Relation between tangent bundle of a manifold and it’s covering manifold

Question

Let $M$ be a smooth (or complex) manifold, $N$ — topological manifold and $p: M \to N$ covering map. Consider the smooth (complex) structure on $N$ obtained by the well-known procedure making $p$ into smooth (holomorphic) map.

Is that true that from triviality of the (holomorphic) tangent bundle $TM$ follows triviality of $TN$? Is there any relation between those two whatsoever?

Answer 1

Suppose that the dimension of $N$ is $n$. The tangent space of $TN$ is trivial if and only if there exists $n$-vector fields $X_1,...,X_n$ such that for each $x\in N, X_1(x),..,X_n(x)$ are linearly independent. Since $p:M\rightarrow N$ is a local diffeomorphism, there exists $Y_1,...,Y_n$ vector fields of $M$ such that for every $y\in M$, $T_yp(Y_i(y))=X_i(p(y))$, this implies that $Y_1,...,Y_n$ is a trivialization of the tangent space of $M$, so the tangent space of $Y$ is trivial.

Answer 2

Triviality of $TM$ does not imply that of $TN$. For example, let $M=\mathbb R^2$, and let $\mathbb Z$ act on $M$ by $$ n\cdot (x,y) = (x+n,(-1)^n y). $$ Let $N$ be the quotient space $M/\mathbb Z$, and let $p:M\to N$ be the quotient map. Then $p$ is a smooth covering map, and $N$ is diffeomorphic to the open Möbius band, which is not even orientable, let alone parallelizable.