$$E^n = \{x \in \mathbb{R}^n : |x| \leq 1\}$$
$$U^n = \{x \in \mathbb{R}^n : |x| < 1\}$$
$$S^{n-1} = \{x \in \mathbb{R}^n : |x| = 1\}$$
In this section, we assume \(X^*\) is Hausdorff, and that \(X\) is a closed subset of \(X^*\) such that \(X^* - X\) is the disjoint union of open subsets \(e^n_\lambda, \lambda \in \Lambda\); each \(e^n_\lambda\) is assumed to be homeomorphic to \(U^n\), and is called an n-cell. Finally, it is assumed that each n-cell is attached to \(X\) by a characteristic map, that is for each \(\lambda\) there exists continuous map:
$$f_\lambda : E^n \to \bar{e}^n_\lambda$$
such that \(f_\lambda\) maps \(U^n\) homeomorphically onto \(e^n_\lambda\) and \(f_\lambda(S^{n-1}) \subset X\). If the number of \(n\)-cells is finite, then no additional requirements are necessary. However, in the case of an infinite family of \(n\)-cells, then we also require that a subset \(A\) of \(X^*\) is closed if and only if \(A \cap X\) and \(f_\lambda^{-1}(A)\) are closed for all \(\lambda \in \Lambda\).
The question we want to know now is, if we know what \(X\) looks like, can we figure out what \(X^*\) looks like? Turns out we can.
Theorem: Under the above hypothesis, \(H_q(X^*,X) = 0\) for all \(q \neq n\). For each \(\lambda \in \Lambda\), the characteristic map \(f_\lambda\) induces a monomorphism of relative homology groups \(f_{\lambda *} : H_n(E^n,S^{n-1}) \to H_n(X^*,X)\) and \(H_n(X^*,X)\) is the direct sum of the image subgroups. Thus, \(H_n(X^*,X)\) is a free abelian group with basis in one to one correspondence with the set of cells \(\{e^n_\lambda | \lambda \in \Lambda\}\)
Corollary: The homomorphism \(i_* : H_q(X) \to H_q(X^*)\) is an isomorphism except possibly for \(q = n\) and \(q = n-1\); the only nontrivial part of the homology sequence of the pair \((X^*,X)\) is the following:
$$0 \to H_n(X) \xrightarrow{i_*} H_n(X^*) \to H_n(X^*,X) \to H_{n-1}(X) \xrightarrow{i_*} H_{n-1}(X^*) \to 0$$
A structure of CW-complex is prescribed on a space \(X\) (which is always assumed to be Haudorff) by the prescription of an ascending sequence of closed subspaces:
$$X^0 \subset X^1 \subset X^2 \subset \cdot\cdot\cdot$$
which satisfy the following conditions:
X^0 has the discrete topology
For \(n > 0\), \(X^n\) is obtained from \(X^{n-1}\) by adjoining a collection of \(n\)-cells in an appropriate manner (i.e. homeomorphing the interior of the cell, and the boundary gets mapped to the skeleton \(X^{n-1}\)).
\(X\) is the union of the subspaces \(X^i\) for \(i \geq 0\)
The space \(X\) and the subspaces \(X^q\) all have the weak topology: A subset \(A\) is closed if and only if \(A \cap \bar{e}^n\) is closed for all \(n\)-cells, \(e^n\), \(n = 0, 1, 2, ...\)
The Homology Groups of a CW-Complex
Let \(K = \{K^n | n = 0,1,2,...\}\) denote a CW-structure imposed on topological space \(X\). We know that $$H_q(K^n,K^{n-1}) = 0$$ for \(q \neq n\) and that $$H_n(K^n,K^{n-1})$$ is a free abelian group generated by the \(n\)-cells of \(K\). We also know that
$$H_q(K^n) = 0$$
for all \(q > n\). Now we define:
$$C_n(K) = H_n(K^n, K^{n-1})$$
$$j_{n-1} \circ \partial_* = d_n : C_n(K) \to C_{n-1}(K)$$
$$H_n(K^n,K^{n-1}) \xrightarrow{\partial_*} H_{n-1}(K^{n-1}) \xrightarrow{j_{n-1}} H_{n-1}(K^{n-1},K^{n-2})$$
where \(\delta_*\) is the boundary operator and \(j_{n-1}\) is the homomorphism induced by the inclusion map. We can now define \(Z_n(K), B_n(K), H_n(K)\) appropriately as the kernel of \(d\), the image of \(d\), the quotient of \(Z\) by \(B\).
Now consider:
$$H_n(X) \xleftarrow{k_n} H_n(K^n) \xrightarrow{j_n} H_n(K^n,K^{n-1}) = C_n(K)$$
Here \(j_n\) and \(k_n\) are homomorphisms induced by inclusion maps. In the above diagram, \(k_n\) is an epimorphism, \(j_n\) is a monomorphism, image \(j_n = Z_n(K)\), kernel \(k_n = j_n^{-1}(B_n(K))\).
We can understand these results 'intuitively'. \(k_n\) represents inclusion from \(K^n\) to \(X\), and intuitively this can only make the \(n\)-dimensional structure 'simpler' (you can't introduce more complexity with higher dimensional manifolds!! Think about a circle going to a sphere; you've simplified one-dimensional group by heaps), so its a surjection. \(j_n\) on the other hand is a mapping which quotients out the skeleton \(K^{n-1}\), which can only make the structure more complicated, (imagine three points connected with edges, and the face in the middle filled in. Now quotient those three points together. Before you had \(\mathbb{Z}\), now you have the free group on three generators!) Hence it is a monomorphism.
Then \(j_n \circ k_n^{-1}\) defines an isomorphism
$$\theta : H_n(X) \to H_n(K)$$
The Euler Characteristic of a CW-Complex
For any abelian group \(A\), let \(r(A)\) denote the rank of \(A\). Let \(K\) be a finite CW-complex on the space \(X\). Then the euler characteristic satisfies the following equation:
$$\chi(K) = \sum_n (-1)^n r(H_n(X))$$
and is invariant on what choice of CW-complex we choose to impose on \(X\).
Let \(K = \{K^n\}\) be a CW-complex on the space \(X\). For each \(n\)-cell \(e^n_\lambda\), define \(\bar{e}^n_\lambda\) as the closure, and \(\dot{e}^n_\lambda\) as the border of the \(n\)-cell. Notice that \(H_n(\bar{e}^n_\lambda,\dot{e}^n_\lambda)\) is infinite cyclic, and hence there are two ways to choose a generator. We can then define \(a^n_\lambda\) as our choice of generator for each \(n\)-cell. We know that \(H_n(K^n, K^{n-1})\) is a free abelian group generated by the set of \(n\)-cells; hence we can then define \(b^n_\lambda\) as our choice of generator \(a^n_\lambda\) under the inclusion map \(l_{\lambda *} : H_n(\bar{e}^n_\lambda, \dot{e}^n_\lambda) \to H_n(K^n,K^{n-1})\) (it's kind of like a choice of basis in linear algebra). Notice that \(\{b^n_\lambda\}\) forms a basis for \(C_n(K) = H_n(K^n, K^{n-1})\).
The above describes the situation for \(n > 0\). Notice however that for \(n = 0\); we have a canonical isomorphism from \(H_0(e^0_\lambda) \to \mathbb{Z}\), there isn't any choice to be made. Canonically we choose "having the point" to be \(1\) over "not having the point".
Recall that we have boundary homomorphisms:
$$d_n : C_n(K) \to C_{n-1}(K)$$
defined by the composition of a boundary homomorphism and an inclusion homomorphism:
$$H_n(K^n,K^{n-1}) \xrightarrow{\partial_*} H_{n-1}(K^{n-1}) \xrightarrow{j} H_{n-1}(K^{n-1},K^{n-2})$$
Since \(\{b^n_\lambda\}\) forms a basis for \(C_n(K)\), boundary homomorphisms \(d_n\) are completely determined by where \(d_n\) takes the basis elements. So we have:
$$d_n(b^n_\lambda) = \sum_{\mu} [b^n_\lambda : b^{n-1}_{\mu}] b^{n-1}_\mu $$
where \([b^n_\lambda : b^{n-1}_{\mu}]\) is some constant that we call the incidence number of the cells \(e^n_\lambda\) and \(e^{n-1}_\mu\) (w.r.t. the chosen orientations).
Properties of the Incidence Numbers
For any \(n\)-cell \(e^n_\lambda\), \([b^n_\lambda : b^{n-1}_{\mu}] = 0\) for all but a finite number of \((n-1)\)-cells.
For any \(n\)-cell \(e^n_\lambda\) and \((n-2)\)-cell \(e^{n-2}_\nu\):
$$\sum_{\mu}[b^n_\lambda : b^{n-1}_{\mu}][b^{n-1}_{\mu} : b^{n-2}_\nu] = 0$$
For any \(1\)-cell \(e^1_\lambda\), \(\sum_\mu [b^1_\lambda : b^{0}_{\mu}] = 0\)
If the cell \(e^{n-1}_\mu\) is not contained in the closure of the cell \(e^n_\lambda\), then \([b^n_\lambda : b^{n-1}_{\mu}] = 0\)
Let \(f : (X,A) \to (Y,B)\) be a map of pairs which is homotopic to a map \(g : (X,A) \to (Y,B)\) such that \(g(X) \subset B\). Then theinduced homomorphism
$$f_* : H_n(X,A) \to H_n(Y,B)$$
is zero for all \(n\).
A CW-complex is regular if for each cell \(e^n\), \(n > 0\), there exists a characteristic map \(f : E^n \to \bar{e}^n\) which is a homeomorphism.
Regular CW - complexes have these properties:
If \(m < n\) and \(e^m\) and \(e^n\) are cells such that \(e^m \cap \dot{e}^n \neq \emptyset \), then \(e^m \subset \dot{e}^n\)
For any \(n\)-cell \(e^n\), \(n \geq 0\), \(\bar{e}^n\) and \(\dot{e}^n\) are the underlying spaces of sub-complexes. Also \(\dot{e}^n\) is the union of closures of \((n-1)\)-cells.
If \(e^m \subset \bar{e}^n\) we say that \(e^m \leq e^n\) and that \(e^m\) is a face of \(e^n\). We say that \(e^m < e^n\) is a proper face if additionaly \(e^m \neq e^n\). Let \(e^n\) and \(e^{n+2}\) be cells of a regular cell complex such that \(e^n\) is a face of \(e^{n+2}\). Then there are exactly two \((n+1)\)-cells \(e^{n+1}\) such that \(e^n < e^{n+1} < e^{n+2}\).
Properties of regular CW-Complexes
The incidence numbers \([b^n_\lambda : b^{n-1}_\mu]\) in a regular CW-complex satisfy the following conditions:
If \(e^{n-1}_\mu\) is not a face of \(e^n_\lambda\), then \([b^n_\lambda : b^{n-1}_\mu] = 0\)
If \(e^{n-1}_\mu\) is a face of \(e^n_\lambda\), then \([b^n_\lambda : b^{n-1}_\mu] = \pm 1\)
If \(e^0_\mu\) and \(e^0_\nu\) are the two vertices which are faces of the 1-cell \(e^1_\lambda\), then
$$[b^1_\lambda : b^0_\mu] + [b^1_\lambda : b^0_\nu] = 0$$
Let \(e^n_\lambda\) and \(e^{n-2}_\rho\) be cells such that \(e^{n-2}_\rho < e^n_\lambda\); let \(e^{n-1}_\mu\) and \(e^{n-1}_\nu\) denote the unique \((n-1)\)-cells \(e^{n-1}\) such that \(e^{n-2}_\rho < e^{n-1} < e^n_\lambda\), then
$$[b^n_\lambda : b^{n-1}_\mu][b^{n-1}_\mu : b^{n-2}_\rho] + [b^n_\lambda : b^{n-1}_\nu][b^{n-1}_\nu : b^{n-2}_\rho] = 0$$
Knowing this; we can now assert the following theorem.
Theorem: Let \(K\) be a regular CW-complex on the topological space \(X\). For each pair \((e^n_\lambda,e^{n-1}_\mu)\) consisting of an n-cell and a (n-1)-cell of \(K\), let there be given an integer \(\alpha^n_{\lambda\mu} \in \{-1,0,1\}\) such that it satisfies the properties that \([b^n_\lambda : b^{n-1}_\mu]\) does above. Then there exists a unique way to choose orientations \(b^n_\lambda\) such that \(\alpha^n_{\lambda\mu} = [b^n_\lambda : b^{n-1}_\mu]\). In other words; the orientations are predetermined by the structure, you have no freedom of choice??? No. In other words, different orientations lead to different incidence numbers; and different incidence numbers lead to different orientations. In other words; provide me with a set of (compatible) incidence numbers, and I'll provide you with your orientations.
An n-dimensional pseudomanifold is an \(n\)-dimensional finite, regular CW-complex which satisfies the following three conditions
Every cell is a face of some \(n\)-cell
Every \((n-1)\)-dimensional cell is a face of exactly two \(n\)-cells.
Given any two \(n\)-cells, \(e^n\) and \(e'^n\), there exists a sequence of \(n\)-cells
$$e^n_0, e^n_1, e^n_2,...,e^n_k$$
such that \(e^n_0 = e^n\), \(e^n_k = e'^n\) and each two consecutive \(n\)-cells shares a common \((n-1)\)-dimensional face.
Let \(K\) be an \(n\)-dimensional pseudomanifold, and let \(e^n_1\) and \(e^n_2\) be \(n\)-cells of \(K\) which have a common \((n-1)\) dimensional face \(e^{n-1}\). We say that the orientations for \(e^n_1\) and \(e^n_2\) are coherent w.r.t. their common shared face if:
$$[e^n_1:e^{n-1}] + [e^n_2:e^{n-1}] = 0$$
\(K\) is orientable if all \(n\)-cells sharing \((n-1)\)-faces are oriented coherently.
Properties of pseudomanifolds
If \(K\) is an orientable n-dimensional pseudomanifold, then \(H_n(K)\) is infinite cyclic; if \(K\) is nonorientable then \(H_n(K) = 0\).
Let \(K\) be an n-dimensional pseudomanifold. If \(K\) is orientable, then \(H_{n-1}(K)\) is torsion-free. If \(K\) is nonorientable, then the torsion subgroup of \(H_{n-1}(K)\) is cyclic of order 2.