differmorphism and homeomorphism for manifolds

Question

For two abstract manifolds that are differmorphic, why are they always homeomorphic?

Why does differentiability imply continuity for abstract manifolds? (for $R^n$ this is certainly clear)

Answer 1

To check continuity is a local matter, so your remark that it's "certainly clear" in $\Bbb R^n$ suffices. The whole point of manifolds (whether topological or smooth) is that local properties are checked in a coordinate chart (and the definition — in terms of compatibility of charts — tells you that this is a well-defined process). If you have a smooth map $f\colon M\to N$, this means that if we choose charts $\phi\colon U\to \Bbb R^n$ and $\psi\colon V\to\Bbb R^n$ around $p\in M$ and $f(p)\in N$, respectively, then $f$ translates into a smooth map $F=\psi\circ f\circ\phi^{-1}\colon \phi(U)\to \psi(V)$. $F$, in turn, is continuous, and so, $f\big|_U = \psi^{-1}\circ F\circ\phi\colon U\to V$ is as well.

In terms of your remark, continuity can always be checked locally, as to prove the inverse image of an arbitrary open set is open, it suffices to show the preimage of a basis element (which can be chosen in $V$) is open. To do this, you need to find a basis element $U'\subset U$ around $p\in f^{-1}(V)$ so that $f(U')\subset V$.

Answer 2

It should be remarked that the definition of a smooth map $f$ from $M$ to $N$ requires the continuity of $f$. Indeed, one defines smoothness by saying that $f$ is smooth in some charts, but in order to do this, we have to fix a chart $V$ in $N$ and a chart $U$ in $M$ such that $f(U) \subset V$. And you cannot do this if you don't assume continuity first.