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

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.

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.
• Can I replace $Y$ by $M$ and $Z$ by $X'$ then use the tranversality homotopy , to show there is a homotopy between $X$ and $X'$? Commented Oct 20, 2014 at 19:04