Immersion in $\mathbb R^3$ of a Klein bottle with Morse-Bott height function without centers Can the Klein bottle be immersed in $\mathbb R^3$ so that the associated height function be of Morse-Bott type and have no centers?
That is, the height function would have only Bott-type extrema and saddle singularities. A Bott-type singularity is a non-degenerate singular circle: a circle where the derivative is zero with the function being quadratic on transverse curves. A center is a Morse-type local extremum: an isolated singularity around which the function is $\pm(x^2_1+x^2_2)$ in some local coordinates.
My intuition is that no. I think such function cannot have (Morse) singularities other than Bott-type extrema (because they would increase the genus), and I cannot see how to connect an (even) number of immersions of Bott-type extrema (circles) by tubes in a non-orientable way without additional singularities (this should follow from the Whitney–Graustein theorem).
For a torus, such an immersion (embedding) is a doughnut lying flat on the table. However, I can't see how this can be done for the Klein bottle. The answer here does not seem to do the trick because it also increases the genus.
 A: I wanted to prove that this is impossible but instead proved that this is possible... 
Unfortunately, it is a bit hard to draw the picture but I'll try to explain how this should look like. 
Construction. In this construction the Klein bottle will be included between the planes $z=0$ and $z=1$. The curves $\{z=1\}\cap K$ and $\{z=0\}\cap K$ are both the eight figure curve (with rotation index $0$). And both are Bott circles. Let us call the first curve $S_1$ and the second $S_0$. 
Now, the function $z$ restricted to $K\setminus S_0\cup S_1$ has no critical points. And $K\setminus S_0\cup S_1$ is the immersed image of two cylinders $C$ and $C'$, both propagating in $\mathbb R^3$ from the plane $z=0$ to the plane $z=1$. The intersection of $C$ and $C'$ with any plane $z=c$ (where $c\in [0,1]$) is a figure eight curve.
The last detail is two explain how $C$ and $C'$ look like. So we will take as $C$ just the direct product of a vertical interval with the figure eight curve. To construct $C'$ we need to do something a bit trickier. Namely to construct it we start from $S_0$ in $z=0$ and then start to rotate it so that it the plane $z=t$ it is the figure eight curve rotated by $\pi t$. Thus, for $t=1$ it will be rotated by $\pi$. 
Now, one can easily check that if we rotate a figure eight by $\pi$, it changes its orientation! So if we glue $C$ with $C'$, we get the Klein bottle indeed.
To finish the construction one just needs to smoothen out the described surface at $S_0$ and $S_1$. But this is not hard to do.
