# Can the Lawvere fixed point theorem be used to prove the Brouwer fixed point theorem?

The Lawvere fixed point theorem asserts that if $X, Y$ are objects in a category with finite products such that the exponential $Y^X$ exists, and if $f : X \to Y^X$ is a morphism which is surjective on points in the sense that the induced map $\text{Hom}(1, X) \to \text{Hom}(1, Y^X)$ is surjective, then $Y$ has the fixed point property: for every morphism $g : Y \to Y$ there exists a point $y : 1 \to Y$ such that $g \circ y = y$.

The Brouwer fixed point theorem asserts that the closed $n$-disks, all of which I will denote by $D$ for ease of notation, have the fixed point property as objects of $\text{Top}$.

Seeing these two theorems together, it is tempting to try to prove the latter from the former by finding a topological space $X$ such that the exponential $D^X$ exists, together with a surjective continuous map $X \to D^X$. Does there in fact exist such an $X$?

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I guess I should mention, in passing, that Lawvere's is much more constructive than Brouwer's. So, if there is an implication that way, the space $X$ will have to contain a lot of non constructive data. (What is needed is essentially a choice of a point from every nonempty compact subspace of $\mathbb{R}$.) –  François G. Dorais Jul 12 '13 at 20:54
Hey! Next, let's prove the measure-theoretic Fubini theorem from the Category theory Fubini theorem! –  Gerald Edgar Jul 12 '13 at 21:08
Category theory seems to have the universal property of eliciting snarky remarks from those who don't have much taste for it. –  Todd Trimble Jul 13 '13 at 11:20
@user36938 : well, I see your point, but I think maybe you're being a little harsh ("does not belong on MO"). Let's say the question might have been a shade offhand (e.g., maybe Qiaochu hadn't considered the constructivity aspects before posting), but that's true of so many questions here; it's okay to kick it around for a few minutes before deciding that it's naive. It all comes under the heading of exploring. For a comparison: there are a lot of naive questions about logic here. But they usually don't elicit the same kinds of cracks, do they? No "logic corrupting the youth of today" trope. –  Todd Trimble Jul 15 '13 at 6:22
Well, I'm not sure. All I can say is that there are a lot more wise-cracks about category theory, and a lot of people think they understand it well enough to make pronouncements on (and jokes about) it. Happily, such jokes are becoming more and more pass\'e, as more and more people see that category theory is seriously useful stuff, and that professional category theorists are not in the business of "trying to get something from nothing". (Still, I can't think of any other field where people make such jokes. Can you?) –  Todd Trimble Jul 15 '13 at 17:37

In my experience it is worth considering variants of Lawvere fixed point theorem. In the present case, I would split things up as follows, in order to circumvent the non-constructive nature of Brouwer's fixed point theorem.

Also, let me point out that we need not worry about exponentials too much, even though they do not exist in the category of topological spaces, unless the exponent is nice enough. We can move over to a cartesian-closed subcategory, such as teh compactly generated spaces, or to a cartesian closed supcategory, such as equilogical spaces.

Theorem: [Approximate Lawvere] Suppose $(B, d)$ is a metric space and $e : A \to B^A$ is a continuous map, such that for every continuous map $g : A \to B$ and $\epsilon > 0$ there is $a \in A$ such that $d(e(a)(a), g(a)) < \epsilon$. Then every continuous map $f : B \to B$ has approximate fixed points: for every $\epsilon > 0$ there is $b \in B$ such that $d(b, f(b)) < \epsilon$.

Proof. Given any $f$ and $\epsilon$ consider the map $g(a) = f(e(a)(a))$. there is $a \in A$ such that $d(e(a)(a), g(a)) < \epsilon$ and then $b = e(a)(a)$is an $\epsilon$-approximate fixed point of $f$. QED.

One way to use the theorem is via the sup metric (allowing infinite distance):

Corollary: If $e : A \to B^A$ has a dense image in the sup metric on $B^A$ then every endomap on $B$ has approximate fixed points.

Suppose we could apply the previous theorem to the closed ball $D^n$. Then we would know (constructively!) that every endomap on $D^n$ has approximate fixed points. then we just have another easy step, which contains all the classical reasoning needed:

Theorem: Suppose $X$ is compact and $f : X \to X$ has an $\epsilon$-approximate fixed point for every $\epsilon > 0$. Then $f$ has a fixed point.

Proof. By countable choice, for every $n$ there is $x_n \in X$ such that $d(x_n, f(x_n)) < 1/n$. Because $X$ is compact, $x_n$ has a subsequence converging to some $y \in X$. It is now easy to see that $y$ is a fixed point of $f$. QED.

But I do not see how to apply the approximate Lawvere to the closed ball, if that is even possible.

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This is not an answer, but I will try to explain why I think that it is unlikely for such a space $X$ to exist.

If we replace $D$ by say, a sphere, then (using Lawvere's fixed point theorem and the fact that spheres do have fixed-point-free self-maps) such a space $X$ does not exist for the sphere. Now, I really don't see how to possibly use the fact that the disc is a disc in constructing the space $X$. So I am tempted to believe that your $X$ does not exist.

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Why can't we do an "Eilenberg swindle" where we take X to be a countable iterated exponential of disks? Something must go wrong because if that worked for disks, it would work for spheres too. Does such an iterated exponential not make sense? –  Chris Schommer-Pries Jul 12 '13 at 20:56
If such a swindle worked, it would work in the category of sets as well. It seems to me the basic problem is that there are no canonical maps between $X$ and $D^X$, so there's no obvious way to take a (co)limit of finite iterated exponentials to get an "infinite iterated exponential". –  Eric Wofsey Jul 12 '13 at 21:19
@Eric: this doesn't quite sink us. For example, as it turns out there is a canonical map $D \to D^D$ (constant functions) as well as a canonical map $D^{D^D} \to D^{D^{D^D}}$ (apply the functor $X \mapsto D^X$ to the previous map twice). These are the odd-numbered maps in a hypothetical diagram of shape $D \to D^D \to D^{D^D} \to ...$ that we would want. But it turns out that there are no canonical candidates for the even-numbered maps (in a precise sense; there are no such maps in the free cartesian closed category on $X$). We can "fix" this by considering a diagram of shape... –  Qiaochu Yuan Jul 13 '13 at 4:37
$D \to D^{D^D} \to D^{D^{D^{D^D}}} \to ...$, which does exist but which doesn't give us the kind of fixed point we would want (if the colimit exists and if the double dual functor $X \mapsto D^{D^X}$ preserves it then it's a fixed point of the double dual functor). –  Qiaochu Yuan Jul 13 '13 at 4:38