Topological "Freshman's Dream"

Question

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.

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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!

Answer 1

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 $X$.

Answer 2

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.

Answer 3

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.

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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.