# Hatcher 0.20

Hatcher, Algebraic Topology, Chapter 0

20. Show that the subspace $X\subset\mathbb{R}^3$ formed by a Klein bottle intersecting itself in a circle, as shown in Figure 1 below, is homotopy equivalent to $S^1\vee S^1\vee S^2$.

Figure 1
The space $X\subset\mathbb{R}^3$ described above

Proof. Let $X$ be the figure shown above consisting of a Klein bottle intersecting itself in a circle $C$. The main key to constructing this result is the fact that the circle $C$ of intersection actually yields a disk $D$ on the “outer surface” of the Klein bottle satisfying $C=\partial D$. Because $D$ is homotopy equivalent to a point, the space $X$ can transformed to the space $X_1$ by way of a homotopy equivalence. The space $X_1$ is shown in Figure 1 below. Note that the blue and orange “loops” in Figure 1 are included to help provide some insight into the geometry present, and the “light” segments are meant to be the continuations of the “dark” ones by way of “surface transparency” of $X_1$. Also, the black “disc” located in the center is meant to be the point $x_0$ onto which $D$ deformation retracts.

Figure 1

Before providing a pictorial reference in Figure 2, consider the following verbal description of the remaining homotopy equivalences.

Consider “stretching” the point $x_0$ so that a segment $\ell$ connects it to a second point $x_0'$. This segment is red in Figure 2 below. This maneuver is a homotopy equivalence with inverse homotopy given by “gluing” the points $x_0$ and $x_0'$ to eliminate $\ell$, a process that’s continuous due to the fact that $\ell$ is homotopy equivalent to a closed interval $I$, and hence to a point. The result of this maneuver is the space $X_2$ shown in the leftmost diagram in Figure 2. Finally, consider homotopy on $X_2$ which glues $x_0$ and $x_0'$ by moving clockwise (that is, $x_0'$ moving “away from” $x_0$ instead of towards it). As mentioned previously, this is a homotopy equivalence with the homotopy inverse that “unglues and separates” $x_0$ and $x_0'$.

The result is a topological figure which is homemorphic to a sphere $S^2$ and which has a “wedge point” $x_0=x_0'$ on its “surface”, along with “wedged circles” $C_1\cong S^1$ “outside it” (the red circle in Figure 2) and $C_2\cong S^1$ “inside it” (the light blue circle in Figure 2) connected via the wedge point $x_0$. This is sketched roughly in the second diagram of Figure 2.

Figure 2

Note that the sketches given here are extremely rough. A much clearer – and more detailed – diagram can be found here. The combination of the linked diagram with the verbal descriptions above completely characterize the series of homotopy equivalences needed to transform the original space $X$ into the resultant space which, topologically, is homeomorphic to $S^2\vee S^1\vee S^1$. Hence, the result.    $\square$

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