Let $M_1,M_2 \in \mathbb{R}^n$ be $k$-dimensional manifolds, $ M=M_1 \cup M_2$
It can happen that $M_1 \cap M_2 =\emptyset $ but $M$ is not a $k$-dimensional manifold. Give a counter example.
So I think my example of $M_1= (0,1)\times \{0\}$, $M_2=(1,2)\times \{0\}$ is good as $M$ here doesn't sustain this condition for manifolds:
- There exists a permutation $(i_1,\dots,i_n)$ of $\{1,\dots,n\}$ and a mapping $g:\mathbb{R}^k\to\mathbb{R}^{n-k}$, continuously differentiable near $(x_{0,i_1},\dots,x_{0,i_n})$ such that $$ x\in M \iff g(x_{i_1},\dots, x_{i_k})=(x_{i_{k+1}},\dots,x_{i_n}) \text{ for all $x$ near $x_0$}$$
but my friend here doesn't agree... can anyone help settle this?
EDIT: Can anyone give a counter-example then?