6
$\begingroup$

Let $X$ and $Y$ be CW complexes.

Inside of the space of maps $\mathrm{map}(X,Y)$, we have the subspace $\mathrm{CW}(X,Y)$, consisting of just the cellular maps from $X$ to $Y$. The Cellular Approximation Theorem tells us that the inclusion $\mathrm{CW}(X,Y)\to \mathrm{map}(X,Y)$ induces an isomorphism on $\pi_0$ (in both the pointed and unpointed contexts).

My question: is the inclusion a weak equivalence? (Feel free to use any reasonable topology.)

EDIT: The answer is NO. I'm leaving this here since some interesting ideas are percolating in the comments.

$\endgroup$
  • 3
    $\begingroup$ Let X be a point. (Note that cellular approximation only gives you a surjection on $\pi_0$ - you need to use cellular homotopies $X \times I \to Y$ as well to get the correct answer.) $\endgroup$ – Tyler Lawson Nov 10 '10 at 19:08
  • $\begingroup$ You're right. Oops. $\endgroup$ – Jeff Strom Nov 10 '10 at 19:11
  • 7
    $\begingroup$ I suppose the simplicial set whose $n$-simplices are the cellular maps $X\times \Delta^n\to Y$ models the homotopy type of $\mathrm{map}(X,Y)$. $\endgroup$ – Charles Rezk Nov 10 '10 at 19:37
  • $\begingroup$ > Charles Rezk Does it true for the infinity-dimensional X? $\endgroup$ – Nikita Kalinin Nov 11 '10 at 8:10
  • $\begingroup$ @Nikita. I would think so. It should be enough to prove that $\mathrm{Cell}(X_m,Y)\to \mathrm{Cell}(X_{m-1},Y)$ is a Kan fibration, where $X_m$ is the $m$-skeleton of $X$, and $\mathrm{Cell}$ is the simplicial set of maps I defined, and that looks like it follows from cellular approximation. $\endgroup$ – Charles Rezk Nov 11 '10 at 15:37

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.