It is well-known that if $q:X\to Y$ is a quotient map, then the self-product $q^2:X^2\to Y^2$ need not be a quotient map. For instance, if $X$ is the real line generated by the basic sets $(a,b)$ and $(a,b)\backslash K$, where $K=\{1,1/2,1/3,...\}$, then the quotient map $q:X\to Y=X/K$ exhibits this failure.

Question: Suppose $q:X \rightarrow Y$ is a quotient map such that the product map $q^2:X^2\rightarrow Y^2$ is also a quotient map. Are the maps $q^n:X^n\rightarrow Y^n$ quotient maps for all $n\geq 3$?

I'd like an answer to the above question but I'd also be interested in conditions on $X$ and $Y$ which are sufficient to imply that $q^n$ is quotient for all $n$. Using a Cartesian closed category like the category of sequential (or compactly generated) spaces, it's easy to see that if $X^n$ is sequential (compactly generated), then $q^n$ is quotient if and only if $Y^n$ is sequential (compactly generated). This is interesting but I find it unfortunate that this depends on what happens to arbitrarily high powers of both $X$ and $Y$. So to clarify, I would be interested in sufficient properties which are not implied by the above fact and which do not depend on $n$.

  • $\begingroup$ I think it is true: If q:X-->Y is a quotient map, then U in Y is open iff q<sup>-1</sup> is open in X. If q<sup>2</sup>:X<sup>2</sup>-Y<sup>2</sup> is a quotient map too, then UxU' is open in Y<sup>2</sup> iff (q<sup>-1</sup>(U),q<sup>-1</sup>(U') is open in X<sup>2</sup>. In general : q^n:X^n-->Y^n , then q^n-1(U1xU2x...Un)= q^-1(U1)xq^-1(U2)x...xq^-1(Un) is open in X^n , as the product of open sets. I think the other side follows. in $\endgroup$ – Herb May 4 '10 at 6:11
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    $\begingroup$ Are there examples where $q$ is a quotient map but $q^2$ is not? $\endgroup$ – Sergei Ivanov May 4 '10 at 14:05

If you add the hypothesis that $q$ is an open map, then your conclusion follows for all powers, including infinite powers, and there is no need in this case separately to assume that $q^2$ is a quotient map.

Theorem. If $q:X\to Y$ is a continuous surjective open map, then all powers $q^I:X^I\to Y^I$ are quotient maps.

Proof. The Wikipedia page on quotient maps mentions that it is a sufficient condition (but regrettably not necessary) for a map to be a quotient map that it is continuous, surjective and open. Suppose that $q:X\to Y$ is continuous, surjective and open. It follows using standard methods that the power map $q^I:X^I\to Y^I$ is also continuous, surjective and open. So it is a quotient map.QED

If you drop the open map hypothesis, however, then I'm not sure what happens. (I couldn't manage to make Herb's suggestion work in the converse direction.) Perhaps one can build a crazy counterexample with a non-open quotient map.

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