Prove the following statement about n-Manifolds

[LCSphysicist]

TL;DR Summary

I am trying to prove the statement i will print below, there is not much to talk here.

I am not sure how to start it, but yet, i imagined it:

Be $\phi^{X}, \phi^{Y}$ two charts from an open set $A, B$ to $R^{n}$:
$\phi^{X}: A -> R^{n}$, where $\phi^{X}$ is continuous
$\phi^{Y}: B -> R^{n}$, where $\phi^{Y}$ is continuous​

I believe we need to show that $\phi^{X}+\phi^{Y}: U = A\bigcup B -> R^{n}$, where $\phi^{Y} + \phi^{X}$ is continuous. But i am not sure how.

 

[fresh_42]

Summary:: I am trying to prove the statement i will print below, there is not much to talk here.

View attachment 272149
I am not sure how to start it, but yet, i imagined it:

Be $\phi^{X}, \phi^{Y}$ two charts from an open set $A, B$ to $R^{n}$:
$\phi^{X}: A -> R^{n}$, where $\phi^{X}$ is continuous
$\phi^{Y}: B -> R^{n}$, where $\phi^{Y}$ is continuous​

I believe we need to show that $\phi^{X}+\phi^{Y}: U = A\bigcup B -> R^{n}$, where $\phi^{Y} + \phi^{X}$ is continuous. But i am not sure how.

It is usually written as a direct product and $\phi^{U}:=\phi^{X}\times\phi^{Y}: U = A\times B -> R^{n}$. Not that it makes much difference, but it shows better that we can work with the components. And here is the key: Start with what you have to show in the product space, translate it according to the definition of the topology and the mappings $\phi^{X},\phi^{Y}$. Then you have the situation in each component, which you clue again with the direct product. It is mainly a notation exercise, less a mathematical one. The product topology is made such that it can be split into the components, apply the property of being an $n-$ manifold and put it together again. It works with continuous, smooth, and homeomorphisms.

 

[Office_Shredder]

Fresh, this isn't the product topology though. If M and N are each a line, the object in question here is two lines, not a plane.

In fact I think the point here is to ignore the structure of the other side entirely. If you have a point in $M\cup N$, it's either in $M$ or it's in $N$. Let's suppose it's in $M$. Then we know there's a chart for an open set $U\subset M$ containing our point.

Without any further notation, can you write some open sets in $M\cup N$? They are really dumb and trivial and of the form $U\cup X$ for some set $X$ you already know.

 

[fresh_42]

Fresh, this isn't the product topology though. If M and N are each a line, the object in question here is two lines, not a plane.

Thanks for the clarification. This makes more sense, as it prepares for foliations and covering spaces.

 

[mathwonk]

I don't know if the question is about giving a rigorous mathematical proof or about understanding it. But understanding it should be "easy". I.e. a space is a manifold if it looks locally like euclidean space everywhere.

I.e. stand at any point and look around near you - it should look euclidean. now a disjoint union of 2 manifolds is formed by taking one manifold and then another one a certain distance away and not meeting the first one. imagine two disjoint spheres with space between them. then choose a point. it is either in one sphere or the other, so if you look around very nearby, it looks like euclidean space, because you can only see points of the sphere your point is on.

My point is only that if you have any idea at all what a manifold is, this result should be pretty clear. So if it seems mysterious, please spend some serious time thinking about this.

maybe i am missing the point of confusion, i.e. maybe it was a misunderstanding about what the "disjoint union" topology defines, i.e. it defines the naive disjoint uniion. namely the point is that one can take as the open set in one space just the empty set, as suggested earlier.

 

[LCSphysicist]

I don't know if the question is about giving a rigorous mathematical proof or about understanding it. But understanding it should be "easy". I.e. a space is a manifold if it looks locally like euclidean space everywhere.

I.e. stand at any point and look around near you - it should look euclidean. now a disjoint union of 2 manifolds is formed by taking one manifold and then another one a certain distance away and not meeting the first one. imagine two disjoint spheres with space between them. then choose a point. it is either in one sphere or the other, so if you look around very nearby, it looks like euclidean space, because you can only see points of the sphere your point is on.

My point is only that if you have any idea at all what a manifold is, this result should be pretty clear. So if it seems mysterious, please spend some serious time thinking about this.

maybe i am missing the point of confusion, i.e. maybe it was a misunderstanding about what the "disjoint union" topology defines, i.e. it defines the naive disjoint uniion. namely the point is that one can take as the open set in one space just the empty set, as suggested earlier.

Yes, i was re reading the question... I don't know if there is a more rigorous way to do it, but:
I propose this, a simple answer by a beginner:
Seems now pretty obvious (here is the problem, maybe it is not obvious) that we can chart from the open sets of A to, say, Rm, and A is open. Now, we can chat too from B to Rm, being B open.
Now, If we get the disjoint between both the Manifolds, or the chart will leave from A's opensets, or it will need to leave from B's openset, to Rm, so immediately it is a manifold m dimensional.

 

[Office_Shredder]

To be totally rigorous you can just observe if $x\in U \subset M$ is a chart, then $U\cup \emptyset$ is an open set in $M\cup N$ and hence a chart with the same charting function.