Source: Madsen & Tornehave, From Calculus to Cohomology
Let $M\subset\Bbb R^k$ be an $n$-dimensional topological manifold. Show the inclusion $M\subset\Bbb R^{k+n}$ is locally flat.
I.e. show that $\forall p\in M$ there exists a homeomorphism $h:U\to V$ with $p\in U$ and $U,V\subset\Bbb R^{n+k}$ open, s. t. that $h(U\cap M) = V\cap\Bbb R^n$.
I have been thinking about this for quite a while now. Especially, does anyone know what's the role of the dimension $n$ here? Since if you would change the question in the title to $M\ldots\subset\mathbb{R}^{k+j}\ldots$. where $j\in\Bbb N$ (and still $\dim M=n\;$) the statement would be false, see, e. g., the Alexander's horned sphere
Thanks for any ideas!
Best Answer
Let $p \in M$ be a point, and $(U, \varphi)$ be a chart neighborhood in $M$, $\varphi : U \to \Bbb R^n$ being the chart homeomorphism. Considering the embedding $M \subset \Bbb R^k$, find a neighborhood $V$ of $p$ in $\Bbb R^k$ such that $V \cap M = U$ and define a continuous extension $\widetilde{\varphi} : V \to \Bbb R^n$, so that $\widetilde{\varphi}|_U = \varphi$.
Now consider the embedding $M \subset \Bbb R^k \times \{0\} \subset \Bbb R^{k+n}$ into the first factor of $\Bbb R^{k+n} = \Bbb R^k \times \Bbb R^n$ and let $B$ be a ball around $p$ in $\Bbb R^{k+n}$ such that $B \cap (\Bbb R^k \times \{0\}) = V$. Define $\Phi : B \to \Bbb R^{k+n}$ by
$$\Phi(x, y) = (\varphi^{-1}(y + \widetilde{\varphi}(x)) - x, y + \widetilde{\varphi}(x))$$
Checking that $\Phi$ is a homeomorphism onto image is a triviality. Note that for any $(x, 0) \in U \times \{0\} \subset M$, $\Phi(x, 0) = (\varphi^{-1}(\varphi(x)) - x, \varphi(x)) = (0, \varphi(x)) \in \{0\} \times \Bbb R^n$, so $\Phi(U \times \{0\}) \subset \{0\} \times \Bbb R^n$. Thus, $M \subset \Bbb R^{k+n}$ is a locally flat submanifold at $p$.
I've decided to make an edit to explain the idea behind the map $\Phi$. For illustration it's easier to imagine a smaller dimensional example of a non locally flat embedding than the Alexander horned sphere. I shall use a properly embedded wild arc $A \subset \Bbb R^3$. We embed $A$ horizontally in $\Bbb R^3 \times \Bbb R^1$.
Our plan is to shift $A \subset \Bbb R^3 \times \{0\}$ vertically up along the $\Bbb R^1$-direction in $\Bbb R^3 \times \Bbb R^1$ so that it "snakes upward" diagonally as we go along the arc. The natural way to do it is to observe the arc admits a homeomorphism $\varphi : A \to \Bbb R$, which we extend by Tietze extension theorem to a map $\widetilde{\varphi} : \Bbb R^3 \to \Bbb R$, and use this to construct a shear homeomorphism
$$\Phi_1 : \Bbb R^3 \times \Bbb R^1 \to \Bbb R^3 \times \Bbb R^1, \; \Phi_1(x, y) = (x, y + \widetilde{\varphi}(x))$$
The advantage of using $\varphi$ along the arc while shearing is that it "unwinds" the many twists in $A$ because $\varphi$ by definition topologically simplifies the arc to the straight real line.
$\hskip1.4in$
What was the point of this construction? The point is now $\Phi_1(A) = \{(x, \varphi(x)) : x \in A\}$ is the graph of $\varphi^{-1} : \Bbb R \to A \subset \Bbb R^3$ over the $\{0\} \times \Bbb R^1$ factor, and graphs of functions are always locally flat submanifolds. Indeed, consider a function $f : \Bbb R^n \to \Bbb R^m$ and $\Gamma_f = \{(x, f(x)) : x \in \Bbb R^n\}$ be the graph in $\Bbb R^n \times \Bbb R^m$. Then another shear homeomorphism
$$\Phi_2 : \Bbb R^n \times \Bbb R^m \to \Bbb R^n \times \Bbb R^m, \; \Phi_2(x, y) = (x, y - f(x))$$
satisfies $\Phi_2(\Gamma_f) \subset \Bbb R^n \times \{0\}$, making $\Gamma_f \subset \Bbb R^n \times \Bbb R^m$ locally flat.
The final homeomorphism $\Phi : \Bbb R^4 \to \Bbb R^4$ such that $\Phi(A) \subset \{0\} \times \Bbb R$, making $A$ locally flat, can be obtained from composing the homeomorphisms $\Phi_1$ and $\Phi_2$, where $\Phi_2$ is done with the Euclidean factors reversed. None of this is special to the example I took, and one can check the formula for $\Phi$ obtained in this manner is exactly what I have above.