Is the converse of this true? I was reading this reference and I was wondering if someone can provide a proof of the converse, well if it is true of course, I mean if two smooth manifolds are diffeomorphic then they have to be homeomorphic, and if fact does the inverse of a diffeomorphism has to be a diffeomorphism ?
Thanks a lot in advance.
 A: Both are true. If manifolds $M$ and $N$ are diffeomorphic, there exists a differentiable bijection $f:M\to N$ which has a differentiable inverse (a diffeomorphism). Since differentiability implies continuity, $f$ is a continuous bijection with a continuous inverse, a homeomorphism. Thus, $M$ and $N$ are homeomorphic.
If $f$ is a diffeomorphism, it is a differentiable bijection with a differentiable inverse. The inverse of a bijection is always a bijection, and $(f^{-1})^{-1}=f$, so $(f^{-1})^{-1}$ is also differentiable. Thus, $f^{-1}$ is a diffeomorphism.
Edit:
If your definition of diffeomorphism is $f:M\to N$ is a diffeomorphism if it is a continuous bijection such that $X\subset M$ is a chart if and only if $f(X)$ is a chart in $N$, then both are still true. To get that $f$ is a homeomorphism, we need to show that $f^{-1}$ is continuous. Let $U\subset M$ be open. Then $U$ can be written as a union of charts, $U=\bigcup U_\alpha$. Thus, $f(U_\alpha)=V_\alpha$ is a chart in $N$. We see that $$f(U)=f\left(\bigcup_\alpha U_\alpha\right)=\bigcup_\alpha f(U_\alpha)=\bigcup_\alpha V_\alpha.$$
Thus, $f(U)$ is open in $N$, so $f^{-1}$ is continuous, and $f$ is a homeomorphism.
To show that $f^{-1}$ is a diffeomorphism, we have to show that it is continuous (done), a bijection (done), and $X$ is a chart in $N$ if and only if $f^{-1}(X)$ is a chart in $M$. Let $X$ be a chart in $N$. Let $Y=f^{-1}(X)$. Then $f(Y)=X$ is a chart in $N$, so by assumption $Y$ is a chart in $M$. Now let $X$ be any set in $N$ such that $f^{-1}(X)=Y$ is a chart in $M$. Then $f(Y)=X$ is a chart in $N$.
