Let $\mathcal{M}$ be a compact 2-manifold, and let $\gamma: S^1 \rightarrow \mathcal{M}$ be a continuous map (you can assume piecewise smooth if it is convenient), with the property that the set $A = \{p \in S^1 : \exists \; q \in S^1, \; p \neq q, \text{ s.t. } \gamma(p) = \gamma(q) \}$ is finite. Define an equivalence relation on $\mathcal{M} \setminus \gamma(S^1)$ that identifies points as follows: $x \thicksim y$ iff $x,y$ are path-connected in $\mathcal{M} \setminus \gamma(S^1)$. The equivalence classes are the ``interiors of the faces'' of $\mathcal{M}$ induced by this construction. Notice that each equivalence class is open (this can be done by choosing any Riemannian metric on $\mathcal{M}$, and then each point in $\mathcal{M} \setminus \gamma(S^1)$ is contained in an open ball that does not intersect $\gamma(S^1)$), and so $\mathcal{M} \setminus \gamma(S^1)$ is a disconnected set, which is the union of its connected components (the equivalence classes).

First, can someone help me prove that the number of equivalence classes is also finite?

Now onto the actual research question: One can interpret this construction as a drawing of a graph on $\mathcal{M}$, whose vertices are the finite set of points $\gamma(A)$, and the edges of the graph are the segments of the curves joining two points. So now you have all the ingredients for an "Euler-like" formula: vertices, faces, edges. I want to know if any such formula is known or not, both in the orientable and non-orientable cases. I'd also be interested to know if any special cases are known.

The particular case I'm interested in is when $\mathcal{M} = S^1 \times S^1$, and additionally $\gamma$ has the property that if $\gamma(p) = \gamma(q)$ for distinct $p,q \in S^1$, then $\not \exists \; r \in S^1$ s.t. $\gamma(r) = \gamma(p)$.

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  • $\begingroup$ I do not think there is such Euler-like formula without further assumptions. Such a formula would have to be independent of the topology of the surface, because you could always connect-sum on a handle and increase the (orientable or non-orientable) genus, so it would have to depend only on $V, E, F$ and hold for all graphs. This seems unlikely. The trouble is that you are not making any assumptions about the topology of the faces. If you assume that the faces are all homeomorphic to discs then you can probably get somewhere. $\endgroup$ – dvitek Mar 14 at 1:39
  • $\begingroup$ @dvitek how about an inequality instead of an equality? $\endgroup$ – Rahul Sarkar Mar 16 at 22:31

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