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Assuming you have an atlas $A=\{(U_n, \phi_n)\}$ with $n$ charts, I understand that a transition map is just the bridge between an overlap of the domain ($U$) of two given charts. However, I am a bit confused on how, assuming you just have two charts $(U_1, \phi_1)$ and $(U_2, \phi_2)$, each of which creates it's own manifold-with-boundary (surface). Let's assume that each chart generates a surface that looks something like the quintessential manifold representation:

surface

Now, if you have two charts within an atlas, there isn't really any concept within the information of an atlas that deals with "compatibility" of the range (surfaces) of the charts. I get that each chart is effectively it's own $nD$ Euclidean space, and perhaps connectivity can verify that these two surfaces are not disconnected, but how can you verify that the connection between these two surfaces actually ends up being "smooth"?

I know that the example of using the graph functions as our manifolds can be iffy (as described here), though it will help to get my point across. If you have two functions, $f(x)=x^2$ and $g(x)=-x$, which are two smooth functions, and then we define the graph's of these two functions as our actual chart maps, $\mathcal{G}f=(x, f(x))$ and $\mathcal{G}g=(x, g(x))$. If you put $\mathcal{G}g$ across the interval $(-1,0)$ and $\mathcal{G}f$ across the interval $(0, 1)$, you create a "manifold-with-boundary" that now has two smooth charts and has no transition map (because the domain never overlaps).

graph

Your "manifold" (if we can even call it that, this is merely to get the point across, I care less about the actual usage of graphs of functions), can be proven to be connected, though the point $(0,0)$, the point between two adjacent charts, does not seem to be a smooth connection. However, at the same time if we try to assume that the two charts will be equal at some point, we run into issues with the identity theorem which makes for an odd dilemma.

Maybe I am missing something here, or focusing too much on the graphs of functions, but I just can't seem to wrap my head around how the range of two adjacent charts can create a "smooth geometric object" without being equal at their edges (like the circle or the sphere).

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    $\begingroup$ But open sets don't have edges. Either two charts $U, V$ don't intersect, or they intersect in another open set. In your example, the origin is not an intersection. Also, when $U, V$ do intersect in a non empty set, the fact that they are part of the same atlas forces the transition map to be smooth (or rather, that's the definition of an atlas). $\endgroup$ Commented Nov 8, 2020 at 5:41
  • $\begingroup$ Yeah totally valid, I overlooked that with this example. But that is the source of my confusion. That means that you can't have two ranges overlap on some point (because that would mean closed sets), than you have to have the ranges overlap on some interval (a, b). But this seems to be in direct violation of the identity theorem. Of course you can bypass this by using non-analytic functions (bump functions), but how can you do it without bumps? $\endgroup$
    – Tug Witt
    Commented Nov 9, 2020 at 22:03
  • $\begingroup$ Perhaps my confusion is in the fact that the identity theorem is saying that some domain interval (a,b) across two functions f and g, where f(x)=g(x) | x ∈ D means that f = g. But with manifolds we don't have to have the same domain. A good example would be f(x) = x and g(x) = x+1. They both have some domain that creates the range (1,2) but a different domain makes it for both f and g. Thus the identity theorem is not violated because its a different domain. $\endgroup$
    – Tug Witt
    Commented Nov 9, 2020 at 22:07
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    $\begingroup$ The charts overlap on open sets, but we don't require the functions/coordinate maps to agree on the overlap, just that the transition (i.e., a change of coordinates) should be continuous/smooth/holomorphic etc. So it doesn't contradict the identity theorem because the functions need not be identical on the overlap, see the answer to your other question. $\endgroup$ Commented Nov 10, 2020 at 2:10

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