# Topological "Freshman's Dream"

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!

• 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. Nov 16, 2021 at 21:14

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 $$X$$.

• $\Bbb Q{/}\Bbb Z$ is not even metrisable. Nor is $\{0,1\}^{\Bbb N}{/}Y$ for many $Y$. Nov 10, 2021 at 6:38
• I require $Y$ to be finite in those examples, though. Nov 10, 2021 at 6:39
• I think discrete and indiscrete are the only easy options. Nov 10, 2021 at 6:39
• 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.) Nov 10, 2021 at 6:41
• 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! :) Nov 10, 2021 at 7:03

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.

• 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. Nov 11, 2021 at 6:36

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.

• 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. Nov 19, 2021 at 1:27
• @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. Nov 19, 2021 at 15:14