There is a general principle that, for finite simply-connected CW complexes, things that are true rationally are usually true once you localize away from a finite list of primes. 


I'm interested in the possibility that a wedge $S^a \vee S^b$ might be a retract of a space, and I'll be satisfied if there is such a retract only after localizing away from a finite list of primes. (This is the case, 
for example, for any suspension $\Sigma X$ of a finite complex $X$.)


My question is this: 



> 
> Suppose $X$ and $Y$ are finite simply-connected complexes, and neither
>  of them has a wedge of two spheres as a retract, even after inverting
>  any finite set of primes. Is it possible that there is a map $f:X\to Y$
>  whose homotopy fiber $F\_f$ *does* have a wedge of two spheres as a
>  retract?
> 
> 
> 



Retracting a wedge of spheres off a homotopy fiber
Take the homotopy fibre of the inclusion $f:\mathbb{C}P^2\vee\mathbb{C}P^2\hookrightarrow \mathbb{C}P^2\times\mathbb{C}P^2$. Both of these spaces are simple connected and neither has a wedge of spheres as a retract, even after inverting any finite collection of primes (unless I've overlooked something). The homotopy fibre $F$ of the map $f$ has the homotopy type


$F\_f\simeq \Omega\mathbb{C}P^2\ast\Omega\mathbb{C}P^2\simeq(S^1\times\Omega S^5)\ast (S^1\times\Omega S^5)\simeq S^3\vee\Sigma^2\Omega S^5\vee \Sigma^2\Omega S^5\vee\dots$


Hopefully there should be enough spheres for you here. Already we have an $S^3$ retracting off, and using the James splitting we get $\Sigma^2\Omega S^5\simeq \bigvee^\infty\_{i=1} S^{4i+2}$. So (at least) $S^3\vee S^6$ indeed retracts off of $F\_f$, and this is integral. In fact $F\_f$ has the integral homotopy type of an infinite wedge of spheres.


