Draft lesson.

Exact bifunctors and the canonical field Künneth map

Original text public domain (CC0).

An exact functor preserves the cycle and boundary objects of a complex. Applied in either variable of a bifunctor, this gives a quasi-isomorphism of finite diagonal totals. Over a field, tensors of cohomology classes give a canonical isomorphism on total cohomology. We construct that map before choosing any splittings, then prove its invertibility by an explicit contraction and verify its graded swap sign.

Use Finite diagonal totals and signed double truncations, Theorem 5.1 and its transposed version, and Signed bifunctors and sum and product total complexes, Sections 3–6 and 8, for the actual total differential and the separately additive tensor construction. Cohomology in abelian categories with translation, Section 5, fixes ordinary cohomology, and Hom, tensor and the exactness of module limits, Section 1, supplies tensor exactness over a field. Work in a fixed universe and use choice for vector-space bases, as in Universes and small categories.

1. Exact functors preserve the actual cohomology quotient

Let \(G:\mathcal A\to\mathcal B\) be an additive exact functor between abelian categories. Exactness here means preservation of short exact sequences. For a complex \(X\), write

Lemma 1.1. There is a canonical natural isomorphism

for every integer \(i\). In particular \(G\) sends a quasi-isomorphism to a quasi-isomorphism.

Proof. Factor any map \(a:U\to V\) as its image epimorphism followed by its image monomorphism. The two short exact sequences

remain exact under \(G\). Thus the factorization of \(G(a)\) has an epimorphism onto \(G(\operatorname{im}a)\) and a monomorphism from it into \(G(V)\). The abelian coimage-to-image comparison identifies that object, with its two specified arrows, with \(\operatorname{im}G(a)\). The first row of (1.3) identifies \(G(\ker a)\) with \(\ker G(a)\), since the final image inclusion is monic. The second identifies \(G(\operatorname{coker}a)\) with the corresponding cokernel.

These identifications are natural in a commuting square of maps. For kernels and cokernels this follows from uniqueness of the universal factors; for images it follows by factoring the square through the image epimorphisms and cancelling the image inclusions. Apply the identifications to the two adjacent differentials in (1.1). They carry the actual inclusion \(B_X^i\hookrightarrow Z_X^i\) to the boundary inclusion inside the cycle object of \(GX\): both inclusions compose to the same map into \(G(X^i)\), so cancellation of its cycle inclusion identifies them. Finally apply \(G\) to \(0\to B_X^i\to Z_X^i\to H^i(X)\to0\). Its quotient is the cycle/boundary quotient of \(GX\), giving (1.2). The preceding universal comparisons also commute with every chain map, proving naturality. A chain map which induces isomorphisms on all \(H^i\) therefore retains that property after applying \(G\). \(\square\)

If \(G=F(-,V)\) is one variable of a bifunctor, (1.2) is natural in \(V\) too. A map \(V\to V'\) gives a natural transformation between those fixed-variable functors; the same universal kernel, image and quotient equations commute with its components. The identical assertion holds for \(F(U,-)\), natural in \(U\). This extra naturality will supply the other differential of a cohomology complex.

2. Both variables of an exact bifunctor

Let \(F:\mathcal A\times\mathcal B\to\mathcal D\) be separately additive and exact in each variable, with all three categories abelian. For complexes \(X,Y\), its commuting double complex has terms and differentials

The functor square makes these differentials commute, and separate additivity makes their squares zero. Set \(F^\bullet(X,Y)=\operatorname{Tot}F(X,Y)\), with differential \(d'+(-1)^id''\) on the \((i,j)\) summand.

Theorem 2.1. Suppose that \(f:X\to I\) and \(g:Y\to J\) are quasi-isomorphisms of bounded-below complexes in their respective categories. Then

is a quasi-isomorphism.

Proof. Choose lower bounds \(a,b\) common to \(X,I\) and to \(Y,J\), respectively. Every possibly nonzero term on diagonal \(n\) has

The interval is finite and is empty for \(n<a+b\). A zero entry in either variable gives zero because its fixed-variable functor is additive. Thus all three doubles needed for the two-step comparison have finite diagonal support; their totals require only finite biproducts in \(\mathcal D\).

First fix \(Y\). For each second degree \(j\), apply Lemma 1.1 to \(F(-,Y^j)\). It gives

Naturality in \(Y^j\to Y^{j+1}\) identifies the induced second differentials with \(F(1,d_Y^j)\). Thus (2.4) identifies actual complexes in degree \(j\), not just their separate objects. Under these identifications, \(H_I^p(F(f,1_Y))\) is the complex map with components \(F(H^p(f),1_{Y^j})\). Each component is invertible, so the map of complexes is invertible: its componentwise inverses satisfy the chain equations by cancellation. Consequently \(F(f,1_Y)\) induces an isomorphism on \(H_{II}H_I\). Theorem 5.1 of the prerequisite makes its total a quasi-isomorphism.

Next fix \(I\). Applying Lemma 1.1 to \(F(I^i,-)\) gives complexes in degree \(i\)

whose first differentials are induced by \(F(d_I^i,1)\). The map \(F(1_I,g)\) induces the componentwise isomorphisms \(F(1_{I^i},H^q(g))\), so it induces an isomorphism on \(H_IH_{II}\). This time use the transposed comparison theorem, with its proved total multiplier \((-1)^{ij}\). Its total is a quasi-isomorphism as well. On every double-complex component the functor law gives

The total has that same factorization, and the composition of the two cohomology isomorphisms proves (2.2). No splitting, injective resolution or derived-functor assumption has entered the proof. \(\square\)

The same proof applies to two bounded-above input pairs. If their common upper bounds are \(a,b\), the finite interval becomes \(n-b\le i\le a\), with total zero for \(n>a+b\). Both mixed doubles in (2.6) retain those bounds. This variant will also explain the finite diagonal condition in the field formula below.

3. The canonical map from classes over a field

Let \(k\) be a commutative field and let \(X,Y\) be bounded-above complexes of \(k\)-vector spaces. If they vanish above degrees \(a,b\), respectively, then each diagonal of \(X^i\otimes_kY^j\) has the finite interval \(n-b\le i\le a\). Its total differential is

The construction and the square-zero check are the tensor specialization of (2.1). Tensor is exact in each variable here: every short exact vector-space sequence splits, by extending a basis of its kernel to a basis of its middle space and using the remaining vectors as a complement. Every additive functor preserves a split sequence, since it preserves the identities giving the inclusion, projection and their splitting. Tensor is separately additive by its bilinear universal property. Hence both fixed-variable tensor functors are exact, also for infinite-dimensional spaces. This is the exact interface of the module prerequisite cited above.

For cycles \(x\in Z_X^i\), \(y\in Z_Y^j\), their tensor is a total cycle. Its class depends only on the two cohomology classes. If \(x\) is changed by \(d_Xu\), with \(u\in X^{i-1}\), then

If \(y\) is changed by \(d_Yv\), with \(v\in Y^{j-1}\), then

Both use that the other factor is a cycle. Changes in both variables may be made successively, since the changed factors remain cycles. This also accounts for the mixed boundary term. Addition and scalar multiplication of cycle representatives therefore give a well-defined bilinear map on the two cohomology quotients. The tensor universal property and the finite diagonal direct sum give a canonical linear map

This sum is finite: the cohomology of \(X\) and \(Y\) also vanishes above \(a,b\). The empty sum is zero. A pair of chain maps carries cycles to cycles and boundaries to boundaries, and its tensor map sends \(x\otimes y\) to the tensor of their images. Evaluating on these generators proves full two-variable naturality of (3.4). Its definition has used no chosen representatives or splittings beyond the well-defined class construction.

4. An explicit contraction proves invertibility

Theorem 4.1. The map (3.4) is an isomorphism for every integer \(n\).

Proof. We construct a homotopy equivalence realizing exactly this map. In each degree choose complements

They exist by basis extension, with no finite-dimensional assumption. Make the choices for every integer degree using choice. The quotient identifies \(\widetilde H_X^i\) with the actual object \(H^i(X)\). The differential kills the first two summands and restricts to an isomorphism

Its kernel on \(L_X^i\) is zero by the complement to cycles, and its image is all of \(B_X^{i+1}\) because the other two summands have zero differential.

Let \(H_X\) be the complex with \(H_X^i=H^i(X)\) and zero differential. The chosen quotient sections and their complementary projections give chain maps

The section lands in cycles; the projection kills boundaries. Put \(E_X=s_Xp_X\). Define \(h_X^i:X^i\to X^{i-1}\) to be zero on \(\widetilde H_X^i\oplus L_X^i\), and on \(B_X^i\) the inverse of (4.2) in degree \(i-1\). Then \(d_Xh_X\) is the projection onto \(B_X\), and \(h_Xd_X\) is the projection onto \(L_X\). Consequently

Construct \(H_Y,s_Y,p_Y,E_Y,h_Y\) identically. The cohomology complexes remain bounded above with the same bounds. Tensoring the chain maps gives

between \(\operatorname{Tot}(H_X\otimes H_Y)\) and \(\operatorname{Tot}(X\otimes Y)\). All involved diagonal sums are finite. Define a degree-minus-one map on homogeneous tensors in the latter total by

Each term is bilinear and lands in total degree \(i+j-1\), so the tensor and finite biproduct universal properties define this map on the whole total.

For the first term of (4.6), the first-direction terms of \(DK+KD\) give \((d_Xh_X+h_Xd_X)x\otimes y\). The mixed second-direction terms have coefficients \((-1)^{i-1}\) and \((-1)^i\) and cancel. Thus its contribution is \((1-E_X)x\otimes y\). For the second term, its two second-direction coefficients are \((-1)^i(-1)^i=1\), giving \(E_Xx\otimes(d_Yh_Y+h_Yd_Y)y\). The first-direction terms cancel with coefficients \((-1)^i\) and \((-1)^{i+1}\), since \(E_X\) is a chain map; indeed both are already zero because \(d_XE_X=E_Xd_X=0\). Its contribution is \(E_Xx\otimes(1-E_Y)y\). Adding them gives

Hence (4.5) is a homotopy equivalence, with the specified homotopy and one composite literally identity. Its source has zero differential, so its degree-\(n\) cohomology is exactly the direct sum on the left of (3.4). The induced map \(H^n(S)\) sends \(a\otimes b\) to the class of \(s_X(a)\otimes s_Y(b)\). Those sections represent precisely the classes \(a,b\), so this map is \(\Phi_{X,Y}^n\), not a different isomorphism obtained from arbitrary splittings. This proves the theorem.

The chosen chain maps need not be natural. The map they realize on cohomology is the canonical map defined in Section 3, so it is independent of every complement and is natural by that section. Its inverse is natural as the unique inverse of a natural isomorphism. All integers \(n\), including degrees with an empty diagonal, are covered. \(\square\)

5. The ordinary flip and its Koszul coefficient

For vector spaces, the ordinary flip

is the map induced by the bilinear function \((m,n)\mapsto n\otimes m\). It respects the scalar balance because \(k\) is commutative. The flip in the opposite order is its inverse: the two composites fix every simple tensor, and simple tensors generate. The same generator check proves naturality in both vector spaces.

The corresponding chain isomorphism of tensor totals is instead

For the chain equation, applying \(\sigma\) to the two terms of (3.1) gives coefficients \((-1)^{(i+1)j}\) on \(y\otimes d_Xx\), and \((-1)^i(-1)^{i(j+1)}=(-1)^{ij}\) on \(d_Yy\otimes x\). Applying the target differential to (5.2) gives \((-1)^{ij+j}\) and \((-1)^{ij}\) on those same two terms. They agree. The inverse is \(\sigma_{Y,X}\), since the combined coefficient is \((-1)^{2ij}=1\), and the maps are natural by the ordinary flip and the degree-preserving chain-map actions.

Let \(t^n\) send the \((i,j)\) summand of the source of (3.4) to the \((j,i)\) summand by \((-1)^{ij}v_{H^i(X),H^j(Y)}\). Then

On a generator \([x]\otimes[y]\), both sides are \((-1)^{ij}[y\otimes x]\). These generators span the tensor summands and the finite direct sum, proving the equation everywhere. Taking the inverses of the two \(\Phi\)'s gives the equivalent commuting square whose horizontal arrows run from total cohomology to the sums of tensor products of classes. The ordinary module flip has no scalar sign by itself; the displayed sign arises from the two complex degrees. All formulas also hold in characteristic two, where the minus signs become plus signs.

6. Four graded exercises with complete solutions

Exercise 1 (introductory: why exactness is required). In abelian groups, let \(X\) have \(\mathbb Z,\mathbb Z,\mathbb Z/2\mathbb Z\) in degrees \(0,1,2\), with differentials multiplication by two and reduction modulo two. Let \(Y=\mathbb Z/2\mathbb Z\) in degree zero. Compare \(0\to X\) and its tensor total with \(Y\).

Solution. Multiplication by two is injective. Its image is \(2\mathbb Z\), the kernel of reduction modulo two, and the reduction map is surjective. Thus \(X\) is acyclic and \(0\to X\) is a quasi-isomorphism of bounded-below complexes. Tensoring each term with \(\mathbb Z/2\mathbb Z\) gives that same cyclic group in all three degrees. The first differential is zero, since two acts as zero; the second is identity, since the reduction map sends the tensor generator to the tensor generator. With \(Y\) concentrated in degree zero, these are exactly the total differentials. The total has \(H^0=\mathbb Z/2\mathbb Z\), and \(H^1=H^2=0\). Therefore its map from zero is not a quasi-isomorphism. This does not contradict Theorem 2.1: the tensor functor with \(\mathbb Z/2\mathbb Z\) fails to preserve the injection \(\mathbb Z\xrightarrow{2}\mathbb Z\), hence is not exact in that variable.

Exercise 2 (intermediate: the two finite intervals). If \(X,Y\) vanish below \(-2,3\), give the possible first degrees on total diagonal \(n\). Do the same when they vanish above \(2,-1\). Show why one lower bound and one upper bound alone do not guarantee finite diagonals.

Solution. In the first case, \(i\ge-2\) and \(j=n-i\ge3\) give \(-2\le i\le n-3\). This is empty for \(n<1\) and finite otherwise. In the second, \(i\le2\) and \(n-i\le-1\) give \(n+1\le i\le2\), empty for \(n>1\) and finite otherwise. For the mixed case, put \(X^i=k\) for every \(i\ge0\), \(Y^j=k\) for every \(j\le0\), and zero elsewhere, with zero differentials. On diagonal zero there is a nonzero \(k\otimes_kk\) for every pair \((i,-i)\), \(i\ge0\). This diagonal is infinite. Vector spaces have the direct sum needed to construct that example's sum total, but the finite-biproduct theorem in an arbitrary abelian target cannot be applied from these mixed bounds alone.

Exercise 3 (advanced: compute the canonical field bases). Over \(\mathbb Q\), let \(X^0=X^1=\mathbb Q^2\) and \(Y^0=Y^1=\mathbb Q^2\), with other terms zero and

Give bases of all total cohomology groups through (3.4), and verify that these classes are independent.

Solution. With standard basis vectors \(e_1,e_2\) for \(X\), put \(a=e_1-e_2\) in degree zero and \(c=e_2\) in degree one. The kernel of \(d_X\) is \(\mathbb Qa\), and its degree-one image is \(\mathbb Qe_1\), so \([c]\) spans \(H^1(X)\). With basis \(f_1,f_2\) for \(Y\), put \(b=f_2\) in degree zero and \(d=f_2\) in degree one. The kernel of \(d_Y\) is \(\mathbb Qb\), and its image is \(\mathbb Q(f_1+f_2)\), so \([d]\) spans \(H^1(Y)\). Formula (3.4) gives

All are cycles because their representatives are cycles in their respective degrees. Define chain projections to the zero-differential cohomology complexes by \(p_X^0(u,v)=-v\), \(p_X^1(u,v)=v\), \(p_Y^0(u,v)=v\), and \(p_Y^1(u,v)=v-u\). The degree-one projections kill the corresponding differential images. Each projection sends its displayed representative to one. Their tensor total kills boundaries and sends (6.2) to the respective standard bases of dimensions \(1,2,1\), so the classes are independent. Theorem 4.1 proves they span as well. Directly, the total term dimensions are \(4,8,4\); the first differential has rank three, since its kernel is the one-dimensional tensor of the two degree-zero cycle lines. The last differential has image \(\operatorname{im}d_X\otimes Y^1+X^1\otimes\operatorname{im}d_Y\), of dimension \(2+2-1=3\). Hence total cohomology has dimensions \(1,8-3-3,1\), verifying the same bases by rank computation.

Exercise 4 (advanced: two ways to detect the graded flip). First take two one-dimensional zero-differential complexes, each concentrated in degree one, over \(\mathbb Q\). Compute the swap on total cohomology. Then take \(X=(\mathbb Q\xrightarrow{1}\mathbb Q)\) in degrees zero and one, and \(Y=\mathbb Q\) in degree one. Show that an unsigned flip is not even a chain map in the second example.

Solution. In the first example, the tensor total is one-dimensional in degree two and \(i=j=1\). Thus (5.2) sends \([x\otimes y]\) to \(-[y\otimes x]\). On the Künneth summand the ordinary flip sends it to the positive class; multiplying that flip by \((-1)^{1\cdot1}\) gives exactly the negative class required by (5.3). Over \(\mathbb Q\) the two classes differ.

In the second example, the source tensor total has one copy of \(\mathbb Q\) in degrees one and two, with differential positive identity, from the first differential of \(X\). In the reversed tensor total, the first factor \(Y\) has degree one, so the second-factor differential has coefficient \(-1\); its differential is negative identity. An unsigned flip acts as positive identity in both degrees, and its chain equation would assert \(-1=1\), which is false over \(\mathbb Q\). The actual graded flip acts as positive identity in degree one, where the degrees are \((0,1)\), and negative identity in degree two, where they are \((1,1)\). Its chain equation is \((-1)\cdot1=(-1)\cdot1\), so it is a chain isomorphism. Both latter totals are acyclic, while the first example detects the same sign on nonzero cohomology.

References

• The linked prerequisite lessons give the exact tensor, ordinary cohomology and finite diagonal total interfaces used above.
• The Stacks Project, Section 15.59, Tensor products of complexes, for the signed tensor total and graded flip conventions. The natural class map and its complete field splitting/contraction proof are supplied in Sections 3–4 here.