Homeomorphism induces differentiable structure Let $M$ be a differentiable manifold and $f:M\to N$ a homeomorphism. I want to show that there is exactly one differential structure on $N$ that makes $f$ a diffeomorphism. 
I have to show that there is a maximal smooth atlas $(V_i, k_i)_{i\in I}$ such that for every chart $h:U\to U^{\prime}\subseteq\mathbb{R}^m$ around $p\in M$ and every chart $k:V\to V^{\prime}\subseteq\mathbb{R}^n$ around $f(p)\in N$ the composition $k\circ f\circ h^{-1}$ is differentiable.
I tried to "tranfer" the charts on $M$ to $N$ using the fact that $f$ is continuous. But I got confused with the possibly different topologies on $M$ and $N$ and the fact that there already is an atlas for $N$. Can someone please help me out?
 A: A differentiable structure on a topological space is the same thing as a maximal atlas on that space.  Let $\mathcal A$ be a maximal atlas of $M$.
For each chart $(U, \phi)$ of $M$ in the atlas $\mathcal A$, define a chart $(f(U), \phi \circ f^{-1})$ of $N$.  Show that all these charts on $N$ are compatible and the set $\mathcal B$ of all these charts is a maximal atlas on $N$.  This defines a differentiable structure on $N$ for which $f: M \rightarrow N$ is a diffeomorphism.
Suppose we have another differentiable structure on $N$ given by another maximal atlas $\mathcal C$, such that $f: M \rightarrow N$ is a diffeomorphism when $N$ is given this differentiable structure.  Let $(W, \psi)$ be a chart of $\mathcal C$.  It suffices to show that $(W,\psi) \in \mathcal B$; this will show that $\mathcal B \subseteq \mathcal C$, and since $\mathcal B$ is maximal, this will imply $\mathcal B = \mathcal C$.
Let $U = f^{-1}(W)$ and $\phi = \psi \circ f$.  Since $f$ is a diffeomorphism, the chart $(U,\phi)$ of $M$ lies in the atlas $\mathcal A$ of $M$.  Then by definition, the chart $(f(U), \phi \circ f^{-1})$ of $N$ lies in the atlas $\mathcal B$ of $N$.  But $$(f(U), \phi \circ f^{-1}, \phi \circ f^{-1}) = (W, \psi)$$
so we are done.
A: I have a rather natural idea for a construction of a differentiable atlas on $N$. But I don't know how to prove that it's unique. I call the atlas on $M$ "$A_M$".
Define the following atlas on $N$:
$A_N = \{ (u,\phi)\ |\ (f^{-1}(u),\phi \circ f) \in A_M \}$
To prove that this set is indeed an atlas, as well as differentiable, we start by showing that it 'covers' all of $N$:
Say $x\in N$. Then  there will exist a chart $(v,\pi)$ in the atlas $A_M$ of $M$ containing $f^{-1}(x)\in M$. We claim that $(f(v), \psi\circ f^{-1})$ is an element of $A_N$. This is clearly the case, as $(f^{-1}(f(v)), \psi\circ f^{-1}\circ f)$ is an element of $A_M$. Hence the atlas contains a chart that has our arbitrarily chosen $x\in N$ in its domain. It therefore 'covers' all of $N$.
Let's show that the chart transition maps of $A_N$ are differentiable.
Assume $(u,\phi)$ and $(v,\psi)$ are charts in $A_N$ with $u\cup w\neq\emptyset$. Denote $u\cup w$ by $v$. Is the transition map
$t=\psi\circ\phi^{-1},\quad t:\phi(v)\rightarrow\psi(v)$
differentiable? Well, we know that $(\psi \circ f) \circ {(\phi \circ f)}^{-1}$ is differentiable, as $\psi \circ f$ and $\phi \circ f$ are chart maps in $A_M$. But we have:
$(\psi \circ f) \circ {(\phi \circ f)}^{-1} = (\psi \circ f) \circ (f^{-1} \circ \phi^{-1}) = \psi \circ (f \circ f^{-1}) \circ \phi^{-1} = \psi\circ\phi^{-1} = t$
Therefore $t$ is differentiable. So $A_N$ is a differentiable structure on $N$.
It is customary that the chart domains of an atlas are open sets. If we look at the definition of the set $A_N$ this is seen trivially for $A_N$: As $(f^{-1}(u),\phi \circ f)$ in $A_M$, $f^{-1}(u)$ is open in $M$. Since $f$ is homeomorphic, $u$ is therefore open in N. As this is the only instance were we rely on the fact that $f$ is homeomorphic, we would already be able to construct a differentiable structure on $N$ if we were only given the data of a continuous function $g:A_N\rightarrow A_M$.
Now, is $f:(M,A_M)\rightarrow (N,A_N)$ differentiable? Let $x\in M$ and $(u,\phi)$ be a chart in $A_M$ containing $x$. Then $(f(u),\phi\circ f^{-1})$ is a chart in $A_N$ containing $f(x)$. So, if we look at $f$ as a real function through these charts, is it differentiable? I.e., is $(\phi)\circ f\circ ({(\phi\circ f)}^{-1})$ differentiable? Well,
$(\phi)\circ f\circ ({(\phi\circ f)}^{-1}) = (\phi)\circ f\circ (f^{-1}\circ\phi^{-1}) = \phi\circ (f\circ f^{-1})\circ\phi^{-1} = \phi\circ\phi^{-1} = id_{\phi(u)}$.
Therefore $f$ is differentiable as a function between the differentiable manifolds $(M,A_M)$ and $(N,A_N)$.
