Universal Coefficient Theorem – When Does the Sequence Split?

abstract-algebrahomological-algebrahomology-cohomology

$\newcommand{\tor}{\operatorname{Tor}}$It is left as an exercise in Weibel $3.6$ to conclude that:

If $C$ is a complex of right $R$-modules such that each $C_n$ and each boundary submodule $B_n$ is flat and $D$ is any complex of left $R$-modules, then there is a short exact sequence:

$$0\to\bigoplus_{p+q=n}H_p(C)\otimes_R H_q(D)\to H_n(C\otimes_R D)\to\bigoplus_{p+q=n-1}\tor^R_1(H_p(C),H_q(D))\to0$$

If $R$ is nice (he takes $R=\Bbb Z$) and $C_n$ is free for all $n$ then this sequence splits.

I can prove there is this exact sequence. I am thoroughly unconvinced that it should split in general; at least, I'm unconvinced that the same arguments he uses (somewhat carelessly?) in the special case of $D$ being concentrated in degree zero will work in general. About "$R$ is nice", for the case where $D$ is just one module it suffices to consider $R$ a hereditary ring; then the sequence will split.

What's my gripe? Well, the standard arguments conclude that $Z_\bullet$ and $B_\bullet$ (cycles and boundaries of $C$) are projective and flat and we say $0\to Z_n\to C_n\to B_{n-1}\to0$ splits, with some splitting $r:C_n\twoheadrightarrow Z_n$. We run some argument and deduce the sequence exists, where the first map has an easy description.

However, to create a projection map $H_n(C\otimes_R D)\to\bigoplus_{p+q=n}H_p(C)\otimes_R H_q(D)$ is hard, since the cycle subgroup of $C\otimes_R D$ is mysterious. Moreover, it should not be true that $Z_p\otimes D_q$ is a subgroup of the $n$-cycles of $C\otimes_R D$; we can only expect this to be true if, say, the differentials of $D$ are trivial (as they are in the $D$-is-a-single-module special case). But the only thing I can know how to do is consider the map $r\otimes1:C_p\otimes_R D_q\twoheadrightarrow Z_p\otimes_R D_q$ and there is absolutely no reason for this map to induce a map on homology. There is also no reason to expect we can magically extend this by some $Z_p\otimes_R D_q\overset{??}{\twoheadrightarrow}H_p(C)\otimes_R H_q(D)$.

I think that there is perhaps an error and we should assume something of $D$ too, if we expect this to split – such as assuming $D_q$ is free for every $q$, because then there is a splitting $D_q\to Z(D)_q$ and we can make everything work using a generalised version of the case $D$ is just one module, in particular using the fact $Z(C)_p\otimes_R Z(D)_q$ is a subgroup of $Z(C\otimes_R D)_n$. Note that in topology this hypothesis holds because $C,D$ will be the free singular complexes.

Am I right, or am I just overthinking this? If I'm wrong, I would really appreciate some help constructing the splitting $H_n(C\otimes_R D)\twoheadrightarrow\bigoplus_{p+q=n}H_p(C)\otimes_R H_q(D)$.

Best Answer

This answer is a work in progress. There's a lot to write up and I'll finish the job hopefully sometime soon.

$\newcommand{\tor}{\operatorname{Tor}}\newcommand{\ext}{\operatorname{Ext}}\newcommand{\hom}{\operatorname{Hom}}$There's a lot to say here. If I don't explicitly say "cocomplex" then you must assume I am discussing a chain complex of homological type, of degree $-1$ in the differential. I have later learned (but it's too late for me) that Rotman's homological algebra contains many of these results. Even later have I learned there are spectral sequence results. These are stronger in the sense of being more generally applicable, but suffer from being far less computable, from failing to recover all the data except in lucky cases and also from failing to recover the splitting claims. It is still good to have a systematic algorithm for recovering pieces though; this is good enough, e.g., to at least deduce claims to the tune of "if $f:A\to B$ is a quasiisomorphism then $f\otimes1:A\otimes_RC\to B\otimes_RC$ is also a quasiisomorphism for any complex $C$". I haven't yet checked in what generality that actually holds, though; to be continued.

For "maximal" practical generality (if anyone knows stronger results than the ones I state, please do say) we will require the existence of "good" resolutions of unbounded complexes. I have only just learned about Cartan-Eilenberg resolutions and I believe these actually suffice for the purposes of demonstrating most if not all of the UCTs, but it is again "too late" for me. I originally researched semiprojective and semiinjective resolutions for the purpose of proving the more general UCTs (thanks to @Thorgott); these have stronger lifting properties than Cartan-Eilenberg resolutions but are by far less elementary in construction.

I cite the results of this and this; the former source discusses some properties and definitions of semi-(free,flat,projective,injective) resolutions and constructs the first three. The former source, when discussing semiinjectivity, implicitly relies on the existence of certain kinds of module; it's not that hard to show but it is crucial - "I" settled that matter in this self-answer. For a construction of semiinjective resolutions: this takes more work and is sketched in the second source. However it can be done in detail. I have no idea if this is written down correctly and in full anywhere, but I will summarise the idea (iirc the linked second paper used cochain notation and one must beware of this):

The idea is that we can resolve complexes $M$ which are bounded above by a complex $I$ which is bounded above by the same bound, where every object of $I$ is injective and an injective quasiisomorphism $M\to I$ exists (this is the "resolution"): as a sketch of this, suppose we have formed the first few objects of $I$ and we have some partial map $(M_a\to M_{a-1}\to M_{a-2}\to\cdots\to M_n)\to(I_a\to I_{a-1}\to\cdots\to I_n)$. To form $I_{n-1}$ and the map $M_{n-1}\to I_{n-1}$ you quotient by boundaries and let $X$ be the pushout of: $$I_n/B_n(I)\leftarrow M_n/B_n(M)\to M_{n-1}$$

Then you embed $X$ into an injective object which we take to be $I_{n-1}$ and take $M_{n-1}\to X\to I_{n-1}$ as the $(n-1)$th map. You then check everything you need to check and proceed to the next inductive stage

Now say $M$ is unbounded. Following the ideas of the second link, it is possible to show (if you can put up with gory details and very, very careful sign handling) that there is a family of bounded above, by $n+1$, complexes of injectives $I^{(n)}$ and injective quasiisomorphisms $\tau_{\le n}(M)\hookrightarrow I^{(n)}$ (where $\tau$ is the "good, soft" truncation) which are compatible with given surjections $I^{(n+1)}\twoheadrightarrow I^{(n)}$ and by several uses of the derived inverse limit functor of a tower, the Mittag-Leffler condition and some mumbling of words which are close in spirit to arguments presented in both links, $I:=\varprojlim I^{(n)}$ does the job; it is a complex of injectives (this is not obvious from some kind of general result as far as I know; it instead follows from careful inspection of the explicit construction), it is genuinely semiinjective and there is a genuine injective quasiisomorphism $M\hookrightarrow I$.

You check the $\tau_{\le n}M\hookrightarrow I^{(n)}$ are quasiisomorphisms by very carefully comparing the mapping cone of this map with a different one, namely a $g:C_f[+1]\to I$ which is a quasiisomorphism (and $f:\tau_{\le(n+1)}M\to I^{(n)}$ is an extension of $\tau_{\le n}(M)\to I^{(n)}$) so the cone of $g$ is acyclic hence the cone of $\tau_{\le(n+1)}(M)\to I^{(n+1)}$ is acyclic (since they magically turn out to be the same) and the map is a quasiisomorphism.

A variety of interesting generalisations of the UCT theorems presented by Weibel can be found in the textbook of Spanier, though Spanier rarely confesses that he means to work in a PID with positive complexes so this should be beared in mind. I have personally checked that we can always further extend these to unbounded complexes.

"Spanier's" (the idea is not originally his) minimal models, or acylic models, are special cases of semifree resolutions. However, his "free resolutions of finite type" are important for technical reasons and cannot be deduced from the general existence of semifree resolutions and thus play an important, separate role in the . The extensions aren't unreasonably hard though, the idea is always the same: take your complex $Q$, maybe its objects aren't free or injective or whatever, and resolve $Q$ by something which you know to be nicer. Then use a previously established UCT for this nicer complex, and pass back to $Q$. The "passing back" step needs to be examined carefully, as the quasiisomorphism of $Q$ with its resolution cannot in general be expected to be preserved. Then perhaps an argument involving some semiflatness condition or some other information like applicability of the $5$ lemma to an exact sequence is necessary to successfully "pass back". For the very final class of UCT I present, the issue actually lies in the "pass forward"; it is not necessarily true (I mean, I don't have a counterexample, but it's "obvious" that we can't expect it to be true and it doesn't follow from anything I already know to be true) that semifree resolutions of unbounded complexes of finite type are again of finite type. As previously remarked this can be arranged - and Spanier proves this - in the bounded below case but I doubt its truth in general. It is also necessary for some of these to check that chain homotopic maps remain chain homotopic upon taking the total tensor product or total hom complex. You also need to be a touch careful creating the splitting in the best possible case where both complexes are (projective/injective), at least in the $\otimes$ case, to ensure we really do get induced maps on homology.

The "$R$ is hereditary" hypothesis is necessary since it's applied to a mysterious extraneous object such as a semifree resolution; the question of whether or not its submodules are projective we cannot hope to answer in general since the complex itself is difficult to understand.

There is also an asymmetric bias for right modules and right hereditary rings. By dualising your ring, this does not matter; swapping all instances of "left" with "right" will give you a "new", correct, theorem.

Theorem $0$: Let $R$ be a right hereditary ring and $C$ a complex of right $R$-modules. If every object of $C$ is projective then every boundary submodule (and every cycle submodule) of $C$ is projective. If every object of $C$ is injective then (consider $B_n=C_{n+1}/Z_{n+1})$ every boundary submodule is injective.

UCT theorems for the homology of a tensor product - these work their way up, and note theorem $1.3$ proves $1.2.1$ as a special case however it requires $1.2.1$ to be proven first:

Theorem $1.1$: Let $R$ be any ring and $P$ any (unbounded) complex of flat right $R$-modules whose every boundary submodule is flat. Then if $Q$ is any (unbounded) complex of left $R$-modules, we get a natural (non-split) short exact sequence: $$0\to\bigoplus_{p+q=n}H_p(P)\otimes_RH_q(Q)\to H_n(P\otimes_RQ)\to\bigoplus_{p+q=n-1}\tor^R_1(H_p(P), H_q(Q))\to0$$Where the first map is the obvious one.

If you are so lucky that both $P$ and $Q$ are complexes of projective modules whose every boundary submodule is projective then the sequence splits unnaturally. If this fails then you need to assume $R$ is right hereditary and appeal to one of the below theorems.

Theorem $1.2.1$: Let $R$ be a (noncommutative, left) hereditary ring, $P$ any (unbounded) complex of right projective $R$-modules whose every boundary submodule is projective. If $Q$ is any (unbounded) complex of left $R$-modules then the exact sequence of theorem $1.1$ exists, is natural and splits unnaturally.

Theorem $1.2.2$: Let $R$ be any ring, $P$ any (unbounded) complex of projective right $R$-modules whose every boundary submodule is projective and $M$ any left $R$-module. Then there is a natural short exact sequence: $$0\to H_n(P)\otimes_RM\to H_n(P\otimes_R M)\to\tor^R_1(H_n(P),M)\to0$$Where the first map is the obvious one. The sequence splits naturally in $M$ (but unnaturally in $P$).

Theorem $1.3$: Let $R$ be a (noncommutative, both left and right) hereditary ring, $P$, $Q$ (unbounded) complexes of right, left $R$-modules such that the $\tor^R_1(P,Q)$ complex - the $\oplus$-totalisation of the obvious double complex on the $\tor^R_1(P_\bullet,Q_\bullet)$ modules whose vertical differentials alternate by a sign akin to how the tensor product of complexes is defined - is acylic, then there is a (canonical up to unique isomorphism) natural short exact sequence: $$0\to\bigoplus_{p+q=n}H_p(P)\otimes_RH_q(Q)\to H_n(P\otimes_RQ)\to\bigoplus_{p+q=n-1}\tor^R_1(H_p(P), H_q(Q))\to0$$Where the first map is the obvious one. The sequence splits unnaturally.

In particular, this is applicable if, say, every object of $Q$ is a flat module.

UCT theorems for cohomology of a hom complex, adapted for Weibel's convention for the $\hom_R$-cocomplex and using the abbreviation $H^\ast(A; B)$ for the cohomology of $\hom_R(A,B)$ - these again work their way up somewhat, and theorem $2.3$ proves (but relies on) theorems $2.2.1.1,2.2.1.2$:

Theorem $2.1.1$: Let $R$ be any ring and $P$ any (unbounded) complex of projective right $R$-modules with projective boundary submodules, $Q$ any (unbounded) cocomplex of right $R$-modules, then there is a natural (non-split) short exact sequence: $$0\to\prod_{p+q=n-1}\ext^1_R(H_p(P),H_q(Q))\to H^n(P; Q)\to\prod_{p+q=n}\hom_R(H_p(P),H_q(Q))\to0$$Where the final map is the obvious one.

If you are so lucky that $Q$ is a complex of injective modules and that every boundary submodule of $Q$ is also injective, then the sequence splits unnaturally. If this fails, you need to assume $R$ is right hereditary and appeal to one of the below theorems.

Theorem $2.1.2$: Let $R$ be any ring and $Q$ any (unbounded) cocomplex of injective right $R$-modules whose every boundary submodule is injective. If $P$ is any (unbounded) complex of right $R$-modules, then there is a natural (non-split) short exact sequence: $$0\to\prod_{p+q=n-1}\ext^1_R(H_p(P),H_q(Q))\to H^n(P; Q)\to\prod_{p+q=n}\hom_R(H_p(P),H_q(Q))\to0$$Where the final map is the obvious one.

If you are so lucky that $P$ is a complex of projective modules and that every boundary submodule of $P$ is also projective, then the sequence splits unnaturally. If this fails, you need to assume $R$ is right hereditary and appeal to one of the below theorems.

Theorem $2.2.1.1$: Let $R$ be any (noncommutative, right) hereditary ring and $P$ any (unbounded) complex of projective right $R$-modules. If $Q$ is any (unbounded) cocomplex of right $R$-modules then the exact sequence of theorem $2.1$ exists, is natural and splits unnaturally.

Theorem $2.2.1.2$: Let $R$ be any (noncommutative, right) hereditary ring and $Q$ any (unbounded) cocomplex of injective right $R$-modules. If $P$ is any (unbounded) complex of right $R$-modules then the exact sequence of theorem $2.1$ exists, is natural and splits unnaturally.

Theorem $2.2.2.1$: Let $R$ be any ring and $P$ any (unbounded) complex of projective right $R$-modules whose every boundary submodule is projective. If $M$ is any right $R$-module, then there is a natural short exact sequence: $$0\to\ext_1^R(H_n(P),M)\to H^n(P; M)\to\hom_R(H_n(P),M)\to0$$Where the last map is the obvious one. The sequence splits naturally in $M$ (but unnaturally in $P$).

Theorem $2.2.2.2$: Let $R$ be any ring and $Q$ any (unbounded) cocomplex of injective right $R$-modules whose every boundary submodule is injective. If $M$ is any right $R$-module, then there is a natural short exact sequence: $$0\to\ext_1^R(M,H_n(Q))\to H^n(M; Q)\to\hom_R(M,H_n(Q))\to0$$Where the last map is the obvious one. The sequence splits naturally in $M$ (but unnaturally in $P$).

Theorem $2.3$: Let $R$ be any (noncommutative, right) hereditary ring, $P,Q$ (unbounded) complexes of right $R$-modules such that the $\ext_1^R(P,Q)$ complex - the $\prod$-totalisation of the obvious double complex on the $\ext_1^R(P_\bullet,Q^\bullet)$ modules with differentials that alternate in sign as per Weibel's convention for the $\hom_R$ double complex - is acyclic, then there is a (canonical up to unique isomorphism) natural short exact sequence: $$0\to\prod_{p+q=n-1}\ext^1_R(H_p(P),H_q(Q))\to H^n(P; Q)\to\prod_{p+q=n}\hom_R(H_p(P),H_q(Q))\to0$$Where the last map is the obvious one. The sequence splits unnaturally.

The UCT of the cross product: this mildly extends Theorem $1.3$ by introducing coefficient modules. I use coefficient notation $H(X; Y):=H(X\otimes Y)$.

Theorem $3$: Let $R$ be a commutative hereditary ring, $P$, $Q$ (unbounded) complexes of flat $R$-modules. If $M$, $M'$ are two $R$-modules with $\tor_1^R(M,M')=0$ then there is a natural short exact sequence which splits unnaturally: $$0\to\bigoplus_{p+q=n}H_p(P;M)\otimes_RH_q(Q;M')\to H_n(P\otimes_R Q;M\otimes_RM')\\\to\bigoplus_{p+q=n-1}\tor_1^R(H_p(P;M),H_q(Q;M'))\to0$$Where the first map is the obvious one.

Theorem $3'$: Let $R$ be a (noncommutative, both left and right) hereditary ring, $P,Q$ (unbounded) complexes of torsionfree Abelian groups. If $M$, $M'$ are right, left $R$-modules with $\tor_1^R(M,M')=0$ then there is a natural short exact sequence which splits unnaturally: $$0\to\bigoplus_{p+q=n}H_p(P;M)\otimes_RH_q(Q;M')\to H_n(P\otimes_{\Bbb Z} Q;M\otimes_RM')\\\to\bigoplus_{p+q=n-1}\tor_1^R(H_p(P;M),H_q(Q;M'))\to0$$Where the first map is the obvious one.

To follow: UCT theorems for the cross product in cohomology with coefficients.

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