Show that the central circle $X$ in the open Mobius band has mod 2 intersection number $I_2(X,X)=1$

Question

Show that the central circle $X$ in the open Mobius band has mod 2 intersection number $I_2(X,X)=1$

Like in this picture

http://i58.tinypic.com/2dkjwug.png

Boundary Theorem: suppose that $X$ is the boundary of some manifold $W$ and $g: X \to Y$ is a smooth map. If $g$ may be extended to all of $W$ then $I_2 (g,Z)=0$ for any closed submanifold $Z$ in $Y$ of complementary dimension.

Tranversality Homotopy theorem: For any smooth map $f:X \to Y$ and any boundarless submanifold $Z$ of boundaryless manifold $Y$, there exist a smooth map $g: X\to Y$ homotopic to $f$ such that $ g$ transversal to $Z$ and $\partial g$ tranversal to $Z$

Somehow I need to show that when the ends of the strip is glued together with a twist, $X'$ becomes a manifold that is a deformation of $X$

Let $f:X\to Y$ be a smooth map of a compact manifold $X$ into a connected manifold $Y$.

From the picture I can see that $dim (X)= \frac {1}{2} dim (Y)$ so clearly, I can't use the mod 2 degree theorem. I'm thinking of using the boundary theorem, but can't find any way to apply it. Any help would be most welcome.

Answer 1

First, you need the fact that $I_2$ is invariant under homotopy. Then $I_2 (X,Y)$ is the number of intersections of $X$ and $Y$ when $X$ is transversal to $Y$. In your case, dim$X$+dim$X'$=dim$M$, where $M$ is the Mobius band, so you just need to notice that the picture you linked as one point of intersection, since $X$ and $X'$ are transversal.

If you really want to write something explicit down for the homotopy between $X$ and $X'$, I would suggest making $X'$ a straight line with slope just above zero. Keep it intersecting $X$ at the midpoint and the ends will line up when you glue the sides of your square together.

I am not sure what you mean by the mod 2 degree theorem, but I don't think that the boundary theorem is necessary. Hope this helps.