In general, we might ask when we can find interesting spaces $X, Y$ such that $X$ is homeomorphic to $[X, Y]$. By the Lawvere fixed point theorem $Y$ must have the fixed point property. Happily, $Y = [0, 1]$ does in fact have the fixed point property (e.g. by the intermediate value theorem), so this is not an obstruction.

I know that computer scientists have constructed spaces $X$ such that $X$ is homeomorphic to $X^X$ in domain theory, but I don't know enough about it to tell whether the techniques are relevant here.

  • 1
    Functions = continuous functions, I assume? – Igor Rivin Mar 17 '17 at 2:06
  • Yes, I suppose that's worth clarifying. Edited. – Qiaochu Yuan Mar 17 '17 at 2:36
  • 1
    I guess you are considering the compact open topology on $[X,Y]$? Or is the question is also about finding the topology on $[X,Y]$? – Cusp Mar 17 '17 at 5:20
  • 3
    There are two candidates one might consider as a first approximation: the direct limit of $C^k(I)\to C^{k+2}(I)\to C^{k+4}(I)\to...$ or the inverse limit of $...\to C^{k+4}(I)\to C^{k+2}(I)\to C^k(I)$ (the latter is $C(\_)$ of the former) – მამუკა ჯიბლაძე Mar 17 '17 at 5:52
  • 3
    I'd pause before decreeing it should be the compact-open topology; I'd say it should be the exponential topology first and foremost, i.e., the topological structure such that there is a natural isomorphism $Top(-, [X, Y]) \cong Top(- \times X, Y)$. This need not be the compact-open topology (although it will be if $X$ is say locally compact Hausdorff). – Todd Trimble Mar 17 '17 at 12:18
up vote 26 down vote accepted

There cannot be such a homeomorphism, because Lawvere's fixed point theorem would give us something too constructive: a continuous map $[0,1]^{[0,1]} \to [0,1]$ that assigns a fixed point to each continuous map $[0,1] \to [0,1]$.

Following the proof of the Lawvere fixed point theorem, given a homeomorphism $f: X \cong [0,1]^X$ and a map $g: [0,1] \to [0,1]$, we construct a map $h: X \to [0,1]$ by $h(x) = g(f(x)(x))$, and then we get an element $x^* = f^{-1}(h)$ of $X$. Then,

$$ f(x^*)(x^*) = h(x^*) = g(f(x^*)(x^*)), $$

so we have found a fixed point of $g$. This construction is continuous in $g$, since we're working with the exponential topology on $[0,1]^{[0,1]}$, so $g \mapsto f(x^*)(x^*)$ gives a continuous fixed-point-finding map $[0,1]^{[0,1]} \to [0,1]$.

There is a lot of work on the nonconstructivity of Brouwer's fixed point theorem for different senses of constructivity. To give a quick visual proof that no continuous map of the type considered here exists, it is sufficient to consider a path through $[0,1]^{[0,1]}$ along which such a map cannot be continuously defined. Consider linearly deforming green to red to purple in the below diagram; a fixed point (i.e. a point where the curve crosses the identity depicted in blue) cannot be continuously chosen.

Three curves showing Brouwer is nonconstructive

It's interesting that this works because we have a homeomorphism $X \cong [0,1]^X$ rather than just a continuous surjection. A continuous surjection $X \to [0,1]^X$ would be enough to deduce the intermediate value theorem using Lawvere's fixed point theorem, but not in a constructive enough way to similarly rule out.

  • Aha, nice construction! – Qiaochu Yuan Mar 22 '17 at 8:08
  • 3
    In fact for this argument to work, we don't need that $[0, 1]^X$ is homeomorphic to $X$, just that it is a retract of $X$. This observation is put to use in an answer on the related thread, here: mathoverflow.net/a/267603/2926 – Todd Trimble Apr 19 '17 at 12:07

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.