5.2 Homological Algebra · Homology & Cohomology

The unknown thing to be known appeared to me as some stretch of earth or hard marl, resisting penetration… the sea advances insensibly in silence, nothing seems to happen, nothing moves, the water is so far off you hardly hear it… yet it finally surrounds the resistant substance.

— Alexander Grothendieck, Récoltes et Semailles, pp. 552–3, translated by Colin McLarty

Several times in this book a computation did something odd, and we checked it and moved on. Working mod 2, the projective plane acquired an $H_{2}$ that it does not have over the integers (§3.4 Computing Homology). In cohomology, the torsion of $H_{1}$ turned up one degree higher, in $H^{2}$ (§4.2 Cohomology Groups). And simplicial, singular and cellular homology, three quite different definitions, kept giving the same answers. None of these is a quirk of a particular space. They are all consequences of the algebraic machine that homology runs on — chain complexes, kernels modulo images, exact sequences — and this chapter takes that machine out of its case to see how it works.

You have already met many of the parts. Chain complexes and chain maps came in §3.2 Chains and the Boundary Operator and §3.5 Maps, Invariance, and First Triumphs; exact sequences and the connecting map in §3.6 Exact Sequences and Mayer–Vietoris; the universal coefficient theorem for cohomology, with its mysterious $Ext,$ in §4.2 Cohomology Groups. Now, with the language of §5.1 Categories and Functors in hand, we can see how the pieces fit, and meet the two functors, $Tor$ and $Ext,$ that measure what is lost when you change coefficients — plus spectral sequences, the great computing engine, and the five axioms that say what “a homology theory” is.

Chain complexes, as a category

Let us restate the definitions from Part III in the abstract, without any space in sight.

Chain complexes and chain maps form a category, $Ch$ (composition is done degree by degree; the identity is the identity in each degree). And homology is a functor on it, $H_{n} : Ch \to Ab :$ a chain map sends cycles to cycles ( $\partial f z = f \partial z = 0)$ and boundaries to boundaries ( $f \partial c = \partial f c),$ so it induces $f_{*} [z] = [f z],$ and $(g \circ f)_{*} = g_{*} \circ f_{*} .$ The homology of a simplicial complex was the composite of two functors: first chains, then homology.

Here is the sentence that organises this entire chapter. A complex is called exact at $C_{n}$ if $im \partial_{n + 1} = ker \partial_{n}$ — exactly the condition met in §1.4 Abelian Groups and Formal Sums. In a chain complex we always have $im \partial_{n + 1} \subseteq ker \partial_{n},$ and the homology $H_{n} (C)$ is the quotient of the second by the first. So:

Chain homotopy

In §3.5 Maps, Invariance, and First Triumphs we proved that homotopic maps induce the same map on homology. The algebraic heart of that proof is worth isolating, because it will reappear below.

A chain homotopy

Figure 5.2.1 Two chain maps

f, g

(vertical) between the rows

C

and

D,

and the diagonal maps

s

that raise degree by one. For spaces,

s

is the “prism” swept out by a homotopy from §3.5 Maps, Invariance, and First Triumphs: the boundary of the prism over a chain is the difference of the two images, plus the prism over the chain’s boundary.

Geometry enters only once: the homotopy between two maps of spaces gives a prism, and the prism gives $s .$ After that, everything is algebra. This division of labour — a geometric input, then an algebraic machine — is the theme of the chapter.

The snake lemma and long exact sequences

The single most useful tool of homological algebra is a lemma about one commutative diagram. It manufactures a map out of nothing, and that map is the connecting homomorphism you used to build every long exact sequence in §3.6 Exact Sequences and Mayer–Vietoris.

Before the proof, read the picture. The kernels of the three vertical maps sit in a row above the diagram, the cokernels in a row below. The connecting map $δ$ joins the right end of the top row to the left end of the bottom row; drawn in, it winds through the diagram like a snake, which is where the name comes from. The figure follows one element along the snake, in a concrete example where every group is $Z$ or $Z / m$ and you can check every step with arithmetic.

Chasing the snake Step through with ‹ › · change m, n and the starting element

0 \to Z \times 4 Z mod 4 Z /4 \to 0

Both rows are exact, and every square commutes (multiplying by 6 commutes with everything). Kernels on top, cokernels below.

The diagram

m 4

n 6

start at

x \in ker c

Figure 5.2.2 The rows are the short exact sequence

0 \to Z \times m Z \to Z / m \to 0

twice, and every vertical map multiplies by

n .

Start at an element

x

ker c

— an

x \in Z / m

with

n x \equiv 0

— then lift it to

B,

push it down with

b,

pull it back to

A^{'}

and read off its class in

coker a = Z / n .

The gold trail is the diagram chase; the gold snake is the map

δ

it defines. Step 7 shows that a different choice of lift gives the same answer.

Proof

Defining $δ .$ Let $x \in ker c .$ Since $p$ is onto, choose $y \in B$ with $p (y) = x .$ The element $b (y) \in B^{'}$ satisfies $p^{'} (b (y)) = c (p (y)) = c (x) = 0,$ because the right square commutes. By exactness of the bottom row at $B^{'},$ $b (y) = i^{'} (z)$ for some $z \in A^{'},$ unique because $i^{'}$ is injective. Set $δ (x) = [z],$ the class of $z$ in $coker a = A^{'} / im a .$

It does not depend on the choice. Another lift $y^{'}$ of $x$ has $p (y^{'} - y) = 0,$ so $y^{'} - y = i (w)$ for some $w \in A$ (exactness of the top row). Then $b (y^{'}) - b (y) = b (i (w)) = i^{'} (a (w)),$ so the new $z$ differs from the old one by $a (w) \in im a$ — the same class in $coker a .$ A similar check shows $δ$ is a homomorphism.

Exactness. Each of the four exactness claims is a short chase of the same kind; for instance, if $x = p (y)$ with $y \in ker b,$ we may use that $y$ as the lift, so $b (y) = 0$ gives $z = 0$ and $δ (x) = 0 .$ The two exactness claims next to $δ$ are Exercise 7, and the other two (at $ker b$ and $coker b)$ are shorter chases of the same kind — the best possible practice in diagram chasing. Weibel’s text has the whole proof (Weibel 1994, Lemma 1.3.2).

From short exact sequences of complexes to long ones

A short exact sequence of chain complexes $0 \to A i B p C \to 0$ is a pair of chain maps that is short exact in every degree: $0 \to A_{n} \to B_{n} \to C_{n} \to 0$ exact for each $n .$ You met two in §3.6 Exact Sequences and Mayer–Vietoris: the chains of a subspace, of the whole space, and of the pair; and the Mayer–Vietoris sequence of chains of $U \cap V,$ of $U$ and $V,$ and of $U \cup V .$

The connecting map $\partial$ is a snake. Take a class $[c] \in H_{n} (C),$ lift the cycle $c$ to $b \in B_{n},$ take its boundary $\partial b \in B_{n - 1};$ it maps to $\partial c = 0$ in $C_{n - 1},$ so it comes from a unique $a \in A_{n - 1},$ which turns out to be a cycle. Then $\partial [c] = [a] .$ Up, across, down, across: the same zig-zag as in Figure 5.2.2. (In fact the theorem follows by applying the snake lemma to the diagram whose rows are $A_{n} / im \partial \to B_{n} / im \partial \to C_{n} / im \partial \to 0$ and $0 \to ker \partial_{n - 1}^{A} \to ker \partial_{n - 1}^{B} \to ker \partial_{n - 1}^{C},$ with the vertical maps induced by $\partial :$ their kernels are $H_{n}$ and their cokernels are $H_{n - 1} .)$ Hatcher proves the theorem by exactly this zig-zag (Hatcher 2002, Thm 2.16).

Hom and tensor: two ways to combine groups

Coefficients have appeared in this book in two different ways. Chains with coefficients in $G$ were formal sums of simplices with coefficients from $G$ (§3.4 Computing Homology). Cochains with values in $G$ were functions from chains to $G$ (§4.2 Cohomology Groups). Each is a way of combining two abelian groups into a third, and each has a name.

Hom

For abelian groups $A$ and $G,$ the set $Hom (A, G)$ of all homomorphisms $A \to G$ is itself an abelian group: add homomorphisms pointwise, $(f + g) (a) = f (a) + g (a) .$ Computing it is a matter of asking where the generators can go.

$Hom (Z, G) ≅ G .$ A homomorphism out of $Z$ is decided by where it sends $1,$ and $1$ may go anywhere: $f \mapsto f (1)$ is the isomorphism. (This is the universal property of the free group on one generator.)
$Hom (Z / n, Z) = 0 .$ If $f (1) = x$ then $n x = f (n \cdot 1) = f (0) = 0$ in $Z,$ so $x = 0 .$ An element of finite order has nowhere to go in a group without torsion.
$Hom (Z / n, Z / m) ≅ Z / g cd (m, n) .$ Now $f (1) = x$ only needs $n x \equiv 0 (mod m),$ and there are exactly $g cd (m, n)$ such $x$ in $Z / m .$ For instance $Hom (Z /6, Z /4) ≅ Z /2 :$ $1$ can go to $0$ or to $2 .$
$Hom (A \oplus A^{'}, G) ≅ Hom (A, G) \oplus Hom (A^{'}, G) :$ a map out of a direct sum is a pair of maps — the coproduct property from §5.1 Categories and Functors.

As a functor of its first argument, $Hom (-, G)$ is contravariant (§5.1 Categories and Functors): a homomorphism $f : A \to B$ gives $f^{*} : Hom (B, G) \to Hom (A, G),$ $φ \mapsto φ \circ f .$ Applied to chains, it produces cochains: $C^{n} (X; G) = Hom (C_{n} (X), G),$ and the coboundary is $δ = \partial^{*} .$

Tensor product

The other combination is less familiar by name, but it is the first of the two ways coefficients appeared above: a chain mod 2 is a sum of simplices, each “times” an element of $Z /2 .$ In general we want a group whose elements are such “formal products” $a \otimes g$ of an element of $A$ with an element of $G,$ obeying the distributive laws and nothing else.

From the relations: $0 \otimes g = 0$ (apply the first with $a = a^{'} = 0),$ and $(na) \otimes g = n (a \otimes g) = a \otimes (n g)$ for every integer $n$ — integers can be moved freely across the $\otimes$ sign. The key property is universal (§5.1 Categories and Functors again): a homomorphism $A \otimes G \to H$ is the same thing as a function $A \times G \to H$ that is additive in each variable separately (bilinear). Now the examples, which have some surprises.

$Z \otimes G ≅ G,$ via $n \otimes g \mapsto n g .$ Tensoring with $Z$ changes nothing.
$Z / m \otimes Z / n ≅ Z / g cd (m, n) .$ Every $a \otimes b$ equals $ab (1 \otimes 1),$ so the group is generated by $1 \otimes 1 .$ Its order divides $m,$ since $m (1 \otimes 1) = m \otimes 1 = 0 \otimes 1 = 0,$ and likewise divides $n,$ so it divides $g cd (m, n) .$ The bilinear map $(a, b) \mapsto ab mod g cd (m, n)$ shows the order is exactly $g cd (m, n) .$
$Z /2 \otimes Z /3 = 0 .$ Watch it happen: $1 \otimes 1 = 3 (1 \otimes 1) - 2 (1 \otimes 1) = (1 \otimes 3) - (2 \otimes 1) = (1 \otimes 0) - (0 \otimes 1) = 0 .$ Two nonzero groups whose tensor product is zero.
$Q \otimes Z / n = 0 :$ $q \otimes x = \frac{q}{n} n \otimes x = \frac{q}{n} \otimes n x = \frac{q}{n} \otimes 0 = 0 .$ Fractions let you divide by $n,$ and $n$ kills everything in $Z / n .$

Exactness, and how it fails

Here is the problem that the rest of the chapter solves. Take a short exact sequence of abelian groups $0 \to A f B g C \to 0$ and apply one of our two operations. Is the result still exact? Only partly.

One example shows both failures. Start from the short exact sequence

0 \to Z \times 2 Z \to Z /2 \to 0.

Apply $Hom (-, Z) .$ Since $Hom (Z, Z) ≅ Z$ and $Hom (Z /2, Z) = 0,$ we get $0 \to 0 \to Z \times 2 Z .$ The last map is multiplication by 2, which is not onto: its cokernel is $Z /2 .$ In words: a homomorphism defined on the subgroup $2 Z$ — say the one sending $2$ to $1$ — cannot be extended to all of $Z,$ because $1$ has no half in $Z .$ The cokernel $Z /2$ measures exactly these non-extendable maps.

Apply $- \otimes Z /2 .$ Since $Z \otimes Z /2 ≅ Z /2,$ we get $Z /2 \times 2 Z /2 \to Z /2 \to 0 .$ But multiplication by 2 is the zero map on $Z /2,$ so the first map is not one-to-one: its kernel is $Z /2 .$ In words: the element $1$ of the first $Z$ was perfectly nonzero, but after reducing mod 2 it maps to $2 \otimes 1 = 1 \otimes 2 = 0 .$ Reducing mod 2 forgot that 2 was not zero.

In both cases the failure is itself an abelian group, here $Z /2 .$ The next section gives these failures their names, $Ext$ and $Tor,$ and a way to compute them.

Resolutions, Tor and Ext

Free resolutions

Free abelian groups are the easy case: if $F$ is free, then $Hom (F, -)$ and $- \otimes F$ preserve all exactness, because $F$ is a sum of copies of $Z,$ and both operations do nothing to $Z .$ So here is the plan: describe any group by free groups — its generators and its relations — and do the measuring there. The failure of exactness then shows up as the homology of a small complex built from those free groups, which we know how to compute.

For abelian groups the story is short. Choose generators of $A :$ that gives a free group $F_{0}$ mapping onto $A .$ The kernel $F_{1}$ of that map — the relations among the generators — is a subgroup of a free abelian group, and a standard theorem (a close cousin of the classification in §1.4 Abelian Groups and Formal Sums) says that such a subgroup is itself free. So every abelian group has a resolution of length one, $0 \to F_{1} M F_{0} \to A \to 0,$ and for finitely generated groups $M$ is the relation matrix whose Smith normal form you met in §1.5 Linear Algebra. The basic example:

0 \to Z \times n Z \to Z / n \to 0.

Two things deserve to be said at once. First, the answer does not depend on which resolution you choose: any two free resolutions of $A$ are chain homotopy equivalent (once more, a chain homotopy saves the day), so they give isomorphic $Tor$ and $Ext .$ We shall take this on trust; Hatcher proves it in a page (Hatcher 2002, Lemma 3.1). Second, the definitions are not a trick: they are exactly “the homology of the complex you get by applying the functor to a resolution”, which is how all derived functors are defined.

Now compute, using the resolution $0 \to Z \times n Z \to Z / n \to 0 .$

$Tor (Z / n, G)$ is the kernel of $G \times n G :$ the elements of $G$ killed by $n .$ So $Tor (Z / n, Z) = 0,$ and $Tor (Z / m, Z / n) ≅ Z / g cd (m, n)$ (the elements $x \in Z / n$ with $m x \equiv 0).$ For example, $Tor (Z /4, Z /6) ≅ Z /2 .$
$Ext (Z / n, G)$ is the cokernel of $G \times n G,$ that is $G / n G .$ So $Ext (Z / n, Z) ≅ Z / n,$ $Ext (Z / n, Z / m) ≅ Z / g cd (m, n),$ and $Ext (Z / n, Q) = 0$ (every rational number is divisible by $n).$
Free groups contribute nothing: $Tor (Z, G) = Ext (Z, G) = 0,$ since $Z$ is its own resolution $0 \to 0 \to Z \to Z \to 0 .$
Sums split: $Tor (A \oplus A^{'}, G) ≅ Tor (A, G) \oplus Tor (A^{'}, G),$ and the same for $Ext .$ With the classification $A ≅ Z^{r} \oplus Z / d_{1} \oplus \dots \oplus Z / d_{k}$ of §1.4 Abelian Groups and Formal Sums, these rules compute everything.

A calculator for ⊗, Tor, Hom and Ext Build A and G from ℤ and ℤ/n pieces · or try a preset

A = Z /4

copies of ℤ in A 0

tap a summand to remove it; +n adds ℤ/n

G = Z /6

copies of ℤ in G 0

tap a summand to remove it; +n adds ℤ/n

A \otimes G

= Z /2

Z /4 \otimes Z /6 = Z /2

Tor (A, G)

= Z /2

Tor (Z /4, Z /6) = Z /2

Hom (A, G)

= Z /2

Hom (Z /4, Z /6) = Z /2

Ext (A, G)

= Z /2

Ext (Z /4, Z /6) = Z /2

Where the numbers come from: a free resolution of

A

0 \to Z M Z \to A \to 0, M = (4)

Tensor with

G = Z /6

(right exact)

G M \otimes G G \to A \otimes G \to 0

The left map should be injective for exactness, and it is not: it multiplies by $4$ on copies of $G$ . Its kernel is $Tor (A, G) = Z /2$ .

Apply

Hom (-, G)

(left exact)

0 \to Hom (A, G) \to G M^{T} G

The right map should be surjective, and it is not: its cokernel, made of the pieces $G / d G$ for each $Z / d$ in $A$ , is $Ext (A, G) = Z /2$ .

try:

Figure 5.2.3 Each operation splits over the cyclic pieces of

A

and

G,

and the cards show every piece. Below, the free resolution of

A :

after tensoring with

G

the first map stops being injective, and its kernel is

Tor (A, G);

after applying

Hom (-, G)

the last map stops being surjective, and its cokernel is

Ext (A, G) .

Notice that

Tor

and

Ext

vanish unless

A

has torsion — free groups resolve themselves.

Where the names come from

Tor is short for torsion. One way to see it: $Tor (A, Q / Z)$ is isomorphic to the torsion subgroup of $A$ — for instance $Tor (Z / n, Q / Z)$ consists of the elements of $Q / Z$ killed by $n,$ which form a copy of $Z / n .$ Tor detects the torsion in $A,$ as seen through $G .$

Ext is short for extension. An extension of $C$ by $A$ is a short exact sequence $0 \to A \to B \to C \to 0 :$ a bigger group $B$ built from a subgroup $A$ and a quotient $C .$ It turns out that the extensions (up to the obvious equivalence) correspond exactly to the elements of $Ext (C, A),$ with the split extension $B = A \oplus C$ corresponding to $0 .$ Take $A = C = Z /2 :$ there are two ways to build a group of order 4 from two copies of $Z /2,$ namely $Z /2 \oplus Z /2$ and $Z /4$ (in which the subgroup ${0, 2} ≅ Z /2$ has quotient $Z /2).$ Correspondingly, $Ext (Z /2, Z /2) ≅ Z /2$ has two elements. (The count is of extensions, not of groups. $Ext (Z /3, Z /3)$ has three elements, but only two middle groups occur, $Z /3 \oplus Z /3$ and $Z /9 :$ the two non-split extensions both use $Z /9,$ glued to the quotient by different maps.)

The universal coefficient theorems

Now the payoff. If you know the integral homology $H_{n} (X) = H_{n} (X; Z)$ of a space, do you know its homology and cohomology with every other coefficient group? Yes — and $Tor$ and $Ext$ are exactly the correction terms.

The second formula is the one stated in §4.2 Cohomology Groups; now you can see where its $Ext$ comes from. In both formulas the correction term is built from the degree below. The proof is a well-chosen short exact sequence of chain complexes and the long exact sequence it produces — homological algebra feeding on itself. We will not write it out; Hatcher does so in a few pages for each version (Hatcher 2002, Thm 3.2, Thm 3A.3). The cohomology version is the coincidence that Eilenberg and Mac Lane set out to explain in 1942, inventing functors along the way (§5.1 Categories and Functors).

Universal coefficients Choose a space and a coefficient group

cells of

RP^{2}

: word aa: ∂(face) = 2a

chains

0 \to Z \times 2 Z 0 Z \to 0

cochains in

Z /2

0 \to Z /2 0 Z /2 \times 2 Z /2 \to 0

	$n = 0$	$n = 1$	$n = 2$
$H_{n} (X; Z)$ integral homology	$Z$	$Z /2$	$0$
$H_{n} (X; Z /2)$ homology with coefficients	$Z /2$ $Z /2$ ⊕ $0$	$Z /2$ $Z /2$ ⊕ $0$	$Z /2$ $0$ ⊕ $Z /2$ from Tor of $H_{1}$
$H^{n} (X; Z /2)$ cohomology	$Z /2$ $Z /2$ ⊕ $0$	$Z /2$ $Z /2$ ⊕ $0$	$Z /2$ $0$ ⊕ $Z /2$ from Ext of $H_{1}$

Each entry is

H_{n} \otimes G \oplus Tor (H_{n - 1}, G)

(homology) or

Hom (H_{n}, G) \oplus Ext (H_{n - 1}, G)

(cohomology); the rose part comes from torsion one degree down. Every entry agrees with a direct computation from the chain complex above.

Figure 5.2.4 Pick a space and coefficients. The table computes

H_{n} (X; G)

and

H^{n} (X; G)

from integral homology by the universal coefficient theorems; the rose part of each entry is the

Tor

Ext

term coming from torsion one degree down. Every entry is checked against a direct computation from the cell chain complex shown at the top (the book’s test suite also checks it on the simplicial models of Part III). Compare

RP^{2}

with

Z /2,

Z /3

and

Q .

Careful Split, but not naturally

The exact sequences in the theorem are natural — a map of spaces gives a map between them. The splittings are not. Collapse the “equator” $RP^{1}$ of $RP^{2}$ to a point: the result is a sphere, and the quotient map $q : RP^{2} \to S^{2}$ induces zero on integral homology in degrees 1 and 2 (the groups are $Z /2 \to 0$ and $0 \to Z).$ A natural splitting of $H_{2} (-; Z /2)$ into a $\otimes$ part and a $Tor$ part would force $q_{*}$ to be zero on $H_{2} (-; Z /2)$ too. But mod 2, $q$ carries the sum of all the triangles of $RP^{2}$ onto the sphere, and $q_{*} : Z /2 \to Z /2$ is an isomorphism (Hatcher 2002, Example 2.51, §3.A). This is the “isomorphic but not naturally isomorphic” phenomenon of §5.1 Categories and Functors in the wild. In practice: the theorem tells you what the groups are, but to compute an induced map you should use the exact sequence, not the splitting.

Two consequences get used constantly. With coefficients in $Q$ (or $R)$ all torsion disappears: when the homology is finitely generated, $H_{n} (X; Q) ≅ Q^{b_{n}},$ and cohomology has the same dimensions as homology. With coefficients in $Z / p$ for a prime $p,$ torsion of order prime to $p$ disappears, while each summand $Z / p^{k}$ of $H_{n}$ is seen twice, as a $Z / p$ in degree $n$ and another in degree $n + 1 .$ That is why the mod-2 Betti numbers of $RP^{2}$ are $1, 1, 1$ although its rational Betti numbers are $1, 0, 0 .$

The Künneth theorem: homology of products

The torus is a product of two circles, $T^{2} = S^{1} \times S^{1} .$ Its homology $Z, Z^{2}, Z$ looks like it ought to be “the product” of the circle’s $Z, Z$ with itself — and the Künneth theorem, glimpsed in §4.5 The Cup Product as the Künneth formula, says exactly how.

Start with cells. If $X$ and $Y$ are cell complexes, then $X \times Y$ is one too: its cells are products $e^{i} \times e^{j}$ of an $i$ -cell of $X$ and a $j$ -cell of $Y,$ of dimension $i + j .$ The circle has a 0-cell and a 1-cell, so the torus gets cells of dimension $0 + 0,$ $0 + 1,$ $1 + 0$ and $1 + 1 :$ one vertex, two edges, one face, the familiar square with opposite sides glued. The cellular chain complex of the product is the tensor product of the two chain complexes, with boundary given by a Leibniz rule, $\partial (a \otimes b) = \partial a \otimes b + (- 1)^{i} a \otimes \partial b (a an i -cell) .$ So computing $H_{n} (X \times Y)$ is a purely algebraic question about the homology of a tensor product of complexes, and the answer is:

Hatcher proves it, for any principal ideal domain of coefficients, in his §3.B (Hatcher 2002, Thm 3B.6). For the torus: $H_{0} = Z \otimes Z = Z;$ $H_{1} = (H_{0} \otimes H_{1}) \oplus (H_{1} \otimes H_{0}) = Z^{2};$ $H_{2} = H_{1} \otimes H_{1} = Z .$ No torsion, no $Tor$ terms. A pleasant way to remember torsion-free cases: multiply the Poincaré polynomials $\sum_{n} b_{n} t^{n} .$ The circle has $1 + t,$ so the torus has $(1 + t)^{2} = 1 + 2 t + t^{2},$ and the $k$ -dimensional torus has $(1 + t)^{k},$ with binomial coefficients as Betti numbers. With coefficients in a field, the $Tor$ term disappears and this multiplication rule is exact.

The $Tor$ term is a genuine correction, and it is the same phenomenon as in the UCT: torsion in two factors combines to produce something one degree higher. For $RP^{2} \times RP^{2},$ the classes of order 2 in the two copies of $H_{1}$ give $Tor (Z /2, Z /2) = Z /2$ in $H_{3} .$ The figure lets you try products.

Homology of a product Choose the two factors

H_{i} (RP^{2}) \to

H_{j} (RP^{2}) ↑

H_{0} = Z

H_{1} = Z /2

H_{2} = 0

H_{2} = 0

n = 2

0

n = 3

0

n = 4

0

H_{1} = Z /2

n = 1

Z /2

n = 2

Z /2

Tor = Z /2

→ n = 3

n = 3

0

H_{0} = Z

n = 0

Z

n = 1

Z /2

n = 2

0

H_{0} (RP^{2} \times RP^{2})

= Z

Z

⊕

0

H_{1} (RP^{2} \times RP^{2})

= (Z /2)^{2}

(Z /2)^{2}

⊕

0

H_{2} (RP^{2} \times RP^{2})

= Z /2

Z /2

⊕

0

H_{3} (RP^{2} \times RP^{2})

= Z /2

0

⊕

Z /2

H_{4} (RP^{2} \times RP^{2})

= 0

0

⊕

0

Each anti-diagonal $i + j = n$ (same colour) adds up to $H_{n}$ ; the rose Tor terms jump one diagonal up. Agrees with the homology of the product cell complex.

X =

Y =

Figure 5.2.5 Cell

(i, j)

holds

H_{i} (X) \otimes H_{j} (Y);

the cells along an anti-diagonal

i + j = n

(one colour) add up to

H_{n} (X \times Y),

together with any

Tor

terms (rose) that arrive from the diagonal below. For

RP^{2} \times RP^{2},

H_{3}

consists of a single

Tor .

Every total is checked against the homology of the product cell complex, computed directly. Keep this grid in mind: a spectral sequence is a grid like this one that has not yet settled down.

Spectral sequences: computing in stages

Many spaces are built in stages. A cell complex is built skeleton by skeleton; a fibre bundle is built from copies of a fibre spread over the cells of a base; a double complex (a grid of groups with two commuting boundary maps) can be cut into columns. In each case there is an obvious plan: compute the homology of each stage separately, then glue. The difficulty is that the stages interact. A cycle that looks new at stage $p$ may have a boundary that lies in an earlier stage — or several stages earlier — and then it is not really a cycle of the whole. A spectral sequence is the bookkeeping device that makes the plan work, correcting the first approximation in a sequence of steps.

Here is the setting. A filtered complex is a chain complex $C$ with a nested sequence of subcomplexes $F_{0} C \subseteq F_{1} C \subseteq \dots \subseteq F_{k} C = C :$ the space after stage 0, after stage 1, and so on. Draw a grid with columns $p = 0, 1, 2, \dots$ for the stages and rows $q,$ and put each generator of degree $n$ that is added at stage $p$ at the position $(p, q)$ with $q = n - p .$ Total degree is constant along the anti-diagonals $p + q = n,$ just as in the Künneth grid.

Page $E^{1}$ records the homology of each stage relative to the stages before it: $E_{p, q}^{1} = H_{p + q} (F_{p} C / F_{p - 1} C) .$ It is the first approximation: the new cycles at each stage, ignoring how stages attach.
The differential $d^{1} : E_{p, q}^{1} \to E_{p - 1, q}^{1}$ records how stage $p$ attaches to stage $p - 1 :$ the boundary of a new cycle, seen in the previous stage. Taking homology gives the next page, $E^{2} = H (E^{1}, d^{1}) .$
The differential $d^{2} : E_{p, q}^{2} \to E_{p - 2, q + 1}^{2}$ records attachments that skip a stage; and in general $d^{r} : E_{p, q}^{r} \to E_{p - r, q + r - 1}^{r}$ moves $r$ columns left and $r - 1$ rows up, with $E^{r + 1} = H (E^{r}, d^{r}) .$
Convergence. With finitely many stages the differentials eventually have nowhere to go, the pages stop changing, and the final page $E^{\infty}$ spells out the homology: over a field, $dim H_{n} (C) = \sum_{p + q = n} dim E_{p, q}^{\infty} .$

Turning the pages Choose a page · switch examples and coefficients

Two disks on a segment

stage 0: two points v, w and a loop c at v
stage 1: a segment t from v to w, and a loop b at w
stage 2: a disk e filling c, a disk f filling b

v

dim 0, stage 0

\partial v = 0

w

dim 0, stage 0

\partial w = 0

c

dim 1, stage 0

\partial c = 0

t

dim 1, stage 1

\partial t = w - v

b

dim 1, stage 1

\partial b = 0

e

dim 2, stage 2

\partial e = c

f

dim 2, stage 2

\partial f = b

On page

E^{1}

, the differential

d^{1}

has slope

(- 1, 0)

. The 2 gold arrows pair off generators that cancel; turn the page to see them vanish.

Figure 5.2.6 Each dot is a generator, placed at (stage, degree − stage). Two disks on a segment: on

E^{1}

two

d^{1}

arrows cancel

t

against

w

and

f

against

b;

E^{2}

d^{2},

of slope

(- 2, 1),

cancels the disk

e

against the loop

c

it fills, two stages earlier;

E^{3} = E^{\infty}

leaves one dot, so the space has the homology of a point. Torus by cells: nothing ever cancels. $RP^{2}$ by cells: over

Q

d^{1}

(multiplication by 2) cancels

s

against

a;

over

Z /2

it is zero and every dot survives — the universal coefficient story again.

The most famous example has a fibration as input. When a space $E$ is fibred over a base $B$ with fibre $F$ — like the Hopf fibration of §5.3 Horizons, where $S^{3}$ is fibred over $S^{2}$ by circles — the Serre spectral sequence has $E_{p, q}^{2} = H_{p} (B; H_{q} (F))$ (when $B$ is simply connected) and converges to $H_{p + q} (E) .$ Its $E^{2}$ page is a Künneth grid, and the differentials measure how far $E$ is from being the product $B \times F .$ Jean-Pierre Serre built it in his 1951 thesis and used it to prove that the homotopy groups of spheres are finitely generated (Serre 1951); two years later he showed that almost all of them are finite (Serre 1953a).

Spectral sequences are still the main tool for computing homotopy groups of spheres, and one of the oldest problems about them was settled only recently, by a spectral sequence computed partly by machine.

Problem The Kervaire invariant one problem

Solved in 2024 Posed by Michel Kervaire and William Browder, 1960–69; proved by Michael Hill, Michael Hopkins and Douglas Ravenel (2009), and Weinan Lin, Guozhen Wang and Zhouli Xu, 2024

In which dimensions is there a closed framed manifold, a manifold with a consistent choice of directions at every point, whose Kervaire invariant is $1,$ so that it cannot be simplified by surgery into a sphere?

The problem began as a question about manifolds and became one about a spectral sequence: whether certain entries on an early page survive every differential and reach the last page. In dimension $126$ the answer needed a computer to calculate a large part of the spectral sequence.

How it was settled. Browder showed in 1969 that only the dimensions $2^{j + 1} - 2,$ that is $2, 6, 14, 30, 62, 126, 254, \dots,$ are possible, and examples were known in dimensions $2, 6, 14, 30$ and $62 .$ In 2009 Hill, Hopkins and Ravenel showed with equivariant homotopy theory that there are none beyond $126 .$ In 2024 Lin, Wang and Xu showed, with a large computer calculation of the Adams spectral sequence, that dimension $126$ has one. The answer: exactly the dimensions $2, 6, 14, 30, 62$ and $126 .$

Before and after. Framed manifolds, up to cobordism, are the same thing as the stable homotopy groups of spheres, so the problem is about which classes survive a spectral sequence to its last page. As of October 2026 the Lin–Wang–Xu paper is a preprint, widely accepted by experts.

(Hill, Hopkins & Ravenel 2016; Lin, Wang & Xu 2024) Status as of October 2026 All open problems

The Eilenberg–Steenrod axioms

This book has met several homology theories: simplicial homology of simplicial complexes, singular homology of all spaces, cellular homology of cell complexes, and (in §4.7 Sheaves and Čech Cohomology) Čech cohomology. Why do they agree whenever they can be compared? In 1945 Samuel Eilenberg and Norman Steenrod answered: because they all satisfy the same short list of properties, and those properties alone determine homology (Eilenberg & Steenrod 1945). Notice that the list is phrased entirely in the language of §5.1 Categories and Functors: functors, natural transformations, exactness.

Definition Homology theory (Eilenberg–Steenrod)

A homology theory assigns to each pair of spaces $(X, A)$ (with $A \subseteq X)$ abelian groups $H_{n} (X, A),$ one for each integer $n,$ functorially in maps of pairs, together with natural connecting homomorphisms $\partial : H_{n} (X, A) \to H_{n - 1} (A)$ (where $H_{n} (A)$ means $H_{n} (A, \emptyset)),$ such that:

Homotopy. Homotopic maps of pairs induce the same homomorphisms.
Exactness. Every pair gives a long exact sequence $\dots \to H_{n} (A) \to H_{n} (X) \to H_{n} (X, A) \partial H_{n - 1} (A) \to \dots .$
Excision. If the closure of $U$ lies in the interior of $A,$ the inclusion $(X ∖ U, A ∖ U) \to (X, A)$ induces isomorphisms in every degree.
Dimension. For a one-point space, $H_{n} (pt) = 0$ when $n \neq = 0 .$
Additivity. For a disjoint union $X = ⨆_{α} X_{α},$ the inclusions induce an isomorphism $⨁_{α} H_{n} (X_{α}) ≅ H_{n} (X) .$

The group $G = H_{0} (pt)$ is called the coefficient group of the theory.

Each axiom is a theorem you have seen for singular homology: homotopy invariance (§3.5 Maps, Invariance, and First Triumphs), the long exact sequence of a pair and excision (§3.6 Exact Sequences and Mayer–Vietoris), the homology of a point (§3.3 Homology Groups), and additivity, which for finitely many pieces follows from the others and was added for infinite unions by John Milnor in 1962 (Milnor 1962). What is new is the claim that nothing else is needed. The figure computes the homology of every sphere using only the five axioms.

Spheres from the axioms Step through · the axioms used in each step light up

Homotopy

f ≃ g \Rightarrow f_{*} = g_{*}

Homotopic maps induce the same homomorphism.

Exactness

\dots \to H_{n} (A) \to H_{n} (X) \to H_{n} (X, A) \partial \dots

Every pair gives a long exact sequence, with natural ∂.

Excision

H_{n} (X ∖ U, A ∖ U) ≅ H_{n} (X, A)

Cutting out a set deep inside A changes nothing.

Dimension

H_{n} (pt) = 0 for n \neq = 0

A point has homology only in degree 0, where it is G.

Additivity

H_{n} (⨆ X_{α}) ≅ ⨁ H_{n} (X_{α})

A disjoint union has the direct sum of homologies.

H ~_{k} (S^{n}) = ?

We want the homology of every sphere, using only the five axioms — no chains, no simplices, no matrices.

The question

Figure 5.2.7 The computation that drives the uniqueness theorem. A disk has the homology of a point; the long exact sequence of the pair

(D^{n}, S^{n - 1})

shifts degree by one; collapsing the rim of the disk gives a sphere one dimension up. Iterating down to the 0-sphere — two points — leaves

H ~_{k} (S^{n}) ≅ G

for

k = n

and

0

otherwise. Here

H ~

is reduced homology (§3.3 Homology Groups), which removes one copy of

G

from

H_{0} .

The idea of the proof is the figure, run inside an induction on cells. Attaching an $n$ -cell changes homology only through the pair $(D^{n}, S^{n - 1}),$ whose homology the axioms pin down; the long exact sequence and the five lemma (a cousin of the snake lemma) carry the isomorphism from one stage of the construction to the next. In fact the argument shows more: any theory satisfying the axioms can be computed by the cellular chain complex. So simplicial, singular and cellular homology agree on complexes — they all satisfy the axioms with coefficient group $Z .$ Hatcher states the uniqueness theorem in his §2.3 and proves it, for all CW pairs, in Chapter 4 (Hatcher 2002, §2.3, Thm 4.59).

Dropping an axiom

The dimension axiom looks the most innocent: of course a point has no interesting homology in degree 5. But it is the one axiom that refers to a specific space, and if you drop it you get a wider class of generalized (co)homology theories that still satisfy homotopy, exactness, excision and additivity — and so still have long exact sequences, Mayer–Vietoris sequences and spectral sequences — but in which a point can have nonzero groups in many degrees. Complex K-theory, built from vector bundles, assigns to a point the group $Z$ in every even degree; cobordism, built from manifolds, assigns to a point the groups of all closed manifolds up to cobordism. These are among the destinations of §5.3 Horizons.

A little history From the axioms to the book

Norman Steenrod 1910 – 1971

With no fellowship after graduating, he spent a year at home in Dayton, Ohio, working alone on problems from Raymond Wilder’s topology course, the one mathematics course he had taken at Michigan, until his first paper brought offers from Harvard, Princeton and Duke (O’Connor & Robertson 2000). He and Eilenberg announced their axioms in a four-page note in the Proceedings of the National Academy of Sciences in 1945, and developed them into Foundations of Algebraic Topology (1952) (Eilenberg & Steenrod 1952). The note pointed out at once that singular and Čech homology both satisfy the axioms, and so must agree on all finite complexes (Weibel 1999). The dimension axiom went in because every homology theory then known obeyed it; within a few years topologists were studying bordism, whose groups of a point are nonzero in infinitely many dimensions (Hatcher 2002, §2.3). The generalized theories of today keep the other four axioms, Milnor’s additivity among them, and simply drop the dimension axiom.

Exercises

Exercise Finishing the snake

In the snake lemma, prove exactness at $ker c$ — that the kernel of $δ$ is exactly the image of $ker b \to ker c$ — and at $coker a .$

Hint

For “ $δ (x) = 0$ implies $x$ comes from $ker b ”:$ if $z = a (w),$ replace the lift $y$ by $y - i (w) .$ For exactness at $coker a,$ chase an element $[z]$ with $i^{'} (z) \in im b$ back up to $ker c .$

Solution

At $ker c .$ If $x = p (y)$ with $b (y) = 0,$ use $y$ as the lift: then $z = 0$ and $δ (x) = 0 .$ Conversely, suppose $δ (x) = 0,$ i.e. the element $z$ with $i^{'} (z) = b (y)$ lies in $im a,$ say $z = a (w) .$ Put $y_{0} = y - i (w) .$ Then $p (y_{0}) = p (y) = x$ (since $p \circ i = 0),$ and $b (y_{0}) = b (y) - b (i (w)) = i^{'} (z) - i^{'} (a (w)) = 0 .$ So $y_{0} \in ker b$ maps to $x .$

At $coker a .$ The composite $ker c \to coker a \to coker b$ sends $x$ to the class of $i^{'} (z) = b (y),$ which is zero in $coker b = B^{'} / im b .$ Conversely, if $[z] \in coker a$ maps to zero in $coker b,$ then $i^{'} (z) = b (y)$ for some $y \in B .$ Let $x = p (y) .$ Then $c (x) = c (p (y)) = p^{'} (b (y)) = p^{'} (i^{'} (z)) = 0,$ so $x \in ker c,$ and by construction (with lift $y)$ $δ (x) = [z] .$

Summary

Recap

Chain complexes and chain maps form a category $Ch;$ homology is a functor on it, and it measures the failure of exactness. Chain homotopic maps induce the same map on homology.
The snake lemma turns a commutative diagram with exact rows into a six-term exact sequence; its connecting map is found by a diagram chase. It yields the long exact sequence of a short exact sequence of complexes.
$Hom (-, G)$ is left exact and $- \otimes G$ is right exact. Examples: $Z / m \otimes Z / n ≅ Z / g cd (m, n),$ $Hom (Z / n, Z) = 0 .$
Applying them to a free resolution measures the failure of exactness: $Tor (Z / m, Z / n) ≅ Z / g cd (m, n),$ $Ext (Z / n, Z) ≅ Z / n;$ free groups give zero. Tor detects torsion; Ext classifies extensions.
The universal coefficient theorems: $H_{n} (X; G) ≅ H_{n} \otimes G \oplus Tor (H_{n - 1}, G)$ and $H^{n} (X; G) ≅ Hom (H_{n}, G) \oplus Ext (H_{n - 1}, G)$ — torsion is felt one degree up. The splittings are not natural.
Künneth: $H_{n} (X \times Y) ≅ ⨁_{i + j = n} H_{i} \otimes H_{j} \oplus ⨁_{i + j = n - 1} Tor (H_{i}, H_{j}) .$
A spectral sequence computes the homology of a filtered complex page by page, with $d^{r}$ of slope $(- r, r - 1);$ cellular homology is the simplest case, and the Serre spectral sequence of a fibration the most famous.
The Eilenberg–Steenrod axioms — homotopy, exactness, excision, dimension, additivity — determine homology on cell complexes. Dropping the dimension axiom gives generalized theories such as K-theory and cobordism.

Homological Algebra

Chain complexes, as a category

Chain homotopy

The snake lemma and long exact sequences

From short exact sequences of complexes to long ones

Hom and tensor: two ways to combine groups

Hom

Tensor product

Exactness, and how it fails

Resolutions, Tor and Ext

Free resolutions

Where the names come from

The universal coefficient theorems

The Künneth theorem: homology of products

Spectral sequences: computing in stages

The Eilenberg–Steenrod axioms

Dropping an axiom

Exercises

Summary

Further reading

References