Chapter 04 - Maps and Variations of Homology

Maps and homology

Theory - Simplicial maps, chain maps

Simplicial maps, chain maps

Simplicial map

A map $f:\mathcal{K}\to \mathcal{L}$ between simplicial complexes is simplicial if it maps vertices of $\mathcal{K}$ to vertices of $\mathcal{L}$, and it maps simplices of $\mathcal{K}$ linearly to simplices of $\mathcal{L}$.

For example:

• Inclusions of one simplicial complex into another.
• Reduction of a simplicial complex to lower dimension, e.g. $[v_0,v_1]\to [v_0]$.

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Chain map

A chain map $f$ is a map between chain complexes, which consists of a collection $f_n,\dots ,f_0$ of homomorphisms:

$$f_i:C_i(\mathcal{K})\to C_i(\mathcal{L})$$

of abelian groups which commute with the boundary operators:

$$f_{i-1}\circ {\partial }_i^{\mathcal{K}}={\partial }_i^{\mathcal{L}}\circ f_i\text{for all}i.$$

Such a chain map makes the following diagram commute:

• The homomorphisms can be collected under a single map $f$, and the commutativity condition expressed simply:

$$f\circ {\partial }^{\mathcal{K}}={\partial }^{\mathcal{L}}\circ f$$

Theorem: Induced chain map

A simplicial map $f:\mathcal{K}\to \mathcal{L}$ determines a map of chain complexes $f:C(\mathcal{K})\to C(\mathcal{L})$.

The simplicial map $f$ is extended linearly from simplices to chains (from generators to linear combinations).

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Theory - Induced maps on homology

Induced maps on homology

Theorem: Induced map on homology

A chain map $f:\mathcal{K}\to \mathcal{L}$ determines a map $f_{\ast }$ of homology groups which splits by degrees:

$${f_{\ast }}_i:H_i(\mathcal{K})\to H_i(\mathcal{L})$$

A chain map $f$ sends cycles to cycles and boundaries to boundaries, so it descends to a map of quotient groups.

Proof of Theorem: Induced map on homology

It is sufficient to check that $f$ maps cycles to cycles and boundaries to boundaries:

• By mapping boundaries to boundaries, $f$ is well-defined on homology groups.
• By mapping cycles to cycles, $f$ maps elements of homology to elements of homology.

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Checking $f$ maps cycles to cycles:

If $\tau \in \mathrm{Im}{\partial }^{\mathcal{K}}$, meaning that $\tau$ is a cycle, then $\tau ={\partial }^{\mathcal{K}}(\sigma )$ for some $\sigma$, whence:

$$\tau =(f\circ {\partial }^{\mathcal{K}})(\sigma )\gg \gg \tau =({\partial }^{\mathcal{L}}\circ f)(\sigma ),$$

but this implies $\tau \in \mathrm{Im}{\partial }^{\mathcal{L}}$.

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Checking $f$ maps boundaries to boundaries:

If ${\partial }^{\mathcal{K}}(\tau )=0$, then:

$$\begin{array}{c}{\partial }^{\mathcal{L}}(f(\tau ))\gg \gg ({\partial }^{\mathcal{L}}\circ f)(\tau )\gg \gg (f\circ {\partial }^{\mathcal{K}})(\tau )\\ \\ \gg \gg f({\partial }^{\mathcal{K}}(\tau ))\gg \gg f(0)\gg \gg 0\end{array}$$

but this means the image of $\tau$ is a cycle.

Proposition: Homotopic maps are the same on homology

If two maps of simplicial complexes $f,g:\mathcal{K}\to \mathcal{L}$ are homotopic, then they induce the same maps on homology groups, $f_{\ast }=g_{\ast }$:

In words: continuous deformations do not change the map induced on homology.

Proposition: Homology respects composition of maps

Given a composition of simplicial maps:

$$\mathcal{K}\stackrel{f}{⟶}\mathcal{L}\stackrel{g}{⟶}\mathcal{M},$$

the map induced by the composite agrees with the composite of the induced maps:

$$(g\circ f{)}_{\ast }=g_{\ast }\circ f_{\ast }:H_i(\mathcal{K})\to H_i(\mathcal{M})$$

The property of respecting composition of maps is called functoriality.

Corollary: Homotopy equivalence implies homology isomorphism

Homotopy equivalent simplicial complexes have isomorphic homology groups:

$$\mathcal{K}\simeq \mathcal{L}⟹H_i(\mathcal{K})\cong H_i(\mathcal{L})\text{for all}i$$

Proof of Corollary

Suppose we have a homotopy equivalence:

$$\begin{array}{c}\mathcal{K}\stackrel{f}{⟶}\mathcal{L},\mathcal{L}\stackrel{g}{⟶}\mathcal{K}\\ \\ g\circ f\simeq {\mathrm{Id}}_{\mathcal{K}},f\circ g\simeq {\mathrm{Id}}_{\mathcal{L}}\end{array}$$

Now apply functoriality to the induced maps on homology:

$$\begin{array}{c}{g}_{\ast }\circ f_{\ast }\gg \gg (g\circ f{)}_{\ast }\gg \gg ({\mathrm{Id}}_{\mathcal{K}}{)}_{\ast }={\mathrm{Id}}_{H_i(\mathcal{K})}\\ \\ f_{\ast }\circ g_{\ast }\gg \gg (f\circ g{)}_{\ast }\gg \gg ({\mathrm{Id}}_{\mathcal{L}}{)}_{\ast }={\mathrm{Id}}_{H_i(\mathcal{L})}\end{array}$$

Therefore $g_{\ast }$ is the inverse of $f_{\ast }$ as homomorphisms of abelian groups.

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Illustration

Contraction to a point

Contraction to a point

The map $f:{\mathbb{R}}^n\to \{p\}$ for some point $p\in {\mathbb{R}}^n$ is a homotopy equivalence, so it induces an isomorphism on homology. Therefore:

$$H_i({\mathbb{R}}^n)=\{\begin{array}{ll}0& i>0\\ \mathbb{Z}& i=0\end{array}$$

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Dunce Hat

Dunce Hat

The Dunce Hat is formed by identifying all sides of a triangle with non-circular identifications:

The Dunce Hat is contractible, but it is not collapsible. The contraction can be performed by embedding the Hat in a 3-ball in ${\mathbb{R}}^4$.

Because it is contractible, the Dunce Hat has the homology of a point.

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Homology of a cylinder

Homology of a cylinder

Recall that the cylinder is homotopic to the circle (§06):

A circle is homeomorphic to a triangle.

We have computed the homology of a triangle (§11 of Part I).

Therefore the homology of the cylinder is:

$$H_i(X)=\{\begin{array}{ll}0& i>2\\ \mathbb{Z}& i=1\\ \mathbb{Z}& i=0\end{array}$$

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Homology variations

Theory - Relative homology

Relative homology

Suppose a simplicial complex $\mathcal{K}$ is embedded within another simplicial complex $\mathcal{L}$. The homology of $\mathcal{L}$ relative to $\mathcal{K}$ is given by zeroing (quotienting out) the cycles fully within $\mathcal{K}$. Only cycles that inescapably involve $\mathcal{L}$ outside of $\mathcal{K}$ contribute to the relative homology.

More precisely, suppose that $\mathcal{K}\subset \mathcal{L}$, so vertices and simplices of $\mathcal{K}$ are also vertices and simplices of $\mathcal{L}$.

Then observe that chains in $\mathcal{K}$ are automatically chains in $\mathcal{L}$, and thus:

$$\mathcal{K}\subset \mathcal{L}⟹C_i(\mathcal{K})<C_i(\mathcal{L})\text{for all}i$$

(A subgroup: because a sum of chains in $\mathcal{K}$ is still a chain in $\mathcal{K}$.)

Relative chain complex

Suppose $\mathcal{K}\subset \mathcal{L}$ is an inclusion of simplicial complexes.

Define $C_i(\mathcal{L},\mathcal{K})$, the relative chain group, as the quotient:

$$C_i(\mathcal{L},\mathcal{K})=\frac{C_i(\mathcal{L})}{C_i(\mathcal{K})}$$

Proposition: Boundary maps descend to relative chain groups

The boundary maps $\partial :C_i(\mathcal{L})\to C_{i-1}(\mathcal{L})$ send chains in $\mathcal{K}$ to chains in $\mathcal{K}$ and thus descend to the quotient groups, giving a relative chain complex:

$$0\to C_n(\mathcal{L},\mathcal{K})\stackrel{{\partial }^{\mathcal{L},\mathcal{K}}}{⟶}\cdots \stackrel{{\partial }^{\mathcal{L},\mathcal{K}}}{⟶}{C}_0(\mathcal{L},\mathcal{K})\to 0$$

(Here ${\partial }^{\mathcal{L},\mathcal{K}}$ is just the map induced by $\partial$ on the quotient groups.)

Proof of Proposition: Boundary maps descend to relative chain groups

It is sufficient to establish that boundaries from $\mathcal{K}$ must also reside in $\mathcal{K}$, i.e. that:

$$\partial (C_i(\mathcal{K}))\subset C_{i-1}(\mathcal{K})$$

But this is clear because $\mathcal{K}$ is a simplicial complex, so it is closed under taking boundaries.

Relative homology

Suppose $\mathcal{K}\subset \mathcal{L}$ is an inclusion of simplicial complexes.

Define $H_k(\mathcal{L},\mathcal{K})$, the relative homology group, as the homology group of the relative chain complex:

$$H_k(\mathcal{L},\mathcal{K})=\frac{\mathrm{Ker}\left({\partial }_k^{\mathcal{L},\mathcal{K}}\right)}{\mathrm{Im}\left({\partial }_{k+1}^{\mathcal{L},\mathcal{K}}\right)}$$

Remark: Relativizing before homology

It is tempting to try relativizing the homology groups directly.

The inclusion $\mathcal{K}\hookrightarrow \mathcal{L}$ does induce maps $H_k(\mathcal{K})\to H_k(\mathcal{L})$. We could study the groups $H_k(\mathcal{L})/H_k(\mathcal{K})$.

These quotient groups are not what relative homology is after! See the illustrations.

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Illustration

Homology of a simplex relative to its boundary

Homology of a simplex relative to its boundary

Consider a simple 2-simplex $\sigma$ and the simplicial complex $\mathcal{L}$ that it determines:

Let $\mathcal{K}$ be the boundary 1-simplex (the edges only from $\mathcal{L}$).

There are no 2-simplices in $\mathcal{K}$, so:

$$C_2(\mathcal{L},\mathcal{K})=C_2(\mathcal{L})\cong ⟨\sigma ⟩\cong \mathbb{Z}$$

All 0- and 1-simplices in $\mathcal{L}$ are also in $\mathcal{K}$, so:

$$C_1(\mathcal{L},\mathcal{K})=0,C_0(\mathcal{L},\mathcal{K})=0$$

The relative chain complex:

$$0\to C_2(\mathcal{L},\mathcal{K})\to C_1(\mathcal{L},\mathcal{K})\to C_0(\mathcal{L},\mathcal{K})\to 0$$

becomes:

$$0\to \mathbb{Z}\to 0\to 0\to 0$$

It follows that the relative homology is:

$$H_i(\mathcal{L},\mathcal{K})=\{\begin{array}{ll}0& i>3\\ \mathbb{Z}& i=2\\ 0& i=1\\ 0& i=0\end{array}$$

Compare this to the homology of a sphere!

Homology of a sphere:

$$H_i({\mathbb{S}}^2)=\{\begin{array}{ll}0& i>3\\ \mathbb{Z}& i=2\\ 0& i=1\\ \mathbb{Z}& i=0\end{array}$$

Notice that ${\mathbb{S}}^2$ is homeomorphic to the quotient space $\mathcal{L}/\mathcal{K}$.

This comparison of homologies is apt; we just need to explain $H_0({\mathbb{S}}^2)\cong \mathbb{Z}$ versus $H_0(\mathcal{L},\mathcal{K})\cong 0$.

Compare this against the quotient of homology groups!

The homology of $\mathcal{K}$ is:

$$H_i(\mathcal{K})=\{\begin{array}{ll}0& i>1\\ \mathbb{Z}& i=1\\ \mathbb{Z}& i=0\end{array}$$

The homology of $\mathcal{L}$, because it is contractible, is:

$$H_i(\mathcal{L})=\{\begin{array}{ll}0& i>0\\ \mathbb{Z}& i=0\end{array}$$

The quotient of homology groups is:

$$H_i(\mathcal{L})/H_i(\mathcal{K})\cong 0\text{for all}i$$

• Relative homology treats $\mathcal{K}$ as a single point.
• The quotient of homology annihilates generators (“holes”) in $\mathcal{L}$ which are also generators (“holes”) in $\mathcal{K}$.
• Different results for different constructions!

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Theory - Cochain complex

Cochain complex

Recall that the chain group $C_k(\mathcal{K})$ is the free abelian group of formal linear combinations of $k$-simplices in $\mathcal{K}$:

$$C_k(\mathcal{K})=\left\{\sum _{i\in I\text{finite}}{n}_i{\sigma }_i|n_i\in \mathbb{Z},{\sigma }_i\in \mathcal{K},\dim ({\sigma }_i)=k\right\}$$

One should think of the chains $\sum n_i{\sigma }_i$ as assignments of integers $n_i$ to the various $k$-simplices in $\mathcal{K}$.

Thinking in this way, the boundary map ${\partial }_k:C_i\to C_{i-1}$ is given on chains by simple linear extension of its definition for individual simplices. Thus ${\partial }_k\left(n_i{\sigma }_i\right)=n_i{\partial }_k({\sigma }_i)$.

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Cohomology is created by interpreting the integers $n_i$ not as coefficients, but as the outputs of linear functions $C_k\to \mathbb{Z}$.

To notate this effectively, define indicator functions ${\phi }_i$ for each of the $k$-simplices:

$$\begin{array}{c}{\phi }_i:\{\text{oriented k-simplices in K}\}\to \mathbb{Z}\\ \\ {\sigma }_i\mapsto 1,{\sigma }_j\mapsto 0\text{for}j\ne i\end{array}$$

Cochains

The space of linear maps $C_k\to \mathbb{Z}$ is isomorphic to the free abelian group of formal linear combinations of indicator functions:

$$C^k(\mathcal{K})=\left\{\sum _{i\in I}{m}_i{\phi }_i|m_i\in \mathbb{Z},{\phi }_i\text{indicator of}{\sigma }_i\right\}$$

Elements of this space, written generically as formal sums $m_1{\phi }_1+m_2{\phi }_2+\cdots$, are called cochains.

(For cochains it is not necessary that $I$ be finite.)

Since there is a bijective correspondence of sets between oriented simplices and their indicator functions, we see that $C^k(\mathcal{K})\cong C_k(\mathcal{K})$ as abelian groups; we have simply reinterpreted the chains as assignments of integers $m_i$ to the indicators of $k$-simplices in $\mathcal{K}$.

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In both cases, the data of a chain / cochain consists of the assignment of a number ($n_i$ or $m_i$) to each of the $k$-simplices in $\mathcal{K}$. But the difference in interpretation of this data allows a difference in the construction of boundary operators:

Coboundary operator

Define the coboundary operator thus:

$$\begin{array}{rl}{\delta }^k:C^k(\mathcal{K})& \to C^{k+1}(\mathcal{K})\\ & \\ {\phi }_i& \mapsto {\phi }_i\circ {\delta }^{k+1}(\text{extended linearly})\end{array}$$

In words: the coboundary of a simplex indicator indicates that simplex in the boundary of a higher simplex.

For example, in the triangular 2-simplex:

If ${\phi }_{[v_0,v_2]}$ is the indicator of $[v_0,v_2]$, then the coboundary ${\delta }^1\left({\phi }_{[v_0,v_2]}\right)$ indicates the presence of $[v_0,v_2]$ in the boundaries of 2-simplices. It returns $-1$ for the triangle itself, $[v_0,v_1,v_2]$, since the boundary of the triangle includes the term $-[v_0,v_2]$.

Thus, in general, ${\delta }^k\left(m_i{\phi }_i\right)=m_i({\phi }_i\circ {\partial }_k)$, in contrast to ${\partial }_k\left(n_i{\sigma }_i\right)=n_i{\partial }_k({\sigma }_i)$ that we saw for ${\partial }_k$.

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Witness a simple computation, applying ${\delta }^{k+1}\circ {\delta }^k$ to an arbitrary indicator:

$$\begin{array}{c}({\delta }^{k+1}\circ {\delta }^k)({\phi }_i)\gg \gg {\delta }^{k+1}({\delta }^k({\phi }_i))\\ \\ \gg \gg {\delta }^{k+1}({\phi }_i\circ {\partial }_k)\gg \gg {\phi }_i\circ {\partial }_k\circ {\partial }_{k+1}\gg \gg 0\end{array}$$

Cochain complex

Given a simplicial (or $\Delta$-) complex $\mathcal{K}$ of maximal dimension $n$, we define the cochain complex of $\mathcal{K}$ as the sequence of morphisms:

$$0⟵C_n\stackrel{{\delta }^{n-1}}{⟵}{C}_{n-1}\stackrel{{\delta }^{n-2}}{⟵}{C}_{n-2}⟵\cdots ⟵C_1\stackrel{{\delta }^0}{⟵}{C}_0⟵0$$

This sequence is a complex because:

$${\delta }^{k+1}\circ {\delta }^k=0\text{for every}k$$

The image and kernel groups are named in the obvious way:

• $B_k:=\mathrm{Im}({\delta }^{k-1})$ — called the coboundary cochains
• $Z_k:=\mathrm{Ker}({\delta }^k)$ — called the cocycles

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Theory - Cohomology

Cohomology

Cohomology

The $k^{\text{th}}$ cohomology of $\mathcal{K}$ is defined as the quotient group of cocycles modulo coboundary chains:

$$H^k(\mathcal{K})=\frac{\mathrm{Ker}({\delta }^k)}{\mathrm{Im}({\delta }^{k-1})}$$

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Cohomology is an upgrade from homology because it has an additional product structure, so cohomology actually gives a collection of rings.

Cup product on cochains, for simplicial complexes

Suppose we are given a simplicial complex $\mathcal{K}$, together with a chosen ordering $O=(v_1,\dots ,v_N)$ of all vertices of $\mathcal{K}$.

There are maps of cochain groups $C^k\times C^{\ell }\to C^{k+\ell }$ defined as follows:

For any pair of cochains $\phi \in C^k$ and $\psi \in C^{\ell }$, and any vertices $w_0<\cdots <w_{k+\ell }$ ordered according to $O$, with $[w_0,\dots ,w_{k+\ell }]$ the $(k+\ell )$-simplex they form, we have:

$$(\phi ⌣\psi )([w_0,\dots ,w_{k+\ell }])=\phi ([w_0,\dots ,w_k])\cdot \psi ([w_k,\dots ,w_{k+\ell }])$$

Note: the oriented simplices here may be written with various orderings, but the splitting into $k$- and $\ell$-simplices must follow the order of $O$.

Cup product using indicators

Let $\phi \in C^k$ be the indicator of an oriented simplex $[u_{i_0},\dots ,u_{i_k}]$, and let $\psi \in C^{\ell }$ be the indicator of another oriented simplex $[v_{i_k},\dots ,v_{i_{k+\ell }}]$.

Let $(w_0,\dots ,w_{k+\ell })$ be the vertices of the total list

$$u_{i_0},\dots ,u_{i_k},v_{i_k},\dots ,v_{i_{k+\ell }}$$

reordered in the ordering of $O$ (allowing repetition), so $w_0\le w_1\le \cdots \le w_{k+\ell }$.

Then $\phi ⌣\psi$ is defined as follows: it is the indicator of $[w_0,\dots ,w_{k+\ell }]$ provided:

• (i) the $w_i$ are all distinct,
• (ii) $u_p\le v_q$ for all $p,q$;

and $\phi ⌣\psi =0$ if (i) and (ii) do not hold.

The definition is extended bilinearly to all of $C^k\times C^{\ell }$.

Vertex orderings and the cup product

Instead of choosing a total ordering $O$ of all vertices of $\mathcal{K}$, it is sufficient to choose a partial ordering of the vertices $v_1,\dots ,v_N$ of $\mathcal{K}$ having the property that $O$ induces a total ordering of each simplex of $\mathcal{K}$.

A set of compatible orderings of the vertices in each simplex is used to define a ‘front half’ and ‘back half’ of each $(k+\ell )$-simplex $\sigma$, thereby cutting $\sigma$ into one $k$-simplex and one $\ell$-simplex.

The cutting method applied to a given simplex must be compatible with the method applied to every other simplex. A global vertex ordering ensures this compatibility.

Cup product on cohomology

The cup product has the following relation to the coboundary operator, a type of signed Leibniz rule:

$$\delta (\phi ⌣\psi )=\delta \phi ⌣\psi +(-1{)}^k\phi ⌣\delta \psi$$

when $\phi \in C^k(\mathcal{K})$.

This formula implies that if $\phi$ and $\psi$ are cocycles (or coboundaries), then $\phi ⌣\psi$ is also a cocyle (coboundary). Therefore the cup product descends to cohomology:

$$H^k(\mathcal{K})\times H^{\ell }(\mathcal{K})\to H^{k+\ell }(\mathcal{K})$$

The cup product on cohomology provides a ring structure to the direct sum ${⨁}_i{H}^i(\mathcal{K})$. The product, and the ring structure, do not depend on the choice of ordering used to define the product on cochains.

(Note: The cup product is always zero when $k+\ell$ is greater than the maximal simplex dimension.)

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Illustration

Some cohomology classes on the 2-torus

Some cohomology classes on the 2-torus

Consider the 2-torus $T$:

The red (poloidal) and pink (toroidal) curves together generate the degree one homology, $H_1(T)\cong {\mathbb{Z}}^2$.

Now recall a simplicial complex representation of the torus:

Let $\phi$ be the cochain which assigns 1 to every edge in the middle column, and zero to all other edges. (So it is 1 on the edges directly below the $\phi$ label, including the diagonals, but excluding all verticals.) Let $\psi$ be the cochain which assigns 1 to every edge in the middle row, and zero to all other edges. (So it is 1 on the edges directly to the right of the $\psi$ label, including the diagonals, but excluding all horizontals.) Accounting now for orientations, let $\phi$ give $+1$ on edges directed rightwards, and $\psi$ give $+1$ on edges directed upwards.

One can check directly, going around the boundaries of each 2-simplex, that $\delta \phi =0$ and $\delta \psi =0$: The sum of the numbers on any triangle, accounting for orientations, is zero. So these cochains are both cocycles.

In order to see that $\phi$ and $\psi$ are not boundaries, and thus represent nontrivial homology classes, and also to see that they don’t represent the same homology class, we devise a method of representing coboundaries of 0-cochains.

A 0-cochain $\eta$ is the assignment of integers to each of the vertices. The condition $\delta \eta =0$, which means $\eta \circ {\partial }_1\equiv 0$, states that these numbers cannot change as one moves across an edge from vertex to vertex. (Applying ${\partial }_1$ to any edge $[u,v]$ gives $[v]-[u]$. By linearity, $\eta ({\partial }_1([u,v]))=\eta ([v])-\eta ([u])$. Therefore $\eta ([v])=\eta ([u])$ when the condition holds.)

From this it follows that $H^0(T)\stackrel{\sim }{⟶}\mathbb{Z}$. The isomorphism is given by sending $\eta$ to the integer it gives on a vertex.

Return now to $H^1(T)$. A boundary in $H^1(T)$ is a cochain $\chi$ assigning numbers to directed edges such that there is a vertex labeling cochain $\eta \in C^1$ with $\chi ([u,v])=\eta ([v])-\eta ([u])$. In other words, that it is possible to derive the edge numbers as the differences of a single vertex numbering scheme.

Now one can see that $\phi$ is not a coboundary. Vertex integers must increase by one as one passes across the middle column of edges, and stay fixed going across all other edges. This is impossible because the left-most and right-most edges are identified. Similarly $\psi$ is not a coboundary, and $\phi -\psi$ is not a coboundary.

(To witness a contrasting case, let $\xi$ be a 1-cochain giving 1 only on the non-vertical edges in the right-most column. This is $\phi$ shifted to the right. Now $\phi -\xi$ is a coboundary: assign 1 on vertices in the middle-right column, and 0 on all other vertices.)

Much more work would be required to show that $\phi$ and $\psi$ have no relations in homology, and that they jointly generate homology.

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A cup product on the 2-torus

A cup product on the 2-torus

We study the cup product $\phi ⌣\psi$:

We must provide an ordering $O$ of all the vertices of the torus. Order them as in a book: top-to-bottom, then left-to-right.

Let us see how to verify the values on the green and yellow 2-simplices. Label the vertices of the inner square as in the figure; the index ordering is compatible with $O$.

For the green 2-simplex $[v_0,v_1,v_2]$, we have $\phi ([v_0,v_1])=+1$ and $\psi ([v_1,v_2])=-1$, so:

$$(\phi ⌣\psi )([v_0,v_1,v_2])=(+1)\cdot (-1)=-1$$

For the yellow 2-simplex $[v_1,v_2,v_3]$, we have $\phi ([v_1,v_2])=-1$ and $\psi ([v_2,v_3])=0$, so:

$$(\phi ⌣\psi )([v_1,v_2,v_3])=(+1)\cdot (0)=0$$

A calculation with more details (illustrating bilinearity) can be performed in terms of the indicator functions. Write:

$$\phi ={\phi }_1+\cdots +{\phi }_6,\psi ={\psi }_1+\cdots +{\psi }_6$$

where ${\phi }_i$ is the indicator for the edge in the support of $\phi$ which is in the $i^{\text{th}}$ place from the top (oriented rightwards), and analogously ${\psi }_j$ is for the one for $\psi$ in the $j^{\text{th}}$ place from the left (oriented upwards).

Applying bilinearity to expand the cup product:

$$\begin{array}{c}\phi ⌣\psi \gg \gg ({\phi }_1+\cdots +{\phi }_6)⌣({\psi }_1+\cdots +{\psi }_6)\\ \gg \gg \sum _{i=1}^6\sum _{j=1}^6{\phi }_i⌣{\psi }_j\end{array}$$

The edges with indicators ${\phi }_i$ and ${\psi }_j$ could span one of the 2-simplices only when $i=3,4,5$ and $j=3,4,5$. And in fact the pairs $(i,j)=(3,5),(5,3)$ can be ruled out as well. The only terms to consider are these six:

$$\begin{array}{c}\sum _{i=1}^6\sum _{j=1}^6{\phi }_i⌣{\psi }_j\gg \gg \begin{array}{r}{\phi }_3⌣{\psi }_3+{\phi }_3⌣{\psi }_4+{\phi }_4⌣{\psi }_3\\ +{\phi }_4⌣{\psi }_5+{\phi }_5⌣{\psi }_4+{\phi }_5⌣{\psi }_5\end{array}\end{array}$$

• Now ${\phi }_3$ is the indicator for $[v_0,v_1]$ and ${\psi }_3$ is the indicator for $[v_0,v_2]$, so ${\phi }_3⌣{\psi }_3=0$ because $v_1\nleqq v_0$ so the total list $v_0,v_1,v_0,v_2$ does not split according to the ordering of $O$.
• Then ${\psi }_4$ is the indicator for $-[v_1,v_2]$, so ${\phi }_3⌣{\psi }_4$ is the indicator for $-[v_0,v_1,v_2]$, meaning it returns $-1$ on the green 2-simplex.
• Then ${\phi }_4$ is the indicator for $-[v_1,v_2]$, so ${\phi }_4⌣{\psi }_3=0$.
• Then ${\psi }_5$ is the indicator for $-[v_1,v_3]$, so ${\phi }_4⌣{\psi }_5=0$ because $v_2\nleqq v_1$.
• Then ${\phi }_5$ is the indicator for $[v_2,v_3]$, so ${\phi }_5⌣{\psi }_4=0$ because $v_2\nleqq v_1$.
• Then ${\phi }_5⌣{\psi }_5=0$ as well, also because $v_2\nleqq v_1$.

By adding these results, we see that $\phi ⌣\psi$ is the indicator for the central 2-simplex, oriented clockwise.

This cochain in $C^2$ is automatically a cocycle because $C^3$ is empty.

Is it a coboundary?

To see that the answer is “no,” we can devise another scheme to represent coboundaries, now for $2$-simplices. We give here only a sketch. Give a 1-cochain $\xi$ which assigns integers to the edges. Draw oriented paths crossing the edges, with the number of paths crossing $e_i$ matching $\xi (e_i)$. Then $\delta \xi (t_j)$ for a triangle $t_j$ counts the total flow of paths into or out from $t_j$; it is a ‘sink’ or ‘source’ number describing extra paths originating or terminating inside $t_j$. The condition $\delta \xi (t_j)=0$ would mean that the paths are conserved in $t_j$, none created or destroyed.

With this perspective, it is obvious that the cochain $\phi ⌣\psi$ depicted in the figure cannot be a coboundary. Every 2-cochain which is a coboundary describes a collection of oriented paths that are created or destroyed only in triangles with nonzero assignments, according to that assignment. So $\phi ⌣\psi$ would describe a single path terminating in the green triangle, but originating nowhere, since there is no other sink or source.

Another observation is worth making about $\phi ⌣\psi$. It is the generator of $H^2(T)$.

Note that the full cohomology of $T$ is given by:

$$H^i(T)\cong \{\begin{array}{ll}0& i>3\\ \mathbb{Z}& i=2\\ {\mathbb{Z}}^2& i=1\\ \mathbb{Z}& i=0\end{array}$$

Here is a fact: All indicator functions on 2-simplices determine the same cohomology class. To see this, take the difference of any two indicators. Draw a simple oriented path out of one of them into the other. This path will define a 1-cochain by assigning $\pm 1$ on any edge it crosses on the way (according to their mutual orientations), and 0 on all edges it does not cross. The coboundary of this cochain gives $+1$ on the triangle from which the path emanates, and $-1$ on the triangle in which the path terminates, and zero on all other triangles. That is exactly the difference of indicator 2-simplices we started with.

Since $H^2(T)$ is generated by indicator 2-simplices, and they are all equivalent, we deduce that it is generated by $\phi ⌣\psi$.

To see that it is in fact infinite and isomorphic to $\mathbb{Z}$, we must exclude relations among the multiples of $\phi ⌣\psi$. A relation $n(\phi ⌣\psi )=m(\phi ⌣\psi )$ is equivalent to the relation $\ell (\phi ⌣\psi )=0$ for $\ell =n-m$. But we can exclude any such relation by the reasoning used above: a cochain $\ell (\phi ⌣\psi )$ cannot be a coboundary because it would represent $\ell$ oriented paths terminating in the green 2-simplex, and originating nowhere.

Therefore we have proven that:

$$H^2(T)\cong ⟨[\phi ⌣\psi ]⟩\cong \mathbb{Z}$$

The cocycle $[\phi ⌣\psi ]$ is called the fundamental cocycle of $T$.

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Cohomology ring has topological significance

Cohomology ring has topological significance

Let $X={\mathbb{S}}^1\times {\mathbb{S}}^1$, the 2-torus.

Let $Y={\mathbb{S}}^1\vee {\mathbb{S}}^1\vee {\mathbb{S}}^2$, a 2-sphere with two circles attached at one common point.

The cohomology groups are isomorphic:

$$H^i(X)\cong H^i(Y)\cong \{\begin{array}{ll}0& i>2\\ \mathbb{Z}& i=2\\ \mathbb{Z}\oplus \mathbb{Z}& i=1\\ \mathbb{Z}& i=0\end{array}$$

However, the cup product operations differ, so the ring structures differ.

• In $X$, the cup product of the two generating cocycles is the fundamental coclass of the 2-torus.
• In $Y$, the cup product of the two generating cocycles (dual to the two copies of ${\mathbb{S}}^1$), is zero.

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Theory - Coefficients

Coefficients

It is possible to define homology and cohomology with other coefficients. Given an abelian group $G$, one defines the chain groups thus:

$$C_k(\mathcal{K},G)=\left\{\sum _{i\in I\text{finite}}{g}_i{\sigma }_i|g_i\in G,{\sigma }_i\in \mathcal{K},\dim ({\sigma }_i)=k\right\}$$

And the cochain groups thus:

$$C^k(\mathcal{K},G)=\left\{\sum _{i\in I}{g}_i{\phi }_i|g_i\in \mathbb{Z},{\phi }_i\text{indicator of}{\sigma }_i\in \mathcal{K},\dim ({\sigma }_i)=k\right\}$$

Homology and cohomology are defined in the usual way from these chain groups.

The most important cases are $G=\mathbb{R}$ (which eliminates torsion), and $G=\mathbb{Z}/2\mathbb{Z}$ (where coefficients track parity only).

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