# category theory: brouwer's fixed point theorem - Functional

This is a discussion on category theory: brouwer's fixed point theorem - Functional ; I'm having trouble with the exercises in Session 10 of Lawvere and Schanuel. Let j : C -&gt; D be the inclusion of a circle into the disc. Suppose we have two continuous maps f : D -&gt; D and ...

1. ## category theory: brouwer's fixed point theorem

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

2. ## Re: category theory: brouwer's fixed point theorem

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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3. ## Re: category theory: brouwer's fixed point theorem

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.

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