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I couldn't find a duplicate, although I think is a very common question.

Given two charts, ($U_{1},φ_{1}$), ($U_{2},φ_{2}$), on a n-dimensional topological manifold M, such that: $U_{1} \cap U_{2}\neq \emptyset$, we get transition maps:

$φ_{1}\circ φ_{2}^{-1} : φ_{2}(U_{1}\cap U_{2}) \rightarrow φ_{1}(U_{1}\cap U_{2})$, and

$φ_{2}\circ φ_{1}^{-1} : φ_{1}(U_{1}\cap U_{2}) \rightarrow φ_{2}(U_{1}\cap U_{2})$

Two charts, as above, are called compatible if the transition maps, as above, are homeomorphisms. If $U_{1} \cap U_{2} = \emptyset$, then they are compatible.

My question is, why do we need this behavior? In addition, if we want to define C$^{\infty}$-compatible charts, why do we need to take transition maps to be smooth, Euclidian mappings?

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    $\begingroup$ I presume by a "chart" on the given topologial manifold $M$ you mean $(U,\phi)$ such that $U \subset M$ is open and $\phi : U \to \mathbb{R}^n$ is a homeomorphism from $U$ onto an open subset of $\mathbb{R}^n$. In which case, it is an exercise that any two charts are compatible. $\endgroup$ – Lee Mosher Feb 21 '16 at 16:39
  • $\begingroup$ Yes, I mean exactly that. I can see why, since φ$_{1}$, φ$_{2}$ are homeomorphisms. But then, why some texts give that like a definition? Not all of them, but there are some that does it. $\endgroup$ – Mathitis Feb 21 '16 at 16:46
  • $\begingroup$ The motivation behind this choice of charts is explained very well in Spivak's book Vol. I $\endgroup$ – Sou Nov 22 '17 at 11:10
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Imagine taking two nice pieces of tissue and gluing them together with the goal to obtain a larger piece of tissue which is still nice. Then you will put part of the second piece of tissue over part of the first one, make it so that there are no wrinkles and then glue. This basically what is happening with manifolds: you are gluing patches of euclidean space together in such a way that the resulting object is "nice", i.e. is again locally euclidean. If you want nicer objects, you will have to take better behaved gluings (you could accept that the resulting piece of tissue forms an angle somewhere, or you could ask for it to be smooth everywhere...)

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Yes, each individual chart $\phi \colon U \to \mathbb{R}^n$ is $C^\infty$, but remember that each chart in only defined on one open neighbourhood $U \subset M$. To allow us to consider behaviour on the whole of the manifold, it's important to be able to patch the $U$ neighbourhoods together nicely, going from one chart $\phi$ to another without having any non-differentiable issues. This is why we need transition maps to be smooth as well as the charts.

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Because we want use the charts to define differentiability. Differentiability depending of the choosing of some specific chart will be a nightmare.

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  • $\begingroup$ No offense, but that's a buzzword. You don't try to explain why we avoid the "choosing of some specific chart" by using this definition. $\endgroup$ – Mathitis Feb 21 '16 at 16:50
  • $\begingroup$ I agree with the point of view of this answer. It is explained, e.g., in the very first pages of Spivak's "A comprehensive introduction to differential geometry", vol. 1. $\endgroup$ – Giuseppe Negro Feb 21 '16 at 17:06
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Take an inclusion of sets $i:U\to M$, where $U$ is an open set in $\mathbb{R}^n$.

We'd like to be able to say that $i$ is a homeomorphism (continuous isomorphism) onto its image—or, for a smooth manifold, to say that $i$ is a smooth isomorphism. Or, for a piecewise-linear manifold, we'd like to say that $i$ is a piecewise linear isomorphism.

In any case, we'd like to check this behavior on charts, since we know how these concepts behave in $\mathbb{R}^n$. But this concept is not well-defined without transition maps with the right property. Otherwise, continuity/smoothness/piecewise linearity would be dependent on which charts you used to verify it, rather than a property of the manifold.

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  • $\begingroup$ So the key is the well defined, in the sense of the property that we want to have. If the charts are not compatible, this concepts might be as we want them, with some charts and not with some others, although each chart is well-defined and homeomorphic to $\mathbb{R}^{n}$. $\endgroup$ – Mathitis Feb 21 '16 at 17:04
  • $\begingroup$ @Mathitis Yes. Speaking somewhat generally: if we are interested in property P, we can define a map of manifolds to have property P when it has property P on a cover by charts. But this is well-defined only when the transition maps have property P. $\endgroup$ – Slade Feb 21 '16 at 18:05

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