Suppose I have a smooth orientable surface $Q$ and a compact region $R$ of $Q$. Suppose there is a closed curve $C$ that divides R into two connected components $R_1,R_2$ but does not divide Q into separate regions. Suppose there is a (smooth) curve $f$ with one endpoint in $R_1$, the other in $R_2$ and is otherwise disjoint from $R$. Must there it be possible to find a closed curve in Q not intersecting R, that does not divide Q into two separate regions? Colloquially must it be possible to find a handle we can "cut" from our surface?

Figure below: Region $R$ in orange is a subset of our surface which is the torus. The curve $C$ in blue divides $R$ into $R_1$ and $R_2$. The red curve has as two endpoints on the boundaries of $R_1$ and $R_2$ respectively. The green curve, disjoint from $R$ does not divide $Q$\ $R$ into two disjoint regions.

enter image description here

Also anyone recommend a basic reference for learning how to prove this kind of thing?(smooth 2dimensional surfaces only focusing on genus and surface operations) Basic topology by Armstrong is understandable but doesn't provide enough tools for this kind of thing. Other textbooks go too far and go beyond 2 dimensions or are a bit too abstract.

  • $\begingroup$ I'm not sure if I understood your question tbh. Are you asking whether there exists an arc $f$ in a closed surface $\Sigma$ of genus $1$ whose ends join a compact subsurface without separating it? Just pick an annulus $R$ inside $Q$ and an arc $f$ that joins the right boundary component to the left boundary component. $\endgroup$
    – Zest
    Nov 18, 2023 at 20:57
  • $\begingroup$ One idea I have is perform surgery along C to yield a surface $Q'$ of smaller genus. Now since $R_1$ and $R_2$ are disjoint, there is a way to "cut" the surface $Q'$ into two smaller surfaces $Q_1$ and $Q_2$ whose genus's when summed up is at most that of $Q'$. Now glue $Q_1$ and $Q _2$ at $C$. Since $Q_1$ and $Q_2$ are disjoint, this produces a surface $Q"$ of genus at most the sum of the genus's of $Q_1$ and $Q_2$. I feel this is intuitively true but I don't know how to rigorously prove that this can work. $\endgroup$
    – Hao S
    Nov 18, 2023 at 23:16
  • $\begingroup$ @Zest in general the genus need not be 1. Second the annulus must not intersect the region $R$. Even for genus 1 idk how to prove such an annulus exists. $\endgroup$
    – Hao S
    Nov 18, 2023 at 23:18
  • $\begingroup$ What do you mean by "the annulus must not intersect the region $R$"? Just take the annulus as $R$. It's a compact subsurface of $Q$. You will always find an arc from one boundary component (which is just a circle) of $R$ to the other boundary component of $R$ since your surface $Q$ is connected. $\endgroup$
    – Zest
    Nov 19, 2023 at 0:37
  • $\begingroup$ @Zest I am looking for a closed curve not intersecting the region R which does not divide the surface Q into two components. Also the R is just a compact subsurface not necessarily an annulus. $\endgroup$
    – Hao S
    Nov 19, 2023 at 0:47

1 Answer 1


No, this need not always be possible.

View the torus $Q$ as a square with opposite sides identified appropriately. Let $R$ be the shaded region in the picture below. Note that $R$ is connected and compact.

Torus Q with shaded region R

Let the closed curve $C$ be the bottom edge (as well as top edge) of the square in the above picture. Then, $Q - C$ is a connected region (homeomorphic to an annulus), but $R - C$ is disconnected into two pieces.

Q - C

Clearly, there is a curve $f$ that connects the shaded part $R_1$ on the top with the shaded part $R_2$ on the bottom that lies completely in the unshaded part of $Q$ (except for its end points). However, the unshaded part (i.e., the region $Q - R$) is homeomorphic to an open disc. So, any closed curve in this region separates $Q$ into two components.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .