Does a coarser topology lead to a non-Hausdorff topological manifold? [closed]

Question

Take a topological manifold $M$. Suppose one considers a strictly coarser topology than the manifold topology. Can such topology result in a non-Hausdorff topological manifold?

NOTE: PLEASE avoid the coarsest/finest topology over the manifold.

Answer 1

Your hypothesis (modifying the topology to produce a non-Hausdorff topological manifold from a Hausdorff one) is impossible, if you assume the non-Hausdorff manifold has dimension equal or less than the original; hence the answer to the question you asked is vacuously "yes".

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Let $(M,\tau)$ be your topological manifold, and denote by $\tau'$ the coarser topology, which you asserted is locally Euclidean.

First we remark that the identity map, regarded as a function from the topological space $(M,\tau)\to (M,\tau')$ is continuous; this is a property of $\tau' \subseteq \tau$ (your coarser assumption).

Given $p\in M$, by the locally Euclidean assumption we may choose an open (w.r.t $\tau'$) neighborhood $O'$ that is homeomorphic to $E'$, some Euclidean space. Note that $O'$ is also open w.r.t. $\tau$.

Similarly there exists another open (w.r.t $\tau$) neighborhood $O$ that is homeomorphic to $E$, another Euclidean space.

Now $O' \cap O$ is an intersection of open sets in the $\tau$ topology, and so is a relatively open subset of $O$, and hence is homeomorphic to some open set $\bar{E}\subseteq E$ in an Euclidean space.

Consider the chain of mappings (for convenience I specify the topology for the middle two spaces)

$$ \bar{E} \to (O'\cap O,\tau) \to (O',\tau') \to E' $$

with the first map the homeomorphism of $O$ and $E$ and the final map the homeomorphism of $O'$ and $E'$. The middle map is the identity map on $M$.

This chain gives an injective continuous map from $\bar{E}$ to $E'$, hence by invariance of domain it is also a homeomorphism to its image. (Technically invariance of domain as commonly stated requires $E$ and $E'$ to have the same dimension. When $\dim(E) > \dim(E')$, however, an upgrade of the theorem shows that there can be no injective continuous maps.)

In particular, this shows the following:

Given any $p\in M$, there exists an open neighborhood (w.r.t $\tau$) $\Omega$ of $p$ such that every open subset $U\subseteq \Omega$ (w.r.t $\tau$) is also open w.r.t. $\tau'$.

Now, assume that $\tau$ is Hausdorff. Given $p\neq q\in M$ let $\Omega_p$ and $\Omega_q$ be as in the previous result. Let $U_p$ and $U_q$ be disjoint open sets (w.r.t. $\tau$) that separate $p$ and $q$. Then $U_p\cap \Omega_p$ and $U_q\cap \Omega_q$ are still disjoint, separates $p$ and $q$, and are both open in the $\tau'$ topology. This shows that $\tau'$ must also be Hausdorff.

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For completeness (mostly for myself), let me spell out the upgrade to invariance of domain.

Corollary to Invariance of Domain If $U\subseteq \mathbb{R}^n$ is open and $f:U\to \mathbb{R}^m$ is injective with $m < n$, then $f$ cannot be continuous.

Proof: WLOG we can assume $U$ is the unit ball at the origin. Let $V$ be unit ball at the origin of in $\mathbb{R}^m$. There is an obvious embedding of $V$ into $U$, which we denote by $g$.

Assume for contradiction that $f$ is continuous. Then $f\circ g$ is a continuous injective map from $V$ to $\mathbb{R}^m$ which by invariance of domain is an open map. So $f\circ g(V)$ is open in $\mathbb{R}^m$, and by the assumed continuity of $f$ we have $g(V)$ is open in $U$. But the standard embedding of $V$ into $U$ does not have open image.