# Projection map is smooth

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?

• 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). – Dog_69 May 25 '18 at 22:56
• @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 – Quang Dao May 25 '18 at 23:13
• 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 – 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.
• @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 $y$. – Stefan4024 May 26 '18 at 0:37
• @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? – HerrWarum Feb 2 at 13:15