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.

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.

exponentialtopology 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). $\endgroup$ – Todd Trimble♦ Mar 17 '17 at 12:182more comments