# $\mathbb{R}\cong X\times Y$ (homeomorphic) implies $X$ or $Y$ is a point.

The following is Problem 2.1.1 from Tammo tom Dieck's Algebraic Topology:

Suppose $$\mathbb{R}\cong X\times Y$$ (homeomorphic). Then $$X$$ or $$Y$$ is a point.

FWIW, the section 2.1 is about (path) connected components (i.e., $$\pi_0$$) and the notion of homotopy.

I tried to use the usual trick by removing one point and consider the number of components, but failed. The spaces $$X,Y$$ are arbitrary, and so I don't know how to handle this.

Any hints will be appreciated!

• – user608030 Dec 22 '18 at 5:29

This comes from the following theorem:

Theorem: Suppose $$X$$ and $$Y$$ are path connected and each have at least two points. Given any point $$(x_0, y_0) \in X \times Y$$, the space $$X \times Y -\{(x_0,y_0)\}$$ is still path connected.

In your case, $$X \times Y$$ must be path connected since $$\mathbb{R}$$ is. Hence, each of $$X$$ and $$Y$$ are path connected, so the above result is applicable. But upon removing a point we get into trouble, since $$\mathbb{R}-pt$$ is no longer path connected, but $$X \times Y -\{(x_0,y_0)\}$$ is. So all you have to do is believe in the theorem.

Sketch Proof. Let $$(a,b)$$ and $$(c,d)$$ be any two points in $$Z=X \times Y -\{(x_0,y_0)\}$$. We must show there exists a path between these two points, the path living in $$X \times Y -\{(x_0,y_0)\}$$ (so, avoiding $$(x_0,y_0)$$). The argument comes in several cases.

First, we could have $$a=c\neq x_0$$. Use the fact that $$\{a\} \times Y\cong Y$$ is path connected and build the path in that slice. That clearly gives a path in $$Z$$. The same argument works if $$b=d\neq y_0$$.

Next, the case $$a=c=x_0$$. It follows that $$b \neq y_0$$ and $$d \neq y_0$$. Here's where you need the assumptions that $$|X|, |Y| \geq 2$$. We can choose a point $$x'\in X$$ which is different from $$x_0$$. Now create the needed path in three segments:
$$(a,b) \to (x',b) \to (x',d) \to (c,d)$$ working in the obvious coordinate slices in each leg (see the argument for the first case). This builds a path $$(a,b) \to (c,d)$$ in $$Z$$. The same thing works if $$b=d=y_0$$.

Finally, there is the case $$a \neq c$$ and $$b \neq d$$. Construct two different three-legged paths $$(a,b) \to (c,b) \to (c,d)$$ and $$(a,b) \to (a,d) \to (c,d).$$ It is not hard to check that $$(x_0,y_0)$$ cannot live on both of these, so one of these gives the escape. Hence we're done.

• The Theorem you cite doesn't imply (on its own) that $X$ and $Y$ must be path-connected, as far as I can tell, so I don't know why you say "Hence, each of $X$ and $Y$ are path connected." (I don't disbelieve the statement, itself, but the statement "all you have to do is believe in the theorem" suggests that said theorem somehow proves that if a product of two spaces is path-connected, then so are the two spaces.) – Cameron Buie Dec 23 '18 at 22:59
• @CameronBuie: The image of a path-connected space under a continuous map is again path-connected. Now, consider the projections of $X\times Y$ to its factors. – Moishe Kohan Dec 27 '18 at 12:48
• @MoisheCohen: I'm aware of that. I'm merely pointing out that the answer is misleading/incomplete. – Cameron Buie Dec 27 '18 at 13:12
• I think that piece is reasonably clear to someone reading tom Dieck and is not at all the crux move in the argument. Feel free to downvote. – Randall Dec 27 '18 at 13:22

Hint: By assumption, $$X\times Y$$ is path-connected, hence so are $$X$$ and $$Y$$. Now you should be able to proceed with your usual trick, e.g. by assuming that both $$X$$ and $$Y$$ have at least two distinct points and then proving that $$X\times Y$$ minus a point must still be path-connected.