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The kunneth formula gives that $H^k(X \times Y) = \bigoplus_{i+j = k} H^{i}(X) \otimes H^{j}(Y)$, where $X$ and $Y$ are both manifolds. I wonder whether this is true on the cochain level. More specifically we have $\pi_1$, $\pi_2$ projections onto $X$ and $Y$, is it true that every $k$ form on $X\times Y$ is of the form $\sum\pi_1^{\ast}(w_1) \land \pi_2^{\ast}(w_2)$?

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No, this is false. The easiest case to see it is $k=0$: then you would be claiming that every smooth function $f$ on $X\times Y$ can be written as a finite sum of functions $(g_i\circ\pi_1)\cdot (h_i\circ\pi_2)$ for smooth functions $g_i$ and $h_i$ on $X$ and $Y$ respectively. But this is false; for instance, writing $f_x(y)=f(x,y)$, this implies that the functions $f_x$ span a finite-dimensional vector space (since they are all linear combinations of the $h_i$), which is not true for all $f$ (exercise: find an explicit smooth function $f$ on $\mathbb{R}\times\mathbb{R}$ such that infinitely many of the functions $f_x$ are linearly independent).

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  • $\begingroup$ But in Bott-Tu's algebraic topology page 34, they claim that every form on $\mathbb R^n \times \mathbb R$ is a uniquely a linear combination of $\pi^{\ast} \phi f$ and $\pi^{\ast} \phi fdt$ where $\pi$ is the projection onto $\mathbb R^n$ and $w$ is a form on $\mathbb R^n$, $f$ a function on $\mathbb R^n \times \mathbb R$ $\endgroup$ – Keith May 25 at 4:38
  • $\begingroup$ The statement you asked about originally includes no such function $f$. Note that crucially $f$ is a function on $\mathbb{R}^n\times\mathbb{R}$, rather than a function on just one of the coordinates. $\endgroup$ – Eric Wofsey May 25 at 4:45
  • $\begingroup$ Would you please explain that why it would be true with such a function? $\endgroup$ – Keith May 25 at 4:52
  • $\begingroup$ A form on $\mathbb{R}^{n+1}$ is just a sum of smooth functions times expressions of the form $dx_{i_1}dx_{i_2}\dots dx_{i_k}$. Each of those terms is either of the form $\pi^*\phi f$ or $\pi^*\phi f dt$, depending on whether $dt$ is one of the $dx_{i_j}$. $\endgroup$ – Eric Wofsey May 25 at 5:16
  • $\begingroup$ It's not clear what Bott and Tu mean when they claim this decomposition is "unique", and it is not actually unique in most obvious senses. But you don't actually need uniqueness: you just need $K$ to be well-defined and linear, and you can achieve that by always using the decomposition given by the coordinate forms as in my previous comment. $\endgroup$ – Eric Wofsey May 25 at 5:21

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