Understanding the Proof of X & Y Connected Topological Spaces: A Deeper Look

[Oxymoron]

If X and Y are two connected topological spaces then so is X \otimes Y.

I want to understand the proof of this theorem but I am having some difficulties. Even though we went over it in class, it is still unclear to me.

The professor constructed this continuous function:

f:X\otimes Y \rightarrow \{0,1\}

Where \{0,1\} is a discrete topological space. Then he shows that f is constant. He then claimed that \{x\} \otimes Y is homeomorphic with Y hence this subspace (\{x\}\otimes Y) is connected - since Y is.

Now this does not make sense to me and I wouldn't be suprised if it didn't make sense to any of you. If you think you have a better way of explaining the proof (it doesn't have to be this one) then I would appreciate the effort.

Im not sure exactly why one would begin by setting up a continuous function which maps points in the product space X \otimes Y to either 0 or 1 in the discrete topological space, then claim that the function constant - then go on to show that \{x\} \otimes Y is homeomorphic to Y.??

 

[Oxymoron]

Where \{0,1\} is a discrete topological space. Then he shows that f is constant. He then claimed that \{x\} \otimes Y is homeomorphic with Y hence this subspace (\{x\}\otimes Y) is connected - since Y is.

Actually come to think of it, I do understand this.

Im just not sure of why this is the correct procedure to proving that the product space is connected.

 

[Oxymoron]

Hmmm, perhaps it is because if one finds a concrete homeomorphism between the product space to a pre-determined connected topological space, then since the homeomorphism preserved topological structure, the product space is therefore connected? Could this be the reason?

 

[HallsofIvy]

I think you may have misunderstood what he was saying. Of course, \{x\} \times Y is homeomorphic to Y: the function f(x,y)= y is clearly continuous with a continuous inverse f^-1(y)= (x,y).
That being true, since Y is connected, so is \{x\} \times Y. A continuous function maps connected sets to connected sets so any continuous function must map \{x\} \times Y to either {0} or {1}- those are the only connected subsets of {0, 1} with the discreet topology.
Now, do same with X \times \{y\}- and observe that any y will be matched with every x, any x with every y: f must map all of X \times Y into a connected set: either {0} or {1} and therefore f is a constant.
Your professor is not first proving that f must be constant and then that f map X \times Y to a connected set, he is first proving that f maps X\times Y to a connected set and then using that to prove that f must be constant.

 

[Oxymoron]

Thanks Halls once again. I understand it now.

However, one small question. Why choose {0,1} as the target space for f? I need motivation for the choice of {0,1}. Why not X or Y or some other space? Is it because {0,1} is easy to work with? I am not sure.

 

[hypermorphism]

Thanks Halls once again. I understand it now.
However, one small question. Why choose {0,1} as the target space for f? I need motivation for the choice of {0,1}. Why not X or Y or some other space? Is it because {0,1} is easy to work with? I am not sure.

The latter reason. He wanted a disconnected space to map to where it was easy to characterize when the image of another space was disconnected.

 

[Oxymoron]

Perfect. Exactly what I thought (I couldn't be sure). Thanks hyper.