Undergrad Properties of the initial topology from a topological manifold

[cianfa72]

TL;DR

About the properties of the initial topology defined on a set though the preimage of a non-injective map

Consider a non-injective map $\pi$ from a set $M$ to a set $N$. $N$ is equipped with a topological manifold structure (Hausdorff, second-countable, locally euclidean).

Take the initial topology on $M$ given from $\pi$ (i.e. a set in $M$ is open iff it is the preimage under $\pi$ of an open set in $N$). Such a topology on $M$ is second-countable, however is it Hausdorff ?

I believe it is not since points in $\pi^{-1} (\{p \}), p \in N$ are always in the same open set in $M$ (let me say there is not enough "resolution" in the initial topology on $M$ to be able to separate its points into disjoint open sets).

What do you think about ? Thanks.

 

[Orodruin]

Indeed, it is not Hausdorff. You should formalize your argument by explicitly stating that you consider two (or more, but two is sufficient) points in $M$ that map to the same point in $N$. There is no open set containing only one of them - it is always both or none - and they can therefore not be separated in the Hausdorff sense.

You should also be able to formally prove that the topology is Hausdorff if instead you have a $\pi$ that is injective.

 

[cianfa72]

You should also be able to formally prove that the topology is Hausdorff if instead you have a $\pi$ that is injective.

Take two points $p, q \in M$ then $m=\pi(p) \neq \pi(q)=s$ assuming $\pi$ to be injective. Since $N$ is Hausdorff there exist disjoint open sets $A,B \subset N$ such that $m \in A$ and $s \in B$.

By injectivity $\pi (\pi^{-1}(A) \cap \pi^{-1}(B)) = \pi(\pi^{-1}(A)) \cap \pi(\pi^{-1}(B))$. Now $\pi(\pi^{-1}(A)) \subseteq A$ and $\pi(\pi^{-1}(B)) \subseteq B$ therefore $$\pi^{-1}(A) \cap \pi^{-1}(B) = 0$$ hence $\pi^{-1}(A)$ and $\pi^{-1}(B)$ are open and disjoint i.e. the initial topology on $M$ is Hausdorff.

 

[cianfa72]

The reason behind the OP is the following: starting from the topological/smooth manifold $M$ consider the definition of tangent bundle $TM$ built on top of it (i.e. $M$ is the base space).

The topology on $TM$ can't be the initial topology from the (surjective) projection map $\pi$ $$\pi: TM \to M$$ since it isn't Hausdorff hence can't be a topological manifold in first place.

 

[martinbn]

The reason behind the OP is the following: starting from the topological/smooth manifold $M$ consider the definition of tangent bundle $TM$ built on top of it (i.e. $M$ is the base space).

The topology on $TM$ can't be the initial topology from the (surjective) projection map $\pi$ $$\pi: TM \to M$$ since it isn't Hausdorff hence can't be a topological manifold in first place.

Why would you think that the topology of $TM$ might be this topology!?

 

[cianfa72]

Why would you think that the topology of $TM$ might be this topology!?

I've been watching Dr. Schuller's lectures. Here at 20:00 starts endowing $TM$ with the coarsest topology such that the (surjective) projection map $$\pi : TM \to M$$ is continuous, i.e. the initial topology on $TM$ from $\pi$. He then shows one can define a smooth atlas on $TM$ turning it into a smooth manifold (by ripping out non-smooth-compatible charts from the topological atlas).

P.s. Does my work in post #3 make sense?

 

[Orodruin]

Why would you think that the topology of $TM$ might be this topology!?

… or in the unforgettable words of one of my math lecturers in first year of university whenever anyone would make an argument that may sound reasonable but ultimately says nothing about the proof in point: ”A common reaction to your argument is ’so what?’.”

 

[martinbn]

I've been watching Dr. Schuller's lectures. Here at 20:00 starts endowing $TM$ with the coarsest topology such that the (surjective) projection map $$\pi : TM \to M$$ is continuous, i.e. the initial topology on $TM$ from $\pi$. He then shows one can define a smooth atlas on $TM$ turning it into a smooth manifold (by ripping out non-smooth-compatible charts from the topological atlas).

P.s. Does my work in post #3 make sense?

Yes, he does say that. May be later on he will correct himself.

 

[cianfa72]

Yes, he does say that. May be later on he will correct himself.

Therefore the underlying $TM$ topology as smooth manifold is that given/induced by the (smooth) atlas he defines later (i.e. a set $A \subset TM$ is defined as open iff for any chart $(U,\varphi)$ in the atlas $\varphi(A \cap U)$ results open in $M$).