# Cup product graded commutative elementary example

I'm trying to get some intuition about the cup product for singular cohomology and was trying to check whether the graded commutativity $$\varphi\smallsmile \psi = (-1)^{kl}\psi \smallsmile \varphi$$ for $$\varphi \in H^k(X)$$ and $$\psi \in H^l(X)$$ is obvious for small $$k,l$$.

The most elementary case is when $$k,l$$ are both zero. Then $$\varphi \smallsmile \psi$$ is simply the pointwise product (where we may view the cocycles $$\varphi$$ and $$\psi$$ as functions on $$X$$) and the formula is immediate.

The next case is the product of a $$0$$-cocycle $$\varphi$$ with a $$1$$-cocycle $$\psi$$. The $$1$$-cocycle $$\varphi \smallsmile \psi$$ assigns to any $$1$$-simplex, which is to say curve, $$f:\sigma_1 \to X$$ the value $$\varphi(f(1,0)) \psi(f)$$. Similarly $$\psi \smallsmile \varphi$$ is the $$1$$-cocycle which assigns $$\varphi(f(0,1))\psi(f)$$ to any curve $$f$$. So we have that $$\varphi \smallsmile \psi = \psi \smallsmile \varphi$$ assigns the value $$(\varphi(f(1)) - \varphi(f(0)))\psi(f)$$ to $$f$$. That $$\psi\smallsmile \varphi - \varphi \smallsmile \psi = 0$$ in cohomology is now equivalent to this value only depending on the endpoints of the curve but I don't see why this is true.

The short answer is that $$f(0)$$ and $$f(1)$$ are homologous via $$f$$, so that $$\varphi(f(0))-\varphi(f(1))$$ is zero for any $$f$$.
Note that the value of a $$1$$-cocycle on a $$1$$-simplex $$f$$ is turned into its opposite if you reverse the orientation of $$f$$. Let's denote by $$-f$$ the simplex $$f$$ with opposite orientation.
$$\begin{eqnarray}(\varphi\smallsmile \psi)(f)&=&-(\varphi\smallsmile \psi)(-f)\\ \varphi(f(0))\psi(f)&=&-\varphi(f(1))\psi(-f)\end{eqnarray}$$ Now use again that $$\psi(-f)=-\psi(f)$$ to conclude.
• I don't understand why it follows from $\psi$ being a cocycle that $\psi(f)$ only depends on the endpoints of $f$. This seems to assume all cocycles are coboundaries. – A. Van Werde Nov 30 '19 at 9:10