In this problem a modified Klein bottle (say $X$) is taken in account which is seen as embedded space in $\mathbb R^3$ (giving subspace topology on the usual self intersecting figure of Klein bottle in $\mathbb R^3$).
I have to prove its fundamental group is $\mathbb{Z}* \mathbb{Z}$. Secondly consider the space $Y$ by removing the open disc enclosed by the circle of self intersection of $X$. Find its fundamental group.
I even don't know how to proceed and apologize for the bad writing style.
The space $X$ deformation retracts onto $S^2\vee S^1 \vee S^1$ (draw pictures) and so $\pi_1(X)=\mathbb{Z}\star \mathbb{Z}$.
For the space $Y$ draw the identification picture of usual Klein bottle with hole removed and give it a CW-structure. It should be easy to get $\pi_1(Y)$ from it, using prop 1.26 of the book. (Hint: add an extra $1$-comlex $c$).
Lastly, note that the space $Y$ can be embedded into $\mathbb{R}^3\setminus Z$ and actually it deformation retracts onto $Y$.
One other way one can do this... I am doing some cheating here by calculating the $\pi_{1}$(Y) first and then from there I'll calculate the fundamental group of X... in that case if we attach a 2-cell to wedge sum of 3 circles by the word $aba^{-1}b^{-1}cbc^{-1}$ we'll get Y where b is denoted the loop of the self intersection circle of X...so by proposition 1.26 of hatcher we can get that $\pi_1(Y) = \langle a,b,c| aba^{-1}b^{-1}cbc^{-1}\rangle$. from here it is easy to construct X as follows...just attach a 2-cell with b loop ...so by proposition 1.26 $\pi_1$ (X) is actually $\mathbb{Z}$ $\star$ $\mathbb{Z}$.
