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!


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.

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  • $\begingroup$ 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.) $\endgroup$ – Cameron Buie Dec 23 '18 at 22:59
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    $\begingroup$ @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. $\endgroup$ – Moishe Kohan Dec 27 '18 at 12:48
  • $\begingroup$ @MoisheCohen: I'm aware of that. I'm merely pointing out that the answer is misleading/incomplete. $\endgroup$ – Cameron Buie Dec 27 '18 at 13:12
  • $\begingroup$ 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. $\endgroup$ – 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.

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