TITLE: Embed the Klein bottle into the 3-manifold $S^{2} \times S^{1}$
QUESTION [9 upvotes]: 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.

REPLY [4 votes]: 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.