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

Section2.4Cup product

Motivation: if \(k\) is a field \begin{align*} H^*(X\times X; k) &\cong (H_*(X\times X; k))^*\\ &\cong (H_*(X; k)\otimes H_*(X; k))^*\\ &\cong H^*(X; k)\otimes H^*(X; k) \end{align*}

Definition2.4.1Cup product
Suppose \(a\in C^i(X)\) and \(b\in C^j(X)\). If \(\sigma\colon \Delta^{i+j} \to X\) then we define \(a\smile b\in C^{i+j}(X)\) by \[(a\smile b)(\sigma) = a(\sigma\circ F_i^+)b(\sigma\circ F_j^-)\] where \begin{align*} F_i^+\colon \Delta^i &\to \Delta^{i+j}\\ (x_0,\ldots,x_i) &\mapsto (x_0,\ldots,x_i,0,\ldots,0) \end{align*} and \begin{align*} F_j^-\colon \Delta^j &\to \Delta^{i+j}\\ (x_0,\ldots,x_j) &\mapsto (0,\ldots,0,x_0,\ldots,x_j). \end{align*}

Subsection2.4.1Digression: De Rahm Cohomology

Let \(M\) be a smooth manifold and \(\omega\in \Omega^k(M)\). If \(\sigma\colon\Delta^k \to M\) is a smooth we define \(\omega(\sigma) = \int_\sigma \omega\). Stokes' theorem then gives us that \(d\omega (\sigma) = \omega (d\sigma)\). Where on the left \(d\colon \Omega^k(M) \to \Omega^{k+1}(M)\) is the differential map and on the right \(d\) is the boundary map in the sense of (smooth) singular chains. I.e. the two \(d\)'s are dual to each other on \(C_*^{\text{smooth}}(M)\).

Example2.4.6
  1. \(H^*(S^n) = \langle 1, X_n\rangle\) where \(\langle X_n\rangle = H^n(S^n)\) and we have \begin{align*} 1\smile X_n &= X_n,\\ X_n\smile 1 &= X_n,\\ 1\smile 1 &= 1,\\ X_n\smile X_n &= 0 \in H^{2n}(S^n) = 0. \end{align*}
  2. For arbitrary \(X,Y\) \[C_*(X\amalg Y) \cong C_*(X) \oplus C_*(Y)\] which implies \[C^*(X\amalg Y) \cong C^*(X) \oplus C_*(Y)\] \[\implies H^*(X\amalg Y) \cong H^*(X) \oplus H_*(Y)\text{ as rings}\] with \[ (\alpha_1 \oplus \alpha_2) \smile (\beta_1 \oplus \beta2) = (\alpha_1 \smile \beta_1)\oplus (\alpha_2 \smile \beta_2).\]
  3. Considering \(X\vee Y\) we have a map \(p\colon X\amalg Y \to X \vee Y\) and so Mayer-Vietoris for \(X \vee Y\) gives \[ \cdots \to H^*(X\vee Y) \xrightarrow{p^*} H^*(X)\oplus H^*(Y) \xrightarrow{i^*} H^*(\text{pt}) \to \cdots\] Now assuming \(X\) and \(Y\) are path connected \(p^*\) is injective and an isomorphism in grading \(\gt 0\). In grading 0 we have \[H^0(X \vee Y) \to H^0(X) \oplus H^0(Y) \to H^0(\text{pt})\] with \(1 \mapsto 1\oplus 1\) and \(a\oplus b \mapsto a-b\). And so as a ring we have \[ H^*(X\vee Y) \cong \ker i^* \subset H^*(X)\oplus H^*(Y)\] where elements in grading \(\gt 0\) look like \((\alpha,\beta)\) and in grading 0 look like \((n,n)\in \ZZ \times \ZZ\).
  4. \(S^2 \vee S^4\), \(H^*(S^2\vee S^4) = \langle 1, X_2, X_4\rangle\) with \(X_2^2 = 0\) since \(X_2^2 = 0\) in \(H^*(S^2)\).

Subsection2.4.2Exterior Product

Let \(\pi_1 \colon X\times Y \to X\) and \(\pi_2 \colon X \times Y \to Y\).

Definition2.4.7Exterior product
If \(\alpha \in H^i(X)\), \(\beta\in H^j(X)\) then their exterior product is \(\alpha\times \beta = \pi_1^*(\alpha) \smile \pi_2^*(\beta) \in H^{i+j}(X\times Y)\).