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I'm thinking to the famous problem of cancellation property in Top, i.e: $$T_1 \times T_2 \cong T_1 \times T_3 \Rightarrow T_2 \cong T_3. $$ Clearly there are many counterexamples like $\prod_{i \in \omega}S_{1_{i}}$ or $ \oplus_{i \in \omega}S_{1_{i}}$ but these counterexamples can be bypassed by giving a definition.

We say that a topological space T is $\Pi$-compact iff $$T \cong \prod_{i\in I}T_i, \ T_i \neq \{e\} \ \Rightarrow |I| < \infty.$$

We say that a topological space T is $\Sigma$-compact iff $$T \cong \oplus_{i\in I}T_i, \ T_i \neq \{e\} \ \Rightarrow |I| < \infty.$$

We say that a topological space is $\times$-compact iff it's $\Pi$ and $\Sigma$ compact.

Actually an easy conjecture is the following.

$$ \ \ \ \ \ T \times\text{-compact} \Rightarrow \text{cancellation property holds}.$$

But I do believe that's false and my plan to prove it is the following: a pair of weeks ago I've posted question about $\times$-compact groups and counterexample is here so if I find a $\times$-compact topological space $\bar{T}$ such that $\pi_1(\bar{T})=$ counterexample I think I'm done.

I'm thinking at this right in these days so probably I'll answer to myselfsoon. I'll note here my doubt and claims.

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R. H. Fox, On a problem of S. Ulam concerning Cartesian products, Fund. Math. 34 (1947) 278-287, constructed nonhomeomorphic compact $4$-dimensional manifolds $A$ and $B$ with $A\times A$ homeomorphic to $B\times B$; it follows that $(A+B)\times A$ is homeomorphic to $(A+B)\times B$.

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  • $\begingroup$ Is the argument a $\pi_1$-argument or they use other ideas? $\endgroup$ – Ivan Di Liberti Dec 5 '14 at 12:59
  • $\begingroup$ Your first link to the Scottish book seems to be broken. $\endgroup$ – Dan Rust Dec 5 '14 at 17:48

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