# Product Spaces

1. Oct 27, 2005

### 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 $Y$.??

2. Oct 27, 2005

### Oxymoron

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.

3. Oct 27, 2005

### 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?

4. Oct 27, 2005

### HallsofIvy

Staff Emeritus
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.

Last edited: Oct 27, 2005
5. Oct 27, 2005

### 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? Im not sure.

6. Oct 27, 2005

### hypermorphism

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.

7. Oct 27, 2005

### Oxymoron

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