Applying Brower's Theorem for Invariance of Domain I'm working through Manifolds and Differential Geometry by Jeffrey M. Lee. In a topological manifold, we have that every point is in an open set which is homeomorphic to $\mathbb{R}^n$ for some $n$. However, I am worried that the dimension $n$ will depend on the point and the open set chosen, or even worse, that we could have homeomorphisms from the same neighborhood to two different dimensional euclidean spaces. To address this, Lee mentions the following theorem:
The image of an open set $U \subset \mathbb{R}^n $ under a continuous injective map $f: U \rightarrow \mathbb{R}^n$ is open and $f$ is a homeomorphism from $U$ to $f(U)$. It follows that if $U \subset \mathbb{R}^n$ is homeomorphic to $V \subset \mathbb{R}^m$ then $m=n$.  
However, I don't see exactly how to apply this theorem to address my concern. 
 A: As I mentioned in my comment, Brouwer's theorem on invariance of domain needs the additional hypothesis that $f$ is injective. With that modification, here's how to use the theorem to rule out the phenomena you're worried about. 
Suppose $M$ is a topological space, $p\in M$, and $U$ and $V$ are neighborhoods of $p$ on which there exist continuous maps $f\colon U\to \mathbb R^n$, $g\colon V\to \mathbb R^m$ that are homeomorphisms onto open subsets. After replacing both $U$ and $V$ by $U\cap V$, we can assume $U=V$.  Assume for contradiction that $m\ne n$; then without loss of generality we may assume $m<n$. Let $\widehat U = f(U)\subset \mathbb R^n$. The following composition of maps is continuous and injective:
$$
\widehat U \overset {f^{-1}} \longrightarrow U 
\overset{g}\longrightarrow \mathbb R^m \hookrightarrow \mathbb R^n.
$$ 
(The rightmost map is $(x^1,\dots,x^m)\mapsto (x^1,\dots,x^m,0,\dots,0)$.) Then the theorem on invariance of domain implies that this composition is an open map so its image is open in $\mathbb R^n$. But since its image is contained in the proper subspace $\mathbb R^m\subset \mathbb R^n$, this is a contradiction.
Your first question, "I am worried that the dimension n will depend on the point and the open set chosen," is actually answered by definition: the dimension $n$ is part of the definition of a topological $n$-manifold.  Otherwise, a disjoint union of a line and a plane would be a manifold, and its dimension would not be well defined.
