Fix $n,m$ non-negative integers such that $m<n$. Show that if $X$ is a manifold of dimension $n$ and $x \in X$ then $X$ contains a submanifold $Y$ of dimension m such that $x \in Y$.
My attempt: Fix $x \in X$. Then $x \in U$ where $(U,f)$ is a chart on $X$. I now attempt to show:
- $(U, \pi \circ f)$ is a chart compatible with $X$, where $\pi : \mathbb{R^n} \to \mathbb{R^m}$ is the standard projection (which omits the first $n-m$ coordinates).
- That $(\pi \circ f)(U \cap Y) = (\pi \circ f)(U) \cap A_{m}$ where $A_{m}$ is an affine subspace.
While trying to prove (1), I run into the problem that $\pi$ is not bijective. But I'm not completely sure. My thinking is that the projection map on an arbitrary open set of $\mathbb{R^n}$ can't possibly be bijective since for example injectivity is violated. Take $\mathbb{R^2}$ for example; $(1,2)$ and $(2,2)$ both have the same image. When looking for similar questions online I read some conflicting answers but that may be me misinterpreting the context of those answers. I just want to make sure.
My questions:
- Is it true that $\pi$ is not bijective?
- If so does that mean that I need to find some other way to do this or is there a way of making this construction work?