# Existence of topological space which has no “square-root” but whose “cube” has a “square-root”

Does there exist a topological space $$XX$$ such that $$X≆Y×YX \ncong Y\times Y$$ for every topological space $$YY$$ but
$$X×X×X≅Z×ZX\times X \times X \cong Z\times Z$$
for some topological space $$ZZ$$ ?

Here $$≅\cong$$ means homeomorphic.

## Answer

Yes.

Here are two pieces of input data.

1) Let $$MM$$ 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 $$MM$$, called the fundamental group at infinity: if you choose a proper (inverse image of compact sets is compact) map $$γ:[0,∞)→M\gamma: [0,\infty) \to M$$, and let $$KnK_n$$ be an increasing compact exhaustion of $$MM$$ (that is, $$Kn⊂Kn+1K_n \subset K_{n+1}$$ and $$⋃Kn=M\bigcup K_n = M$$) so that $$γ(t)∉Kn\gamma(t) \not \in K_n$$ for $$t∈[n,n+1]t \in [n, n+1]$$, then one may write the inverse limit $$π∞1(M,γ):=lim\pi_1^\infty(M,\gamma) := \lim \pi_1(M - K_n, \gamma(n));$$ strictly speaking we have restriction maps $$\pi_1(M – K_n, \gamma(n)) \to \pi_1(M – K_{n-1}, \gamma(n)),\pi_1(M - K_n, \gamma(n)) \to \pi_1(M - K_{n-1}, \gamma(n)),$$ but we may use the path $$\gamma\gamma$$ from $$\gamma(n)\gamma(n)$$ to $$\gamma(n-1)\gamma(n-1)$$ to get a natural isomorphism $$\pi_1(M – K_{n-1}, \gamma(n)) \to \pi_1(M – K_{n-1}, \gamma(n-1))\pi_1(M - K_{n-1}, \gamma(n)) \to \pi_1(M - K_{n-1}, \gamma(n-1))$$ so that we may take the inverse limit over a sequence of maps as above. This is essentially independent of the choice of sequence $$K_nK_n$$. It only depends on the ray $$\gamma\gamma$$ up to a proper homotopy.

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

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

Our strategy, therefore, is to find a noncompact, contractible smooth manifold $$MM$$ of dimension $$n \geq 3n \geq 3$$ with nontrivial fundamental group at infinity. We will argue that $$\pi_1^\infty(M^k, \gamma) = 0\pi_1^\infty(M^k, \gamma) = 0$$ for $$k>1k>1$$, and hence that $$M^k \cong \Bbb R^{nk}M^k \cong \Bbb R^{nk}$$. Because you asked for a square root of $$3n3n$$, we should take $$nn$$ even. At the end we will specify $$n = 4n = 4$$.

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

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

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

What this proves, altogether, is that if $$MM$$ is a compact, contractible manifold of dimension $$n \geq 3n \geq 3$$, and $$\pi_1(\partial M) \neq 0\pi_1(\partial M) \neq 0$$, then $$MM$$ is not homeomorphic to $$\Bbb R^n\Bbb R^n$$, but $$M^kM^k$$ is homeomorphic to $$\Bbb R^{nk}\Bbb R^{nk}$$ for any $$k > 1k > 1$$. What remains is twofold: to show that such $$MM$$ 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 $$n \geq 5n \geq 5$$, if $$\Sigma\Sigma$$ be an $$(n-1)(n-1)$$-manifold which has $$H_*(\Sigma;\Bbb Z) \cong H_*(S^{n-1};\Bbb Z)H_*(\Sigma;\Bbb Z) \cong H_*(S^{n-1};\Bbb Z)$$, it is a theorem of Kervaire that $$\Sigma\Sigma$$ bounds a contractible manifold $$MM$$. If $$\pi_1 \Sigma \neq 0\pi_1 \Sigma \neq 0$$ (which is equivalent to saying “$$\Sigma\Sigma$$ is not homeomorphic to the $$(n-1)(n-1)$$-sphere”, by the higher-dimensional Poincare conjecture), then this gives an example of what we want. (In fact, for $$n \geq 6n \geq 6$$, Kervaire proved that you can even construct such `homology $$(n-1)(n-1)$$-spheres’ with any specified finitely presented fundamental group $$\pi\pi$$, modulo the conditions $$H_1(\pi) = H_2(\pi) = 0H_1(\pi) = H_2(\pi) = 0$$.) So we see there is any such compact manifold $$MM$$, and hence noncompact manifold $$M^\circM^\circ$$, for any dimension $$n \geq 4n \geq 4$$.

If $$M^\circM^\circ$$ were a product $$X \times XX \times X$$ of two spaces, then first, observe $$XX$$ would need to be contractible; second, it is a homology manifold (this is a local condition in terms of the relative homology of $$(X, X – p)(X, X - p)$$ at all points $$pp$$ which ensures duality properties) of dimension $$\dim M/2\dim M/2$$. A homology manifold of dimension $$\leq 2\leq 2$$ 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 $$\dim M = 4\dim M = 4$$ here; then $$XX$$ is a contractible surface, so the classification of compact surfaces implies $$X \cong \Bbb R^2X \cong \Bbb R^2$$ (see eg here). This contradicts $$0 = \pi_1^\infty(\Bbb R^4) \cong \pi_1^\inf(M^\circ) \neq 00 = \pi_1^\infty(\Bbb R^4) \cong \pi_1^\inf(M^\circ) \neq 0$$, and so this is impossible.

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

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

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

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