Find the DeRham's cohomology of the following open sets, and then looking at the product conclude that they are not diffeomorphic.
a) $M=\mathbb{R}^3\setminus (L_1\cup C)$, $L_1=\{x=y=0\}$ and $C=\{x^2+y^2=1,z=0\}$.
b) $N=\mathbb{R}^3\setminus (L_2\cup C)$, $L_2=\{x=3,y=0\}$ and $C=\{x^2+y^2=1,z=0\}$.
Can you give me an idea?
I tried:
Idea I write $$M= \left(\mathbb{R^3}\setminus L_1\right)\cap \left(\mathbb{R^3}\setminus C\right)$$ So I had to find first the cohomology of $\left(\mathbb{R^3}\setminus L_1\right)$ and $\left(\mathbb{R^3}\setminus C\right)$. But I dont how to find them and neither how to use the different positions of the line respecto to the circle
Thank you
Perhaps some visual hints will help, if you'll excuse the hand-drawn examples...
First we see $M$ deformation retracts to a torus. Intuitively, the missing line at $x=y=0$ becomes the hole through the donut, while the missing circle at $x^2+y^2 = 1, z=0$ becomes the hole inside the donut.
Next, we consider $N$. Since the circle and the line aren't interlinked, we may consider them separately, and then glue them together at the end with the Mayer-Vietoris Sequence.
On the left, we have $\mathbb{R}^3 \setminus S^1$, which is "well known" to deformation retract to $S^2 \vee S^1$. Instead of thinking about this as a sphere with a circle on its side, think of it as a sphere with a diameter linking its two poles. Then the missing circle becomes the hollow bit between this diameter and the surface of the sphere... Sorry I don't have a better picture for this. Perhaps the graphics here will help. This can also be found on page 46 of Hatcher.
On the right, we have $\mathbb{R}^3$ without a line. This deformation retracts to the plane without a point (by squishing down), and of course this is well known to be a circle. The missing point becomes the hole in the center of the circle.
Do you think you can use this information (as well as the Mayer-Vietoris Sequence for $N$) to discover the cohomology of these spaces, and thereby deduce they aren't diffeomorphic?
I hope this helps ^_^