category theory: brouwer's fixed point theorem - Functional

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= \ /\ \|
www.mathematik.tu-darmstadt.de/~blumensath /"\ o----|
____________________________________________________________________\___|

Achim Blumensath wrote:
> 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.

Yes, you are correct. I had to more carefully read the previous pages.
The "retraction theorem" (one of three they give) says: There is no
continuous map which is a retraction for j.

[...]

Thank you for your reply. I am still pondering it...

-thant