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I need to show that a diffeomorphism from a differentiable manifold $M$ to a differentiable manifold $N$ is an isomorphism. We have defined a diffeomorphism to be a $C^{\infty}$ homeomorphism with a $C^{\infty}$ inverse.

But, as I am new to pure mathematics, I don't know what it means for something to be an isomorphism(structure-preserving map would be my guess, whatever that means precisely) and I don't know what I need to show in order to prove that the map is an isomorphism.

Any help will be appreciated.

TheQuantumMan
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    This question makes no sense as stated: "Isomorphism" of what kind of a structure? Your tag suggests that this structure is a Riemannian metric, but in this category one says "an isometry", not "an isomorphism", – Moishe Kohan Oct 02 '17 at 21:59
  • @MoisheCohen I just have a map from a differentiable manifold M to another differentiable manifold N. No Riemannian metric. Sorry for the confusion – TheQuantumMan Oct 02 '17 at 22:00
  • The diffeomorphisms are the isomorphisms in the category of smooth manifolds and smooth maps nearly by definition, depending on your definitions. Can you say exactly what your definitions are? If not, you should ask whoever gave you this question for their definitions. – Qiaochu Yuan Oct 02 '17 at 22:03
  • @QiaochuYuan All our definitions are the same as those in Do Carmo's Riemannian geometry. We haven't talked about isomorphisms is the lectures. We just have to prove the above as an exercise. – TheQuantumMan Oct 02 '17 at 22:05
  • By the way, this is related: https://math.stackexchange.com/questions/211254/if-f-is-diffeomorphism-the-linear-map-f-is-an-isomorphism?rq=1 – TheQuantumMan Oct 02 '17 at 22:05
  • Then the question is underspecified. In category theory, in order to say what an isomorphism is you need to specify a category to be working in. If the category is the category of smooth manifolds and smooth maps, then an isomorphism is a smooth map with smooth inverse. That may or may not be your definition of diffeomorphism; if it is, then the statement is true by definition. So what is the question? Can you ask your professor? – Qiaochu Yuan Oct 02 '17 at 22:10
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    @QiaochuYuan Well, our definition of diffeomorphism is a map that is a homeomorphism and $C^{\infty}$. Yes, I will ask my professor, although I originally thought that it would be a good idea to ask here since this question has not been asked before(as far as I know), so others might also have this question. – TheQuantumMan Oct 02 '17 at 22:12
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    Okay, cool, so now there is at least a possible reasonable reading of the question: namely, to prove that a smooth homeomorphism has a smooth inverse. Unfortunately, this statement is false, and in fact that is not a correct definition of diffeomorphism: the map $\mathbb{R} \ni x \mapsto x^3 \in \mathbb{R}$ is a homeomorphism and $C^{\infty}$ but not a diffeomorphism because its inverse $\sqrt[3]{x}$ fails to be differentiable at $x = 0$. – Qiaochu Yuan Oct 02 '17 at 22:13
  • @QiaochuYuan Just to state everything, the professor did mention something about the map being linear. Also, the exercise is the same as the link I posted above. – TheQuantumMan Oct 02 '17 at 22:14
  • The link you posted above is a different question from the question you stated: the question as you wrote it here makes no mention of the differential at a point. (There's still the more important issue that the definition of diffeomorphism you just gave is incorrect, so either you wrote it down incorrectly or your professor's using an incorrect definition.) – Qiaochu Yuan Oct 02 '17 at 22:15
  • @QiaochuYuan Yes, my question is more general. But, I just wanted to state the place where it originated from since it might give you a clue of what's going on! – TheQuantumMan Oct 02 '17 at 22:17
  • Your question is not more general; it is just a different question. – Qiaochu Yuan Oct 02 '17 at 22:17
  • @QiaochuYuan I don't understand why it's not more general. The question is about a general map being an isomorphism. Isn't the differential also a map? Also, yes, my definition is incorrect. We also stated that the inverse must also be $C^{\infty}$ – TheQuantumMan Oct 02 '17 at 22:20
  • The other question asks to show that if some map $f$ is a diffeomorphism then another map, namely the differential, constructed out of $f$ is an isomorphism, presumably in the category of vector spaces. This question asks to show that if $f$ is a diffeomorphism then it, not another map, is an isomorphism (presumably in the category of smooth manifolds although this is still ambiguous). It's just a different question. Neither statement implies the other. – Qiaochu Yuan Oct 02 '17 at 22:22
  • @QiaochuYuan Yes, you are right. Sorry. I just want to know what I need to show in order to prove that a map is an isomorphism. I think that if I actually learn this, I will be able to prove the other question too. – TheQuantumMan Oct 02 '17 at 22:25
  • That is yet another question! By definition, an isomorphism is a map with an inverse (in some category). So the obvious way to show that a map is an isomorphism is to write down its inverse. There are less obvious ways but this is enough to do the linked problem. – Qiaochu Yuan Oct 02 '17 at 22:27
  • @QiaochuYuan Lol, yeah, I turned this question into a carnival! Thanks for the patience! So, for MY exercise, I will prove that an inverse exists and it is a differential that takes tangent vectors from $T_{\phi (p)} N$ to $T_pM$, right? I'll probably have to use somewhere that the map $\phi$ is a diffeomorphism. – TheQuantumMan Oct 02 '17 at 22:32
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    Yes and yes. The statement is certainly not true if $\phi$ is not a diffeomorphism. – Qiaochu Yuan Oct 02 '17 at 22:33
  • @QiaochuYuan Wow, at last! haha! Sorry for being such a pain! In any way, thanks a lot. Appreciate the clarifications. – TheQuantumMan Oct 02 '17 at 22:33

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In differential topology, isomorphism can be taken to mean homeomorphism, ie. Isomorphism between topological spaces. An isomorphism between topological spaces is a continuous bijection with continuous inverse.

mathfan27543
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