Let $S$ be a closed orientable surface embedded in $S^2\times S^1$ assume that the projection of the surface $S$ on $S^1$ is not a point. I am trying to prove that exist a smooth loop $\gamma: S^1\to S^2$ such that $\forall x\in S^1$ we have $(\gamma(x),x)\notin S$, i.e. $(\gamma,id)(S^1)\cap S=\emptyset $.
I tried to use Transversality theorem in the following way: We define $F=id: S^2\times S^1 \to S^2\times S^1$ such that $F(y,x)=(y,x)$ since $F$ is transversal to $S$ we have that exist a dense subset $D\subset S^2$ such that $F_y$ is transversal to $S$ for each $y\in D$. We fix $y\in D$ and we call $\gamma:=F_y$.
Suppose that $\exists x\in S^1$ such that $(\gamma(x),x)\notin S$, since $\gamma$ is transversal we have:
$T_{(\gamma(x),x)}S^2\times S^1= Im\:d_x\gamma \:+\:T_{(\gamma(x),x)}S=0\times T_xS^1\:+\:T_{(\gamma(x),x)}S$,
now I tried to project on the subspace $T_{\gamma(x)}S^2$ and from that we have: $T_{\gamma(x)}S^2=\pi (T_{(\gamma(x),x)}S)$.
And I got stuck, I thought to use the condition on the projection of $S$, but I don't know how to use it.
Thank you for your time.
What you are trying to prove is not true, because counterexamples like $S^2 \times \{z\}$ which project to a single point $z \in S^1$ (and which your hypothesis rules out) can be easily perturbed so that they do not project to a point (and so your hypothesis does not rule them out). So you need stronger hypotheses on your surface $S$ in order to get the conclusion that you want.
Here's an example. Take any nonconstant function $f : S^2 \mapsto (-1/2,1/2)$, such as $f(r,s,t) = t/3$. Define $$S = \{(p, e^{2 \pi i f(p)}) \mid p \in S^2\} \subset S^2 \times S^1 $$ For every smooth loop $\gamma : S^1 \to S^2$, the loop $\{(\gamma(x),x) \mid x \in S^1\}$ has nonempty intersection with the surface $S$.