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Let $U\subset\mathbb{R}^n$, $V\subset\mathbb{R}^m$ and a bijection $f:U\to V$ is a diffeomorphism if $f$ and $f^{-1}$ are differentiable.

I would like to know the intuitive meaning of two open sets being diffeomorphic.

For example, if two spaces are homeomorphic, these spaces share the same topological properties. And we have a clear intuitive idea of two spaces being homeomorphs, like the classic relationship between a donut and a mug.

Is there a similar intuitive idea for diffeomorphism?

Edit: The proprieties that homeomorphism preserves is, for example, if one of them is compact, then the other is as well; if one of them is connected, then the other is as well; if one of them is Hausdorff, then the other is as well; their homotopy and homology groups will coincide.

So, what are the properties preserved by diffeomorphism?

I quoted homeomorphism, to indicate what I meant by an intuitive idea.

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  • $\begingroup$ The diffeomorphisms are smooth so while a half cone is homeomorphic to the plane, it is not diffeomorphic.to it. $\endgroup$ – John Douma Jun 29 '20 at 21:32
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    $\begingroup$ @JohnDouma That example always bugs me. What smooth structure are you endowing the half cone with so as to make meaningful the statement that it's not diffeomorphic to the plane? It does not have a smooth structure naturally induced by its being a subset of $\mathbb{R}^3$, and in fact it does not have one satisfying your criteria at all, as pulling back such a smooth structure to the plane via our homeomorphism would imply the existence of an exotic $\mathbb{R}^2$. $\endgroup$ – jawheele Jun 29 '20 at 21:50
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    $\begingroup$ @GReyes I think you've overcomplicated things a bit-- there's no need to invoke connections or metrics to specify a smooth structure. In fact, it's the other way around. In order to induce a metric on the cone $C$ from the ambient $\mathbb{R}^3$ by restricting it from $T_p \mathbb{R}^3$ to $T_p C$, it is necessary (but not sufficient) to first choose a smooth structure on the cone, as $T_p C$ is a property of the smooth structure. This still cannot be done if the cone is to include its vertex $v$, however, as there is no way to identify $T_v C$ with a subspace of $T_v \mathbb{R}^3$. $\endgroup$ – jawheele Jun 29 '20 at 22:31
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    $\begingroup$ You are absolutely right. I misunderstood the question. It is about the existence of an isomorphism as differentiable manifolds... $\endgroup$ – GReyes Jun 29 '20 at 22:38
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    $\begingroup$ As well as the property I mentioned in my answer, another property that a differential $n$- manifold might have is that every compact subset may be enclosed by a smoothly embedded $n-1$ sphere. Clearly $\mathbb{R}^4$ has this property (compact subsets are bounded with respect to the usual metric, so you just need a sphere of large enough radius). However there exist manifolds homeomorphic to $\mathbb{R}^4$ that do not have this property. See planetmath.org/donaldsonfreedmanexoticr4 for a little detail. $\endgroup$ – tkf Jun 30 '20 at 2:13
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Diffeomorphisms are precisely the isomorphisms in the category of differentiable manifolds.

So it might be helpful to think about diffeomorphisms between differentiable manifolds in exactly the same way as you would think about homeomorphisms between topological spaces.

Since we consider spaces that are isomorphic " to be essentially the same" or "indistinguishable", this is what we have in mind in terms of differentiable manifolds whenever we talk about them being diffeomorphic.

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It may help to illustrate in what ways two spaces can be homeomorphic but not diffeomorphic. Consider the half open square $X=(0,1)\times [0,1]\subset\mathbb{R}^2$ with usual differential structure. Topological properties of the space include that it is connected, simply connected, contractible.

What might a property of its differential structure be? One example is the fact that you can draw a smooth curve starting at one edge and ending at the other. In a way this is a bad example, because there is no "exotic" square, which is topologically the same as $X$, but where you cannot draw such a smooth curve.

However in dimension 4 this becomes more meaningful. There is a subset of $\mathbb{R}^4$ which is homeomorphic to $X \times X$, but where you cannot smoothly map a disk, so that the boundary of the disk goes once round the boundary of $X \times X$.

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