# Is it "identity function is diffeomorphism" implies the two smooth structures are equal?

I am reading Lee's Introduction to Smooth Manifolds, 2nd edition, and happen to read a sentence in a proof for Proposition 5.2 goes like this:

With this smooth structure on $S$, the map $F$ is a diffeomorphism onto its image, and this is obviously the only smooth structure with this property.

Perhaps I missed something from previous chapters, but this is not obvious to me at first sight. Maybe I would prove the following:

Say $F:M\to N$ is a homeomorphism between two topological manifolds. We endow $M$ with only one smooth structure, and $N$ with two smooth structures $C_1$ and $C_2$, such that $F$ is a diffeomorphism in both cases. We would show that $C_1=C_2$. Due to symmetric footing, it suffices to show $C_1\subseteq C_2$.

Observe $id=F\circ F^{-1}:(S,C_1)\to (S,C_2)$ is a diffeomorphism. To show $C_1\subseteq C_2$, I think this could be stated and proved more generally:

For a diffeomorphism $f:X\to Y$ and any smooth chart $(U,\varphi)$ on $X$, we have that $(f(U),\varphi\circ f^{-1})$ is a smooth chart on $Y$.

Now let $(U,\varphi)$ be a smooth chart on $X$ and $(V,\psi)$ be a smooth chart on $Y$. We are to show $\psi\circ f\circ\varphi^{-1}:\varphi\circ f^{-1}(f(U)\cap V)\to\psi(f(U)\cap V)$ and $\varphi\circ f^{-1}\circ\psi^{-1}:\psi(f(U)\cap V)\to\varphi\circ f^{-1}(f(U)\cap V)$ are smooth as maps between subsets of $\Bbb R^n,\Bbb R^m$, where $f(U)\cap V$ is nonempty for non-triviality.

For a point $p\in f^{-1}(f(U)\cap V)$, by smoothness of $f$, pick a chart $(U_p,\varphi_p)$ containing $p$, a chart $(V_{f(p)},\psi_{f(p)})$ containing $f(p)$, such that $f(U_p)\subseteq V_{f(p)}$, and $\psi_{f(p)}\circ f\circ\varphi_p^{-1}:\varphi_p(U_p)\to\psi_{f(p)}(V_{f(p)})$ is smooth. Impose also that $U_p\subseteq f^{-1}(f(U)\cap V)$; if not, we can use the chart $(U_p\cap f^{-1}(f(U)\cap V),\varphi|_{U_p\cap f^{-1}(f(U)\cap V)})$ in place of $(U_p,\varphi_p)$, which still satisfies the above conditions.

Now observe that the composition of smooth maps $(\psi\circ\psi_{f(p)}^{-1})\circ(\psi_{f(p)}\circ f\circ\varphi_p^{-1})\circ(\varphi_p\circ\varphi^{-1})$ is smooth. Cancellation gives that the composition is smooth on the domain of definition. We are to check that the domain is a neighbourhood of $\varphi(p)$.

To see this, observe the condition $U_p\subseteq f^{-1}(f(U)\cap V)$ gives $U_p\subseteq U$, so that the transition map $\varphi_p\circ\varphi^{-1}$ has domain and codomain $\varphi(U_p)$ and $\varphi_p(U_p)$ respectively; the conditions $U_p\subseteq f^{-1}(f(U)\cap V)$ and $f(U_p)\subseteq V_p$ gives $f(U_p)\subseteq V_{f(p)}\cap V$, so that the map $\psi_{f(p)}\circ f\circ\varphi_p^{-1}$ has range being a subset of $\psi_{f(p)}(V_{f(p)}\cap V)$; the transition map $\psi\circ\psi_{f(p)}^{-1}$ has domain $\psi_{f(p)}(V_{f(p)}\cap V)$. Hence, $(\psi\circ\psi_{f(p)}^{-1})\circ(\psi_{f(p)}\circ f\circ\varphi_p^{-1})\circ(\varphi_p\circ\varphi^{-1})$ is defined on the whole neighbourhood $\varphi(U_p)$ of $\varphi(p)$. Since this function $\psi\circ f\circ\varphi^{-1}$ is locally smooth around each point $\varphi(p)\in\varphi\circ f^{-1}(f(U)\cap V)$, it is smooth on the whole set $\varphi\circ f^{-1}(f(U)\cap V)$.

Smoothness of $\varphi\circ f^{-1}\circ\psi^{-1}$ is proved similarly: for each point $p'\in f(U)\cap V$, choose a chart $(V_{p'},\psi_{p'})$ containing $p'$, a chart $(U_{f^{-1}(p')},\varphi_{f^{-1}(p')})$ containing $f^{-1}(p')$, such that $f^{-1}(V_{p'})\subseteq U_{f^{-1}(p')}$, and $\varphi_{f^{-1}(p')}\circ f^{-1}\circ\psi_{p'}^{-1}$ is smooth. Impose $V_{p'}\subseteq f(U)\cap V$. We can then prove $\varphi\circ f^{-1}\circ\psi^{-1}$ is locally smooth around $\psi(p')$ by the same arguments as above.

Is the above proof correct? Is there a shorter proof for the above claim? Or is this proved in the book but I overlooked?

Edit: I will include the whole statement of Proposition 5.2 here.

Suppose $M$ is a smooth manifold with or without boundary, $N$ is a smooth manifold, and $F:N\to M$ is a smooth embedding. Let $S=F(N)$. With the subspace topology, $S$ is a topological manifold, and it has a unique smooth structure making it into an embedded submanifold of $M$ with the property that $F$ is a diffeomorphism onto its image.

• It seems like you may be establishing that there is only one smooth structure in a diffeomorphism class on a manifold, which is, I believe, false. .. btw is there any more info on $F$ and $S$ in your example
– user403337
Aug 21, 2017 at 7:17
• @ChrisCuster $F$ is simply required to be a fixed diffeomorphism.
– edm
Aug 21, 2017 at 7:20
• What about $S$?
– user403337
Aug 21, 2017 at 7:30
• @ChrisCuster do you mean the $S$ from the book?
– edm
Aug 21, 2017 at 7:31
• yeah. Is it an arbitrary smooth manifold; or anything more specific?
– user403337
Aug 21, 2017 at 7:34

Here is a possible reason why it could be okay to call it 'obvious'. It is a straight forward combination of three basic observations. Let me use the provisory definition that a topological chart on a topological space $X$ is a pair $(U,\varphi)$ of an open subset $U\subset X$ and a homeomorphism $\varphi\colon U\to U'$ onto any open subset $U'$ of some $\mathbb{R}^n$.

1. Let $U,V,W$ be smooth manifolds. If $f\colon U\to V$ is a diffeomorphism, then a map $g\colon V\to W$ is a diffeomorphism if and only if $g\circ f\colon U\to W$ is a diffeomorphism.

2. The maximal atlas of a smooth manifold $X$ is the set of topological charts $(U,\varphi)$ on $X$ where $\varphi\colon U\to U'$ is a diffeomorphism. (Note that this is somehow recursive since we're taking $U$ with the differentiable structure naturally inherited from $X$; this is exactly the trick here.)

3. Let $f\colon X\to Y$ be a homeomorphism between topological manifolds. For each topological chart $(U,\varphi)$ on $X$ we get a topological chart $f_*(U,\varphi):=(f(U),\varphi\circ f^{-1}|_{f(U)})$ on $Y$ and each topological chart on $Y$ is of this form since the process can be inverted by mapping a topological chart on $Y$, $(V,\psi)$, to the topological chart $f^*(V,\psi) := (f^{-1}(V),\psi\circ f|_{f^{-1}(V)})$ on $X$.

In combination, we see that if $f\colon X\to Y$ is a diffeomorphism, then we can recover the maximal atlas $A_Y$ (hence, the smooth structure) of $Y$ from the maximal atlas $A_X$ of $X$ as follows. A topological chart $(V,\psi)$ on $Y$ is actually a chart, i.e., $(V,\psi)\in A_Y$, if and only if $\psi$ is a diffeomorphism; but this is the case if and only if $\psi\circ f|_{f^{-1}V}$ is a diffeomorphism and this in fact precisely means $f^*(V,\psi)\in A_X$. Thus, we get a pair of inverse bijections $f_*\colon A_X\to A_Y$ and $f^*\colon A_Y\to A_X$. In particular, the smooth structure on $Y$, that is, $A_Y =\{f_*(U,\varphi)\mid(U,\varphi)\in A_X\}$, is uniquely determined by $(X,A_X)$ and $f$.

The embedded submanifold $S$ inherits a unique smooth structure from the manifold $M$. With this structure the inclusion $\mathcal i:S \to M$ is an embedding. Note, however, there certainly can be more than one atlas defining this structure. These atlases are called compatible...

It's clear that if we took a different smooth structure, that is, one defined by atlases not smoothly equivalent to the above, $F$ would not be a diffeomorphism anymore. ..

So, say $\phi_\alpha \circ \psi_\beta^{-1}$ isn't a diffeomorphism .. then $F\circ \psi_\beta^{-1}$ isn't either. .. because, say $g=F\circ \psi_\beta^{-1}$ is a diffeomorphism. .. then $\phi_\alpha \circ F^{-1}\circ g=\phi_\alpha \circ \psi_\beta^{-1}$ would be a diffeomorphism. ..

And I now think you're right, he didn't mean up to diffeomorphism. Incompatible maps can define diffeomorphic smooth structures, as in @Ben's example. .. ($(\mathbb R, id)$ and $(\mathbb R, x\to x^3)$)

• I am aware that the uniqueness of smooth structure is under the assumption that $F$ has to be a diffeomorphism. What is not obvious is the statement "all smooth structures on the image $F(S)$ that make $F$ a diffeomorphism must be equal to each other".
– edm
Aug 8, 2017 at 5:03
• I don't think the author means "up to diffeomorphism". He wants to make a fixed mapping $F$ to be a diffeomorphism.
– edm
Aug 21, 2017 at 7:32
• I'm aware of the function $F$. I meant diffeomorphic atlases (smooth structures), which would make $F$ a diffeomorphism. ..
– user403337
Aug 21, 2017 at 9:41
• An easy example for the claim that there can be incompatible but diffeomorphic atlases: Take $\mathbb{R}$ once with the most obvious chart $(\mathbb{R},\mathrm{id})$ and once with the chart $(\mathbb{R},\varphi)$ where $\varphi\colon\mathbb{R}\to \mathbb{R}$ is the map $x\mapsto x^3$. The two charts are incompatible but there is still a diffeomorphism between them, namely, $\varphi\colon (\mathbb{R},\varphi)\to (\mathbb{R},\mathrm{id})$.
– Ben
Aug 21, 2017 at 10:16
• @Ben This is getting further and further away from what I am asking... To respond to ChrisCuster's addendum, I don't think I want to use results that comes from latter pages because the author does not seem to postpone the proof as he writes "...is obviously the only smooth structure...".
– edm
Aug 21, 2017 at 10:51