All mapping space between CW complexes is a CW complex? Let $\mathrm{Map}(X,Y)$ denote the (unbased) cellular mapping space from $X$ to $Y$.
If $X$ and $Y$ are finite CW complexes, is $\mathrm{Map}(X,Y)$ a CW complex?
Can we know the cell structure of $\mathrm{Map}(X,Y)$?
For example, what is the cell structure of $\mathrm{Map}(S^n,S^k)$ for $n \geq k$?
Please recommend related papers and textbooks.
 A: I think I should be posting this as a comment, but as i don't have enough reputation to comment I take the liberty to post this as an answer.  
I am here just elaborating on Eric Wofsey's answer.
Let's take the case, say, $X=\mathbb{N}$ and $Y=I=[0,1]$.  Then 
$Map(X,Y)$ is Hilbert cube.
On the other hand,  the definition of a CW-complex says that a CW complex has to be the union of finite skelta, and 
for reasons explained in Eric Wofsey's answer, this is not the case.
A: A mapping space $\mathrm{Map}(X,Y)$ between two finite CW-complexes never admits a cell structure if both $X$ and $Y$ are positive-dimensional.  If you use the compact-open topology, this essentially follows from this answer: the mapping space is a complete metric space and thus satisfies the Baire category theorem, but if every component of $Y$ is positive-dimensional then it's easy to see that any open subset of $\mathrm{Map}(X,Y)$ is infinite-dimensional.  If you use the compactly generated topology, the Baire category theorem may not hold in $\mathrm{Map}(X,Y)$ itself, but you can find a Hilbert cube $H$ embedded in it.  Since $H$ is compact, its topology does not change when you pass to the compactly generated topology, and any open subset of $H$ is still infinite-dimensional.  Thus by the Baire category theorem, $H$ cannot be the union of its intersections with finite-dimensional skeleta.
However, $\mathrm{Map}(X,Y)$ does always have the homotopy type of a CW-complex.  For a quite general version of this, see Milnor's paper "On spaces having the homotopy type of a CW-complex".
