$T^2\times S^n$ is parallelizable This is taken from a UCLA Geometry/Topology qualifying exam.
How would one prove that $T^2\times S^n$ is parallelizable for all $n\geq 1$? Is there a way to find $n+2$ linearly independent vector fields? I am trying to think of the simplest case $n=2$ where $S^2$ is not parallelizable, but $T^2\times S^2$ has to be in some way. I would appreciate a general strategy to treat such problems.
 A: There is a similar post a few days ago asking why
$$
T(\mathbb{S}^{2}\times \mathbb{S}^{1})
$$
is trivial. The idea is to use the extra dimension coming from $T(\mathbb{S}^{1})$ to fill in the place of the normal bundle. For your question I think it is similar but more subtle, as you have to consider carefully how the pull back bundle splits and how to place the tangent vectors. I am not sure if a long exact sequence+Whitehead's thoerem type of argument would be useful, but you can also give a try. 
A: I just want to point out that this is elementary. Consider $S^n \times [1,2] \subseteq \mathbb{R}^{n+1}$ let the vector fields be defined by $\mathbf{e}_i$ the constant fields in the coordinate directions. Now identify the points $(x,1)$ and $(x,2)$ this gives $S^n \times S^1$ the vector fields are equal at the identified points so this gives $n+1$ fields on $S^n \times S^1$
A: I may elaborate the reason. Since $\mathbb{S}^1$ has trivial tangent bundle. Denote $\xi$ to be one-dimensional trivial bundle, and thus
$$T_{\mathbb{T}^2}=T_{\mathbb{S}^2}=\pi_1^* T_{S^1}\oplus\pi_2^* T_{S^1}=\pi_1^*\xi\oplus\pi_2^*\xi=\xi^2.$$
Also use the same formula, note that the tangent bundle $\tau_n$ os $\mathbb{S}^n$ we have $\tau_n\oplus \xi =\xi^{n+1}$. Thus
$$T_{\mathbb{S}^n\times\mathbb{T}^2}=\pi_1^*T_{\mathbb{S}^n}\oplus\pi_2^*\xi\oplus\pi_3^*\xi=\pi_1^*(\tau_n\oplus\xi)\oplus\xi=\pi_1^* \xi^{n+1}\oplus\xi=\xi^{n+2}.$$
