Existence of a Global Trivialization of Pullback Bundle Good evening!
Let us assume that we have a $n$-dimensional compact Riemannian manifold $(M,g)$, and a smooth cylinder $u\in C^\infty(\mathbb{R}\times\mathbb{S}^1,M), (s,t)\longmapsto u(s,t)$. My question then is: Can the pullback tangent bundle $u^* TM$ always be globally trivialized along $u$?
I would argue as follows: Take any point $(s_0,t_0)\in\mathbb{R}\times\mathbb{S}^1$, and define a local orthonormal basis $(Z_1(s_0,t_0),...,Z_n(s_0,t_0))$ of $T_{u(s_0,t_0)}M$. Then, via parallel transport (w.r.t. the Levi-Civita connection) along $t\longmapsto u(s_0,t)$, one gets an orthonormal basis $(Z_1(s_0,t),...,Z_n(s_0,t))$ of $T_{u(s_0,t)}M$ for every $t\in\mathbb{S}^1$ (since parallel transport leaves $g(Z_i(t),Z_j(t))$ constant) that depends smoothly on $t$.
Now, to reach the other values of $s\in\mathbb{R}$, one parallel transports (for fixed $t\in\mathbb{S}^1$)the basis  $(Z_1(s_0,t),...,Z_n(s_0,t))$ along $s\longmapsto u(s,t)$.
Thus, the map $\Theta:\mathbb{R}\times\mathbb{S}^1\times \mathbb{R}^n\longrightarrow u^*TM, \Theta(s,t)e_i = Z_i(s,t)$ (where $(e_1,...,e_n)$ is the canonical basis of $\mathbb{R}^n$) is a global trivialization of $u^*TM$.
Of course, this gets difficult if $u$ intersects itself at some point (i.e. $u(s,t)=u(s',t')$ for $(s,t)\neq(s',t')$), since then one would have two different trivializations for $T_{u(s,t)} = T_{u(s',t')}$. However, I am not sure how big of a problem this is, since the map $(s,t)\longmapsto (Z_1(s,t),...,Z_n(s,t))$ should still be well-defined.
I would like to know what has to be corrected to make this rigorous, or what additional assumptions I should put on $M$.
 A: Questions about the triviality of bundles are typically answered in the language of algebraic topology, so I hope you are familiar with this terminology. I'll use the first Stiefel-Whitney class $w_1\in H^1(M,\mathbb{Z}/2)$. Firstly, we can deformation retract the cylinder onto $S^1$ and then ask whether the pullback of a tangent bundle to $S^1$ is trivial (or not). Consider the embedding $u:S^1=\mathbb{RP}^1\to\mathbb{RP}^2$. We know that $w_1(T\mathbb{RP}^2)\neq 0$, i.e. $\mathbb{RP}^2$ is not orientable. By naturality, we have $w_1(u^*T\mathbb{RP}^2)=u^*w_1(T\mathbb{RP}^2)$.
Noting that the cohomology ring with $\mathbb{Z}/2$-coefficients for $\mathbb{RP}^n$ is the polynomial ring $\mathbb{Z}/2[x]/(x^{n+1})$, the element $w_1(T\mathbb{RP}^2)\in H^1(\mathbb{RP}^2,\mathbb{Z}/2)$ must be the generator $x$, which pulls back to the generator of the cohomology ring of $\mathbb{RP}^1$, and is thus also a non-trivial element. Consequently, $w_1(u^*T\mathbb{RP}^2)\neq 0$ which implies that $u^*T\mathbb{RP}^2$ is non-trivial.
On the other hand, if you consider maps $u:S^1\to M$ where $M$ is an orientable manifold, then $w_1(TM)=0$ and hence $w_1(u^*TM)=0$. Therefore, if $M$ is orientable, the pullback of its tangent bundle to a cylinder will always be trivial. This is because any vector bundle $E$ over $S^1$ with $w_1(E)=0$ can be trivialised. For a discussion of this fact, see here.
