# If $X\times Y$ is homotopy equivalent to a finite-dimensional CW Complex, are $X$ and $Y$ as well?

Is there a space $X$ that is not homotopy equivalent to a finite-dimensional CW complex for which there exists a space $Y$ such that the product space $X\times Y$ is homotopy equivalent to a finite-dimensional CW complex? If so, how might we construct an example?

A first consideration could be where $X$ has infinitely many nontrivial homology groups, in which case $X$ is not homotopy equivalent to a finite-dimensional CW complex. Yet, in this case it follows from the Künneth Formula that no suitable $Y$ exists. Of course, this only rules out spaces with infinitely many nontrivial homology groups.

Another consideration could be where $X$ is an Eilenberg–Maclane space for which $\pi_1(X)$ has non-trivial torsion, in which case $X$ is, again, not homotopy equivalent to a finite-dimensional CW complex. In this case, if $\pi_1(X)$ is abelian then $X$ is homotopy equivalent to $L\times Z$ for some other space Z and an infinite-dimensional lens space $L$, and hence $X$ has infinitely many nontrivial homology groups, which implies, again, that no suitable $Y$ exists. This has led me to wonder if there exists a space $Y$ and a noncommutative group $G$ with non-trivial torsion such that $Y\times K(G,1)$ is homotopy equivalent to a finite-dimensional CW complex, where $K(G,1)$ denotes an Eilenberg–Maclane space.

Any insight into approaching this problem is greatly appreciated.

• A triviality: Y can be the empty space – Thomas Rot Apr 19 '17 at 22:21
• Not exactly what you wanted, but there are pointed CW complexes $X$ that are not finite, but such that $S^1\wedge X$ is finite, for example $BG$ for $G$ an infinite acyclic group. – Denis Nardin Apr 19 '17 at 22:55
• If $Y$ is nonempty then $X$ is a homotopy retract of a CW complex, therefore homotopy equivalent to a CW complex. If $X$ is a homotopy retract of a finite-dimensional CW complex, then $X$ has no cohomology above some dimension $d$, for any coefficient system. This in turn implies (if $d\ge 3$) that $X$ is weakly, therefore strongly equivalent, to a $d$-dimensional complex. – Tom Goodwillie Apr 20 '17 at 0:52
• Just to add references to Tom's answer. The last sentence it is theorem E (p.63) of C.T.C. Wal, "Finiteness conditions for CW complexes", see math.uchicago.edu/~shmuel/tom-readings/…. The statement that a homotopy retract of a CW-complex is a CW complex is e.g. in Hatcher's "Algebraic Topology", proposition A.11. – Igor Belegradek Apr 20 '17 at 1:22
• A postscript to Igor Belegradek's comment: The argument in Proposition A.11 of my book shows that a space dominated by a CW complex of dimension $n$ is homotopy equivalent to a CW complex of dimension $n+1$, where the extra dimension arises from a mapping telescope construction. The argument, which is due to J.H.C.Whitehead if I remember correctly, uses only elementary homotopy theory. Using cohomology and Wall's work one obtains the stronger result that the increase in dimension is not really necessary, at least in dimensions 3 and greater. – Allen Hatcher Apr 20 '17 at 2:53