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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 $S^1 \vee S^1 \vee S^2 = Y$.

                                           enter image description here

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 $H_2(X)=0$.

Pete L. Clark
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Dávid Natingga
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    By the way, "homotopic" and "homotopy equivalent" are different. Maps can be homotopic; spaces can be homotopy equivalent. – Zev Chonoles Apr 19 '13 at 05:51
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    The point is that you are considering an "immersion" (it isn't actually an injective map) of the Klein bottle in $\Bbb R^3$, which is orientable and self intersecting. A Klein bottle can be properly embedded in $\Bbb R^n$ only for $n\geq 4$. – A.P. Apr 19 '13 at 06:20

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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 $\begin{smallmatrix} 1 & 2\\ 3 & 4\\ 5 & 6\end{smallmatrix}$.

enter image description here enter image description here

enter image description here enter image description here

enter image description here enter image description here

Zev Chonoles
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    Thanks! I drew them when I had this problem in my algebraic topology class at Brown, so they were all set to go on my computer when I saw this post :) – Zev Chonoles Apr 19 '13 at 07:03
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    So the immersion of the Klein bottle (although the Klein bottle itself in $\mathbb{R}^4$ not) has inside and outside. This is the point I was missing - I thought that at the intersection $D^2$ region was cut from the surface of the original $K$ in order to make a "tunnel" from inside to outside and make these ambient subspaces (forming the whole ambient space) connected. – Dávid Natingga Apr 19 '13 at 15:20
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    The fact that the mentioned $D^2$ is present on the surface of $X$ is clearly visible from the deformation retraction of $D^2$ in the homotopy equivalence between the 1st and the 2nd picture. Thanks! – Dávid Natingga Apr 19 '13 at 15:26