Technical question about the fiber of tangent bundle over a smooth manifold

Question

I am attempting to self-study differential geometry because it is relevant to my profession, but I have run into a technical question that I can't find addressed in the many textbooks out there.

Let $M$ be a manifold of degree $n$ and let $TM$ be the tangent bundle. Since I don't know how to do commutative diagrams here, I'd like to refer to diagram (1) in this Wikipedia link: Fiber Bundle.

With regard to that diagram, I am completely clear on the right hand side. My question concerns the left hand side. I know from various textbooks that a fiber $F$ of $TM$ is a vector space (isomorphic to $T_pM$). Furthermore, I know that $F=\pi^{-1}(p)$ at a point $p \in M$.

My question is what does $\pi^{-1}(p)$ look like? Note that this is before the homeomorphism to the Cartesian product $\{ p \} \times F$ is applied so it really can't be $(p,v)$. Furthermore, what is $\pi$? Most books call it a projection map, but it projects what onto what? It's clear from the diagram I cited that $\text{proj}_1$ is the projection map onto the first "slot" in the Cartesian product: $(p,v) \rightarrow p$. So what does $\pi$ project?

Finally, on the left hand side, what encodes information about $p$ so that the homeomorphism "knows" there is a $p$ to place in the first slot?

Answer 1

The fiber at $p$ is not isomorphic to $T_pM$; it is the tangent space $T_pM$. To get a better feeling for things, suppose $M\subset \Bbb R^N$. Then $$TM = \{(p,v)\in M\times\Bbb R^N: v\in T_pM\subset\Bbb R^N\},$$ and, yes, $\pi\colon TM\to M$ is literally given by $\pi(p,v) = p$.

Answer 2

a point, $p,$ in the tangent bundle is a point in the manifold together with a a vector, $v,$ which expresses a tangent distance and direction. $\pi$ expresses the action of dropping $v$ so you just have $p.$ The fibre at $p$ is the set of all the vectors that can be attached at $p$ so it is essentially a copy of $R^n.$ The subtleties lie in how the copies of $R^n$ join together.