Does there exist a topological space X such that X≆Y×Y for every topological space Y but

X×X×X≅Z×Z

for some topological space Z ?Here ≅ means homeomorphic.

**Answer**

Yes.

Here are two pieces of input data.

1) Let M be any noncompact space. Beyond the usual invariants like homology and homotopy groups, there is a further invariant of the homeomorphism type (or proper homotopy type) of M, called the fundamental group at infinity: if you choose a proper (inverse image of compact sets is compact) map γ:[0,∞)→M, and let Kn be an increasing compact exhaustion of M (that is, Kn⊂Kn+1 and ⋃Kn=M) so that γ(t)∉Kn for t∈[n,n+1], then one may write the inverse limit π∞1(M,γ):=lim strictly speaking we have restriction maps \pi_1(M – K_n, \gamma(n)) \to \pi_1(M – K_{n-1}, \gamma(n)), but we may use the path \gamma from \gamma(n) to \gamma(n-1) to get a natural isomorphism \pi_1(M – K_{n-1}, \gamma(n)) \to \pi_1(M – K_{n-1}, \gamma(n-1)) so that we may take the inverse limit over a sequence of maps as above. This is essentially independent of the choice of sequence K_n. It only depends on the ray \gamma up to a proper homotopy.

(Similarly, there is a notion of the set of ends of a space – this is the inverse limit over \pi_0(M – K_n). This is the set we choose \gamma from, in the sense that we choose a connected component for the usual fundamental group.)

2) If M is a smooth, connected, noncompact manifold of dimension n \geq 5, a theorem of Stallings (the piecewise linear structure of Euclidean space, here) says that if M is both contractible and \pi_1^\infty(M,\gamma) = 0 for the unique end \gamma of M, then M \cong \Bbb R^n.

Our strategy, therefore, is to find a noncompact, contractible smooth manifold M of dimension n \geq 3 with nontrivial fundamental group at infinity. We will argue that \pi_1^\infty(M^k, \gamma) = 0 for k>1, and hence that M^k \cong \Bbb R^{nk}. Because you asked for a square root of 3n, we should take n even. At the end we will specify n = 4.

Here is a helpful tool in constructing such noncompact manifolds. If M is a compact manifold with boundary, then its ends of its interior M^\circ are in bijection with \pi_0(\partial M), and the fundamental group at infinity is equal to \pi_1(\partial M). (Take the ray to extend to a map [0, \infty] \to M, and let the basepoint in \partial M be \gamma(\infty); if [0,1) \times \partial M \subset M is a collar of the boundary, let the compact exhaustion be the complement of [0, 1/n) \times \partial M.)

In this situation above, the product M \times M is a compact topological manifold with boundary (it has “corners”, but these are topologically the same as boundary points). The boundary is homeomorphic to (\partial M \times M) \cup_{\partial M \times \partial M} (M \times \partial M). If \partial M is connected and M is simply connected, the Seifert van Kampen theorem dictates that the fundamental group of the result is \pi_1\partial M *_{\pi_1 \partial M \times \pi_1 \partial M} \pi_1 \partial M = 0.

Therefore, if M is simply connected with connected boundary, M \times M has simply connected boundary; and hence (M \times M)^\circ = M^\circ \times M^\circ.

What this proves, altogether, is that if M is a compact, contractible manifold of dimension n \geq 3, and \pi_1(\partial M) \neq 0, then M is not homeomorphic to \Bbb R^n, but M^k is homeomorphic to \Bbb R^{nk} for any k > 1. What remains is twofold: to show that such M exist; and to find one that is itself not a square.

First, existence. In dimension 3 there are none of these of interest: a compact contractible 3-manifold is homeomorphic to the 3-ball by the solution of the Poincare conjecture. In dimension 4 these are called Mazur manifolds and come in great supply. In dimension n \geq 5, if \Sigma be an (n-1)-manifold which has H_*(\Sigma;\Bbb Z) \cong H_*(S^{n-1};\Bbb Z), it is a theorem of Kervaire that \Sigma bounds a contractible manifold M. If \pi_1 \Sigma \neq 0 (which is equivalent to saying “\Sigma is not homeomorphic to the (n-1)-sphere”, by the higher-dimensional Poincare conjecture), then this gives an example of what we want. (In fact, for n \geq 6, Kervaire proved that you can even construct such `homology (n-1)-spheres’ with any specified finitely presented fundamental group \pi, modulo the conditions H_1(\pi) = H_2(\pi) = 0.) So we see there is any such compact manifold M, and hence noncompact manifold M^\circ, for any dimension n \geq 4.

If M^\circ were a product X \times X of two spaces, then first, observe X would need to be contractible; second, it is a homology manifold (this is a local condition in terms of the relative homology of (X, X – p) at all points p which ensures duality properties) of dimension \dim M/2. A homology manifold of dimension \leq 2 is a manifold (this seems to be well-known, but the only reference I could find was Theorem 16.32 in Bredon’s sheaf theory), so let’s take \dim M = 4 here; then X is a contractible surface, so the classification of compact surfaces implies X \cong \Bbb R^2 (see eg here). This contradicts 0 = \pi_1^\infty(\Bbb R^4) \cong \pi_1^\inf(M^\circ) \neq 0, and so this is impossible.

In fact, with some more work you can show that this M may not even be decomposed into a product at all.

**EDIT:** Thanks to Moishe Cohen’s answer here we can prove that if M is a compact contractible manifold of dimension n \geq 4 for which \pi_1 \partial M \neq 0, then M does not admit a square root. For if it did, X \times X = M, the space X would be a contractible homology manifold of dimension at least 2; by Moishe’s answer, it must have one end. Using the decomposition \text{End}(X \times X) \sim \text{End}(X) * \text{End}(X) of end-spaces of a product, we see that \pi_1^\inf(X \times X) = \pi_1^\inf(X) *_{\pi_1^\inf(X) \times \pi_1^\inf(X)} \pi_1^\inf(X) = 0, exactly as in the case of manifolds with boundary. Thus M admits no square root.

This method thus produces some M that admits no square root but whose nth power admits a kth root, for any pair of positive integers (n,k) with n > 1. It has no power to find spaces for which X^j is similarly un-rootable for j in some range; it is unique to j=1 that this works.

**Attribution***Source : Link , Question Author : user521337 , Answer Author : Community*