这是一份southampton南安普敦大学MATH3080W1-01作业代写的成功案例
To define the cup product we consider cohomology with coefficients in a ring $R$, the most common choices being $\mathbb{Z}, \mathbb{Z}_{n}$, and $\mathrm{Q}$. For cochains $\varphi \in C^{k}(X ; R)$ and $\psi \in C^{\ell}(X ; R)$, the cup product $\varphi \smile \Psi \in C^{k+\ell}(X ; R)$ is the cochain whose value on a singular simplex $\sigma: \Delta^{k+\ell} \rightarrow X$ is given by the formula
$$
(\Phi \smile \Psi)(\sigma)=\Phi\left(\sigma \mid\left[v_{0}, \cdots, v_{k}\right]\right) \psi\left(\sigma \mid\left[v_{k}, \cdots, v_{k+\ell}\right]\right)
$$
where the right-hand side is the product in $R$. To see that this cup product of cochains induces a cup product of cohomology classes we need a formula relating it to the coboundary map:
MATH3080W1-01 COURSE NOTES ：
The cup product formula $(\varphi \sim \psi)(\sigma)=\varphi\left(\sigma \mid\left[v_{0}, \cdots, v_{k}\right]\right) \psi\left(\sigma \mid\left[v_{k}, \cdots, v_{k+\ell}\right]\right)$ also gives relative cup products
$H^{k}(X ; R) \times H^{\ell}(X, A ; R) \longrightarrow H^{k+\ell}(X, A ; R)$
$H^{k}(X, A ; R) \times H^{\ell}(X ; R) \longrightarrow H^{k+\ell}(X, A ; R)$
$H^{k}(X, A ; R) \times H^{\ell}(X, A ; R) \stackrel{\smile}{\longrightarrow} H^{k+\ell}(X, A ; R)$
since if $\varphi$ or $\psi$ vanishes on chains in $A$ then so does $\varphi \cup \psi$. There is a more general relative cup product
$$
H^{k}(X, A ; R) \times H^{\ell}(X, B ; R) \longrightarrow H^{k+\ell}(X, A \cup B ; R)
$$