MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

4 added 704 characters in body
and $f:X\to X$ is given by $f(x,y):= (x,y + u(x))$, using that EM-spaces are abelian groups). There are also self-maps of finite simply connected complexes that are the identity on homology, but not on homotopy, see Diarmuid Crowleys answer to http://mathoverflow.net/questions/11364/cohomology-of-fibrations-over-the-circle-how-to-compute-the-ring-structure/55609#55609

• Construct two simply-connected CW complexes $X$ and $Y$ such that $H^* (X;F) \cong H^* (Y;F)$ for any field, as rings and modules over the Steenrod algebras, but which are not homotopy equivalent. EDIT: Appropriate Moore spaces do the job, see Eric Wofseys answer.

• Let $f: X \to Y$ be a map of CW-complexes. Assume that $[T,X] \to [T;Y]$ is a bijection for each finite CW complex $T$ ($[T,X]$ denotes free homotopy classes). What assumptions are sufficient to conclude that $f$ is a weak homotopy equivalence? EDIT: the answer has been given by Tyler Lawson, see below.

• Do there exist spaces $X,Y,Z$ and a homotopy equivalence $X \times Y \to X \times Z$, without $Y$ and $Z$ being homotopy equivalent? Can I require these spaces to be finite CWs? EDIT: without the finiteness assumptions, this question was ridiculously simple.

• Do you know examples of fibrations $F \to E \to X$, such that the integral homology of all three spaces is finitely generated (so that the Euler numbers are defined) and such that the Euler number is not multiplicative, i.e. $\chi(E) \neq \chi(F) \chi(X)$? Remark: is $X$ is assumed to be simply-connected, then the Euler number is multiplicative (absolutely standard). Likewise, if $X$ is a finite CW complex and $F$ is of finite homological type (less standard, but a not so hard exercise). So any counterexample would have to be of infinite type. The above fibration $BSL_2 (Z) \to BZ/12$ is a counterexample away from the primes $2,3$, but do you know one that does the job in all characteristics?. Of course, the ordinary Euler number is the wrong concept here.