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*}
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
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\(i: A\cap B \hookrightarrow A\) induces \(i^*: H_*(A\cap B) \to H_*(A)\)
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\(j: A\cap B \hookrightarrow B\) induces \(j^*: H_*(A\cap B) \to H_*(B)\)
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\(l: A \hookrightarrow A\cup B\) induces \(l^*: H_*(A) \to H_*(X)\)
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\(r: B \hookrightarrow A\cup B\) 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*}
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
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\(H_{n}(X/A) \cong \tilde H_{n}(X, A)\) when \(A\subset X\) has a neighborhood that deformation retracts onto it.
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LES of a pair
- \((A \hookrightarrow X) \mapsto (A, X, X/A)\)
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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*}
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\(H_{n}(X, A) \cong_? H_{n}(X/A, {\operatorname{pt}})\)
Exercises
Show that \(S^2 \times{\mathbf{RP}}^4 \not\simeq S^4 \times{\mathbf{RP}}^2\).
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 \({\mathbb{B}}^{n+1}\).
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 \({\mathbf{Z}}\to 0\).