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By the Theorem of Milnor in his paper "On spaces having the homotopy type of a CW-complex", the function space $\operatorname{Hom}(X,Y)$ (with the compact-open topology) is homotopy equivalent to a CW complex (say $Z$) when $X$ is compact.

Do we know anything about the cell structure (of $Z$) in this case? [When $X$ is $S^n$, I believe there is a well-known construction, and I would like to know more about it as well.]

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    $\begingroup$ These mapping spaces do not admit CW structures. They are (when $X$ is compact) homotopy equivalent to CW complexes. $\endgroup$
    – Tyrone
    Commented Dec 13 at 8:11
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    $\begingroup$ Oh yes, I meant that I would like to know the cell structure of the space $Z$ that is homotopy equivalent to $\operatorname{Hom}(X,Y)$ when $X$ is compact (also when it is the $n$-sphere). Thanks for the clarification! :) $\endgroup$
    – May
    Commented Dec 13 at 8:43
  • $\begingroup$ I have edited my question. ^^ $\endgroup$
    – May
    Commented Dec 13 at 8:45
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    $\begingroup$ You can make a simplicial set of the right homotopy type by letting the $n$-simplices be the maps $\Delta^n\times X \to Y$. $\endgroup$
    – Dave Benson
    Commented Dec 13 at 8:52
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    $\begingroup$ Related question mathoverflow.net/q/309542 There is a well known construction when $X=S^n$ and $Y=S^n Z$, with $Z$ connected. The construction generalises to the case when $X$ is a parallellisable n-dimensional manifold. $\endgroup$
    – Gregory Arone
    Commented Dec 13 at 9:35

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