Quotient map on a product space

[Tom1992]

suppose q:M -> M/R is a quotient map.

i've asked my dad what is the quotient map from MxM to (M/R)x(M/R)?

he told me it is qxq: MxM -> (M/R)x(M/R) defined by

(qxq)(x,y) = (q(x), q(y)),

but there are some conditions to be met, but he could not remember what those conditions are. i went through all 6 or so of my topology textbooks and could not find it.

does anybody know what the conditions are for qxq: MxM -> (M/R)x(M/R) defined by

(qxq)(x,y) = (q(x), q(y)),

to be a quotient map if q:M -> M/R is a quotient map?

 

[Gib Z]

It seems like no one wants to answer your question :S

In fact, the only reason I am posting is so this thread goes to the top of the list so someone might notice it, because I sure as hell don't know it.

Good Luck

 

[Tom1992]

thanks a lot gib, i accepted already that apparently no one knows. the munkres topology book gives a counterexample of where a product of quotient maps is not a quotient map and i can see why it fails in the counterexample but can't figure out the general condition for failure. the counterexample munkres gives is so weird that i assume that qxq: MxM -> (M/R)x(M/R) is a quotient map in most cases except for the bizzarre ones.

thanks buddy for watching out for me. i took your advice about learning math for the sake of discovering what i really like, rather than just trying to scale up mount everest as quickly as possible. in fact, today i reverted to rereading the proofs to basic set theory to make sure i master the basics before jumping to 4th year courses.

 

[Hurkyl]

Well, we can look at the pieces.

As a function on the underlying sets, qxq is certainly a surjective function, and "is" thus a quotient map.

As a map between topolocial spaces, qxq is clearly a continuous function.

So, the only possibility is that the topology on (M/R)x(M/R) is not the one induced by the map qxq, right?

...