28
$\begingroup$

When one learns about quotient and product spaces in topology for the first time, it is perhaps natural to expect that they would behave like mutual inverses:

Topological Freshman's Dream (TFD). For a space $X$ and subspace $\emptyset \neq Y\subseteq X$, the spaces $(X/Y)\times Y$ and $X$ are homeomorphic.

It is not too hard to see that TFD is not true even for very simple spaces. For example, pick $X=[0,1]$ and $Y=\{0,1\}$, then $X/Y\times Y$ is the disjoint union of two copies of $S^1$, obviously not the same as $X$.

There are two trivial cases when TFD does hold, when $Y$ is a single point and when $Y$ is the whole space $X$.

Q. Is there any nontrivial example when TFD holds?

I've tried for a while to construct such an example without success.


Some incomplete observations:

  • If $X$ is connected, then $Y$ must be as well. Otherwise, $(X/Y)\times Y$ would be disconnected.
  • We can apply the tools of algebraic topology to see, for example, that TFD implies $\pi_n(X)\cong\pi_n(X/Y)\times\pi_n(Y)$. This condition is quite hard to satisfy since it implies that the homotopy groups of the quotient space $X/Y$ are simpler than that of $X$, which generally fails quite spectatularly. A similar idea can also be applied to the homology and cohomology groups.
  • A special case of the above point implies that if $X$ is simply connected, then both $Y$ and $X/Y$ are simply connected (take the fundamental group $\pi_1$).

Any input is appreciated!

$\endgroup$
1
  • 2
    $\begingroup$ In light of the answers it would be very interesting to see whether there can be any "nice" (say Hausdorff) connected space with this property. $\endgroup$
    – M. Winter
    Nov 16, 2021 at 21:14

3 Answers 3

17
$\begingroup$

There are many examples where both $X/Y\cong X$ and $X\times Y\cong X$, from which it follows that $X/Y\times Y\cong X$. Probably the easiest is if $X$ is an infinite discrete space and $Y$ is any nonempty subspace of $X$ such that $|X\setminus Y|=|X|$. For a less trivial example, you could take $X=\mathbb{Q}$ or $X=\{0,1\}^\mathbb{N}$ and $Y$ any nonempty finite subset of $X$ (for these it takes some work to show that $X/Y\cong X$).

Here's a nontrivial example along these lines where $X$ is a CW complex. Let $X\subset\mathbb{R}^2$ be the union of all the circles of radius $1$ centered at points of the form $(2a,3b)$ where $a,b\in\mathbb{Z}$. This is a disjoint union of infinitely many infinite "chains of circles" attached at single points. Let $Y=\{(0,1),(0,-1)\}$. Then $X/Y\cong X$, since this quotient just takes one of the circles and pinches it together to form two circles (so its chain of circles remains an infinite chain of circles), and $X\times Y\cong X$ since that product just doubles the already infinite number of chains of circles in $X$.

$\endgroup$
10
  • $\begingroup$ $\Bbb Q{/}\Bbb Z$ is not even metrisable. Nor is $\{0,1\}^{\Bbb N}{/}Y$ for many $Y$. $\endgroup$ Nov 10, 2021 at 6:38
  • $\begingroup$ I require $Y$ to be finite in those examples, though. $\endgroup$ Nov 10, 2021 at 6:39
  • $\begingroup$ I think discrete and indiscrete are the only easy options. $\endgroup$ Nov 10, 2021 at 6:39
  • 1
    $\begingroup$ I don't see anything in the question which indicates that. (That would be an interesting different question though! Though, you need to at least require $Y$ to be nonempty to have any hope.) $\endgroup$ Nov 10, 2021 at 6:41
  • 4
    $\begingroup$ Hi, thanks for both of your responses and the discussion in the comments! My original intention was indeed to allow a specific choice of $Y$, not for it to hold for all nonempty subspaces $Y$ in fixed $X$. Sorry if the phrasing was unclear! :) $\endgroup$
    – YiFan Tey
    Nov 10, 2021 at 7:03
8
$\begingroup$

A potential class of examples: If $X$ is infinite discrete, say $|X| = \kappa \ge \aleph_0$.

Then any subspace $Y$ is also discrete and so is $X{/}Y$. A product of two discrete spaces is also discrete so it comes down to sizes, as discrete spaces are homeomorphic iff they have the same cardinality:

If $Y$ is finite, TFD holds as $X{/}Y$ has the same size as $X$ and $|X| = |X| \times n$ for $\kappa$ infinite and $n$ finite.

If $\lambda:=|Y| < |X|$ and is also infinite then $|X{/}Y| = \kappa$ still and indeed $\kappa \times \lambda = \kappa$ so TFD holds.

If $|Y| = |X|$ there are some cases depending on $|X\setminus Y|$, check it out.

Another potential class: indiscrete spaces, as all subspaces and quotients by subspaces are indiscrete and homeomorphism just depends on size.

$\endgroup$
1
  • $\begingroup$ If $X$ is discrete of size 4 and we identify a two point set $Y$ then $X{/}Y$ is discrete of size $3$ and as $3 \times 2 \neq 4$ we don't have the isomorphism. So finite ones must be excluded. $\endgroup$ Nov 11, 2021 at 6:36
4
$\begingroup$

Here is a connected compact Hausdorff space with this property: the Hilbert cube $X:=[0,1]^\omega$. We can choose $Y:=\{x\in [0,1]^\omega\mid x_1=0\}\subset X$.

Admittably, this is still an "Eric Wofsey type example": it satisfies $X/Y\cong X$ and $X\times Y\cong X$, and now I wonder whether there are example that satisfy neither.


Proving $X/Y\cong X$

The Hilbert cube can be identified with the following convex compact subset of $ \ell^2$:

$$X:=[0,1]\times[0,1/2]\times[0,1/3]\times\cdots.$$

I believe that the quotient $X/Y$ is homeomorphic to the following subset of $X$:

$$X/Y\cong \Big\{x\in X \;\Big\vert\; x_i\le \frac{x_1}i\text{ for all $i\ge 2$}\Big\}.$$

I furthermore believe that this is still a convex compact subset of $\ell^2$ of infinite dimension. Wikipedia state that every infinite-dimensional convex compact subset of $\ell^2$ is homeomorphic to the Hilbert cube.

$\endgroup$
2
  • $\begingroup$ Is there any easy proof that $X/Y\cong X$ in this case? I believe it should be true but it looks messy to write down a homeomorphism. $\endgroup$ Nov 19, 2021 at 1:27
  • $\begingroup$ @EricWofsey I admit I believed this to be easier. I had to change the definition of $Y$, but I added some sketch for how I believe one can show $X/Y\cong X$. There are still a lot of "believes" in there, and I make use of a claim from Wikipedia. $\endgroup$
    – M. Winter
    Nov 19, 2021 at 15:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .