A path connected product of path connected spaces

[radou]

I just went through and exercise which asks if a product of two path connected spaces is path connected.

There seems no reason to believe it is in general, for the only thing we know is that if two spaces X, Y are path connected, then they are connected, and their product X x Y is connected.

So I thought of this Lemma:

Let X, Y be two path connected spaces. Their product X x Y is path connected if the domains of the paths fx in X and fy in Y coincide.

Assume the hypothesis is true, then for X and Y there exist continuous functions fx : [a, b] --> X and fy : [a, b] --> Y, such that, for any pair of points x1, x2 of X and y1, y2 of Y fx(a) = x1, fx(b) = x2 and fy(a) = y1, fy(b) = y2.

Let f : [a, b] --> X x Y be given with f(x) = (fx(x), fy(x)). Since fx and fy are continuous, f is continuous too, and for any pair of points (x1, x2), (y1, y2) of X x Y f(a) = (x1, y1), f(b) = (x2, y2).

I think this should work, any comments?

 

[Landau]

I have no idea what your lemma is saying: "The domains of the path fx and fy coincide"...what are fx and fy?

The statement is true. Suppose X and Y are path-connected. Let (x1,y1) and (x2,y2) be two points in X x Y. There is a path f:[0,1]->X from x1 to x2. There is a path g:[0,1]->Y from y1 to y2. But then then path h:[0,1]->X x Y given by h(t):=(f(t),g(t)) is a path from (x1,y1) to (x2,y2).

It looks like your proof is trying something like this, but I can't follow you. Perhaps your definition of "path in X" is any continuous function [a,b]-> X and you are worrying about the fact that the domain [a,b] can differ? That is no problem because you can just compose with a homeomorphism between [0,1] and [a,b]. Or you are in fact considering arc-connected spaces instead of path-connected ones?

 

[radou]

Perhaps your definition of "path in X" is any continuous function [a,b]-> X and you are worrying about the fact that the domain [a,b] can differ? That is no problem because you can just compose with a homeomorphism between [0,1] and [a,b]. Or you are in fact considering arc-connected spaces instead of path-connected ones?

Yes, that was exactly the thing I was worrying about - that the domains differ in general.

The idea about the homeomorphism didn't occur to me, thanks a lot.

So, we construct, for every function a homeomorhpism from [0, 1] to the "domain", and then compose these continuous functions in order to obtain the final function (i.e. path), right?

 

[radou]

By the way, does this generalize for an arbitrary product of topological spaces? i.e. if every space is path-connected, then their product is, too? Or does it depend on the topology of the product space? I know there's a theorem which states that a function f : A --> ∏ Xj (j is from J, and J is an index set) given with f(a) = (fj(a)), fj is a function fj : A --> Xj if and only if every function fj is continuous, and ∏ Xj is given the product topology. Now in our case, it is "<==" (since we know that the coordinate functions are continuous), does this hold only for the product topology, or can the product have some other topology?

 

[Landau]

So, we construct, for every function a homeomorhpism from [0, 1] to the "domain", and then compose these continuous functions in order to obtain the final function (i.e. path), right?

Let F:[0,1]\to[a,b] be a homeomorphism. If f:[a,b]\to X is a path from x1 to x2, then fF:[0,1]\to X is as well. In other words, we can restrict ourselves to paths with domain [0,1]. That's why many authors define a path to have domain [0,1]: the possibility of having [a,b] as domain does not add anything new.

By the way, does this generalize for an arbitrary product of topological spaces? i.e. if every space is path-connected, then their product is, too? Or does it depend on the topology of the product space?

Well, obviously it depends on the topology. Path-connected is a topological property, concerning certain continuous maps. It is usually understood that the product of a collection of spaces has the 'product toplogy'.

I know there's a theorem which states that a function f : A --> ∏ Xj (j is from J, and J is an index set) given with f(a) = (fj(a)), fj is a function fj : A --> Xj if and only if every function fj is continuous, and ∏ Xj is given the product topology.

Now in our case, it is "<==" (since we know that the coordinate functions are continuous), does this hold only for the product topology, or can the product have some other topology?

It holds for the product toplogy. The product toplogy is the unique (up to isomorphism) topology with the universal property: for any top. space Y and continuous functions f_i:Y\to X_i, there is a unique continuous function f:Y\to ∏ Xi such that p_if=f_i for all i, where p_i:∏ Xi\to Xi is the projection.

I don't know whether it is an interesting question to consider other topologies on the product.

 

[radou]

OK, thanks.

 

[lavinia]

I just went through and exercise which asks if a product of two path connected spaces is path connected.

There seems no reason to believe it is in general, for the only thing we know is that if two spaces X, Y are path connected, then they are connected, and their product X x Y is connected.

So I thought of this Lemma:

Let X, Y be two path connected spaces. Their product X x Y is path connected if the domains of the paths fx in X and fy in Y coincide.

Assume the hypothesis is true, then for X and Y there exist continuous functions fx : [a, b] --> X and fy : [a, b] --> Y, such that, for any pair of points x1, x2 of X and y1, y2 of Y fx(a) = x1, fx(b) = x2 and fy(a) = y1, fy(b) = y2.

Let f : [a, b] --> X x Y be given with f(x) = (fx(x), fy(x)). Since fx and fy are continuous, f is continuous too, and for any pair of points (x1, x2), (y1, y2) of X x Y f(a) = (x1, y1), f(b) = (x2, y2).

I think this should work, any comments?

Though it is a little confusing I think you have a good idea. I think of a path that first connects the x coordinates keeping one of the y's fixed then connects the y's keeping the x endpoint fixed.