Suppose $M$ is a topological manifold and $(U,\phi)$ a local chart around $p\in M$.
Is it always possible to find a chart $(U,\psi)$ such that $\psi(U)=B$ where $B$ is, say, the unit ball in $\mathbb{R}^n$ ?
I'm led to think that this is the case. I know one can have $\psi(U)\supset B$ by composing $\phi$ with a scalar multiplication, but I don't see how to get equality. I think one would have to modify $\phi$ if $\phi(U)$ isn't connected to get a connected image, and I wonder if any two open connected sets in $\mathbb{R}^n$ are homeomorphic (which seems plausible).
Even if $U$ is connected, this won't work. For example, $B\setminus \{0\}$ is a connected open subset of $\Bbb R^n$ ($n>1$), but it isn't homeomorphic to $B$ (because it isn't contractible).
However, as Daniel Fischer mentioned in the comments, it is always possible to find some $V \subset U$ open such that $\psi(V)=B$.