I have the following questions:
Let $p:\Bbb{T}^2\to \Bbb{S}^1$ be a circle bundle. Is $p$ trivial ?
Before I show you what I attempted, I just want to let you know that I don't have any knowledge of fiber bundles. I just base my intuition on the little I know about vector bundles, and from what I quiclky read on Wikipedia I think that I'm allowed to do what follows.
Take the quotient map $q:[0,1]\to S^1$. We can pull back the bundle $p$ via $q$. Because $[0,1]$ is contractible, this pulled back bundle is trivial so we have $$\begin{array} {ccc} [0,1]\times \Bbb{S}^1 & \stackrel{h}{\longrightarrow} & E=\Bbb{T}^2\\ p' \Big\downarrow & & \Big\downarrow p\\ [0,1] & \stackrel q {\longrightarrow} & S^1 \end{array}$$
Then I guess I should glue the sides of the cylinder $[0,1]\times \Bbb{S}^1$ by factorizing $h$ but I don't know how to do it formally (I don't see how $f$ factors). Also should I have taken the universal cover $\mathbb{R}\to \Bbb{S}^1$ instead of $q$?
Second question: I am looking for a (short) introductory book to learn about the theory of fiber bundles. I don't know much about it so I can't really be specific in my request, but I'd like something which is "differentiable manifolds related". From what I have read I think that I'm looking for something like the 3rd chapter of Dale Husemoller, Fiber Bundles. I also know about Norman Steenford The topology of fibre bundles, but I'm not sure if this one is exactly what I'm looking for. Do you have any advices? (I've had a course of differential topology already but we didn't study fiber bundles)
Thanks in advance!
This is true.
Sketch : Let $x \in S^1$, $I = S^1 \backslash \{x\} \cong (0,1)$ and $U = p^{-1}(I)$. The map $p_{|U} : U \to I$ is a trivial fiber bundle because $I$ is contractible, i.e it's the projection $S^1 \times I \to I$.
Moreover, let us extend $p_{|U}$ to a map $\overline p : \overline{I} := [0,1] \times S^1 \to [0,1]$. By construction one can reconstruct $p$ from $\overline p$ by identifying $0 \times S^1$ and $1 \times S^1$ together.
The map should be a diffeomorphism, so it should either reverse or preserve the orientation. It's easy to see that you can only get the Klein bottle in the first case or the torus in the second case.
EDIT : here are more details. With OP's notation, it's clear that $h$ has the same values on $0 \times S^1$ and $1 \times S^1$ since the diagram is commutative. Thus $h$ is a surjection and since $[0,1] \times S^1$ is compact we get an homeomorphism $[0,1] \times S^1 / \sim \cong T^2$. This homeomorphism is determined by a continuous bijection $S^1 \to S^1$. Up to homotopy there is only two such maps : $z \mapsto z$ and $z \mapsto \overline{z}$. In the second case we would obtain the Klein bottle which is not possible, so the gluing map is homotopic to $z \mapsto z$ which means that the trivialisation extends to the whole space i.e the fiber bundle was trivial.