Cellular homology
Cellular homology provide an efficient way to compute homology groups of CW complexes, based on degree calculations.
Let $X$ be a CW complex; then:
- $H_k(X^n, X^{n - 1})$ is zero for $k \neq n$, and free abelian otherwise, with basis in one-to-one correspondence with the n-cells of $X$. This follows from the observation that $(X^n, X^{n - 1})$ is a good pair, and $X^n/X^{n - 1}$ is the wedge sum of $n$-spheres, one for each $n$-cell of $X^n$.
- $H_k(X^n) = 0$ for $k \gt n$.
- The inclusion $i : X^n \hookrightarrow X$ induces the isomorphism $i_* : H_k(X^n) \rightarrow H_k(X)$ if $k \lt n$, since $H_k(X^n) \approx H_k(X^{n + 1}) \approx \ldots H_k(X^{n + m})$ for all $m \geq 0$, as in (b).
To prove (b), consider the l.e.s of the pair $(X^n, X^{n - 1})$:
If $k \neq n, n - 1$, then the outer two groups are zero, and we get $H_k(X^{n - 1}) \approx H_k(X^n)$. If $k > n$, $H_k(X^n) \approx H_k(X^{n - 1}) \approx \ldots \approx H_k(X^0)$.
Pairs $(X^{n + 1}, X^n), (X^n, X^{n - 1}), (X^{n - 1}, X^{n - 2})$ fit into the l.e.s:
$d_{n + 1}, d_n$ are relativizations of the boundary map $\partial_{n + 1}, \partial_n$, and the composition $d_n d_{n + 1}$ is zero. The horizontal row in the diagram is termed cellular chain complex in $X$. Since $H_n(X^n, X^{n - 1})$ is free with basis in one-to-one correspondence with the $n$-cells, it can be thought of as linear combinations of $n$-cells of $X$. The homology groups of this cellular chain complex are called cellular homology groups.
$H_n^\text{CW}(X) \approx H_n(X)$. From the diagram, $H_n(X)$ can be identified with $H_n(X^n)/\text{im } \partial_{n + 1}$. Further, since $j_n$ is injective, it maps $\text{im } \partial_{n + 1}$ isomorphically onto $\text{im }(j_n \partial_{n + 1}) = \text{im } d_{n + 1}$, and $H_n(X^n)$ onto $\text{im } j_n = \text{ker } \partial_n$. Since $j_{n - 1}$ is injective, $\text{ker } d_n = \text{ker } \partial_n$. Thus, $j_n$ induces a isomorphism of the quotient $H_n(X^n)/\text{im } \partial_{n + 1}$ onto $\text{ker } d_n/\text{im } d_{n + 1}$.
- $H_n(X) = 0$ if $X$ is a CW complex with no $n$-cells.
- If $X$ is a CW complex with $k$ $n$-cells, $X$ is generated by atmost $k$ elements; $H(X^n, X^{n - 1})$ is free abelian on $k$ generators, so $\text{ker } d_n$ must be generated by at most $k$ elements, hence also the quotient $\text{ker } d_n/\text{im } d_{n + 1}$.
- If $X$ is a CW complex with no two cells in adjacent dimensions, then it is free abelian with basis in one-to-one correspondence with the $n$-cells of $X$. This is because the cellular boundary maps $d_n$ are zero.
$\mathbb{C}P^n$ has a CW-structure with each cell of each even dimension, where $\mathbb{C}P^n$ is the complex projective $n$-space of complex lines through the origin in $\mathbb{C}^{n + 1}$.
As another example, $S^n \times S^n$ has a product CW-structure, with one $0$-cell, two $n$-cells, and one $2n$-cell.
To compute cellular boundary maps $d_n$, note that $d_1 : H_1(X^1, X^0) \rightarrow H_0(X^0)$ is the same as the simplicial boundary map $\Delta_1(X) \rightarrow \Delta_0(X)$. If $X$ is connected, and has only one $0$-cell, $d_1$ must be zero, otherwise $H_0(X)$ would not be $\mathbb{Z}$. When $n \gt 1$, $d_n$ can be computed in terms of degrees:
Cellular boundary formula: $d_n(e_\alpha^n) = d_{\alpha \beta} e_\beta^{n - 1}$ where $d_{\alpha \beta}$ is the degree of the map $S_\alpha^{n - 1} \rightarrow X^{n - 1} \rightarrow S_\beta^{n - 1}$ that is the composition of attaching map $e_\alpha^n$ with the quotient map collapsing $X^{n - 1} - e_\beta^{n - 1}$ to a point. The cellular boundary formula can then be obtained from the following diagram: