Jtdylktuy

I'm familiar with the idea in topology that shapes that can be continuously deformed into one another are considered "equivalent". I read about the Euler Characteristic as being Vertices-Edges+Faces. Thinking of this number as related to the genus of an object, and the genus as the number of holes it has (which is probably where I'm going wrong), I began wondering what the Euler number or genus of a Y-shaped pipe (see below) would be, having three openings that all converge together.

Y-shaped pipe:

Thanks in advance for any help!

This can be answered with some knot theory. The openings can be thought of as an unlink on three components. This just means three circles that are not "linked" together. This Y shaped pipe can be deformed to a sphere with three holes missing from it. Just think about the pipe as if it were a weird ballon. If we would fill these holes, we get a sphere, which means that it has genus zero. Then, we know that the Euler characteristic of sphere is two. (For a genus $g$ "shape," the Euler characteristic is $2-2g$. )

But we added those disks, which each count as a face. So, to get back to our Y shaped pipe, we need to subtract those disks (faces), which means we have $2-3=-1$.

Or, we could have used this: From Rolfsen's Knots and Links, we have a relation which says $$ g(M)=1-frac{chi(M)+b}{2} $$ where $g(M)$ is the genus of a manifold $M$ (our Y shaped pipe), $chi(M)$ is the Euler characteristic, and $b$ is the number of boundary components, which means the holes. We know the genus, and the number of holes, all you have to do is solve for $chi(M)$.

The easy way to find the Euler characteristic is to note that we can deformation retract the pair of pants (that's the name topologists normally give this space) onto a subset which is homeomorphic to a circle with an interval glued at one end to the north pole or the circle and the other end glued to the south pole of the circle. This in turn is homotopy equivalent to a wedge of two circles by shrinking the extra interval to a point. This space has one vertex and two edges so we get $chi=1-2=-1$. As Euler characteristic is a homotopy invariant, it follows that this is also the Euler characteristic of the pair of pants.

It's actually pretty easy to put a cellular decomposition straight onto the pair of pants without worrying about homotopy equivalences. Note that the pair of pants is homeomorphic to a closed disk with two smaller open disks removed from its interior. We can decompose this into cells by placing an edge between the boundaries of the removed disks and also an edge from each boundary of the removed disks to the boundary of the large disk. This gives us a cellular decomposition consisting of $6$ vertices on the boundary, $3$ edges in the interior, $6$ edges on the boundary, and $2$ faces in the interior. This gives $chi=6-(3+6)+2=-1$ as expected.