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category theory: brouwer's fixed point theorem
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| Thant Tessman 2007-08-15, 7:11 pm |
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I'm having trouble with the exercises in Session 10 of Lawvere and
Schanuel.
Let j : C -> D be the inclusion of a circle into the disc. Suppose we
have two continuous maps f : D -> D and g : D -> D, and that g satisfies
g o j = j. Use the retraction theorem to show that there must be a point
x in the disk at which f(x) = g(x). Hint: The fixed point theorem is the
special case g = 1D.
The retraction theorem says: Given f : A -> B, a retraction for f is a
map r : B -> A for which r o f = 1A.
The retraction in our case is presumably:
r : D -> C
And by the retraction theorem we have:
r o j = 1C
I think I know what all these words mean, but I have no clue where to
take it from here, and this feels like an important point. I'd be
grateful for a clue.
-thant
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| Achim Blumensath 2007-08-16, 4:31 am |
| Hello,
Thant Tessman wrote:
> Let j : C -> D be the inclusion of a circle into the disc. Suppose we
> have two continuous maps f : D -> D and g : D -> D, and that g satisfies
> g o j = j. Use the retraction theorem to show that there must be a point
> x in the disk at which f(x) = g(x). Hint: The fixed point theorem is the
> special case g = 1D.
>
> The retraction theorem says: Given f : A -> B, a retraction for f is a
> map r : B -> A for which r o f = 1A.
This seems to be the definition of a retraction, not the retraction
theorem. My guess would be that the "retraction theorem" is the
statement:
There exists no retraction from D to C, i.e., j has no left inverse.
> I think I know what all these words mean, but I have no clue where to
> take it from here, and this feels like an important point. I'd be
> grateful for a clue.
Assuming that f(x) <> g(x), for all x, try to construct a retraction
from D to C.
Achim
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Achim Blumensath \O/ \___/\ |
TU Darmstadt =o= \ /\ \|
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