Topology problem

  • Thread starter kakarukeys
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mathwonk said:
let me make my objection to your proof as simple as possible:

just because h = f+g, it does not follow that h(U) = f(U)+g(U), which is what you are apparently claiming in your argument. this is what hurkyl's counter example shows.

i.e. h(U) consists only of all points of form f(x)+g(x) for all x in U, while f(U)+g(U) consits of the much larger collection of points of form f(x)+g(y) for all x,y in U.

I.e. f(U) +g(U) is the image of the product set UxU under the map f+g, while h(U) is the image only of the diagonal.
Well, now that you put it that way, I see the problem. I tried to eliminate that problem by writing the addends in the form of h(x,y)=f(x)g(y), but now I see that the sum of them does not quite conform to the way I wanted it to.
 

mathwonk

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here is simpler conundrum for you: your type of argument would also prove that det^2 is an open mapping whereas clearly it is not, even for n=1.
 

mathwonk

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ok here's a quicky solution, inspired by hurkyl's:

we must show that in every nbhd of any matrix A with det = 0, there are matrices whose determinants have both signs.

first of all, note that most polynomials of degree n have n distinct roots (those that do not lie on the hypersurface where the discriminant equals zero).

hence in every open set around A there are matrices which are diagonalizable. now it is trivial that on every nbhd of a diaonalizable matrix of det zero, there are matrices with det of boths signs. thus we are done.

i.e. if A is diagonalizable, the for some invertible P, P^-1 AP = D is diagonal. Hence there are matrices near D with determinants of both signs, and thus applying
P(.)P^-1 to them yields matrices near A with determinants of both signs.
 

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