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Assume that $X$ is a path connected hausdorff topological space. Let $\alpha\in C^{1}(X,\mathbb{R})$ and $\beta\in C^{n}(X,\mathbb{R})$ be cochains in real singular cohomology. Asume that $\alpha \smile \beta=0$.

Is there a $n-1$ cochain $\gamma$ with $\beta=\alpha \smile \gamma$ ?

This is motivated by the similar situation, with affirmative answer, in De Rham cohomology and differential forms. In fact this smooth version is used to define the Godbillon Vey invariant for a codimension one foliation.

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    $\begingroup$ What if $\alpha=0$. In fact, it's completely unclear what the relation between the two conditions is as, in general, $\alpha\smile\alpha\ne0$ for a $1$-cochain (or even cocycle) $\alpha$. $\endgroup$
    – Alex Degtyarev
    Commented Jan 28, 2015 at 8:18
  • $\begingroup$ @AlexDegtyarev befor this post I was thinking to singular cohomological version of "Non vanishing one form" in term of cochain. Is it a good idea to define $\alpha$ vanish at a point $z\in X$ if for every curve passing $z$, the pull bach cochain is vanishing at a point of interval?(with an appropriate definition, for the intervalcase). In fact My main motivation was the following: Let $\alpha$ be a non vanishing 1-cochain on compact n dimensional space $X$(not necessarilly manifold) with $\alpha \smile \sigma \alpha=0$. can we divid $X$ to disjoint union of n-1 dim subspace $L_{t}$.... $\endgroup$
    – Ali Taghavi
    Commented Jan 28, 2015 at 8:40
  • $\begingroup$ such that the restriction of $\alpha$ to each $L_{t}$ is identically zero? $\endgroup$
    – Ali Taghavi
    Commented Jan 28, 2015 at 8:42
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    $\begingroup$ The corresponding question in cohomology also has a negative answer. If $X$ is the plane with $n$ points removed, then $H^*(X)$ is freely generated over $\mathbb{Z}$ together with classes $\alpha_1,\dotsc,\alpha_n$ satisfying $\alpha_i\alpha_j=0$ for all $i$ and $j$. $\endgroup$
    – Neil Strickland
    Commented Jan 28, 2015 at 10:37
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    $\begingroup$ @AlexDegtyarev: note OP is taking coefficients in $\mathbb{R}$. $\endgroup$
    – Thomas Rot
    Commented Jan 28, 2015 at 11:42

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