Manifolds do not satisfy Topological Freshman’s Dream

Recently, I asked this question on MSE:

Let the “Topological Freshman’s Dream” (TFD) be the property, for a topological space $X$ and a nonempty subspace $Y \subseteq X$, that $Y\times (X/Y) \cong X$.
When $Y$ is a single point, and when $Y$ is the whole space $X$, TFD holds. These are the “trivial” examples.
Does there exist any non-trivial example where TFD holds?

The reason why this is interesting is of course that the quotient and product, which one might expect to be inverse operations, fail so spectacularly to actually be inverses in the category of topological spaces. (By the way, it’s interesting to ask why: certainly in algebraic categories like $\mathsf{Mod}_R$ quotients and products are indeed inverse, so maybe there’s a way to figure out what categorical property allows this to be true?)

The answer to the above question, as it turns out, is yes. Several different cool constructions are provided in the linked MSE question, by Eric Wofsey, the late Henno Brandsma, and M. Winters.

Here, let me provide an explanation for the phenomenon on the opposite side, namely the curious observation that all “nice” spaces that one might initially try in the construction don’t seem to work. Here is a proof sketch that all topological manifolds $X$ cannot contain a nontrivial subspace that satisfies TFD. (My original intention was to formulate a rigorous proof, but I got lazy… I hope the missing details aren’t bad enough to derail the idea, though to be honest I’m not completely sure :P)

Claim. Let $X$ be a connected topological manifold, and $Y$ a subspace, which is not a single point, with $X-\bar{Y}$ nonempty. Then $(X/Y)\times Y$ and $X$ are not homeomoprhic.

Proof Sketch. Pick some point $x\in X-\overline Y$. Being the complement of a closed set, this set is open, so there exists a neighbourhood $U$ of $x$ contained in $X-\overline Y$. The projection map $p\colon X\to X/Y$ restricts to a homeomorphism of $U$ with $p(U)$, since $U$ does not contain any points affected by the identification of $Y$. On the other hand, $U$ can be chosen to be homeomorphic to some Euclidean space $\mathbb R^n$, so the same is true for $p(U)$. Therefore, a neighbourhood of $p(x)\times y$ for any $y\in Y$ is homeomorphic to $\mathbb R^n\times V$ where $V$ is some open set in $Y$ containing $y$.

Suppose now that $(X/Y)\times Y$ is homeomorphic to $X$. Then every point in $(X/Y)\times Y$, including $p(x)\times y$ in particular, has a neighbourhood homeomorphic to $\mathbb R^n$. Thus we must have $\mathbb R^n\times V\cong\mathbb R^n$. Well, $Y$ should probably be sufficiently nice such that this can never occur: if, for instance, $V$ is locally homeomorphic to $\mathbb R^m$, invariance of dimension would give a contradiction. But here is where I am not totally sure how to make my argument rigorous.