How to use intersections to deduce facts about the cohomology ring of $S_1 \vee S_2$

Question

Let $[i]$ and $[j]$ be the inclusions maps of the homology fundamental classes of $S_1$ and $S_2$ respectively inside $S_1 \vee S_2$.

I want to show that the cup product of a cocycle that is zero on $C_*(S_1) \hookrightarrow C_*(S_1 \vee S_2)$ and a cocycle that is zero on $C_*(S_2) \hookrightarrow C_*(S_1 \vee S_2)$, is zero.

Is there a way to show this by showing that the intersection of any simplex contained in $S_1$ and $S_2$ is degenerate?

Answer 1

Let $a$ be a $1$-cocycle on $S^1 \vee S^2$ that is zero on $C_*(S^1)$, and let $b$ be a $1$-cocycle on $S^1 \vee S^2$ that is zero on $C_*(S^2)$. I do not think it is true that $a \cup b$ is necessarily zero. The reason is that a $2$-chain $c$ in the wedge sum need not have its front face in $S^1$ and its back face in $S^2$. It could be that $c$ has its front face in $S^2$ and its back face in $S^1$. Then $(a \cup b)(c)$ is not zero.

Answer 2

The cup product on relative cohomology looks like $$ \smile: H^k(X, A) \times H^l(X, B) \to H^{k+l}(X, A \cup B).$$ If you have cocylces $\alpha \in H^k(S^1 \vee S^2, S^1)$ and $\beta \in H^l(S^1 \vee S^2, S^2)$, then $\alpha \vee \beta$ lives in $H^{k+l}(S^1 \vee S^2, S^1 \vee S^2) = 0$.

See Hatcher, page 209.