If I have already known a topological space $N$ is homeomorphic to a smooth manifold $M$ then will it be a smooth manifold? The atlas of $N$ is the preimage of the atlas of $M$ and the coordinate map is the composition of homeomorphism composites the coordinate map?

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    $\begingroup$ You can even do it even if $M$ is just a set bijective with $N$, by transporting the topology as well as the differential structure. $\endgroup$
    – Arthur
    Jan 22, 2018 at 15:32
  • $\begingroup$ @Arthur thank you, I got that :) $\endgroup$
    – xxyshz
    Jan 22, 2018 at 15:35

3 Answers 3


Let me emphasize something that is perhaps not readily apparent in the other answers. "Smooth" is not something that a topological space is. You can't formally say "this topological space is a smooth manifold." It is not a property of a topological space. It is extra structure that you add on top of the already existing space.

Indeed, you have to choose an atlas such that all the coordinate change maps are smooth. If you choose an atlas at random, chances are, it won't be smooth, even if you know otherwise that the space is homeomorphic to a smooth manifold.

However there is a property here: can the space be attributed a smooth manifold structure? That is a possible property, it's a yes/no question. In this case the answer is yes! As the other answer points out, you can use the homeomorphism in a manner called transport of structure to choose an atlas that you know will be smooth. But there may be others! For example the $7$-sphere $S^7$ has famously several different, non-equivalent smooth structures (the ones other than the standard one are called exotic).

Therefore it's important to distinguish property and structure. When you ask "Is this space a smooth manifold?" you're implicitly saying that "being a smooth manifold" is a property, which it isn't. And in this particular case it may help you clear up the question: when you wonder "can this space be made into a smooth manifold", then you see immediately that you need to add some stuff on top of what you already have, namely, an atlas. You can't just look at the space and ask yourself "is this smooth" in the same way that you look at a car and ask "is this red". And since you now know what you have to do, it's certainly easier to actually do it. (See also this other answer of mine or this nLab article.)

  • $\begingroup$ I think the coordinate map may not have to be smooth, it has to be homeomorphic to the open set of $\mathbb{R}$, am I right? $\endgroup$
    – xxyshz
    Jan 22, 2018 at 14:30
  • $\begingroup$ Thanks a lot. I got the point :) $\endgroup$
    – xxyshz
    Jan 22, 2018 at 14:39

Yes, you can transport the structure; see jabo's answer. However, the smooth structure you induce in this way may not be the natural one. I have in mind the example of a square in $\mathbb R^2$. The square is homeomorphic to a circle, which is a smooth manifold. However, the square is not a smooth submanifold of $\mathbb R^2$, because it has corners.

Therefore, the smooth structure you import from the circle is not the natural one that you expect on subsets of $\mathbb R^2$.

Let me try to illustrate the main point with a toy model from linear algebra. Consider the vector space $(\mathbb R^2, +, \cdot )$. One may ask: does there exist a vector space structure on
$$A=\left\{ (x, y)\in \mathbb R^2\ |\ x=1\right\},$$ which is a subset of $\mathbb R^2$? The answer is affirmative, because the map $$ \phi\colon A\to \mathbb R,\qquad \phi(x, y)=y, $$ is a bijection and so you may transport the vector space structure of $\mathbb R$ on $A$.$^{[1]}$

However, a more natural question would be whether $A$ is a vector subspace of $\mathbb R^2$. And this is not true, as you surely know.

Summarizing, the structure you transported on $A$ is completely unrelated to the structure of the bigger space $\mathbb R^2$, and so it is less likely to be useful. This is a general fact: if you transport structures, you might well end up with something crazy and potentially useless.

[1] This means that if $(x, y), (x',y')\in A$ you can define $$(x,y)+_\phi(x', y')=\phi^{-1}(y+y'),$$ and if $\lambda\in \mathbb R$ you can define $$\lambda\bullet_\phi(x, y)=\phi^{-1}(\lambda y)$$ and $(A, +_\phi, \bullet_\phi)$ is a vector space.

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    $\begingroup$ It's good to point this out, maybe it also helps to state this in more generality: Assume $f\colon S\hookrightarrow M$ is an embedding of topological manifolds and $M$ has a smooth structure. Then even though $S$ might admit smooth structures, there is no reason to assume that also the embedding $f$ will be smooth with respect to all of them or even a single one. This phenomenon occurs both for the square embedded into $\mathbb{R}^2$ or for $\text{id}\colon S^7 \rightarrow S^7$, where one of the spheres carries an exotic smooth structure. $\endgroup$
    – Jan Bohr
    Jan 22, 2018 at 10:29
  • $\begingroup$ Why the square is not a smooth manifold, if I choose identification map to be the coordinate map, will it work? $\endgroup$
    – xxyshz
    Jan 22, 2018 at 14:27
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    $\begingroup$ @R.Sherlock: I didn't say that. I said that the square is not a smooth submanifold of $\mathbb R^2$. A submanifold of $\mathbb R^n$ is, roughly speaking, a set that can be locally described as the graph of a smooth function. The square cannot be described as such, because you will never find a smooth function whose graph has a corner. $\endgroup$ Jan 22, 2018 at 14:35
  • $\begingroup$ @GiuseppeNegro I'm sorry missing that ...Since I have not met the concept of submanifold... $\endgroup$
    – xxyshz
    Jan 22, 2018 at 14:38
  • $\begingroup$ No need to be sorry, that's actually a slippery notion (that was the first question I posed on this site, btw). $\endgroup$ Jan 22, 2018 at 14:40

Sure this works and is sometimes called transport of structure.

  • $\begingroup$ I know that, I feel clearer than before ! Thank you ~~ $\endgroup$
    – xxyshz
    Jan 22, 2018 at 14:26

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