**EDIT:** The answer is trivially positive; the question arose from my misunderstanding of the figure below.

Can a non-orientable closed surface of **odd genus** 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][1]_ 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 $\pm(x_1^2+x_2^2)$ in some local coordinates.

My intuition is that no. (I've asked this [question][2] on math.SE but did not get any answer.)

Consider the projective plane $\mathbb RP^2$ as the Boy surface (left) [**EDIT:** wrong, both figures are not immersions, and the left figure is not the Boy surface] and the Klein bottle $K^2$ (right):

[![|Fig. 1][3]][3]

(image from the [book][4]). The surfaces shown are images of immersions (i.e., locally smooth embeddings) in the complements of the singular points, shown in white. The vertical line in the right-side figure is a homologically non-trivial cycle.

For even genera $g$ (except $g=2$, which is a different story), it is easy to do: e.g., connect the top and bottom of $K^2$ (right) by a tube (as if you drill a wormhole along the vertical axis), which will form a surface of genus $g=4$ immersed [**EDIT:** wrong, this is not an immersion] with two Bott-type extrema (circles) and two Morse-type saddles. (You can get any even genus $g\ge4$ by adding more handles.)

However, adding such a handle to the $\mathbb RP^2$ (left) seems not possible. Suppose you add such a handle connecting the bottom to the top of the figure (left). There must be a singularity on that handle. Indeed, consider the evolution of the level sets from the bottom to the top along this handle. The level sets at its endpoints are circles $S^1$ immersed in the plane: O-shaped at the bottom and 8-shaped at the top, which are not regularly homotopic by the Whitney–Graustein theorem. Therefore, there must be a singularity in between.

My intuition is that the singularity will be similar to the saddle shown on the picture (roughly speaking, not preserving the parity of the total turning number below and above). Though the singular contour can be more complicated (e.g., connecting more handles), it will effectively convert the left-side picture into the right-side one: it would cause an additional cycle passing through the first singularity (like the cycle between the two singularities shown on the right), thus making the genus $g$ even.

I think this argument would generalize to a surface with more handles, as soon as any cycle exists between the "bottom" and "top" of the singular level of the type shown in the figure (left): in the absence of centers, for each saddle "that changes the orientation," along the evolution of the level sets (passing transparently the Bott-type extrema along the handle) there will occur another similar saddle adding a second cycle.

Unfortunately, I lack the skill to convert this into a formal proof, and even if I could do it for this particular type of immersion [**EDIT:** wrong, this is not an immersion] of $\mathbb RP^2$ (Boy surface), it would not prove the claim in the general case. Could you provide such a proof, or point to sources where a proof can be found? Detailed explanations would be greatly appreciated, since I am not an expert.


  [1]: https://en.wikipedia.org/wiki/Morse_theory#Morse%E2%80%93Bott_theory
  [2]: https://math.stackexchange.com/questions/3389004/immersion-of-non-orientable-surface-in-mathbb-r3-with-conditions-on-the-heig
  [3]: https://i.sstatic.net/cGgSw.jpg
  [4]: https://www.researchgate.net/publication/220692408_Computational_Topology_An_Introduction