Two spaces, each a retract of the other, which are not homotopy equivalent Hoping to find a simple example of spaces $X$ and $Y$ such that each is a retract of the other (i.e there exist continuous $i : X \to Y, p : Y \to X$ with $p\circ i = \mathrm{id}_X$ and also continuous $j : Y \to X, q : X \to Y$ with $q \circ j =\mathrm{id}_Y$) but such that $X$ and $Y$ are not homotopy equivalent.
I think there should be a fairly simple example, since I don't care if the spaces $X$ and $Y$ are connected. A connected example would be nice, of course...
Added: Actually, I just noticed it's not very hard to do this if we don't mind using some kind of infinite swindle. Lots of examples are possible along these lines.... Let $T_n$ denote the space which is the wedge of $n$ circles. Observe $T_n$ is a retract of $T_m$ when $m \geq n$. Thus,  $X = T_1 \sqcup T_3 \sqcup T_5 \sqcup \ldots$ and $Y =T_2 \sqcup T_4 \sqcup T_6 \sqcup \ldots$ are retracts of each other, but they shouldn't be homotopy equivalent.
In light of this addition, let me change the question to:

Revised Question: Can we find a nice example of two connected spaces, each a retract of the other, such that they are not homotopy equivalent?

 A: Here is an obstruction to getting an example that is too "nice".  Suppose you want $X$ and $Y$ to be simply connected and have the homotopy type of CW-complexes.  Then since $i:X\to Y$ is not a homotopy equivalence, there is some $n$ such that $i_*:H_n(X)\to H_n(Y)$ is not an isomorphism.  We then see that $j_*i_*:H_n(X)\to H_n(X)$ is an inclusion of a proper direct summand.  No finitely generated abelian group can be isomorphic to a proper direct summand of itself, and so $H_n(X)$ cannot be finitely generated.  In particular, for instance, $X$ cannot have the homotopy type of a finite CW-complex.
(So morally, this says any example must either be "infinite" or involve exploiting weird nonabelian stuff with $\pi_1$.  I don't know whether you can actually get examples using weird nonabelian stuff.  For instance, I don't at the moment see whether it is possible to have a finitely generated group that is isomorphic to a proper retract of itself.)
A: $X=S^1 \vee T^2 \vee T^2 \vee ...$, $Y=T^2\vee T^2 \vee T^2 ...$. But I do not know any examples which are finite CW complexes.
