Is the Kunneth formula for de Rham cohomology true on the cochain level

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 $$\sum\pi_1^{\ast}(w_1) \land \pi_2^{\ast}(w_2)$$?

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).
• 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 $\mathbb R^n \times \mathbb R$ – Keith May 25 at 4:38
• 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. – Eric Wofsey May 25 at 4:45
• 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 $dx_{i_j}$. – Eric Wofsey May 25 at 5:16
• 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. – Eric Wofsey May 25 at 5:21