5
$\begingroup$

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 $\mathbb{R}^{m+n}$.

$\endgroup$
  • 6
    $\begingroup$ You know that $\mathbb{R}^{n+m}$ is (homeomorphic to) $\mathbb{R}^n \times \mathbb{R}^m$ with the product topology? $\endgroup$ – Daniel Fischer Oct 23 '15 at 8:57
7
$\begingroup$

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 $F^{-1}(x,y) = (f^{-1}(x), g^{-1}(y))$.

$\endgroup$
  • $\begingroup$ 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 $E^{m}$. $\endgroup$ – jackwo Oct 23 '15 at 13:59
  • $\begingroup$ @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 $\bigcup_{p \in M} U_p = M$? $\endgroup$ – Mnifldz Oct 23 '15 at 17:44
  • $\begingroup$ 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? $\endgroup$ – jackwo Oct 24 '15 at 1:27
  • 1
    $\begingroup$ @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. $\endgroup$ – Mnifldz Oct 24 '15 at 6:29
  • 1
    $\begingroup$ @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 $U$. $\endgroup$ – Mnifldz Oct 26 '15 at 4:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.