# Product of submanifolds is a submanifold

Suppose that $X_1$ is an $n_1$-dimensional submanifold of $\mathbb{R}^{N_1}$, and $X_2$ is an $n_2$-dimensional submanifold of $\mathbb{R}^{N_2}$. Prove that $X_1\times X_2\subseteq \mathbb{R}^{N_1}\times\mathbb{R}^{N_2}$ is an $(n_1+n_2)$-dimensional submanifold of $\mathbb{R}^{N_1}\times\mathbb{R}^{N_2}$.

Suppose $q=(r,s)\in X_1\times X_2$, where $r\in X_1$ and $s\in X_2$.

By definition of a submanifold, there exists a neighborhood $V_r\in\mathbb{R}^{N_1}$, and an open subset $U_1\in\mathbb{R}^{n_1}$ and a diffeomorphism $\phi_1:U_1\rightarrow X_1\cap V_r$. Similarly, there exists a neighborhood $V_s\in\mathbb{R}^{N_2}$, and an open subset $U_2\in\mathbb{R}^{n_2}$ and a diffeomorphism $\phi_2:U_2\rightarrow X_2\cap V_s$.

(Edited to take user103402’s comment into account) Now, I could take the neighborhood $V_r\times V_s\in\mathbb{R}^{N_1}\times\mathbb{R}^{N_2}$ and the open subset $U_1\times U_2\in\mathbb{R}^{n_1}\times\mathbb{R}^{n_2}$. I want to define the map $\phi:U_1\times U_2\rightarrow(X_1\times X_2)\cap (V_r\times V_s)$. Note that $$(X_1\times X_2)\cap (V_r\times V_s)=(X_1\cap V_r)\times(X_2\cap V_s).$$

So the map $\phi$ is simply given by $\phi(u_1,u_2)=(\phi_1(u_1),\phi_2(u_2))$ for $u_1\in U_1,u_2\in U_2$.

Now, why is $\phi$ a diffeomorphism between $U_1\times U_2$ and $(X_1\cap V_r)\times(X_2\cap V_s)$? Even though $\phi_1$ is a diffeomorphism between $U_1$ and $X_1\cap V_r$, and $\phi_2$ is a diffeomorphism between $U_2$ and $X_2\cap V_s$, it does not seem clear to me that $\phi$ should be a diffeomorphism.

#### Solutions Collecting From Web of "Product of submanifolds is a submanifold"

If it is not clear why $\phi$ is a diffeomorphism, we should examine bijectivity and smoothness. Let $\phi(u_1,u_2)=\phi(v_1,v_2)$ for $(u_1,u_2),(v_1,v_2)\in U_1\times U_2$. Then, $$(\phi_1(u_1),\phi_2(u_2))=(\phi_1(v_1),\phi_2(v_2))\Leftrightarrow\phi_1(u_1)=\phi_1(v_1)~\text{and}~\phi_2(u_2)=\phi_2(v_2)$$
and since $\phi_i$, $i=1,2$ is in particular one-to-one it follows that $u_1=v_1$ and $u_2=v_2$, so $(u_1,u_2)=(v_1,v_2)$. To see $\phi$ is onto we again use the fact that $\phi_i$ is onto, that is $\forall x_i\in X_i\cap V_j\exists u_i\in U_i$ s.t. $\phi_i(u_i)=x_i$. Combining this for $\phi_1$ and $\phi_2$, we see that this also holds for all $(x_1,x_2)\in(X_1\cap V_r)\times(X_2\cap V_s)$, i.e. that $\phi$ is onto.

Smoothness is also rather immediate: $\phi$ is smooth since its component functions $\phi_1$ and $\phi_2$ are smooth.

So basically, taking the cartesian product preserves every property for charts.