Understanding attaching space

Question

I have some problem understanding the attaching space when learning topology. I cannot understand these two examples.

Example 1 Let $X=A=S^1$,$Y=I\times S^1$ and $B=\{0\}\times S^1$. Let $h:B\to A$ be the map that sends $B$ twice around $A$, i.e., $h(z)=z^2$ where we think of the circle as the set of complex number of modulus $1$. The attaching space $X\cup_h Y$ is homeomorphic to the Möbius strip.

I have trouble understanding "sends $B$ twice around $A$". I cannot imagine how a cylinder becomes a Möbius strip by this attaching map. I can only imagine rolling a cylinder twice and identify the boundary that touches each other. Is there any method to understand this?

Example 2 $X=A=S^1$,$Y=D^2$ and $B=\partial D^2=S^1$. Using the same attaching mapping, we claim that the attaching space $X\cup_h Y$ is homeomorphic to attaching a two-dimensional sphere to Möbius strip, which is in turn homeomorphic to $P^2(\mathbb{R})$.

How to understand "attaching a two-dimensional sphere to Möbius strip" and why is this homeomorphic to $P^2(\mathbb{R})$?

I'm new to this area, so sorry for the rather dumb-sounding problem. Thanks for your attention!

Answer 1

The key in understanding these two example is to understand what the attaching map does.

The attaching map
$$f \colon S^1 \to S^1$$ defined by $f(z)=z^2$ is the parametrization of the path which turn two times around the circle.

This map cover the image with both the subspaces $\{e^{i\theta} \mid \theta \in [0,\pi]\}$ and $\{ e^{i\theta} \mid \theta \in [\pi,2\pi]\}$, identifying the antipodal points: this map is a quotient map which identifies the antipodal points and in particular is surjective.

So in the first example the attaching space is nothing more than the space obtained attaching the two semicirles of $S^1 \times \{0\}$ along the $S^1$ and so is the quotient space of $S^1 \times I$ obtained identifying in the subspace $S^1 \times \{0\}$ the antipodal points: this quotient is the Moebius strip.

The second example is obtained by attaching both the semicircles of the boundary of $D^2$ on $S^1$ and so the attaching space is the quotient of $D^2$ obtained identifying the antipodal points in the boundary, and this quotient is exactly $P^2(\mathbb R)$.

Hope this helps.

Answer 2

I do not know if your source gives some more explanations, and I do not want to repeat these. I would like to point out though, that the first example is something that you can actually check with a paper model: Take a Möbius strip. Mark the circle in the middle of the strip. (The one that does not cross itself.) Cut along this circle. What you obtain is a cylinder, even though it is unusually embedded in $3$-space. Now see what parts of the boundary of the cylinder you would have to glue back together to obtain the Möbius strip.