Show a subspace formed by a Klein bottle is homotopy-equivalent to $S^1 \vee S^1 \vee S^2$

I am trying to solve Hatcher, 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 the figure, is homotopy equivalent to $S^1 \vee S^1 \vee S^2 = Y$.

                                           

I do not understand how $X$ can be homotopic to $Y$ since I do not think that they are even homologic. Intuitively, $H_2(Y)=\mathbb{Z}$ since a subspace $S^2$ disconnects $\mathbb{R}^3$ into two pieces. But $X$ 'disconnects' $\mathbb{R}^3$ only into one piece so $H_2(X)=0$.

Solution 1:

I'm not sure why you think $X$ disconnects $\mathbb{R}^3$ into "one piece" (I'm not sure what that would even mean); there's an inside and an outside to $X$, just like there is for the sphere.

Here is a (pictoral) proof of the homotopy - of course, you should provide some justification for why each step is a homotopy equivalence:

The order of the pictures is $\begin{smallmatrix} 1 & 2\\ 3 & 4\\ 5 & 6\end{smallmatrix}$.