \beta_U. This uses an extension over the L-scheme U, not an extension over R. The class

therefore has zero coefficient boundary and the same residues as \xi_0. The coefficient exact sequence in Section 2 supplies an integral representative z\in K_2(J) of \eta. Write \partial_q z\in K_1(\kappa(q))=\kappa(q)^\times for its ordinary tame symbol. Reduction of localization residues commutes with coefficient reduction, so (4.4) becomes

In particular, \partial_q z is a fifth power in \kappa(q)^\times at every retained point other than the p_i. We have obtained an ordinary K_2-class with the required control of its tame symbols. It remains to derive the relation that these symbols must satisfy on D.

The horizontal divisor. For each closed point q of C, let R_q be the valuation ring of the unique extension of the valuation of L to \kappa(q), let k_q be its residue field, and let v_q be its normalized valuation. The extension \kappa(q)/L is finite, so R_q is a DVR finite over R. Properness gives the extension

used to define r(q). This map is finite. Indeed, its graph in \Spec R_q\times_R\mathcal X is a closed immersion because \mathcal X/R is separated, and the projection from this fiber product to \mathcal X is finite, being the base change of \Spec R_q\to\Spec R. Set

Smoothness over the regular DVR R makes \mathcal X a regular surface, and its function field is J. The codimension filtration for coherent K-theory gives the following row of the coniveau complex:

The assertion needed here is only that the composite is zero: the first differential d_1 of the coniveau spectral sequence satisfies d_1^2=0 [18, §7, Theorem 5.4 and equation (5.5)]. There is exactly one vertical point in \mathcal X^{(1)}, the generic point of the integral curve D. Its residue \partial_Dz\in k(D)^\times contributes the principal divisor \operatorname{div}_D(\partial_Dz). The other codimension-one points are the closed points q of C.

We compute their contributions without assuming that a horizontal closure is normal. On \Spec R_q, the localization boundary

is the normalized valuation v_q: a unit has boundary zero, and a uniformizer \pi_q has boundary [R_q/\pi_qR_q]=[k_q] [27, V.6.1.2 and V.(6.6)]. Functoriality of coherent localization for the finite map j_q identifies the horizontal boundary on \mathcal X with the pushforward of this DVR boundary [18, §7, paragraphs 2.7–2.8, Proposition 3.2 and paragraph 3.4]. At the generic point this pushforward is the identity on \kappa(q). At the closed point it sends [k_q] to [k_q:\kappa(r(q))]\,[\kappa(r(q))]. Thus the contribution at r(q) is exactly f_qv_q(\partial_qz). There is no additional ramification factor: the boundary already uses the normalized valuation on \kappa(q), and the closed-point pushforward contributes only the residue degree f_q.

Define the horizontal divisor

The sum is finite because the first arrow of (4.6) takes values in a direct sum. All codimension-two points of the proper surface lie on D. The identity d_1^2z=0 therefore says, as an equality of divisors on D,

Consequently

The Picard test. Apply \psi to (4.8). If r(q)\in Z, the corresponding summand vanishes because \psi([r(q)])=0. If r(q)\notin Z, the hypothesis on U retains q. For such a point other than the p_i, (4.5) makes v_q(\partial_qz) divisible by five, so its contribution also vanishes in B. Finally, \kappa(p_i)=L, k_{p_i}=k=\kappa(r_i), and hence f_{p_i}=1 and v_{p_i}=v_L. The remaining congruence in (4.5) gives

The identity \psi([H(z)])=0 is therefore

contrary to (4.1). This proves (4.2).□

Proof. The underlying curve E has four points over \F_5. At s=0 there are two affine points, while at each nonzero s the value s^4+1=2 is a nonsquare. There are also two rational points at infinity, since the leading coefficient of the quartic is a square. Choosing any rational point as origin over \F_5, the Frobenius polynomial is therefore T^2-2T+5, and its roots compute extension-field point counts [19, Theorem V.2.3.1]. If a_m is the sum of the mth powers of its roots, then

It follows that a_4=-14 and

This count is independent of the choice of origin. Hence, with origin o, the finite group E(k_0) has a point Q\notin5E(k_0).

Such a point remains nondivisible by 5 over k. In fact, if Q=5R with R\in E(k), the trace for the quadratic extension k/k_0, using the group law defined over k_0, would give

Multiplication by 2 is invertible on E(k_0)/5E(k_0), a contradiction. Every branch point is 2-torsion and therefore divisible by 5, so Q is not a branch point. A fiber of the degree-two morphism h has a point over an extension of k_0 of degree at most two. Since k is its quadratic extension, we may choose P\in D(k) over Q. The description of h on Z=0 gives Z(P)\ne0.

Finally, on the dense chart Y\ne0 the morphism h\circ g_a fixes s and multiplies w by a^{-2}. It therefore agrees with h when a^2=1 and with [-1]\circ h when a^2=-1. These are identities of morphisms on all of D: they extend from the dense chart because D is integral and the target is separated. This also covers the possibility that Q is a point at infinity. Thus

Each weighted summand in (5.4) equals Q modulo 5. Their sum is 4Q, which is nonzero because multiplication by 4 is invertible on E(k)/5E(k).□

Proof. The regular local ring V[x,y]_{(5,x,y)} has dimension three. The equation 5+x^4+y^4 is congruent to 5 modulo the square of its maximal ideal, so it is part of a regular system of parameters. Its quotient is regular local of dimension two and is a domain [24, Tags 00NQ and 00NP].

Equations (6.1) and (6.3) show that the assignment respects the defining equation. Every denominator inverted in (6.4) has nonzero constant residue at x=y=0, hence maps to a unit of \OO. Thus the homomorphism is defined and local. At least one of X_0,Y_0 is a unit of V, so its image contains t. Surjectivity follows from (6.2).

The element Y_0x-X_0y belongs to the kernel and has nonzero image in \mathfrak m_A/\mathfrak m_A^2. In particular the kernel is nonzero. It is a nonmaximal prime, since the quotient \OO is a DVR, and therefore has height one in the two-dimensional local domain A. Since Y_0x-X_0y is a regular parameter, it generates a height-one prime and hence generates this kernel.□

Proof. The V-module L is flat, since V\to H_0 is a localization and L is a vector space over H_0. The substitution x=tX, y=tY transforms 5+x^4+y^4=0 into X^4+Y^4=1. Tensoring and localizing therefore give the asserted description of T. Its coordinate ring is a localization of the coordinate ring of a smooth geometrically integral affine curve over L; it is regular, integral, and Noetherian, of dimension at most one.

Since 5 is invertible in the second tensor factor, the coordinate ring of the base-changed divisor is

Here \OO[1/5]=L, and the last isomorphism is the usual splitting of a finite Galois extension. The images of x and y on the \delta-factor are \delta(t)X_0 and \delta(t)Y_0, giving (6.6). Every inverted denominator evaluates to a unit of \OO, so these factors indeed define points of T. They are distinct: the ratios \delta(t)/t are the four different fourth roots of unity, and at least one of X_0,Y_0 is a unit. Thus T has closed points and has dimension one.

The coordinates in (6.6) are integral over \OO and satisfy the Fermat equation, so they define sections of \Dcal. Reducing (6.1) gives their stated specializations. The last assertion follows directly from the factor maps in (6.7).□

Proof of Theorem 1.1. Let c\in K_5(S) be the class from Proposition 3.3, and write c_L\in K_5(L) for its restriction along S\to L. There is no map S\to\OO in this construction, since 5 is invertible in S. Instead, integral localization for the DVR \OO gives

where the last equality is Quillen’s computation for finite fields [16]; see also [27, IV.1.13]. The residue field here has characteristic five; the equality is the integral finite-field calculation. Choose an integral lift \widetilde c\in K_5(\OO). The closed immersion i:\Spec\OO\hookrightarrow\Spec A of Lemma 6.1 defines

The equality uses the canonical resolution isomorphism for the regular ring A. Restriction to \operatorname{Frac}A annihilates every coherent module supported on \Spec\OO. Thus \gamma is in the kernel of the canonical augmentation, integrally.

To show that it is nonzero, make the flat base change V\to L. The square

is Cartesian by Lemma 6.2. To check flat base change with the point indices, put B_T=A\otimes_VL. For every finitely generated \OO-module M, scalar extension gives natural isomorphisms of B_T-modules

The \delta-summand is a B_T-module through the quotient defining p_\delta. These are exact functors: B_T is flat over A, and each map \OO\to L is a localization followed by an automorphism. Thus (7.1) proves the required identity on integral coherent K-theory.

Write \iota:S\to L for the fixed embedding. Since \widetilde c restricts to c_L=\operatorname{res}_\iota c, its image on the \delta-factor is

The equality uses the specified identification \Delta=\Gal(L/H_0) and requires no equivariance of the chosen DVR preimage. Consequently, still integrally,

Naturality of coefficient reduction and Proposition 3.3 now give

Here b uses the fixed embedding S\to L, and the character factors come from Proposition 3.3. No choice of the integral lift \widetilde c affects this formula.

Apply Proposition 4.1 to the model \Dcal/\OO, to U=T, and to the set Z_\infty and homomorphism \psi constructed in Section 5. The required specialization condition follows directly from the definition of A: let q be any closed point of C and let \OO_q be the valuation ring of its finite residue-field extension \kappa(q)/L. If q specializes in the chart Z\ne0, the proper extension \Spec\OO_q\to\Dcal lies in that chart, since an open set containing the closed point of a DVR contains its generic point. Hence X/Z and Y/Z belong to \OO_q. The original coordinates x=tX/Z and y=tY/Z have positive valuation. Every denominator inverted outside (5,x,y) has nonzero constant residue and hence is a unit at q. Thus q belongs to T.

All the hypotheses of Proposition 4.1 are now verified: the smooth model and localized chart come from Section 6, the retained-point condition holds for every closed point, and (5.3) says that \psi kills the entire set Z_\infty. The points p_\delta specialize to g_{\chi(\delta)}P and are L-rational. Every \delta u is a uniformizer, so v_L(\delta u)=1. The sum tested by Proposition 4.1 for (7.2) is therefore

by Lemma 5.1. It follows that \overline{\gamma|_T}\ne0. Hence \gamma\ne0 in the original integral group, as required.□

Proof. Take the encoded endomorphisms of X=(1,1)_B

Their actual chain matrices are

In each degree the determinants of U,U',L,L' are, respectively, 1,1-g,1,1-g. All are units, so these are chain isomorphisms. The first two are upper triangular and the last two lower triangular. A direct calculation with (A.2) gives

Equivalently, both degree-one composites are \left(\begin{smallmatrix}1&1\\g&1\end{smallmatrix}\right) and both degree-zero composites are \left(\begin{smallmatrix}1&g\\1&1\end{smallmatrix}\right). Hence the diagram

commutes. Each row is split exact since its first arrow is an isomorphism. Besides that first arrow, all structure arrows of the corresponding S_2-object are zero or identity maps and are therefore upper triangular. The vertical maps (L,L',\operatorname{id}_0) are lower triangular isomorphisms. Thus the diagram is a morphism in the proposed source category.

Both upper-left entries of L,L' are 1, whereas those of U,U' are 1,1-g. Either assignment would therefore send the left square to

Its two composites are 1 and 1-g in both degrees. They are unequal because g\ne0. The proposed image is therefore not a morphism of S_2-diagrams.□