Intersection of topological manifolds. A condition for the intersection of two smooth manifolds to be a smooth manifold is that they intersect transversally. Is this only an obstruction because of the smooth structure?
Question: Is the intersection of two topological manifolds always a topological manifold? If not, are there conditions which can be added to the manifold(Not the smooth+transversal intersection) so that it is true?
 A: No, the general intersection of topological manifolds need not be another topological manifold.
For instance, suppose I have two two-dimensional manifolds $M_1$ and $M_2$. Here $M_1$ is the $xy$-plane in $\mathbb{R}^3$, and $M_2$ is the union of the unit sphere $S_a$ with center $a=(0,0,1/2)$ and of the unit sphere $S_b$ with center $b=(3,0,1)$. If you want $M_2$ to be connected, you can connected $S_a$ to $S_b$ via a tube that doesn't intersect the $xy$-plane.
As you can see, the intersection $M_1\cap M_2$ consists of a circle in the $xy$-plane (in $\mathbb{R}^3$) with center $(0,0,0)$ together with the point $(3,0,0)$. It is some object that is both one- and two-dimensional, which cannot be a manifold.
A: Actually, the intersection of two topological manifolds in a Euclidean space can be almost anything. Here's an example to show how bad things can get. Let $C$ be any closed subset of $\mathbb R^n$ whatsoever, and let $f\colon \mathbb R^n \to \mathbb R$ be the function 
$$
f(x) = \operatorname{dist}(x,C).
$$
Thus $f$ is continuous, and $f(x)=0$ if and only if $x\in C$.
Let $M\subset\mathbb R^{n+1}$ be the graph of $f$, and let $N\subset\mathbb R^{n+1}$ be the graph of the zero function (i.e., $N = \mathbb R^n\times \{0\}$). Then $M$ and $N$ are both topological submanifolds of $\mathbb R^{n+1}$, and $M\cap N = C \times \{0\}$.
