§ Cup product [TODO ]

• We need an ordered simplex, so there is a total ordering on the vertices. This is to split a chain apart at number $k$.
• Can always multiply functions together. This takes a $k$ chain $\xi$ and an $l$ chain $\eta$ and produces $\xi \cup \eta$ which is a $k + l$cochain. The action on a $(k+l)$ chain $\gamma$ acts by $(\xi \cup \eta)(\gamma) \equiv \xi (\gamma_{\leq k}) \cdot \eta (\gamma_{> k})$.
• No way this can work for chains, can only ever work for cochains.
• This cup product "works well" with coboundary. We have $\partial (\xi \cup \eta) \equiv (\partial \xi \cup \eta) + (-1)^k (\xi \cup \partial \eta)$.
• We get cocycle cup cocyle is cocycle.
• Similarly, coboundary cup cocycle is coboundary.
• Simiarly, cocycle cup coboundary is coboundary.
• The three above propositions imply that the cup product descends to cohomology groups.
• The algebra of cohomology (cohomology plus the cup product) sees the difference between spaces of identical homology!
• The space $S^1 \times S^1$ have the same homology as $S^2 \cap S^1 \cap S^1$. Both have equal homology/cohomology.
• However, we will find that it will be zero on the torus and non-zero on other side.
• The cup product measures how the two generators are locally product like. So if we pick two generators on the torus, we can find a triangle which gives non-zero