Useful Facts
H_0(X) is a free abelian group on the set of path components of X. Thus if X is path connected, H_0(X) \cong {\mathbf{Z}}. In general, H_0(X) \cong {\mathbf{Z}}^{{\left\lvert {\pi_0(X)} \right\rvert}}, where {\left\lvert {\pi_0(X)} \right\rvert} is the number of path components of X.
\begin{align*} \tilde H_*(A\vee B) &\cong H_*(A) \times H_*(B) \\ H_{n}\qty{\bigvee_\alpha X_\alpha} &\cong \prod_\alpha H_{n} X_\alpha \end{align*} See footnote for categorical interpretation. 1
\todo[inline]{May need some good pair condition?}
\begin{align*} H_{n}(\bigvee_{k} S^n) = {\mathbf{Z}}^k .\end{align*}
proof (?):
Mayer-Vietoris.
H_{k} \qty{ \prod_ \alpha X_ \alpha} is not generally equal to \prod_ \alpha \qty{ H_{k} X_ \alpha }. The obstruction is due to torsion – if all groups are torsionfree, then the Kunneth theorem 2 yields \begin{align*} H_{k} (A\times B) = \prod_{i+j=k} H_{i} A \otimes H_{j} B \end{align*}
Todo
\todo[inline]{Excision.}
:::{.fact title="Assorted facts}
\envlist
- H_{n}(X) = 0 \iff X has no n{\hbox{-}}cells.
- C^0 X = {\operatorname{pt}}\implies d_{1}: C^1 \to C^0 is the zero map.
:::
Known Homology
\begin{align*} H_{i}(S^n) = \begin{cases} {\mathbf{Z}}& i = 0, n \\ 0 & \text{else}. \end{cases} \end{align*}
\todo[inline]{Homology examples.}
Mayer-Vietoris
Since {\mathbf{Z}} is free and thus projective, any exact sequence of the form 0 \to {\mathbf{Z}}^n \to A \to {\mathbf{Z}}^m \to 0 splits and A\cong {\mathbf{Z}}^{n}\times{\mathbf{Z}}^m.
Mnemonic: X = A \cup B \leadsto (\cap, \oplus, \cup)
Let X = A^\circ \cup B^\circ; then there is a SES of chain complexes \begin{align*} 0 \to C_{n}(A\cap B) \xrightarrow{x\mapsto (x, -x)} C_{n}(A) \oplus C_{n}(B) \xrightarrow{(x, y) \mapsto x+y} C_{n}(A + B) \to 0 \end{align*} where C_{n}(A+B) denotes the chains that are sums of chains in A and chains in B. This yields a LES in homology: \begin{align*} \cdots H_{n}(A \cap B) \xrightarrow{(i^*,~ j^*)} H_{n}(A) \oplus H_{n}(B) \xrightarrow{l^* - r^*} H_{n}(X) \xrightarrow{\delta} H_{n-1}(A\cap B)\cdots \end{align*} where
-
i: A\cap B \hookrightarrow A induces i^*: H_*(A\cap B) \to H_*(A)
-
j: A\cap B \hookrightarrow B induces j^*: H_*(A\cap B) \to H_*(B)
-
l: A \hookrightarrow A\cup B induces l^*: H_*(A) \to H_*(X)
-
r: B \hookrightarrow A\cup B induces r^*: H_*(B) \to H_*(X)
More explicitly,
The connecting homomorphisms \delta_{n} :H_{n}(X) \to H_{n-1}(X) are defined by taking a class [\alpha] \in H_{n}(X), writing it as an n-cycle z, then decomposing z = \sum c_{i} where each c_{i} is an x+y chain. Then {\partial}(c_{i}) = {\partial}(x+y) = 0, since the boundary of a cycle is zero, so {\partial}(x) = -{\partial}(y). So then just define \delta([\alpha]) = [{\partial}x] = [-{\partial}y].
Handy mnemonic diagram: \begin{align*} \begin{matrix} && A\cap B & \\ &\diagup & & \diagdown \\ A\cup B & & \longleftarrow & & A \oplus B \end{matrix} .\end{align*}
H_*(A # B): Use the fact that A# B = A \cup_{S^n} B to apply Mayer-Vietoris.
\begin{align*}H^i(S^n) \cong H^{i-1}(S^{n-1}).\end{align*}
proof:
Write X = A \cup B, the northern and southern hemispheres, so that A \cap B = S^{n-1}, the equator. In the LES, we have:
\begin{align*} H^{i+1}(S^n) \xrightarrow{} H^i(S^{n-1}) \xrightarrow{} H^iA \oplus H^i B \xrightarrow{} H^i S^n \xrightarrow{} H^{i-1}(S^{n-1}) \xrightarrow{} H^{i-1}A \oplus H^{i-1}B .\end{align*}
But A, B are contractible, so H^iA= H^iB = 0, so we have
\begin{align*} H^{i+1}(S^n) \xrightarrow{} H^{i}(S^{n-1}) \xrightarrow{} 0 \oplus 0 \xrightarrow{}H^i(S^n) \xrightarrow{} H^{i-1}(S^{n-1}) \xrightarrow{} 0 .\end{align*}
In particular, we have the shape 0 \to A \to B \to 0 in an exact sequence, which is always an isomorphism.
More Exact Sequences
There exists a short exact sequence \begin{align*} 0 \to \prod_{i+j=k} H_{j}(X; R) \otimes_{R} H_{i}(Y; R) \to H_{k}(X\times Y; R) \to \prod_{i+j=k-1} \operatorname{Tor}_{R}^1(H_{i}(X; R), H_{j}(Y; R)) .\end{align*} If R is a free R{\hbox{-}}module, a PID, or a field, then there is a (non-canonical) splitting given by \begin{align*} H_{k} (X\times Y) \cong \left( \prod_{i+j = k} H_{i} X \oplus H_{j} Y\right) \times\prod_{i+j = k-1}\operatorname{Tor}(H_{i}X, H_{j} Y) \\ \end{align*}
For changing coefficients from {\mathbf{Z}} to G an arbitrary group, there are short exact sequences
\begin{align*} 0\to \operatorname{Tor}_{\mathbf{Z}}^0 (H_{i}(X;{\mathbf{Z}}), A) &\to H_{i}(X;A)\to \operatorname{Tor}_{\mathbf{Z}}^1 (H_{i-1}(X;{\mathbf{Z}}),A)\to 0 \\ & \quad \Downarrow \\ \\ 0 \to H_{i} X \otimes G &\to H_{i}(X; G) \to \operatorname{Tor}_{\mathbf{Z}}^1(H_{i-1}X, G) \to 0 \end{align*} and \begin{align*} 0\to \operatorname{Ext}_{{\mathbf{Z}}}^{1}(H_{i-1}(X; {\mathbf{Z}}),A) &\to H^{i}(X; A)\to \operatorname{Ext}_{{\mathbf{Z}}}^{0}(H_{i}(X; {\mathbf{Z}}),A) \to 0 \\ &\quad \Downarrow \\ \\ 0 \to \operatorname{Ext}(H_{i-1} X, G) &\to H^i(X;G) \to \hom(H_{i} X, G) \to 0 .\end{align*} These split unnaturally: \begin{align*} H_{i}(X;G) &= (H_{iX}\otimes G) \oplus \operatorname{Tor}(H_{i-1}X; G) \\ H^i(X; G) &= \hom(H_{i}X, G) \oplus \operatorname{Ext}(H_{i-1}X; G) \end{align*}
When all of the H_{i}X are all finitely generated (e.g. if G is a field), writing H_{i}(X; {\mathbf{Z}}) = {\mathbf{Z}}^{\beta_{i}} \oplus T_{i} as the sum of a free and a torsionfree module, we have \begin{align*} H^i(X; {\mathbf{Z}}) &\cong {\mathbf{Z}}^{\beta_{i}} \times T_{i-1} \\ H^i(X; A) &\cong \qty{H_i(X; G)} {}^{ \vee }\coloneqq\hom_{\mathbf{Z}}(H_{i}(X; G), G) .\end{align*}
In other words, letting F({-}) be the free part and T({-}) be the torsion part, we have \begin{align*} H^i(X; {\mathbf{Z}}) &= F(H_{i}(X; {\mathbf{Z}})) \times T(H_{i-1}(X; {\mathbf{Z}}))\\ H_{i}(X; {\mathbf{Z}}) &= F(H^i(X; {\mathbf{Z}})) \times T(H^{i+1}(X; {\mathbf{Z}})) \end{align*}
\todo[inline]{Might need assumptions: finite CW complex?}
Relative Homology
-
H_{n}(X/A) \cong \tilde H_{n}(X, A) when A\subset X has a neighborhood that deformation retracts onto it.
-
LES of a pair
- (A \hookrightarrow X) \mapsto (A, X, X/A)
-
For CW complexes X = \left\{{X^{(i)}}\right\}, we have \begin{align*} H_{n}(X^{(k)},X^{(k-1)}) \cong \begin{cases}{\mathbf{Z}}[\left\{{e^n}\right\}]~ &k=n,\\ 0 &\text{otherwise}\end{cases} \qquad\text{ since } X^k/X^{k-1} \cong \bigvee S^k \end{align*}
-
H_{n}(X, A) \cong_? H_{n}(X/A, {\operatorname{pt}})
Exercises
Show that S^2 \times{\mathbf{RP}}^4 \not\simeq S^4 \times{\mathbf{RP}}^2.
solution:
Take coefficients in { \mathbf{F} }_2, apply Kunneth to get \begin{align*} H^*(X_1; { \mathbf{F} }_2) = { {\bigwedge}^{\scriptscriptstyle \bullet}} _{{ \mathbf{F} }_2}{x_2} \otimes_{{ \mathbf{F} }_2} {{ \mathbf{F} }_2 { \left[ \scriptstyle {y_1} \right] } \over \left\langle{ y_1^5 }\right\rangle \\ H^*(X_2; { \mathbf{F} }_2) = { {\bigwedge}^{\scriptscriptstyle \bullet}} _{{ \mathbf{F} }_2}{w_4} \otimes_{{ \mathbf{F} }_2} {{ \mathbf{F} }_2 { \left[ \scriptstyle {z_1} \right] } \over \left\langle{ z_1^3 }\right\rangle \\ .\end{align*}
Now use that any homotopy equivalence induces a graded ring isomorphism f, where f(y_1) = z_1 in H^1, but these have different orders.
Show that S^n is not a strong deformation retract of {\mathbb{B}}^{n+1}.
solution:
If \iota S^n \hookrightarrow{\mathbb{B}}^{n+1} then a strong deformation retract yields a p: {\mathbb{B}}^{n+1} \to S^n with p\circ \iota = \operatorname{id}. Then apply homology to factor the identity map {\mathbf{Z}}\to {\mathbf{Z}} through a constant zero map {\mathbf{Z}}\to 0.