# Prove that the Cartesian product of two topological manifolds is a topological manifold.

I need help on the following problem, any responses would be greatly appreciated:

Let $M$ be a topological $m$-manifold and $N$ be a topological $n$-manifold. Prove that $M \times N$ is a topological $(m + n)$-manifold.

I know how to prove that $M \times N$ is Hausdorff and that it has a countable base for its topology (namely the product topology). However, I am unsure on how to prove that it is locally Euclidean I.e. It has an open cover by sets that are homeomorphic to open subsets of $\mathbb{R}^{m+n}$.

• You know that $\mathbb{R}^{n+m}$ is (homeomorphic to) $\mathbb{R}^n \times \mathbb{R}^m$ with the product topology? – Daniel Fischer Oct 23 '15 at 8:57

Every point $(p,q) \in M\times N$ has a product open set $U\times V$ where $U \subseteq M$ and $V \subseteq N$ are each open with $p \in U, q \in V$. Observe that we can pick $U$ and $V$ such that $U$ is homeomorphic to $\mathbb{R}^m$ and $V$ is homeomorphic to $\mathbb{R}^n$. If $f:U \to \mathbb{R}^m$ and $g:V \to \mathbb{R}^n$ are homeomorphisms, we can define $F: U \times V \to \mathbb{R}^{m+n}$ given by $F(x,y) = (f(x), g(y))$. We see immediately that $F$ is continuous (each component is continuous), and since $f^{-1}$ and $g^{-1}$ exist and are continuous, we find that $F^{-1}$ exists, is continuous, and is given by $F^{-1}(x,y) = (f^{-1}(x), g^{-1}(y))$.
• How would I convert this into a proof using open covers? My definition for locally euclidean is that $M$ is locally $E^{m}$ if it has an open cover by sets that are homeomorphic to open subsets of $E^{m}$. – jackwo Oct 23 '15 at 13:59
• @jackwo Normally I see the definition of locally Euclidean is that each point has a neighborhood homeomorphic to $\mathbb{R}^n$. Note that if $M$ is your manifold (of dimension $n$), $p \in M$ and $U_p$ a neighborhood of $p$ homeomorphic to $\mathbb{R}^n$, then wouldn't you agree that $\bigcup_{p \in M} U_p = M$? – Mnifldz Oct 23 '15 at 17:44
• Yes that makes perfect sense, but why can we pick the open sets $U$, $V$ so that they are homeomorphic to $\mathbb{R}^{m}$ and $\mathbb{R}^{n}$, respectively? – jackwo Oct 24 '15 at 1:27
• @jackwo $M$ and $N$ are manifolds. Let $p \in M, q \in N$. Then by definition of $M$ and $N$ being locally Euclidean there exist open sets $U \subseteq M$ containing $p$, and $V \subseteq N$ containing $q$ such that $U \approx \mathbb{R}^m$ and $V \approx \mathbb{R}^n$. Now $U\times V$ is open in $M\times N$ in the product topology, and since a product of homeomorphisms is again a homeomorphism, we have that $U \times V \approx \mathbb{R}^{m+n}$. Does this answer your question? I'm not sure where you're getting stuck. – Mnifldz Oct 24 '15 at 6:29
• @jackwo Observe that homeomorphisms preserve the number of connected components between topological spaces, thus we can restrict our case to when $U \subseteq M$ is connected. There is a proposition that shows that the map $f: \mathbb{B}^n \to \mathbb{R}^n$ (where $\mathbb{B}^n$ is the open unit ball) given by $f(x) = \frac{x}{1 - ||x||}$ is a homeomorphism. Now if $g:U \to V \subseteq \mathbb{R}^n$ is a homeomorphism, there is an open ball in $V$ such that $g^{-1}$ restricts to a homeomorphism of a neighborhood of the point in question in $U$. – Mnifldz Oct 26 '15 at 4:54