4.2 Cohomology Groups · Homology & Cohomology

We have also isolated the algebra from the topology to show that a large part of cohomology and Hodge theory is nothing more than the linear algebra of matrices satisfying AB = 0.

— Lek-Heng Lim, Hodge Laplacians on Graphs (2020)

In the last chapter, measurements arrived as answers to puzzles: heights on a trail map, climbs on its trails, a staircase that could not exist. Now we build the machine. It will look almost exactly like homology — groups of cochains, a coboundary map, cocycles and coboundaries, and their quotient — and for good reason: it is homology run through a mirror.

If the reflection were perfect, there would be little reason to build it. It is not, and the flaws are the reward. Maps between spaces will act on cohomology backwards. The twisted, finite part of homology — the torsion that made the projective plane strange in Part III — will reappear one floor higher than you would expect. And measurements, unlike places, can be multiplied, which later in this part lets cohomology tell apart spaces whose homology is identical.

Cochain groups: measurements as homomorphisms

A cochain is still what it was in the last chapter: a number on every simplex. This section only dresses that idea in the formal clothes the rest of Part IV expects, starting from the chains of §3.2 Chains and the Boundary Operator. For a simplicial complex $K,$ the group $C_{k} (K)$ consists of formal sums $c = \sum_{i} c_{i} σ_{i}$ of oriented $k$ -simplices with integer coefficients — inventories of places, with multiplicities. Its elements add coefficient by coefficient, and the $k$ -simplices themselves form a basis: $C_{k} (K)$ is a free abelian group, with one copy of $Z$ for each $k$ -simplex.

A measurement assigns a value to each $k$ -simplex. The values can live in any abelian group $G$ — the real numbers $R,$ the integers $Z,$ or the two-element group $Z /2$ of the gear ring — and the choice is called the coefficients. Once we know the value $φ (σ)$ of a measurement on every simplex, there is only one sensible way to measure a chain: add up, with the multiplicities,

φ (i \sum c_{i} σ_{i}) = i \sum c_{i} φ (σ_{i}) .

(Here $c_{i} φ (σ_{i})$ means $φ (σ_{i})$ added to itself $c_{i}$ times, or subtracted if $c_{i}$ is negative; that makes sense in any abelian group.) This rule respects addition, $φ (c + c^{'}) = φ (c) + φ (c^{'}),$ and a function between groups that respects addition is a homomorphism. Conversely, because the simplices form a basis, a homomorphism out of $C_{k} (K)$ can take any values on the simplices, and those values determine it. So “a value on every $k$ -simplex” and “a homomorphism $C_{k} (K) \to G ”$ are two descriptions of the same thing.

Concretely, a $k$ -cochain is a list of $n_{k}$ values in $G,$ one for each of the $n_{k}$ simplices of dimension $k,$ so $C^{k} (K; G) ≅ G^{n_{k}} .$ The hollow triangle has $C^{1} (K; Z) ≅ Z^{3} :$ three integers, one per edge. The simplest cochains are the indicator cochains: for a $k$ -simplex $τ,$ the cochain $1_{τ}$ is $1$ on $τ$ and $0$ on every other $k$ -simplex. Every cochain is a combination of them, $φ = \sum_{τ} φ (τ) 1_{τ} .$

With real coefficients this is a familiar construction: $C^{k} (K; R)$ is the dual space of the vector space $C_{k} (K; R)$ from §1.5 Linear Algebra, its elements are the linear functionals on chains, and the indicator cochains form the dual basis. Cochains are to chains what row vectors are to column vectors.

The value of a cochain on a chain gets its own notation, the pairing of §4.1 Cochains: Measuring Shapes:

⟨ φ, c ⟩ = φ (c) \in G .

Think of it as integrating the measurement $φ$ over the place $c .$ Try it.

Measuring a chain Tap an edge, then ±1 to change its coefficient · try the presets

Edge [2,3] in the chain:

c=[0,1]+[1,2]+[2,3]
⟨φ,c⟩=3+1+(−2)=2
∂c=−[0]+[3] 

Figure 4.2.1 A cochain

φ

is a number on each edge (gold); a chain

c

is a combination of edges (violet, with its coefficients). The pairing

⟨ φ, c ⟩

multiplies and adds. Switch to

φ = δ f :

now the pairing depends only on the boundary of the chain,

⟨ δ f, c ⟩ = ⟨ f, \partial c ⟩,

so it is

0

on every loop.

The coboundary map is a transpose

In §4.1 Cochains: Measuring Shapes we met the coboundary as the discrete gradient and the discrete curl, and noticed that both are “evaluate on the boundary”. Here is the definition in every degree and for every coefficient group.

The matrix of $δ$

Choose the simplices as bases: $k$ -simplices for $C_{k}$ and their indicator cochains for $C^{k} .$ Then $\partial_{k + 1}$ is a matrix with a row for each $k$ -simplex and a column for each $(k + 1)$ -simplex; column $σ$ lists the faces of $σ$ with their signs. What is the matrix of $δ_{k} : C^{k} \to C^{k + 1} ?$ Its column for $τ$ is $δ 1_{τ},$ and its entry in row $σ$ is

(δ 1_{τ}) (σ) = 1_{τ} (\partial σ) = the coefficient of τ in \partial σ = the entry of \partial_{k + 1} in row τ, column σ .

Rows and columns have swapped places:

Read the formula both ways. A column of $\partial$ answers “what are the faces of this simplex?”. The same numbers, read as a row of $δ,$ answer “which values does $δ φ$ on this simplex look at?”. And a column of $δ$ — the coboundary of an indicator $1_{τ}$ — answers “which simplices have $τ$ as a face?”: $δ 1_{τ}$ is $\pm 1$ on the simplices one dimension up that contain $τ,$ its cofaces. The boundary looks down to faces; the coboundary looks up to cofaces.

Daniel Tubbenhauer, in his video What is…cohomology? Or: Reversing arrows, watches this happen on the solid tetrahedron with vertices $0, 1, 2, 3,$ working mod 2 so that every sign disappears. Going down, the tetrahedron has four triangles as its boundary, a triangle has three edges and an edge has two ends. Going up, a vertex reaches the three edges that leave it, an edge the two triangles that contain it, and a triangle the one solid it bounds:

δ 1_{[0]} = 1_{[0, 1]} + 1_{[0, 2]} + 1_{[0, 3]}, δ 1_{[0, 1]} = 1_{[0, 1, 2]} + 1_{[0, 1, 3]}, δ 1_{[0, 1, 2]} = 1_{[0, 1, 2, 3]} .

His slogan fits on one line: “Homology goes down, cohomology goes up.”

One matrix, read two ways Hover a matrix entry · tap a vertex or edge

Tap a vertex to put a 1 on it: its coboundary is ±1 on the edges it is a face of.

δ (1_{1}) = 1_{01} - 1_{12} - 1_{13}

Boundary

\partial_{1}

: columns list faces

	$[0, 1]$	$[0, 2]$	$[1, 2]$	$[1, 3]$	$[2, 3]$
$[0]$	−1	−1	0	0	0
$[1]$	1	0	−1	−1	0
$[2]$	0	1	1	0	−1
$[3]$	0	0	0	1	1

Coboundary

δ_{0} = \partial_{1}^{T}

: rows list faces

	$[0]$	$[1]$	$[2]$	$[3]$
$[0, 1]$	−1	1	0	0
$[0, 2]$	−1	0	1	0
$[1, 2]$	0	−1	1	0
$[1, 3]$	0	−1	0	1
$[2, 3]$	0	0	−1	1

(δ f) ([a, b]) = f (\partial [a, b]) = f (b) - f (a)

Figure 4.2.2 The boundary matrix

\partial

and the coboundary matrix

δ = \partial^{T}

of two triangles sharing an edge. Hover an entry: it lights up the same pair (face, coface) in the picture and the mirrored entry in the other matrix. Tap a vertex (or an edge) to put a

1

on it: its coboundary is

\pm 1

on the edges (or triangles) it borders — a column of

δ,

a row of

\partial .

The coboundary of a coboundary

For every cochain $φ$ and every simplex $σ,$

(δ δ φ) (σ) = (δ φ) (\partial σ) = φ (\partial\partial σ) = φ (0) = 0,

so $δ \circ δ = 0,$ as a direct consequence of $\partial \circ \partial = 0 .$ In matrices it is one line of §1.5 Linear Algebra: $δ_{k + 1} δ_{k} = \partial_{k + 2}^{T} \partial_{k + 1}^{T} = (\partial_{k + 1} \partial_{k + 2})^{T} = 0,$ because the transpose of a product is the product of the transposes in the opposite order.

The tetrahedron shows the same thing without matrices. Climb twice from the vertex $0 :$ first to its three edges, then from each edge to its two triangles. Every triangle at $0$ contains exactly two edges at $0,$ so it is reached twice — $[0, 1, 2]$ by way of $[0, 1]$ and by way of $[0, 2]$ — and mod 2 the two arrivals cancel. (With integers they arrive with opposite signs.) A coface of a coface is always reached along two routes, which is $\partial\partial = 0$ read upside down.

The cochain groups and coboundary maps form a sequence in which each map raises the degree:

0 ⟶ C^{0} (K; G) δ_{0} C^{1} (K; G) δ_{1} C^{2} (K; G) δ_{2} \dots, δ_{k + 1} \circ δ_{k} = 0.

Such a sequence is a cochain complex. Compare the chain complex of Part III, $\dots \to C_{2} \to C_{1} \to C_{0} \to 0,$ whose maps lower the degree: same groups, arrows reversed. Why did homology point its arrows downwards in the first place? “For no good reason,” says Tubbenhauer: as a bare sequence of groups and maps, a cochain complex is a chain complex numbered the other way. The surprises of this chapter come from somewhere else, from the fact that its groups measure chains.

Cocycles, coboundaries and cohomology

Everything is now in place to copy the definition of homology from §3.3 Homology Groups, with every arrow reversed.

Every word has a picture from the last chapter. A 1-cocycle is an edge labelling that passes every local test (zero curl on every filled triangle). A 1-coboundary is a gradient. Two 1-cocycles are cohomologous when one turns into the other by adding bumps at vertices, so a class in $H^{1}$ is a global obstruction with all the irrelevant detail forgotten. In degree 0 there are no coboundaries, so $H^{0} = Z^{0}$ is the group of locally constant functions. In degree 2 the question is “is this measurement on triangles the curl of something?”, and $H^{2}$ collects the answers “no”.

Counting dimensions over a field

When the coefficients form a field $F$ — the real numbers, the rationals, or $Z /2$ — every group above is a vector space and we can count dimensions with rank–nullity from §1.5 Linear Algebra. Writing $n_{k}$ for the number of $k$ -simplices, $dim Z^{k} = n_{k} - rank δ_{k}$ and $dim B^{k} = rank δ_{k - 1},$ so

dim H^{k} (K; F) = n_{k} - rank δ_{k} - rank δ_{k - 1} = n_{k} - rank \partial_{k + 1} - rank \partial_{k} = dim H_{k} (K; F),

because a matrix and its transpose have the same rank. Over a field, cohomology and homology have the same dimensions: the Betti numbers count both. (In fact $H^{k} (K; F)$ is the dual vector space of $H_{k} (K; F),$ as we will see.) In particular the Euler characteristic can be read off either: $χ (K) = \sum_{k} (- 1)^{k} dim H^{k} (K; F) .$ The integers will be more interesting.

First computations: graph, circle, sphere, torus

Graphs

A graph has no triangles, so $δ_{1} = 0$ and every 1-cochain is a cocycle. The last chapter did the rest: $H^{0} = G^{c},$ one value per connected piece, and $H^{1} = C^{1} / im δ_{0} ≅ G^{E - V + c},$ one free value per edge outside a spanning tree — the loop sums of the fundamental loops. With $G = Z :$ $H^{0} ≅ Z^{b_{0}}$ and $H^{1} ≅ Z^{b_{1}} .$

Read the graph case as a verdict on one map. The cochain complex is $0 \to C^{0} δ C^{1} \to 0,$ so $H^{0} = ker δ$ measures how far $δ$ is from being one-to-one (the “ $+ C ”:$ different heights with the same climbs), and $H^{1} = C^{1} / im δ$ measures how far it is from being onto (the climbs that no heights produce: the obstructions). Every cohomology group asks the same question of the $δ$ on either side of it: what is left over that the maps fail to account for? The YouTube channel K-Theory, in What, and why, is coHomology, works through such tiny complexes (one term, two terms, a short exact sequence) and turns to spaces only at the end.

The circle and the torus, step by step

The next figure computes the cohomology of the circle (as a hollow triangle) and of the torus. For the torus we use Hatcher’s small model: a square whose opposite sides are glued, cut along a diagonal into two triangles. All four corners become a single vertex, so this is not a simplicial complex but a Δ-complex (§2.5 Simplicial Complexes); chains, cochains and (co)boundary matrices work exactly the same way, and the answers agree with those of genuine triangulations.

Computing H* step by step Use ‹ › to step · switch between circle and torus

The hollow triangle has three vertices, three edges and no triangles. A 0-cochain is three numbers $f_{0}, f_{1}, f_{2},$ a 1-cochain is three numbers $ψ_{01}, ψ_{02}, ψ_{12},$ and there are no 2-cochains:

$C^{0} ≅ Z^{3}, C^{1} ≅ Z^{3}, C^{2} = 0.$

The cochains

Figure 4.2.3 Cohomology in four moves: write down the cochains, write down

δ

as a transposed boundary matrix, find the cocycles (the kernel) and the coboundaries (the image), and take the quotient. The circle gives

H^{0} = H^{1} = Z;

the torus gives

H^{0} = Z,

H^{1} = Z^{2}

and

H^{2} = Z .

For the circle, $δ_{0}$ has rank 2, its kernel is the constants ( $H^{0} ≅ Z),$ and since there are no triangles, $H^{1} = C^{1} / im δ_{0} .$ The loop sum $ψ \mapsto ψ_{01} + ψ_{12} - ψ_{02}$ vanishes exactly on the image and takes every integer value, so it identifies $H^{1} ≅ Z .$ A generator is the cochain that is $1$ on a single edge: a fence across the loop.

For the torus, the single vertex makes $δ_{0} = 0 .$ Both triangles have boundary $a + b - c,$ so the cocycle condition is $φ (c) = φ (a) + φ (b),$ with two free choices; nothing is divided out, and $H^{1} ≅ Z^{2},$ with basis

α = (1, 0, 1), β = (0, 1, 1) (values on a, b, c) .

In degree 2 the two triangles carry values $p$ and $q;$ the coboundaries are exactly the pairs with $p = q,$ and $(p, q) \mapsto p - q$ identifies $H^{2} ≅ Z .$ The number $p - q$ is the value on $L - U,$ the 2-cycle that represents the whole torus.

The sphere

Take the hollow tetrahedron, with four triangles, six edges and four vertices. It is connected, so $H^{0} ≅ Z .$ Every 1-cocycle is a coboundary: a 1-cocycle passes every local test, and every loop on the sphere is the boundary of a union of triangles, so all its loop sums vanish and the gradient test of §4.1 Cochains: Measuring Shapes makes it a gradient. Hence $H^{1} = 0 .$ In degree 2 every cochain is a cocycle (there are no 3-simplices). Let $[S^{2}] = \partial [0, 1, 2, 3]$ be the 2-cycle made of all four triangles with their signs. The map $φ \mapsto ⟨ φ, [S^{2}]⟩$ kills every coboundary, by Stokes: $⟨ δ ψ, [S^{2}]⟩ = ⟨ ψ, \partial [S^{2}]⟩ = 0 .$ It takes the value $\pm 1$ on each triangle indicator, so it hits every integer. Its kernel is exactly the coboundaries, and you can see why by pushing weight around. Any two triangles of the tetrahedron share an edge, and adding a multiple of the coboundary of that edge’s indicator moves weight between those two triangles and touches no others (the exercise “Two triangles of the sphere” makes one such move). Three moves pile all of a cochain’s weight onto a single triangle without changing its value on $[S^{2}] .$ If that value was $0,$ nothing is left on the pile, so the cochain was a coboundary all along. So

H^{0} (S^{2}) ≅ Z, H^{1} (S^{2}) = 0, H^{2} (S^{2}) ≅ Z,

and any single triangle indicator generates $H^{2} :$ a measurement that puts all its weight on one tiny triangle is, up to coboundaries, the same as one spread evenly over the sphere — only the total $⟨ φ, [S^{2}]⟩$ matters.

Fences: what a 1-cocycle measures

Here is the most useful picture of a 1-cocycle on a surface. Hatcher describes it for coefficients $Z /2 :$ if $ψ$ is a cocycle, then on each triangle it takes the value $1$ on an even number of edges, so we can draw short arcs inside each triangle joining the edges with value $1$ in pairs. The arcs link up into curves — in his words, “a collection $C_{ψ}$ of disjoint curves in $X$ crossing the 1-skeleton transversely, such that the number of intersections of $C_{ψ}$ with each edge is equal to the value of $ψ$ on that edge.” With integer coefficients the curves get a direction, and crossings count with signs.

Call such curves a fence. Then the value of the cocycle on a closed loop is the number of times the loop crosses the fence, counted with sign. And if $ψ = δ f,$ Hatcher notes, the curves are “something like level curves for a function $φ$ with $δ φ = ψ ”:$ $f$ jumps by one every time you step across the fence, so a fence coming from a function must separate the surface into regions at different heights.

Fences on a torus Drag to rotate · choose a fence and a loop · play the wiggle

⟨ α, γ ⟩ = (+ 1) = 1

the loop γ goes 1× around the hole and 0× around the tube

Fence

Loop

straight wiggled

Figure 4.2.4 A 1-cocycle drawn as a fence standing on the torus. Its value on the golden loop is the number of crossings, counted with sign. The fence

α,

across the tube, counts trips around the hole;

β,

along the tube, counts trips around the tube. Wiggling a fence adds a coboundary: crossings may appear, but only in cancelling pairs. A fence around a little disk is a coboundary, and every loop crosses it a net zero times.

The figure is the torus computation made visible. The cocycle $α = (1, 0, 1)$ is $1$ on the edges $a$ and $c$ and $0$ on $b :$ as a fence, it is a circle running across the square in the $b$ direction, which the loop $a$ crosses once and the loop $b$ never crosses. So $⟨ α, a ⟩ = 1$ and $⟨ α, b ⟩ = 0;$ likewise $⟨ β, a ⟩ = 0$ and $⟨ β, b ⟩ = 1 .$ The two classes $α, β$ are exactly the measurements “how many times around in direction $a ”$ and “how many times around in direction $b ”.$

Measuring classes with classes

The figure also shows that wiggling the fence does not change the count on a closed loop, and that moving the loop does not either. Both facts are Stokes’ formula $⟨ δ φ, c ⟩ = ⟨ φ, \partial c ⟩$ from the last chapter.

This is often called the Kronecker pairing. So cohomology classes are measurements of homology classes: each $[φ] \in H^{k}$ gives a homomorphism $H_{k} (K) \to G .$ For the torus, $α$ and $β$ are the “dual basis” to the loops $a$ and $b,$ and every class in $H^{1} (T^{2})$ is determined by its two values $⟨ φ, a ⟩$ and $⟨ φ, b ⟩ .$ Whether a class is always determined by its values on cycles is the question the Universal Coefficient Theorem answers, at the end of this chapter: almost, but not quite.

Arrows reverse: pulling measurements back

In §3.5 Maps, Invariance, and First Triumphs a map of spaces $f : X \to Y$ pushed chains forward: a simplex of $X$ is sent to its image in $Y,$ giving a chain map $f_{♯} : C_{k} (X) \to C_{k} (Y)$ and an induced map $f_{*} : H_{k} (X) \to H_{k} (Y) .$ Places travel in the direction of the map.

Measurements travel the other way. Your drive to work is a map from the half-hour of the journey into the country; a weather map is a measurement on the country. The weather map gives you a temperature for every minute of the drive — read it off wherever you happen to be — but no record of the drive will ever produce a weather map. A map $X \to Y$ carries measurements on $Y$ back to $X,$ and not the other way. In symbols: if $φ$ measures chains of $Y,$ we can measure a chain $c$ of $X$ by first pushing it into $Y$ and then measuring there. That defines a cochain on $X,$

f^{*} φ = φ \circ f_{♯}, (f^{*} φ) (c) = φ (f_{♯} (c)),

called the pullback of $φ .$ Note the direction: $f$ goes from $X$ to $Y,$ but $f^{*}$ goes from cochains on $Y$ to cochains on $X .$ You have seen this reversal twice before. The preimage of §1.1 Sets and Functions takes subsets of $Y$ back to subsets of $X,$ and the transpose of §1.5 Linear Algebra turns a matrix from $R^{m}$ to $R^{n}$ into one from $R^{n}$ to $R^{m} .$ A pullback is both at once: $f^{*} 1_{τ}$ is $\pm 1$ on the simplices that land on $τ$ (the sign records whether the orientation is kept) and $0$ elsewhere, and the matrix of $f^{*}$ is the transpose of the matrix of $f_{♯} .$

A rule that turns spaces into groups and maps into homomorphisms in the same direction, like $H_{k},$ is called covariant; one that reverses the direction, like $H^{k},$ is called contravariant. (The precise language of functors waits for §5.1 Categories and Functors.) Hatcher puts the whole distinction in one sentence: “The basic distinction between homology and cohomology is thus that cohomology groups are contravariant functors while homology groups are covariant.”

Forward and back Choose k · drag the rose fence around the right-hand circle

Wrapping number k

Move the point

f_{*} [S^{1}] = 2 [S^{1}]

f^{*} [φ] = 2 [φ]

⟨ f^{*} φ, [S^{1}]⟩ = 2 = ⟨ φ, f_{*} [S^{1}]⟩

Figure 4.2.5 The map

z \mapsto z^{k} .

The point on the left goes around once while its image on the right goes around

k

times: chains are pushed forward,

f_{*} [S^{1}] = k [S^{1}] .

The fence

φ

on the right pulls back to

∣ k ∣

fences on the left, each crossed with sign

sign k :

measurements are pulled back,

f^{*} [φ] = k [φ] .

The two arrows point in opposite directions and agree on the pairing.

A little history Why “co”, and when

James Waddell Alexander 1888 – 1971

Andrey Kolmogorov 1903 – 1987

Cohomology was found twice in one week. At the first international topology conference, in Moscow in September 1935, James Alexander and Andrey Kolmogorov independently presented what we now call cohomology and its product; Alexander’s talk was titled “On the ring of a complex and the combinatory theory of integration” (Apushkinskaya et al. 2019). The two discoverers were an unlikely pair. Alexander, a Princeton topologist who had inherited a fortune, never drew a salary from the Institute for Advanced Study, and colleagues remembered his fondness for limericks. Kolmogorov had worked as a railway conductor after leaving school, and two years before Moscow he had built probability theory on axioms (O’Connor & Robertson 2000; O’Connor & Robertson 1999). The names came a little later: Hassler Whitney’s 1938 paper “introduced the modern ‘co’ terminology: coboundary (δ) and cocycle”, the “co” standing for the duality with boundaries and cycles (Weibel 1999; Whitney 1938).

Torsion moves up: the projective plane and the Klein bottle

Now the promised surprise. In §3.4 Computing Homology you met the real projective plane $RP^{2},$ a square whose opposite sides are glued with a twist, and found $H_{0} = Z,$ $H_{1} = Z /2$ and $H_{2} = 0 :$ there is a loop that is not a boundary while twice it is — torsion. What does cohomology make of it?

The torsion shift Use ‹ › to step through the computation

Glue opposite sides of a square with a twist: $a$ to $a$ and $b$ to $b,$ with the arrows. The corners become two vertices $v$ and $w;$ the edges are $a, b$ (from $v$ to $w)$ and the diagonal $c$ (from $w$ to $w);$ the faces are two triangles with

$\partial T_{1} = a - b + c, \partial T_{2} = - a + b + c .$

The complex

Figure 4.2.6 The projective plane as a Δ-complex: two vertices

v, w,

edges

a, b, c,

triangles

T_{1}, T_{2} .

Homology has its

Z /2

in degree 1; integer cohomology has

H^{1} = 0

and its

Z /2

in degree 2. The same “2” causes both: it sits in

\partial_{2},

and transposed, in

δ_{1} .

With

Z /2

coefficients every group is

Z /2 .

Here is the computation of the figure in words. The triangles have $\partial T_{1} = a - b + c$ and $\partial T_{2} = - a + b + c .$ A 1-cocycle $φ$ must therefore satisfy $φ (a) - φ (b) + φ (c) = 0$ and $- φ (a) + φ (b) + φ (c) = 0 .$ Adding the two equations gives $2 φ (c) = 0 .$ With integer values that forces $φ (c) = 0,$ and then $φ (a) = φ (b) .$ But those are exactly the coboundaries: $δ f$ has the value $f (w) - f (v)$ on both $a$ and $b$ (each runs from $v$ to $w)$ and $0$ on $c$ (which runs from $w$ to $w).$ So

H^{1} (RP^{2}; Z) = 0.

No integer measurement can detect the twisted loop $c .$ The reason is visible in the equation $2 φ (c) = 0 :$ because $2 c$ is a boundary, any cocycle must give it the value $0,$ and the only integer whose double is $0$ is $0 .$ In algebraic terms, there is no non-zero homomorphism from $Z /2$ to $Z .$

The torsion has not vanished, though. In degree 2 a cochain is a pair $(p, q)$ of values on $T_{1}, T_{2},$ and a coboundary $δ φ = (x - y + z, - x + y + z)$ always has an even total $p + q = 2 z .$ Every pair with an even total is a coboundary, and the parity of $p + q$ can never be changed, so

H^{2} (RP^{2}; Z) ≅ Z /2, H^{0} (RP^{2}; Z) ≅ Z .

Homology had $Z /2$ in degree 1; cohomology has it in degree 2. The torsion has moved up one degree. Notice that $H^{2} \neq = 0$ even though $H_{2} = 0 :$ cohomology is not just homology with a superscript.

With $Z /2$ coefficients

Rerun the computation with values in $Z /2,$ where $1 + 1 = 0 .$ Now $2 φ (c) = 0$ holds for every $φ (c),$ so the cocycle $φ = (1, 0, 1)$ — a fence crossing the twisted loop once — is allowed, and it is not a coboundary (those are $(0, 0, 0)$ and $(1, 1, 0)).$ So $H^{1} (RP^{2}; Z /2) ≅ Z /2,$ and in fact every group of $RP^{2}$ with $Z /2$ coefficients is $Z /2,$ in homology and cohomology alike. Mod-2 measurements can see one-sidedness that integer measurements cannot.

The Klein bottle

The Klein bottle $K$ — a square with one pair of sides glued straight and the other with a flip — has $H_{0} = Z,$ $H_{1} = Z \oplus Z /2,$ $H_{2} = 0$ (§3.4 Computing Homology). The same computation as for the torus, with the twisted boundary formulas $\partial L = a + b - c$ and $\partial U = c + a - b$ of that section, gives cocycles with $φ (a) = 0$ and $φ (c) = φ (b) :$

H^{0} (K; Z) ≅ Z, H^{1} (K; Z) ≅ Z, H^{2} (K; Z) ≅ Z /2.

Again the free part matches homology and the $Z /2$ has climbed from degree 1 to degree 2. With $Z /2$ coefficients the Klein bottle has groups $Z /2, (Z /2)^{2}, Z /2$ — exactly the same as the torus. Cohomology groups alone cannot tell the two surfaces apart mod 2; the cup product will (§4.5 The Cup Product). In general, a closed connected surface has $H^{2} (Σ; Z) ≅ Z$ when it is orientable and $Z /2$ when it is not.

The Universal Coefficient Theorem

The examples suggest a rule: integer cohomology in degree $k$ has the free part of $H_{k}$ together with the torsion of $H_{k - 1} .$ That rule is a theorem, and it says that homology determines cohomology completely.

Hatcher proves it as Theorem 3.2 (Hatcher 2002, §3.1). Let us read the two pieces, for finitely generated groups (the only kind our complexes produce).

$Hom (H_{k} (K), Z)$ is the group of homomorphisms from $H_{k}$ to $Z$ — the “measurements of homology classes” from the pairing. If $H_{k} ≅ Z^{r} \oplus T$ with $T$ finite, a homomorphism to $Z$ must kill $T$ (an element of finite order can only go to the only integer of finite order, $0),$ and it can do anything on $Z^{r} .$ So $Hom (H_{k}, Z) ≅ Z^{r} :$ the free part of $H_{k}$ .
$Ext (H_{k - 1} (K), Z)$ is a new construction, defined properly in §5.2 Homological Algebra. For finitely generated groups it can be computed with three rules: $Ext (Z, Z) = 0,$ $Ext (Z / n, Z) ≅ Z / n,$ and $Ext$ of a direct sum is the direct sum. So $Ext (H_{k - 1}, Z) ≅$ the torsion of $H_{k - 1}$ .

In short:

H^{k} (K; Z) ≅ (free part of H_{k}) \oplus (torsion of H_{k - 1}) .

Check it on $RP^{2} :$ $H^{1} = Hom (Z /2, Z) \oplus Ext (Z, Z) = 0$ and $H^{2} = Hom (0, Z) \oplus Ext (Z /2, Z) = Z /2 .$ On the Klein bottle: $H^{1} = Hom (Z \oplus Z /2, Z) = Z$ and $H^{2} = Ext (Z \oplus Z /2, Z) = Z /2 .$ The “one 2, two places” picture is the reason the formula is true; the proof makes that picture precise.

Other coefficients

The theorem holds for any coefficient group $G,$ with $G$ in place of the second $Z :$

H^{k} (K; G) ≅ Hom (H_{k} (K), G) \oplus Ext (H_{k - 1} (K), G), Ext (Z / n, G) ≅ G / n G .

With $G = Z /2$ and $K = RP^{2} :$ $H^{1} = Hom (Z /2, Z /2) \oplus Ext (Z, Z /2) = Z /2$ and $H^{2} = Hom (0, Z /2) \oplus Ext (Z /2, Z /2) = Z /2,$ as we computed by hand. When the coefficients form a field $F,$ there is also a version of the theorem that uses homology with coefficients in $F$ itself; over a field every vector space is free, the $Ext$ term disappears, and $H^{k} (K; F)$ is the dual vector space of $H_{k} (K; F) :$ same dimension, as we found by counting ranks. (With integer homology the $Ext$ term does not vanish in general: for $RP^{2}$ and $F = Z /2$ it is the whole of $H^{2} .)$

Three kinds of measurement Choose a space

Projective plane one-sided, closed · triangulated with 6, 15, 10 vertices, edges, triangles · χ = 1

	k = 0	k = 1	k = 2
$H_{k} (X; Z)$	$Z$	$Z /2$	$0$
$H^{k} (X; Z)$	$Z$	$0$	$Z /2$
$H^{k} (X; Z /2)$	$Z /2$	$Z /2$	$Z /2$
$H^{k} (X; R)$	$R$	$0$	$0$

Torsion: the $Z /2$ in homology degree 1 reappears in integer cohomology degree 2; the free parts match degree by degree. Euler characteristic from the real dimensions: $χ = 1 - 0 + 0 = 1$ , the same as $V - E + F .$

Figure 4.2.7 Homology with integer coefficients, and cohomology with integer, mod-2 and real coefficients, computed live from a triangulation of each space. Torsion glows rose: wherever homology has a

Z /2

in degree

k,

integer cohomology has it in degree

k + 1 .

Real coefficients see only the free part;

Z /2

coefficients see the twisting in every degree.

Exercises

Exercise Cohomology from homology

Suppose a space $X$ has $H_{0} = Z,$ $H_{1} = Z^{2} \oplus Z /3,$ $H_{2} = Z /2$ and $H_{k} = 0$ for $k \geq 3 .$ Use the Universal Coefficient Theorem to find $H^{k} (X; Z)$ for all $k,$ and the dimensions of $H^{k} (X; Z /2) .$ Check your answer with the Euler characteristic.

Hint

For $Z /2$ coefficients use $Hom (Z / n, Z /2) ≅ Z /2$ if $n$ is even and $0$ if $n$ is odd, and $Ext (Z / n, Z /2) ≅ (Z /2) / n (Z /2) .$

Solution

Integer coefficients, “free part of $H_{k}$ plus torsion of $H_{k - 1} ”:$ $H^{0} = Z,$ $H^{1} = Z^{2},$ $H^{2} = 0 \oplus Z /3 = Z /3,$ $H^{3} = Z /2,$ and $H^{k} = 0$ for $k \geq 4 .$ Mod 2: $H^{0} = Hom (Z, Z /2) = Z /2;$ $H^{1} = Hom (Z^{2} \oplus Z /3, Z /2) = (Z /2)^{2}$ (the $Z /3$ contributes nothing) plus $Ext (Z, Z /2) = 0;$ $H^{2} = Hom (Z /2, Z /2) \oplus Ext (Z^{2} \oplus Z /3, Z /2) = Z /2 \oplus 0,$ because $3$ is invertible mod 2; $H^{3} = Ext (Z /2, Z /2) = Z /2 .$ Dimensions $1, 2, 1, 1 .$ Euler characteristic: over $Q$ the Betti numbers are $1, 2, 0, 0,$ giving $χ = - 1;$ mod 2, $1 - 2 + 1 - 1 = - 1 .$ They agree, as they must.

Summary

Recap

$C^{k} (K; G) = Hom (C_{k} (K), G) :$ a $k$ -cochain is a value in $G$ on each $k$ -simplex, extended to chains by linearity. The pairing $⟨ φ, c ⟩ = φ (c)$ measures a place.
The coboundary $(δ φ) (σ) = φ (\partial σ)$ raises degree; as a matrix $δ_{k} = \partial_{k + 1}^{T};$ and $δ δ = 0$ because $\partial\partial = 0 .$
$H^{k} = Z^{k} / B^{k}$ = cocycles ( $δ φ = 0)$ modulo coboundaries ( $φ = δ g).$ Over a field, $dim H^{k} = dim H_{k} .$
Circle: $Z, Z .$ Sphere: $Z, 0, Z .$ Torus: $Z, Z^{2}, Z,$ with $H^{1}$ generated by two fences dual to the two loops.
A 1-cocycle on a surface is a fence; its value on a loop counts signed crossings. The pairing $H^{k} \times H_{k} \to G$ is well defined, by Stokes.
Maps act backwards: $f : X \to Y$ gives $f^{*} : H^{k} (Y) \to H^{k} (X),$ with $(g \circ f)^{*} = f^{*} g^{*}$ and $⟨ f^{*} φ, c ⟩ = ⟨ φ, f_{*} c ⟩ .$
Torsion moves up: $RP^{2}$ has $H_{1} = Z /2,$ $H_{2} = 0$ but $H^{1} = 0,$ $H^{2} = Z /2;$ the Klein bottle has $H^{*} = Z, Z, Z /2 .$ With $Z /2$ coefficients, both see their one-sidedness.
Universal Coefficient Theorem: $H^{k} (K; Z) ≅ Hom (H_{k}, Z) \oplus Ext (H_{k - 1}, Z)$ = free part of $H_{k}$ ⊕ torsion of $H_{k - 1} .$

Cohomology Groups

Cochain groups: measurements as homomorphisms

The coboundary map is a transpose

The matrix of $δ$

The coboundary of a coboundary

Cocycles, coboundaries and cohomology

Counting dimensions over a field

First computations: graph, circle, sphere, torus

Graphs

The circle and the torus, step by step

The sphere

Fences: what a 1-cocycle measures

Measuring classes with classes

Arrows reverse: pulling measurements back

Torsion moves up: the projective plane and the Klein bottle

With $Z /2$ coefficients

The Klein bottle

The Universal Coefficient Theorem

Other coefficients

Exercises

Summary

Further reading

References

Cochain groups: measurements as homomorphisms

The coboundary map is a transpose

The matrix of δ

The coboundary of a coboundary

Cocycles, coboundaries and cohomology

Counting dimensions over a field

First computations: graph, circle, sphere, torus

Graphs

The circle and the torus, step by step

The sphere

Fences: what a 1-cocycle measures

Measuring classes with classes

Arrows reverse: pulling measurements back

Torsion moves up: the projective plane and the Klein bottle

With Z/2 coefficients

The Klein bottle

The Universal Coefficient Theorem

Other coefficients

Exercises

Summary

Further reading

References

The matrix of $δ$

With $Z /2$ coefficients