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10:40 PM
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A: projective space and torus

Anubhav.KHere I am trying to give an intuitive idea...Let see how far I can make it clear... let $U,V$ be two chart which cover these two spaces, $U\cup V=X$ and $U,V$ is homeomorphic to open disc $\mathbb{D^2}$...now since X is compact so boundary of $U$ should properly contained in $V$ similarly bound...

 
sorry if you think this language is not properly written...
 
why should they be homeomorphic to an open disc, I mean I don't see why $U,V$ have to be connected.
 
they are chart...
 
I don't see where the connectedness enters.
 
I dont understand your question...
 
10:40 PM
I mean, I don't see why a chart can only be defined on connected sets. You clearly assume this, as you say that $U,V$ are homeomorphic to a disc.
 
$U,V$ both are homemorphic with $\mathbb{R^2}$ so homeomorphic with $\mathbb{D^2}$
 
but connectedness is preserved under homeomorphisms, so this assumes that $U$ and $V$ are connected, right?
 
you are right...
 
but this does not have to be the case in general.
 
I dont understand what are you talking about now...
 
10:40 PM
well, I am saying your answer is wrong cause you assume that the chart is defined on connected sets. this does not have to be the case
 
please go and read the definitions in wiki...
and then try to understand those meanings...
 
I mean where do we have in this definition that $U$ is homeomorphic to the open disc?
 
Do you know that open disc and R2 is homeomorphic??
 
yes, I do......
I mean, my question is just. How do you know that $U$ has to be homeomorphic to the open disc or $\mathbb{R}^2$?
 
I define them like that
 
10:40 PM
so your proof is not valid in general, cause you used that restriction?
 
It is valid... You please show this proof to some one
 
well if it is valid, then why don't you tell me why $U$ NEEDS(!) to be homeomorphic to the open disc?
 

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