I want to prove that $\pi: M \times N \rightarrow M$ is smooth where $M, N$ are smooth manifolds. Let $(U \times V, \phi \times \varphi)$ be a chart on $M \times N$, and $(W, \psi)$ be another chart on $M$, then $\psi \circ \pi\circ (\phi \times \varphi)^{-1}$ is smooth. I know that function $\psi \circ \pi\circ (\phi \times \varphi)^{-1} = \psi \circ \phi^{-1}$, which is smooth since $M$ is smooth, but those two have different domains, so how can they be equal?

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    $\begingroup$ The domain of both function is $\mathbb R$. Furthermore you have a mistake: the function indsde the brackets is $\phi\times\varphi$, and not $\phi\circ \varphi$ (such a composition doesn't exist). $\endgroup$ – Dog_69 May 25 '18 at 22:56
  • $\begingroup$ @Dog_69: the domain of $\psi \circ \phi^{-1}$ is $\phi (U \bigcap W)$, while the domain for the other function is $\phi \times \varphi ((U \times V)\bigcap (W \times N))$, which are not equal to me $\endgroup$ – Quang Dao May 25 '18 at 23:13
  • $\begingroup$ Since both $\phi^{-1}$ and $\varphi^{-1}$ are defined over $\mathbb R^n$ I was thinking about the product as the map $x\longmapsto(\phi^{-1}(x),\varphi^{-1}(x))$ but not. The product must be a map defined over $\mathbb R^{2n}$, you're right. Thanks $\endgroup$ – Dog_69 May 26 '18 at 10:09

The problem is that $\psi \circ \pi\circ (\phi \times \varphi)^{-1} = \psi \circ \phi^{-1}$ isn't valid, since as you've noticed the domain don't coincide. What is true is the following:

$$\psi \circ \pi\circ (\phi \times \varphi)^{-1}(x,y) = \psi \circ \pi(\phi^{-1}(x),\varphi^{-1}(y)) = \psi(\phi^{-1}(x)) = \psi \circ \phi^{-1}(x)$$

which is a smooth map from $\mathbb{R}^{m+n} \to \mathbb{R}^m$, as $\psi \circ \phi^{-1}$ is a smooth map itself.

  • $\begingroup$ So you are using the fact that smoothness is local? $\endgroup$ – Quang Dao May 26 '18 at 0:31
  • $\begingroup$ @PatrickLeung From the answer you can prove that the function is smooth at every point and hence smooth on the whole domain. But also you can conclude it immediately, as $\psi \circ \phi$ changes in a smooth manner. In fact the Jacobian of $\psi \circ \pi (\phi \times \varphi)^{-1}$ is $m \times (m+n)$ matrix, whose left $m \times m$ part is the Jacobian of $\psi \cdot \phi^{-1}$ and the right $m \times n$ part is all zeroes, as the value of the function doesn't depend on $y$. $\endgroup$ – Stefan4024 May 26 '18 at 0:37
  • $\begingroup$ So in general, you compute the image of some point in the domain, and if the image is smooth, then the function is smooth? $\endgroup$ – Quang Dao May 26 '18 at 1:16
  • $\begingroup$ The image can't be smooth, as the image is the set of all points the function attains. On the other hand if you think about the right hand side agove, then yes. After all the right hand side is the value of the function, which is what you use to determine whether one is smooth or not $\endgroup$ – Stefan4024 May 26 '18 at 1:19
  • $\begingroup$ @Stefan4024, Could we not have chosen $(W,\psi)$ and $(U,\phi)$ to be the same chart on $M$, since they are charts at the same point? $\endgroup$ – HerrWarum Feb 2 at 13:15

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