Cellular Homology

Cellular homology provides a powerful and efficient method for computing the homology groups of CW complexes. It relies on a chain complex constructed directly from the cells of the CW structure.

The Cellular Chain Complex

Lemma

Let $X$ be a CW complex with $X^n$ as its $n$-skeleton.

1. The relative homology group $H_k(X^n,X^{n-1})$ is given by: \newcommand{\Z}{\mathbb{Z}}H_{k}(X^{n}, X^{n-1}) \cong \begin{cases} \Z\{n\text{-cells of } X\} & \text{if } k=n \\ 0 & \text{if } k \neq n \end{cases} where $\mathbb{Z}\{n\text{-cells}\}$ denotes the free abelian group generated by the $n$-cells.
2. For $k>n$, the homology group $H_k(X^n)=0$.
3. The inclusion map $i:X^n\hookrightarrow X$ induces a map on homology $i_{\ast }:H_k(X^n)\to H_k(X)$ which is an isomorphism for $k<n$ and surjective for $k=n$.

Proof

1. The pair $(X^n,X^{n-1})$ is a good pair, and the quotient space $X^n/X^{n-1}$ is a wedge sum of $n$-spheres, one for each n-cell of $X$. Thus, $H_k(X^n,X^{n-1})\cong {\tilde{H}}_k(X^n/X^{n-1})\cong {\tilde{H}}_k(\bigvee S^n)$, which gives the result.
2. This is shown by considering the long exact sequence for the pair $(X^n,X^{n-1})$ and using induction on $n$.
3. This follows from considering the long exact sequence of the pair $(X,X^n)$. Any cycle in $X$ is compact and can be deformed into some finite skeleton $X^m$, which allows us to relate $H_k(X)$ to the homology of its skeletons.

$\square$

We can now define a chain complex based on the structure of a CW complex $X$.

Cellular Chain Groups

The cellular chain group $C_n^{\mathrm{CW}}(X)$ is the free abelian group generated by the $n$-cells of $X$. By the lemma above, we have an isomorphism: $C_n^{\text{CW}}(X)\cong H_n(X^n,X^{n-1}).$ The cellular boundary map $d_n:C_n^{\text{CW}}(X)\to C_{n-1}^{\text{CW}}(X)$ is defined as the composite of maps from the long exact sequence of the pair $(X^n,X^{n-1})$: $d_n:H_n(X^n,X^{n-1})\stackrel{{\partial }_n}{\to }{H}_{n-1}(X^{n-1})\stackrel{j_{n-1}}{\to }{H}_{n-1}(X^{n-1},X^{n-2})$ where ${\partial }_n$ is the mystery map and $j_{n-1}$ is induced by inclusion.

Lemma

The composition of two consecutive cellular boundary maps is zero, i.e., $d_n\circ d_{n+1}=0$. Thus, $\left(C_{\ast }^{\text{CW}}(X),d_{\ast }\right)$ forms a chain complex.

Proof The composition $d_n\circ d_{n+1}$ involves the segment $H_n(X^n)\stackrel{j_n}{\to }{H}_n(X^n,X^{n-1})\stackrel{{\partial }_n}{\to }{H}_{n-1}(X^{n-1})$. This is a portion of the long exact sequence for the pair $(X^n,X^{n-1})$, so the composition ${\partial }_n\circ j_n=0$. The full map is $d_n\circ d_{n+1}=(j_{n-1}\circ {\partial }_n)\circ (j_n\circ {\partial }_{n+1})$. Since ${\partial }_n\circ j_n=0$, the entire composition is zero. $\square$

Cellular Homology

The $n$-th cellular homology group of $X$ is the homology of this chain complex: $H_n^{\text{CW}}(X)=\frac{\ker (d_n)}{\mathrm{im}(d_{n+1})}.$

The key result is that this new homology theory is naturally isomorphic to the singular homology.

Theorem

For a CW complex $X$, the cellular homology is isomorphic to the singular homology: $H_n^{\text{CW}}(X)\cong H_n(X).$

Proof The situation is summarized by the following commutative diagram: where the arrows with same colors come from the same section of the long exact sequence. Define $\eta :H_n(X)\to H_n^{\text{CW}}(X)$ as follows: Take $x\in H_n(X)$, since $i_n$ is surjective, there exists $y\in H_n(X^n)$ such that $i_n(y)=x$. Define $\eta (x)=[j_n(y)]$. We need to check that $\eta$ is well-defined. Firstly, $j_n(y)\in \ker (d_n)$ because $d_n(j_n(y))=j_{n-1}({\partial }_n(y))=0$. Thus, $[j_n(y)]\in H_n^{\text{CW}}(X)$. Secondly, we need to show that $\eta$ does not depend on the choice of $y$. If there is another $y^{\prime }\in H_n(X^n)$ such that $i_n(y^{\prime })=x$, then $y-y^{\prime }\in \ker (i_n)=\text{im}({\partial }_{n+1})$, so $y-y^{\prime }={\partial }_{n+1}(z)$ for some $z\in C_{n+1}^{\mathrm{CW}}(X)$. Thus, $j_n(y-y^{\prime })=j_n({\partial }_{n+1}(z))=d_{n+1}(z)$, which implies $[j_n(y)]=[j_n(y^{\prime })]$. And clearly that $\eta$ is a homomorphism, so it is well-defined. Next, we show that $\eta$ is injective. If $\eta (x)=[j_n(y)]=0$, then $j_n(y)=d_{n+1}(z)$ for some $z\in C_{n+1}^{\mathrm{CW}}(X)$. Since $j_n$ is injective, $y={\partial }_{n+1}(z)$, which implies $i_n(y)=0$. Thus, $x=0$ and $\eta$ is injective. Finally, we show that $\eta$ is surjective. For any $[w]\in H_n^{\text{CW}}(X)$ with $w\in \ker (d_n)$, so that $j_{n-1}({\partial }_n(w))=0$. Since $j_{n-1}$ is injective, ${\partial }_n(w)=0$, which implies $w\in \text{im}(j_n)$. Thus, there exists $y\in H_n(X^n)$ such that $j_n(y)=w$. Define $x=i_n(y)$, then $\eta (x)=[j_n(y)]=[w]$. So $\eta$ is surjective. $\square$

Cellular Boundary Formula

To compute the boundary maps in practice, we use a formula based on the degrees of maps.

Cellular Boundary Formula

If we denote the $n$-cells by $\{e_{\alpha }^n{\}}_{\alpha \in A}$ and the $(n-1)$-cells by $\{e_{\beta }^{n-1}{\}}_{\beta \in B}$, the boundary map $d_n^{\mathrm{CW}}$ is given by: $d_n^{\mathrm{CW}}(e_{\alpha }^n)={\sum }_{\beta }{d}_{\alpha \beta }{e}_{\beta }^{n-1}.$ The coefficient $d_{\alpha \beta }\in \mathbb{Z}$ is the degree of the composite map: $S_{\alpha }^{n-1}\stackrel{{\phi }_{\alpha }}{\to }{X}^{n-1}\stackrel{q}{\to }{X}^{n-1}/X^{n-2}\cong {\bigvee }_{\beta }{S}_{\beta }^{n-1}\stackrel{p_{\beta }}{\to }{S}_{\beta }^{n-1}$ where ${\phi }_{\alpha }$ is the attaching map of the cell $e_{\alpha }^n$, $q$ is the quotient map, and $p_{\beta }$ is the projection onto the sphere corresponding to the cell $e_{\beta }^{n-1}$.

Proof Observe that the following diagram commute: Following the blue route, $1\in \mathbb{Z}$ will be sent to $d_{\alpha \beta }\in \mathbb{Z}$, while following the red route, $1\in \mathbb{Z}$ will be sent to $p_{\beta }\left(d_n^{\mathrm{CW}}(e_{\alpha }^n)\right)$, which is the coefficient of $e_{\beta }^{n-1}$ in $d_n(e_{\alpha }^n)$. Since the diagram commutes, these two must be equal. $\square$

Comparison with Simplicial Homology

For simplicial homology, the chain group $C_n^{\Delta }(X)$ is the free abelian group on the $n$-simplices of $X$. The coefficients of the boundary map are always $0,1,$ or $-1$.

Example: Real Projective Space

The real projective space $\mathbb{R}{\mathrm{P}}^n$ has a CW structure with exactly one $k$-cell $e_k$ in each dimension $k$ for $0\le k\le n$. The cell $e_k$ is attached to the $(k-1)$-skeleton ${\text{RP}}^{k-1}$ via the standard 2-sheeted covering map ${\varphi }_k:S^{k-1}\to {\text{RP}}^{k-1}$. The cellular chain complex is $C_k^{\text{CW}}({\text{RP}}^n)=\mathbb{Z}$ for $0\le k\le n$ and $0$ otherwise. The boundary map $d_k:C_k^{\text{CW}}\to C_{k-1}^{\text{CW}}$ is determined by the degree of the composite map: $S^{k-1}\stackrel{{\varphi }_k}{\to }{\text{RP}}^{k-1}\stackrel{q}{\to }{\text{RP}}^{k-1}/{\text{RP}}^{k-2}\cong S^{k-1}$Let’s call this composite map $f:S^{k-1}\to S^{k-1}$. To find its degree, we can sum the local degrees over the preimage of a regular point $y\in S^{k-1}$. The preimage under ${\varphi }_k$ consists of two points, which we can identify with $y$ and its antipode $-y$ in the domain $S^{k-1}$.

1. Near $y$, the map is a local homeomorphism that preserves orientation. The local degree is +1.
2. Near $-y$, the map is a composition of the antipodal map $a:S^{k-1}\to S^{k-1}$ (which has degree $(-1{)}^k$) and an orientation-preserving local homeomorphism. The local degree is $(-1{)}^k$.

Therefore, the total degree is $\text{deg}(f)=1+(-1{)}^k$. The boundary map is multiplication by this integer: $d_k=\{\begin{array}{ll}0& \text{if }k\text{ is odd}\\ 2& \text{if }k\text{ is even and }k>0\end{array}$

Homology of ${\text{RP}}^4$

Let’s compute the homology of ${\text{RP}}^4$. The chain complex is: $0\to C_4\stackrel{d_4}{\to }{C}_3\stackrel{d_3}{\to }{C}_2\stackrel{d_2}{\to }{C}_1\stackrel{d_1}{\to }{C}_0\to 0$Substituting $C_k=\mathbb{Z}$ and the computed boundary maps: $0\to \mathbb{Z}\stackrel{\times 2}{\to }\mathbb{Z}\stackrel{\times 0}{\to }\mathbb{Z}\stackrel{\times 2}{\to }\mathbb{Z}\stackrel{\times 0}{\to }\mathbb{Z}\to 0$The homology groups are:

• $H_0({\text{RP}}^4)=\ker (d_0)/\text{im}(d_1)=\mathbb{Z}/\text{im}(\times 0)=\mathbb{Z}$
• $H_1({\text{RP}}^4)=\ker (d_1)/\text{im}(d_2)=\ker (\times 0)/\text{im}(\times 2)=\mathbb{Z}/2\mathbb{Z}$
• $H_2({\text{RP}}^4)=\ker (d_2)/\text{im}(d_3)=\ker (\times 2)/\text{im}(\times 0)=0/0=0$
• $H_3({\text{RP}}^4)=\ker (d_3)/\text{im}(d_4)=\ker (\times 0)/\text{im}(\times 2)=\mathbb{Z}/2\mathbb{Z}$
• $H_4({\text{RP}}^4)=\ker (d_4)/\text{im}(d_5)=\ker (\times 2)/0=0$
• $H_k({\text{RP}}^4)=0$ for $k>4$.

So, we have: $H_k({\text{RP}}^4)=\{\begin{array}{ll}\mathbb{Z}& k=0\\ \mathbb{Z}/2\mathbb{Z}& k=1,3\\ 0& \text{otherwise}\end{array}$

In general, the homology of ${\text{RP}}^n$ is: $H_k({\text{RP}}^n)\cong \{\begin{array}{ll}\mathbb{Z}& k=0\\ \mathbb{Z}/2\mathbb{Z}& \text{for }0<k<n,k\text{ odd}\\ \mathbb{Z}& k=n,n\text{ odd}\\ 0& \text{else}\end{array}$

Example: Surface of Genus g

A standard CW structure for the orientable surface of genus $g$, $M_g$, has:

• 1 0-cell (a point).
• $2g$ 1-cells, denoted $a_1,b_1,\dots ,a_g,b_g$.
• 1 2-cell, $E$, attached along the path ${\prod }_{i=1}^g[a_i,b_i]=a_1{b}_1{a}_1^{-1}{b}_1^{-1}\dots a_g{b}_g{a}_g^{-1}{b}_g^{-1}$.

The cellular chain complex is $0\to \mathbb{Z}\stackrel{d_2}{\to }{\mathbb{Z}}^{2g}\stackrel{d_1}{\to }\mathbb{Z}\to 0$.

• The map $d_1$ is the zero map because the boundary of each 1-cell is the single 0-cell, mapping to $v-v=0$.
• To compute $d_2(E)=\sum m_i{a}_i+\sum n_j{b}_j$, we find the degree of the attaching map projection. For each $a_i$, the attaching map traverses $a_i$ once forward and once backward ($a_i^{-1}$). The corresponding degrees cancel out, so $m_i=0$. The same logic applies to the $b_j$‘s, so $n_j=0$.
• Therefore, both $d_2$ and $d_1$ are zero maps.

The homology groups are then the chain groups themselves: $H_k(M_g)\cong \{\begin{array}{ll}\mathbb{Z}& k=0\\ {\mathbb{Z}}^{2g}& k=1\\ \mathbb{Z}& k=2\\ 0& \text{else}\end{array}$

Example: Complex Projective Space

The complex projective space $\mathbb{C}{\mathrm{P}}^n$ has a CW structure with one cell in each even dimension $2k$ for $0\le k\le n$. The chain complex consists of $\mathbb{Z}$ in even dimensions and 0 in odd dimensions:$0\to \mathbb{Z}\stackrel{d_{2n}}{\to }0\to \mathbb{Z}\to \cdots \to 0\to \mathbb{Z}\to 0$All boundary maps must be zero since they map from or to a zero group. The homology groups are therefore the chain groups themselves: $H_k({\text{CP}}^n)\cong \{\begin{array}{ll}\mathbb{Z}& 0\le k\le 2n,k\text{ even}\\ 0& \text{else}\end{array}$