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Let $(M,\mathfrak{X})$ be a smooth manifold, where $M$ is a topological manifold and $\mathfrak{X}$ a smooth structure on it.

It is commonly mentioned (in introductory resources on smooth manifolds) that there may exist a (finite or infinite) number of other smooth structures $\mathfrak{X}'$ on $M$ such that $(M,\mathfrak{X}')$ is not diffeomorphic to $(M,\mathfrak{X})$; and that in some special cases, some kind of uniqueness does hold (e.g. for the three-dimenional case, cf. this M.O post).

I am wondering about the analogous question in the category of topological manifolds (i.e., about something analogous to the notion of "exotic structures", but for topological manifolds).

Namely:

Let $(X,\mathfrak{T})$ be a topological manifold (where X is a set and $\mathfrak{T}$ is a topology on $X$ that makes it a topological manifold). In general, does there exist another topology $\mathfrak{T}'$ on $X$ such that $(X,\mathfrak{T}')$ is a topological manifold that is not homeomorphic to $(X,\mathfrak{T})$ ? If so, are there more restrictive situations where we still have uniqueness of the topological-manifold structure ?

qubitsandwich
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If the only thing you're preserving is the set $X$, then you can biject it to the set of points in some other manifold and use that manifold's topology.

For example, suppose you start with $(\mathbb{R},\mathfrak{T})$ where $\mathfrak{T}$ is the standard topology on $\mathbb R$. Then set $f:\mathbb{R}^2\to\mathbb{R}$ to be your favorite bijection (perhaps as in MJD's answer to Examples of bijective map from $\mathbb R^3\to\mathbb R$).

If $\mathfrak{T}'$ is the standard topology on the plane $\mathbb R^2$, then we can turn it into a topology $\mathfrak{T}''$ on $\mathbb R$ by using the forward image under $f$:

$$\mathfrak{T}''=\left\{f[U]\mid U\in\mathfrak{T}'\right\}$$

Now $(\mathbb R,\mathfrak{T}'')$ (the plane encoded so the points are technically real numbers) is not homeomorphic to $(\mathbb R,\mathfrak{T})$ (the real line).

(Instead of the plane, I could have used another 1-d manifold like the circle, but I figured changing dimension would be more striking/make it clear how little "$X$ is the same" controls.)

Mark S.
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    As an additional remark, all manifolds of positive dimension have the same cardinality (that of the continuum), so if you fix bijections and use the method of this answer, there are as many topologies on $\mathbb{R}$ that result in pairwise non-homeomorphic manifolds as there are homeomorphism types of positive-dimensional manifolds (and those are also continuum many, if I recall correctly). – Thorgott Apr 02 '23 at 16:54
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    I see, of course. So in fact in general, if one wishes to study the non-homeomorphic topological spaces $(X,\mathfrak{T})$, with all $X$ having the carnality of the reals, and all different topologies $\mathfrak{T}$ on $X$, one could without loss of generality represent them as the topological spaces $(X_0,\mathfrak{T})$, where the $\mathfrak{T}$'s are all different possible topologies on $X_0$, and $X_0$ is some fixed set of that cardinality (e.g. $[0,1]$, $S^1$, $\mathbb{R}^n$,...) . – qubitsandwich Apr 02 '23 at 18:35