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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?

2 Answers 2

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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.

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    This is expected: The information we have is that the intersection of these two cycles must be a point - but this is 'not degenerate enough for one-cycles'. In the setting of my quetsion one expects the the cup product of a 1-cycle supported on $S_1$ and a 2-cycle supported on $S_2$ to be zero because the 'poincare dual' to the intersection of poincare dual cycles, would have to be in degree 2, while the cup product of the two cocycles would be in degree 3., so you would expect both to be zero. – 2017-02-04
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    On the other hand, the cup product of a 1-cycle supported on $S_1$ and a $1-cycle$ supported on $S_2$ is not expected to be zero by this reasoning, because the intersection will be in degree zero, the 'poincare dual' to this will be in degree 2, and this is the degree that the cup product is supposed to be in, so one cannot 'conclude' that the cup product of these one-cycles is zero. – 2017-02-04
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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.

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    Thank you for this answer. I am familiar with the cup product calculation. I am trying to better my intuition on intersections. – 2017-02-04
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    The intersection picture of cohomology relies on Poincare duality, which only applies to manifolds, right? $S^1 \vee S^2$ is not a manifold. Try working on a surface of genus 2 or something. – 2017-02-04
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    Trying to think of a manifold with a boundary that is homotopy equivalent to $S^1 \vee S^2$. – 2017-02-09
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    @user062295 yeah you can do that. Say a solid torus minus an interior open ball. You can try to think about intersections there – 2017-02-10