# 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?

• Since every diffeomorphism is a homeomorphism, you don't have to worry about different topologies on $N$. You can consider the topology on $N$ to be fixed in this problem: a set $W \subset N$ is open if and only if $f^{-1}(W)$ is open in $M$. – D_S May 11 at 21:39

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.

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 $$(N,A_N)$$.

• Refer to @D_S's comment for a proof of uniqueness. – querryman May 11 at 22:37