Can different topological spaces have the same square?












7














For topological spaces $X$ and $Y$, is it possible that $X times X$ and $Y times Y$ are homeomorphic, but $X$ and $Y$ are not homeomorphic?



(I poked around with finite spaces, and manifolds, and the Cantor set, without seeing any examples.)



This was inspired by Existence of topological space which has no “square-root” but whose “cube” has a “square-root”. Cartesian product makes the proper class of topological spaces (up to homeomorphism) into a large abelian monoid, so here's a bonus question: what is known about the structure of this monoid? For example, the dogbone space shows that it is not cancellative. Does it have torsion in the sense that sometimes $X^n notcong X$ but $X^{n + 1} cong X$?










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  • 2




    It's not possible for finite spaces, by my argument at mathoverflow.net/questions/269542/….
    – Eric Wofsey
    5 hours ago










  • Maybe take a look at the Whitehead manifold. See: numdam.org/article/BSMF_1960__88__131_0.pdf and en.wikipedia.org/wiki/Whitehead_manifold
    – Alvin Jin
    5 hours ago






  • 1




    The existence of such spaces $X, Y$ is what's used in Mike Miller 's example, see his answer in the linked question. But Whitehead manifold would also do the job.
    – Moishe Cohen
    5 hours ago






  • 1




    As for the "torsion" examples, see references in mathoverflow.net/questions/10128/when-is-a-isomorphic-to-a3
    – Moishe Cohen
    4 hours ago












  • @MoisheCohen Interesting, for OP's sake I will observe that there exists a functor from groups to spaces, given by constructing a simplicial complex called "$BG$" whose $n$-simplices are labelled by sequences $[g_1, cdots, g_n]$ of elements of $G$ (and the gluings involve multiplications), and $B(G times H) = BG times BH$. In particular, functoriality means that if $G$ and $H$ are homeomorphic, so are $BG$ and $BH$, and conversely (if $BG cong BH$ then $G cong H$ by looking at fundamental groups). So if $A cong A^3 neq A^2$, then $BA cong (BA)^3 neq (BA)^2$.
    – Mike Miller
    4 hours ago
















7














For topological spaces $X$ and $Y$, is it possible that $X times X$ and $Y times Y$ are homeomorphic, but $X$ and $Y$ are not homeomorphic?



(I poked around with finite spaces, and manifolds, and the Cantor set, without seeing any examples.)



This was inspired by Existence of topological space which has no “square-root” but whose “cube” has a “square-root”. Cartesian product makes the proper class of topological spaces (up to homeomorphism) into a large abelian monoid, so here's a bonus question: what is known about the structure of this monoid? For example, the dogbone space shows that it is not cancellative. Does it have torsion in the sense that sometimes $X^n notcong X$ but $X^{n + 1} cong X$?










share|cite|improve this question


















  • 2




    It's not possible for finite spaces, by my argument at mathoverflow.net/questions/269542/….
    – Eric Wofsey
    5 hours ago










  • Maybe take a look at the Whitehead manifold. See: numdam.org/article/BSMF_1960__88__131_0.pdf and en.wikipedia.org/wiki/Whitehead_manifold
    – Alvin Jin
    5 hours ago






  • 1




    The existence of such spaces $X, Y$ is what's used in Mike Miller 's example, see his answer in the linked question. But Whitehead manifold would also do the job.
    – Moishe Cohen
    5 hours ago






  • 1




    As for the "torsion" examples, see references in mathoverflow.net/questions/10128/when-is-a-isomorphic-to-a3
    – Moishe Cohen
    4 hours ago












  • @MoisheCohen Interesting, for OP's sake I will observe that there exists a functor from groups to spaces, given by constructing a simplicial complex called "$BG$" whose $n$-simplices are labelled by sequences $[g_1, cdots, g_n]$ of elements of $G$ (and the gluings involve multiplications), and $B(G times H) = BG times BH$. In particular, functoriality means that if $G$ and $H$ are homeomorphic, so are $BG$ and $BH$, and conversely (if $BG cong BH$ then $G cong H$ by looking at fundamental groups). So if $A cong A^3 neq A^2$, then $BA cong (BA)^3 neq (BA)^2$.
    – Mike Miller
    4 hours ago














7












7








7


1





For topological spaces $X$ and $Y$, is it possible that $X times X$ and $Y times Y$ are homeomorphic, but $X$ and $Y$ are not homeomorphic?



(I poked around with finite spaces, and manifolds, and the Cantor set, without seeing any examples.)



This was inspired by Existence of topological space which has no “square-root” but whose “cube” has a “square-root”. Cartesian product makes the proper class of topological spaces (up to homeomorphism) into a large abelian monoid, so here's a bonus question: what is known about the structure of this monoid? For example, the dogbone space shows that it is not cancellative. Does it have torsion in the sense that sometimes $X^n notcong X$ but $X^{n + 1} cong X$?










share|cite|improve this question













For topological spaces $X$ and $Y$, is it possible that $X times X$ and $Y times Y$ are homeomorphic, but $X$ and $Y$ are not homeomorphic?



(I poked around with finite spaces, and manifolds, and the Cantor set, without seeing any examples.)



This was inspired by Existence of topological space which has no “square-root” but whose “cube” has a “square-root”. Cartesian product makes the proper class of topological spaces (up to homeomorphism) into a large abelian monoid, so here's a bonus question: what is known about the structure of this monoid? For example, the dogbone space shows that it is not cancellative. Does it have torsion in the sense that sometimes $X^n notcong X$ but $X^{n + 1} cong X$?







general-topology






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 5 hours ago









Hew Wolff

2,218716




2,218716








  • 2




    It's not possible for finite spaces, by my argument at mathoverflow.net/questions/269542/….
    – Eric Wofsey
    5 hours ago










  • Maybe take a look at the Whitehead manifold. See: numdam.org/article/BSMF_1960__88__131_0.pdf and en.wikipedia.org/wiki/Whitehead_manifold
    – Alvin Jin
    5 hours ago






  • 1




    The existence of such spaces $X, Y$ is what's used in Mike Miller 's example, see his answer in the linked question. But Whitehead manifold would also do the job.
    – Moishe Cohen
    5 hours ago






  • 1




    As for the "torsion" examples, see references in mathoverflow.net/questions/10128/when-is-a-isomorphic-to-a3
    – Moishe Cohen
    4 hours ago












  • @MoisheCohen Interesting, for OP's sake I will observe that there exists a functor from groups to spaces, given by constructing a simplicial complex called "$BG$" whose $n$-simplices are labelled by sequences $[g_1, cdots, g_n]$ of elements of $G$ (and the gluings involve multiplications), and $B(G times H) = BG times BH$. In particular, functoriality means that if $G$ and $H$ are homeomorphic, so are $BG$ and $BH$, and conversely (if $BG cong BH$ then $G cong H$ by looking at fundamental groups). So if $A cong A^3 neq A^2$, then $BA cong (BA)^3 neq (BA)^2$.
    – Mike Miller
    4 hours ago














  • 2




    It's not possible for finite spaces, by my argument at mathoverflow.net/questions/269542/….
    – Eric Wofsey
    5 hours ago










  • Maybe take a look at the Whitehead manifold. See: numdam.org/article/BSMF_1960__88__131_0.pdf and en.wikipedia.org/wiki/Whitehead_manifold
    – Alvin Jin
    5 hours ago






  • 1




    The existence of such spaces $X, Y$ is what's used in Mike Miller 's example, see his answer in the linked question. But Whitehead manifold would also do the job.
    – Moishe Cohen
    5 hours ago






  • 1




    As for the "torsion" examples, see references in mathoverflow.net/questions/10128/when-is-a-isomorphic-to-a3
    – Moishe Cohen
    4 hours ago












  • @MoisheCohen Interesting, for OP's sake I will observe that there exists a functor from groups to spaces, given by constructing a simplicial complex called "$BG$" whose $n$-simplices are labelled by sequences $[g_1, cdots, g_n]$ of elements of $G$ (and the gluings involve multiplications), and $B(G times H) = BG times BH$. In particular, functoriality means that if $G$ and $H$ are homeomorphic, so are $BG$ and $BH$, and conversely (if $BG cong BH$ then $G cong H$ by looking at fundamental groups). So if $A cong A^3 neq A^2$, then $BA cong (BA)^3 neq (BA)^2$.
    – Mike Miller
    4 hours ago








2




2




It's not possible for finite spaces, by my argument at mathoverflow.net/questions/269542/….
– Eric Wofsey
5 hours ago




It's not possible for finite spaces, by my argument at mathoverflow.net/questions/269542/….
– Eric Wofsey
5 hours ago












Maybe take a look at the Whitehead manifold. See: numdam.org/article/BSMF_1960__88__131_0.pdf and en.wikipedia.org/wiki/Whitehead_manifold
– Alvin Jin
5 hours ago




Maybe take a look at the Whitehead manifold. See: numdam.org/article/BSMF_1960__88__131_0.pdf and en.wikipedia.org/wiki/Whitehead_manifold
– Alvin Jin
5 hours ago




1




1




The existence of such spaces $X, Y$ is what's used in Mike Miller 's example, see his answer in the linked question. But Whitehead manifold would also do the job.
– Moishe Cohen
5 hours ago




The existence of such spaces $X, Y$ is what's used in Mike Miller 's example, see his answer in the linked question. But Whitehead manifold would also do the job.
– Moishe Cohen
5 hours ago




1




1




As for the "torsion" examples, see references in mathoverflow.net/questions/10128/when-is-a-isomorphic-to-a3
– Moishe Cohen
4 hours ago






As for the "torsion" examples, see references in mathoverflow.net/questions/10128/when-is-a-isomorphic-to-a3
– Moishe Cohen
4 hours ago














@MoisheCohen Interesting, for OP's sake I will observe that there exists a functor from groups to spaces, given by constructing a simplicial complex called "$BG$" whose $n$-simplices are labelled by sequences $[g_1, cdots, g_n]$ of elements of $G$ (and the gluings involve multiplications), and $B(G times H) = BG times BH$. In particular, functoriality means that if $G$ and $H$ are homeomorphic, so are $BG$ and $BH$, and conversely (if $BG cong BH$ then $G cong H$ by looking at fundamental groups). So if $A cong A^3 neq A^2$, then $BA cong (BA)^3 neq (BA)^2$.
– Mike Miller
4 hours ago




@MoisheCohen Interesting, for OP's sake I will observe that there exists a functor from groups to spaces, given by constructing a simplicial complex called "$BG$" whose $n$-simplices are labelled by sequences $[g_1, cdots, g_n]$ of elements of $G$ (and the gluings involve multiplications), and $B(G times H) = BG times BH$. In particular, functoriality means that if $G$ and $H$ are homeomorphic, so are $BG$ and $BH$, and conversely (if $BG cong BH$ then $G cong H$ by looking at fundamental groups). So if $A cong A^3 neq A^2$, then $BA cong (BA)^3 neq (BA)^2$.
– Mike Miller
4 hours ago










1 Answer
1






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4














Copying part of my answer to this Math Overflow question:




Whether two (a) topological spaces, (b) metric spaces, or (c) groups $A$ and $B$, with isomorphic squares, are necessarily isomorphic, was Problem 77 (of Ulam) in the Scottish book. The following information is from the commentaries on Problem 77, by Mauldin for (a) and (b) and by Kaplansky for (c), in The Scottish Book: Mathematics from the Scottish Café, R. Daniel Mauldin, ed., 1981.

Mauldin gives a bibliography of 17 items for parts (a) and (b). The first counterexample for part (a), as well as a positive answer for two-dimensional compact manifolds, was given by R. H. Fox, On a problem of S. Ulam concerning Cartesian products, Fund. Math. 34 (1947), 278-287. The first counterexample to part (b) was given by G. Fournier, On a problem of S. Ulam, Proc. Amer. Math. Soc. 29 (1971), 622. Mauldin writes (in 1981):

"However, it is open whether there is an affirmative solution to (b) in the case that $A$ and $B$ are complete metric spaces. In fact, part (b) is open in the case where $A$ and $B$ are assumed to be compact."







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  • I deleted my silly comment, which parsed (b) as "metrizable".
    – Mike Miller
    44 mins ago











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1 Answer
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4














Copying part of my answer to this Math Overflow question:




Whether two (a) topological spaces, (b) metric spaces, or (c) groups $A$ and $B$, with isomorphic squares, are necessarily isomorphic, was Problem 77 (of Ulam) in the Scottish book. The following information is from the commentaries on Problem 77, by Mauldin for (a) and (b) and by Kaplansky for (c), in The Scottish Book: Mathematics from the Scottish Café, R. Daniel Mauldin, ed., 1981.

Mauldin gives a bibliography of 17 items for parts (a) and (b). The first counterexample for part (a), as well as a positive answer for two-dimensional compact manifolds, was given by R. H. Fox, On a problem of S. Ulam concerning Cartesian products, Fund. Math. 34 (1947), 278-287. The first counterexample to part (b) was given by G. Fournier, On a problem of S. Ulam, Proc. Amer. Math. Soc. 29 (1971), 622. Mauldin writes (in 1981):

"However, it is open whether there is an affirmative solution to (b) in the case that $A$ and $B$ are complete metric spaces. In fact, part (b) is open in the case where $A$ and $B$ are assumed to be compact."







share|cite|improve this answer





















  • I deleted my silly comment, which parsed (b) as "metrizable".
    – Mike Miller
    44 mins ago
















4














Copying part of my answer to this Math Overflow question:




Whether two (a) topological spaces, (b) metric spaces, or (c) groups $A$ and $B$, with isomorphic squares, are necessarily isomorphic, was Problem 77 (of Ulam) in the Scottish book. The following information is from the commentaries on Problem 77, by Mauldin for (a) and (b) and by Kaplansky for (c), in The Scottish Book: Mathematics from the Scottish Café, R. Daniel Mauldin, ed., 1981.

Mauldin gives a bibliography of 17 items for parts (a) and (b). The first counterexample for part (a), as well as a positive answer for two-dimensional compact manifolds, was given by R. H. Fox, On a problem of S. Ulam concerning Cartesian products, Fund. Math. 34 (1947), 278-287. The first counterexample to part (b) was given by G. Fournier, On a problem of S. Ulam, Proc. Amer. Math. Soc. 29 (1971), 622. Mauldin writes (in 1981):

"However, it is open whether there is an affirmative solution to (b) in the case that $A$ and $B$ are complete metric spaces. In fact, part (b) is open in the case where $A$ and $B$ are assumed to be compact."







share|cite|improve this answer





















  • I deleted my silly comment, which parsed (b) as "metrizable".
    – Mike Miller
    44 mins ago














4












4








4






Copying part of my answer to this Math Overflow question:




Whether two (a) topological spaces, (b) metric spaces, or (c) groups $A$ and $B$, with isomorphic squares, are necessarily isomorphic, was Problem 77 (of Ulam) in the Scottish book. The following information is from the commentaries on Problem 77, by Mauldin for (a) and (b) and by Kaplansky for (c), in The Scottish Book: Mathematics from the Scottish Café, R. Daniel Mauldin, ed., 1981.

Mauldin gives a bibliography of 17 items for parts (a) and (b). The first counterexample for part (a), as well as a positive answer for two-dimensional compact manifolds, was given by R. H. Fox, On a problem of S. Ulam concerning Cartesian products, Fund. Math. 34 (1947), 278-287. The first counterexample to part (b) was given by G. Fournier, On a problem of S. Ulam, Proc. Amer. Math. Soc. 29 (1971), 622. Mauldin writes (in 1981):

"However, it is open whether there is an affirmative solution to (b) in the case that $A$ and $B$ are complete metric spaces. In fact, part (b) is open in the case where $A$ and $B$ are assumed to be compact."







share|cite|improve this answer












Copying part of my answer to this Math Overflow question:




Whether two (a) topological spaces, (b) metric spaces, or (c) groups $A$ and $B$, with isomorphic squares, are necessarily isomorphic, was Problem 77 (of Ulam) in the Scottish book. The following information is from the commentaries on Problem 77, by Mauldin for (a) and (b) and by Kaplansky for (c), in The Scottish Book: Mathematics from the Scottish Café, R. Daniel Mauldin, ed., 1981.

Mauldin gives a bibliography of 17 items for parts (a) and (b). The first counterexample for part (a), as well as a positive answer for two-dimensional compact manifolds, was given by R. H. Fox, On a problem of S. Ulam concerning Cartesian products, Fund. Math. 34 (1947), 278-287. The first counterexample to part (b) was given by G. Fournier, On a problem of S. Ulam, Proc. Amer. Math. Soc. 29 (1971), 622. Mauldin writes (in 1981):

"However, it is open whether there is an affirmative solution to (b) in the case that $A$ and $B$ are complete metric spaces. In fact, part (b) is open in the case where $A$ and $B$ are assumed to be compact."








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share|cite|improve this answer



share|cite|improve this answer










answered 5 hours ago









bof

50.2k457119




50.2k457119












  • I deleted my silly comment, which parsed (b) as "metrizable".
    – Mike Miller
    44 mins ago


















  • I deleted my silly comment, which parsed (b) as "metrizable".
    – Mike Miller
    44 mins ago
















I deleted my silly comment, which parsed (b) as "metrizable".
– Mike Miller
44 mins ago




I deleted my silly comment, which parsed (b) as "metrizable".
– Mike Miller
44 mins ago


















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