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?
[Math] Projection map is smooth
smooth-manifolds
Related Solutions
$(\Rightarrow)$ If $M^m\times N^n$ is orientable, there is a volume form $\eta\in\Omega^{n+m}(M\times N)$. For any fixed point $q\in N$, take a basis $\{w_1,...,w_n\}$ for $T_qN$. Now define $\omega\in\Omega^m(M)$ by: $$\omega(X_1,...,X_m):=\eta_{(\cdot,q)}\left(X_1,...,X_m,w_1,...,w_n\right)$$
We will prove that $\omega$ is a volume form. For a fixed $p$, take a basis $\{v_1,...,v_m\}$ for $T_pM$. Using the identification $T_{(p,q)}M\times N\equiv T_pM\oplus T_qN$, then $\{v_1,...,v_m,w_1,...,w_n\}$ is a basis for $T_{(p,q)}M\times N$. Since $\eta$ is a volume form, $\omega_p(v_1,...,v_m)=\eta_{(p,q)}(v_1,...,v_m,w_1,...,w_n)\neq 0$, which means $\omega$ is a volume form, so $M$ is orientable. By a similar argument, $N$ is orientable.$_\blacksquare$
$(\Leftarrow)$ If $M,N$ are orientable, there are volume forms $\omega\in\Omega^m(M), \sigma\in\Omega^n(N)$. For the natural projections $\pi_M:M\times N\to M$ and $\pi_N:M\times N\to N$, define: $$\eta:=\pi_M^*(\omega)\wedge \pi_N^*(\sigma)\in\Omega^{n+m}(M\times N)$$
We will prove $\eta$ is a volume form. For a fixed $x=(p,q)\in M\times N$, take basis $\{v_1,...,v_n\}$ for $T_pM$ and $\{w_1,...,w_n\}$ for $T_qN$. Because $\omega$ and $\sigma$ are volume forms, we have $\omega_p(v_1,...,v_m)\neq 0$ and $\sigma_q(w_1,...,w_n)\neq 0$. Since $\{v_1,...,v_m,w_1,...,w_n\}$ is a basis for $T_x(M\times N)$, it's enough to check that $\eta_x(v_1,...,v_m,w_1,...,w_n)\neq 0$. Indeed, noticing that $(d\pi_M)_x(v_i)=v_i$, $(d\pi_M)_x(w_j)=0$, $(d\pi_N)_x(v_i)=0$ and $(d\pi_N)_x(w_j)=w_j$, we have: $$\eta_x(v_1,...,v_m,w_1,...,w_n)=\underbrace{\omega_p(v_1,...,v_m)}_{\neq 0}\,\underbrace{\sigma_q(w_1,...,w_n)}_{\neq 0}\neq 0\,\,_\blacksquare$$
The counterexample just shows that two diffeomorphic smooth structures on the same set $X$ do not need to share a common atlas. However, in any case, two diffeomorphic structures cannot be distinguished in the smooth category, in the sense that every true statement in the smooth category still remains true after replacing one structure with the other one.
This also happens in the topological category: it is possible to have two homeomorphic, but different, topological structures on a set $X$. For instance, the two topologies on $\mathbb{R}$ given by $$\mathcal{T}_1 = \{[a, \, + \infty) \; : \; a \in \mathbb{R} \cup \{-\infty\} \}, \quad \mathcal{T}_2 = \{(- \infty, \, a] \; : \; a \in \mathbb{R} \cup \{+\infty\} \}$$ are clearly different. However, the map $$f \colon (\mathbb{R}, \, \mathcal{T}_1) \to (\mathbb{R}, \, \mathcal{T}_2), \quad x \mapsto -x$$ is a homeomorphism, so from the topological point of view the two spaces have exactly the same properties.
Best Answer
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.