Embed the Klein bottle into the 3-manifold $S^{2} \times S^{1}$ Can the Klein bottle $K$ be embedded into $S^{2} \times S^{1}$?
If so, how does it work? If not, what is the obstruction?
Thanks in advance.  
 A: The Klein bottle can be given by $S^1 \times [0,2\pi] $ with the identificition 
$$(\theta, 0) \sim (-\theta, 2\pi)$$ 
(where I used $\theta$ to parametrizes the circle). Now define 
$$S^1 \times [0,2\pi] \to S^2 \times S^1, \ f(\theta, t)  =\big( a(\theta, t), e^{it} \big),$$ 
where 
$$a(\theta, t ) = \sin\theta (\cos\frac{t}{2} , \sin \frac{t}{2}, 0)+ \cos \theta (0,0,1).$$
Note that 
$$f(-\theta, 2\pi) = (a(-\theta, 2\pi ), e^{i2\pi}) = (a(-\theta, 2\pi ), e^{i0}) $$
and 
\begin{equation}
\begin{split} 
a(-\theta, 2\pi ) &=\sin(-\theta) (\cos \pi , \sin \pi, 0)+ \cos (-\theta) (0,0,1)\\
&= \sin \theta (\cos 0 , \sin 0, 0)+ \cos \theta (0,0,1) \\
&= a(\theta, 0)
\end{split}
\end{equation}
Thus $f(\theta, 0) = f(-\theta, 2\pi)$ and so $f$ descends to a map 
$$\tilde f: K  \to  S^2 \times S^1. $$
Note that $\tilde f$ is injective. 
A: The Klein bottle is the space $S^1 \times I/\sim$, where the equivalence relation identifies $(z,0) \sim (\bar z, 1)$, where $\bar z$ denotes complex conjugation (aka reflection across the x-axis). 
Denote by $\text{rot}_\theta$ the rotation of 3-space which moves the z-axis down theta degrees toward the y-axis (this is a rotation whose fixed set is the x-axis). Write $g: S^1 \to S^2$ for the standard inclusion of the equator. 
$f: S^1 \times I \to S^1 \times S^2$, given by $f(z,t) =(2\pi t, \text{rot}_{\pi t} g(z))$ descends to an embedding of the Klein bottle. 
