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I'm trying to understand the proof of Brouwer's theorem form Lawvere and Schanuel's Conceptual Mathematics:

To prove that the non-existence of a retraction implies that every continuous endomap has a fixed point, all we need to do is to assume that there is a continuous endomap of the disk which does not have any fixed point, and to build from it a continuous retraction for the inclusion of the circle into the disk.

This seems to involve unspoken assumptions that I don't know how to justify. If the non-existence of the retraction proves $ \exists x[f(x) = x] $, then I believe that it must also be true that the lack of fixed points $ \forall x[f(x) \neq x] $ implies the existence of a retraction which shares the same property as the endomap, namely:

$$∀x[(f(x) \neq x)] \to \exists r[r \circ j = 1_C]$$

(Edit note: this expression was originally $∀x[(f(x) \neq x) \to (r(x) \neq x)]$. I mistakenly thought that contradicting $ r(x) \neq x$ was necessary to the argument.)

So my question is twofold:

  • How does the lack of a fixed point in the endomap imply (as opposed to simply making possible) the existence of a retraction, $ r \circ j = 1_C $?
  • Why is it assumed that the retraction must have properties that behave as the endomap lead to a contradiction? In other words, it seems like they could be completely independent so that it's both true that $ \forall x[f(x) \neq x] $ and $ r \circ j = 1_C $.

Here's the proof from the book:

enter image description here

So, let $ j: C \to D $ be the inclusion map of the circle into the disk as its boundary, and let's assume that we have an endomap of the disk, $ f: D \to D $, which does not have any fixed point. This means that for every point $x$ in the disk $D$, $f(x) \neq x.$

From this we are going to build a retraction for $j$, i.e. a map $r: D \> \to C$ such that $r \circ j$ is the identity on the circle. The key to the construction is the assumed property of $f$, namely that for every point $x$ in the disk, $f(x)$ is different from $x$. Draw an arrow with its tail at $f(x)$ and its head at $x$. This arrow will 'point to' some point $r(x)$ on the boundary. When $x$ was already a point on the boundary, $r(x)$ is $x$ itself, so that $r$ is a retraction for $j$, i.e. $rj= 1_c$.

(The book doesn't assume any knowledge about topology nor do I know much about it.)

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You can't construct a retraction if there is a fixed point because the basic ingredient in the construction of the retraction is the displacement direction $f(x)-x$, which can be normalized to get a unit vector. If the displacement is zero then you can't get a unit vector because you can't divide by zero. In more detail, with regard to your question

How does the lack of a fixed point in the endomap imply the existence of a retraction, $ r \circ j = 1_C $?

The answer is that the absence of a fixed point of the endomap is what allows one to define a continuous retraction. With regard to your question

Why is it assumed that the retraction must behave as the endomap? In other words, it seems like they could be completely independent so that it's both true that $ \forall x[f(x) \neq x] $ and $ \exists x [r(x) = x]$

The retraction itself (when it is defined, i.e. so long as the endomap has no fixed points) satisfies $r(x)=x$ for all points $x$ on the boundary, by construction of $r$.

As I mentioned in the comments, the account provided by the authors you cite leaves an essential topological fact unmentioned. This is the fact that the identity map of the circle cannot be extended over the disk.

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  • $\begingroup$ People keep saying that that's the basic ingredient, but I've yet to find anyone explain why. But, even so, my question is about about why is it assumed that a retraction is necessary. How does the lack of a fixed point in the endomap imply the existence of a retraction? $\endgroup$ – User4407 May 3 '17 at 16:37
  • $\begingroup$ The retraction needs to be a continuous map (otherwise you haven't accomplished anything since one can certainly always find a discontinous "retraction"). If there is no fixed point then the trick with normalizing the displacement vector produces a continuous retraction. $\endgroup$ – Mikhail Katz May 3 '17 at 16:38
  • $\begingroup$ I understand, but that's not what I'm asking. $\endgroup$ – User4407 May 3 '17 at 16:40
  • $\begingroup$ Wait a minute, you need to know what the basic strategy of Brouwer's proof is. Modern proofs usually use either a homology argument or a homotopy argument. The point is that the existence of such a retraction would contradict the nonvanishing of such a class. For example, the relative homology class of the disk with respect to its boundary is nonzero. If a retraction existed, it would imply that such a class is zero, which gives a contradiction. $\endgroup$ – Mikhail Katz May 3 '17 at 16:43
  • $\begingroup$ I don't think you understood the question. If $ \neg R \to F $ is a relevant argument, it is so in virtue of $ \neg F \to R $. You keep addressing $ \neg R $ but my question is about the basis for $ \neg F \to R $. $\endgroup$ – User4407 May 3 '17 at 17:14
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The claim made is essentially: suppose we can show that there is no continuous retraction from $B^n$ onto $S^{n-1}$ (you would use homotopy or homology to show that, usually).

Then every continuous $f: B^n \to B^n$ (an endomap) has a fixed point. Reason: if we have any such endomap $f$ without a fixed point, then define $r(x)$ as the unique point of intersection of the ray $x + t(f(x) -x), t \in [0,\infty)$ with $S^{n-1}$ having the smallest $t$, is well-defined (as $f(x) \neq x$ on $B^n$, we always have a non-zero vector as a direction for the ray), continuous (hint here, e.g.) and a retraction from $B^n \to S^{n-1}$, (for $x \in S^{n-1}$ $t=0$ is the first intersection point, so $r(x) = x$) which we have shown to be impossible.

Note that also that having no such retraction is a necessary condition for the fixed point property: if we'd have such an $r$ we could define $f = h\circ r: B^n \to B^n$, where $h$ is a non-trivial rotation of $S^{n-1}$,and this $f$ has no fixed point (points inside map to points on the boundary, so are never fixed, and on the boundary $r$ keeps them fixed but $h$ moves them).

So the fixed point property for $B^n$ (i.e. Brouwer's theorem) is equivalent to the no-retraction theorem, in the sense that one is easily derived from the other.

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  • $\begingroup$ Your reason: "if we have any such endomap $f$ without a fixed point..." only implies that the retraction would be well defined and continuous. It doesn't, however, imply that its existence would be a necessary consequence. If the existence of the retraction is only a trivial possibility, then proving it's non-existence is also trivial. Therefore, there has to be some logical connection between the endomap and the retraction: $ ∀x[(f(x) \neq x) \to (r(x) \neq x)] $. Otherwise it's completely irrelevant to the proof. $\endgroup$ – User4407 May 3 '17 at 18:37
  • $\begingroup$ Yes, if we have a fixed point free endomap, we necessarily have a retraction. The argument I gave shows that. The point is that most proofs will argue the nonexistence of such a retraction. I don't understand your $r(x)\neq x$ remark. Retractions have a lot of fixed points, the whole of $S^{n-1}$. $\endgroup$ – Henno Brandsma May 3 '17 at 18:47
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    $\begingroup$ @PédeLeão I give the construction using the point of intersection with the ray. This $r(x)$ is constructed from such an $f$ $\endgroup$ – Henno Brandsma May 4 '17 at 14:01
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    $\begingroup$ You ask "how does the lack of a fixpoint lead to the existence of a retraction?" That's what the definition of $r$ shows. $\endgroup$ – Henno Brandsma May 4 '17 at 14:23
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    $\begingroup$ The possibility to define it shows its existence. $\neg F \to R$ is only falsified by $\neg F \land \neg R$. We show $F$ because we can define $r$ from a counterexample to $F$. This shows $\neg F \to R$ which is equivalent to $\neg R \to F$ by contraposition. As we supposedly have shown $\neg R$ we conclude $F$. $\endgroup$ – Henno Brandsma May 4 '17 at 18:44
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The claim being made is actually (roughly) $$(\exists r)(r\,\mathrm{is\,a\,retraction})\implies (\forall f)(\exists x)(f(x)=x)$$ which is distinct from the formula you gave. I do not know if this is the source of your confusion, but it's worth pointing out.

For your first question, suppose $f:B^n\to B^n$ has no fixed point. When we may define $r: B^n\to S^{n-1}$ as follows: given $x\in B^n$, consider the unique line intersecting both $x$ and $f(x)$ (this line is unique because $f(x)\neq x$). Then define $r(x)$ to be the point of intersection of this line with the sphere, on the side of $f(x)$. This defines a continuous retraction from the ball to its boundary.

Thus, proving that no retraction exists from the ball to its boundary is sufficient to prove the Brouwer fixed point theorem. For if there were a map with no fixed point, as shown, we could contruct such a retraction. This should clarify your second question as well.

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  • $\begingroup$ The claim that you're making is irrelevant if the retraction need not exist. If an endomap, free of fixed points, could exist without a retraction, it doesn't make sense to posit any arguments about the retraction's properties. $\endgroup$ – User4407 May 3 '17 at 16:33

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