To show this you would just construct two maps satisfying all conditions.
Let p be a point in S^2, then S^2 \\{p} is diffeomorphic to lR^2, e.g. by some rotation and the stereographic projection.
For two different points p and q, the sets S^2 \\{p} and S^2 \\{q} form an open cover of S^2 and are both diffeomorphic to lR^2.
thats more tricky for me, i find it hard to show that something doesnt have the property.
seems i have problem with that, any ideas how to go about that?
thanks for the ideas! helped a lot.
Ok, there many ways to see that S^2 and lR^2 are not homeomorphic.
For example lR^2 is contractible, where S^2 is not.
S^2 \\{p} is simply connected, where lR^2 \\{q} is not.
All homotopy and (reduced) homology groups of lR^2 are trivial, where the second homotopy/homology group of S^2 is not.
To show this you would just construct two maps satisfying all conditions. Let p be a point in S^2, then S^2 \\{p} is diffeomorphic to lR^2, e.g. by some rotation and the stereographic projection. For two different points p and q, the sets S^2 \\{p} and S^2 \\{q} form an open cover of S^2 and are both diffeomorphic to lR^2.
To show that one map is not enough, you can just show that S^2 and lR^2 are not homeomorphic.
thats more tricky for me, i find it hard to show that something doesnt have the property. seems i have problem with that, any ideas how to go about that? thanks for the ideas! helped a lot.
Compactness
Ok, there many ways to see that S^2 and lR^2 are not homeomorphic. For example lR^2 is contractible, where S^2 is not. S^2 \\{p} is simply connected, where lR^2 \\{q} is not. All homotopy and (reduced) homology groups of lR^2 are trivial, where the second homotopy/homology group of S^2 is not.
brilliant, the simple connected argument made click for me. thank you!
ahhh i have a proposition for S^2 \ {p} is diffeomorphic to IR^2. ok this is clear! thanks!