Powers of quotient maps

Suppose $q:X \rightarrow Y$ is a quotient map of topological spaces 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 $n\geq 3$? If not, are there sufficient conditions that make this the case?

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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
There are lots of examples and this failure is directly tied to the greatest deficiency of the category of topological spaces (that it is not Cartesian closed). The easiest example is the real K-topology where the closed set K is identified to a point (I think this appears in Munkres' chapter on quotient spaces). In fact, even when $X$ is a metric space $q^2$ does not have to be a quotient map. –  Jeremy Brazas May 4 '10 at 17:02
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