\( \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\Map}{Map} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\Tor}{Tor} \DeclareMathOperator{\Ext}{Ext} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\id}{id} \DeclareMathOperator{\coker}{coker} \DeclareMathOperator{\CC}{\mathbf{C}} \DeclareMathOperator{\QQ}{\mathbf{Q}} \DeclareMathOperator{\RR}{\mathbf{R}} \DeclareMathOperator{\ZZ}{\mathbf{Z}} \DeclareMathOperator{\RP}{\mathbf{RP}} \DeclareMathOperator{\C}{C_*} \DeclareMathOperator{\U}{\mathcal{U}} \)

Section1.4Homology of a pair

Subsection1.4.1Exact sequences

Definition1.4.1Exact sequence
A sequence \[ \cdots \to A_{n+1} \to A_n \to A_{n-1} \to \cdots \] of \(R\)-modules and \(R\)-linear maps is exact at \(A_n\) if \(\ker f_n = \im f_{n+1}\).
A sequence is exact if it is exact at all \(A_n\), then \((A_*,f)\) is known as a acyclic chain complex (the homology is zero).
Example1.4.2
  1. \(0 \to A \xrightarrow{f} B\) is exact if and only if \(f\) is surjective.
  2. \(B \xrightarrow{g} C \to 0\) is exact if and only if \(g\) is injective.
  3. \(0 \to A \xrightarrow{f} A' \to 0\) is exact if and only if \(f\) is an isomorphism.
  4. \(0 \to A \to B \to C \to 0\) is exact if and only if \(A\subset B\) and \(C\cong B/A\).
Definition1.4.3Short exact sequence
A sequence \[ 0 \to A_* \xrightarrow i B_* \xrightarrow{\pi} C_* \to 0 \] is a short exact sequence of chain complexes if
  1. \(A_*,B_*,C_*\) are chain complexes.
  2. \(i, \pi\) are chain maps.
  3. \[ 0 \to A_n \xrightarrow i B_n \xrightarrow{\pi} C_n \to 0 \] is exact for all \(n\).
Exercise 1.4.6
Suppose we have a map of short exact sequences commuting, then show there is a map of long exact sequences of homology commuting .
Example1.4.7Reduced homology
If \(X\) is a space let \[ \tilde C_*(X) = \begin{cases}C_*(X), &* \ne -1,\\\ZZ, & * = -1,\end{cases} \] then define \(d(e_p)= e\) for \(p \in X\). We have \(d^2e_\gamma = d(e_{\gamma(1)} - e_{\gamma(0)}) = e - e = 0\). So \(\tilde C_*\) is a chain complex.
This construction can be motivated by thinking of \(\Delta^{-1} = \{\}\) and then considering \(\Map(\{\},X) = \{e\}\). If \[ A_* = \begin{cases} \ZZ = \langle e \rangle, & * = -1, \\ 0, & * = -1.\end{cases} \] Then we have a short exact sequence \[ 0 \to A_* \to \tilde C_*(X) \to C_*(X) \to 0. \] We have \(H_*(A) = A_*\) and so the long exact sequence of homology then says that for \(n \ge 0\)\[ \cdots \to H_n(A) = 0 \to H_n(\tilde C(X))\to H_n(C(X)) \to H_{n-1}(A)= 0 \to \cdots \] giving that \(H_n(\tilde C_*(X)) \cong H_n(C_*(X))\) for \(n \gt 0\). We write \(\tilde H_*(X)\) for \(H_*(\tilde C_*(X))\). At \(n = 0\) we get \[ \cdots \to 0 \to \tilde H_0(X)\to H_0(X) \to \ZZ \to 0 \to\cdots \] giving that \(\tilde H_0(X)\) has one fewer copies of \(\ZZ\) than \(H_0(X)\).
Example1.4.8Homology of a pair
Suppose \(A\subset X\), then \(C_*(A) \subset C_*(X)\) is a subcomplex. Define \(C_*(X,A) = C_*(X)/C_*(A)\) and so we have a short exact sequence \[ 0 \to C_*(A) \to C_*(X) \to C_*(X,A) \to 0. \]

Subsection1.4.2Homology of a pair

Definition1.4.9Homology of a pair
\(H_*(X,A) = H_*(C_*(X,A))\) is the homology of the pair \((X,A)\).

From this we obtain:

Definition1.4.10Long exact sequence of a pair
The long exact sequence of the pair \((X,A)\) is \[ \cdots \to H_n(A) \to H_n(X) \to H_n(X,A) \xrightarrow{\partial} H_{n-1} (A) \to \cdots.\]

If \(f\colon (X,A) \to (Y,B)\) is a map of pairs then we get an induced map \(f_\# \colon C_*(X) \to C_*(Y)\) defined by \[ (\sigma\colon \Delta^n \to A) \mapsto f\circ \sigma. \] Observe that \(f_\#(C(A)) \subset C_*(B)\) and so \(f_\#\) descends to a map \[ f_\#\colon C_*(X)/C_*(A) \to C_*(Y)/C_*(B) \] or equivalently \(f_\#\colon C_*(X,A) \to C_*(Y,B)\) this then induces \(f_*\colon H_*(X,A) \to H_*(Y,B)\).

Subsubsection1.4.2.1Visualising relative homology classes

If \(W^{n+1}\) is a connected oriented compact manifold then we'll show that \(H_{n+1}(W,\partial W) \cong \ZZ = \langle [W,\partial] \rangle\) (where the \(\partial\) notation means relative to boundary). So given \(f\colon (W, \partial) \to (X,A)\) we get \(f_*([W,\partial W]) \in H_{n+1}(X,A)\).

Example1.4.12
Let \(X = \RR^3\) and \(A= S^1\) then \(W = D^2\) defines a class in \(H_2(\RR^3,S^1)\) (the boundary of \(W\) lies inside of \(A\)).

Subsection1.4.3Good pairs

Definition1.4.13Good pair
\((X,A)\) is a good pair if
  1. \(\exists U \subset X\) open and \(A \subset U\)
  2. \(\exists \pi \colon U \to A\) with \(\pi|_A = \id_A\).
  3. \(\pi \sim 1_{(U,A)}\) as maps of pairs from \((U,A)\) to itself.
(i.e. \(A\) is a deformation retract of \(U\)).
Example1.4.14Good pairs
  1. \((\text{smooth manifold}, \text{closed submanifold})\).
  2. \((\text{simplicial complex}, \text{subcomplex})\).
Example1.4.15Not good pairs
  1. \((\RR, \QQ)\).
  2. Letting \(H\subset \RR^2\) be the Hawaiian earring then \((\RR^2, H)\) is not a good pair.
Exercise 1.4.17
The composite map \(\phi\) in \[ \tilde H_*(X) \to H_*(X) \to H_*(X,*) \] is an isomorphism. i.e. \[ H_*(X,*) = \begin{cases} H_*(X), &* \gt 0,\\H_0(X)/\ZZ, &* \le 0.\end{cases} \]
Example1.4.20Chains generating \(H_n(S^n, *)\)
  1. Choose \(f\colon \Delta^n \to \Delta^n\) a homeomorphism and let \(e\colon \Delta^n \to \Delta^n\) be the identity map. Then \(a_{n-1} = f_\#(de) \in C_*(S^{n-1})\) and we have \(d a_{n-1} = df_\#(de) = f_\# (d^2 e) = 0\) and so \(a_{n-1}\) is closed.
  2. Choose \(g\colon (D^n/S^{n-1}, S^{n-1}/S^{n-1}) \to (S^n, *)\). Now let \(b_n = g_\#(f_\#(e)) \in C_n(S^n,*)\). Then \(db_n = g_\#(a_{n-1}) \in C_*(*)\) implies \(b_n\) is closed in \(C_*(S^n, *)\).
    For example if \(n = 2\) we are crushing the boundary of the \(2\)-simplex to a point.
Definition1.4.22Wedge product
If \((X_i, p_i)\) are pointed spaces the wedge product\[ \bigvee_{i\in I} (X_i,p_i) \text{ is } \coprod_{i\in I} X_i\Big/\{p_i : i\in I\}. \] If \(X_i\) is such that for any \(p,q\in X_i\) there exists a homeomorphism \(f\colon X_i \to X_i\) with \(f(p) = q\) we can drop \(p_i\) from the notation (for example in the case of \(X_i\) a connected manifold we can do this).
Example1.4.23
Example1.4.25
\[ H_*(S^1 \vee S^2) = \begin{cases}\ZZ, &* = 0,1,2\\0, &\text{otherwise}.\end{cases} \]