# Powers of quotient maps

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 $n$.

• 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 – Herb May 4 '10 at 6:11
• Are there examples where $q$ is a quotient map but $q^2$ is not? – 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