Gluing of Möbius strips Let $X$ be a Möbius strip. Take $n$ copies of $X$ and glue them by their boundaries (where the morphisms $S^1 \to S^1$ are homeomorphisms) and denote the new space by $Y$. What is a homotopy type of $Y$? At least, how to compute $\pi_1 (Y)$?
I think we need to endow $Y$ with a CW-structure but I don't really see how to do this. It seems to me $\pi_1$ doesn't change but I think this is wrong. Also, I would appreciate any references to places where I can find similar problems with solutions.
 A: I am not sure why the comment is upvoted: your meaning is clear, take $M_1 = \cdots = M_n = M$ all copies of the Mobius band; there are inclusion maps $f_i: S^1 \to M_i$ for all $i$ parameterizing the boundary; take the space $X_n = (M_1 \sqcup \cdots \sqcup M_n)/\sim$ where $f_i(x) \sim f_j(x)$ for all $i,j$. Write $f: S^1 \to X_n$ for the map $f_i$ (which is independent of $i$).
Now it is clear that "$\pi_1$ does not change" is preposterous, because for $X_2 = K$ the Klein bottle. However we can try to take an inductive approach. We have $X_{n+1} = X_n \sqcup_{S^1} M_{n+1}$. We will keep track of both the fundamental group and the particular element $[f]$. In what follows I will freely replace $f$ with $f^{-1}$ as appropriate for the cleanest possible examples; it doesn't really matter.
For $n = 1$ we see that the fundamental group is $\Bbb Z$ with $[f] = 2$. For $n = 2$ we see that the fundamental group is $\langle x, y \mid x^2 = y^2\rangle$ with $[f] = y^2$. For $n = 3$ we see that the fundamental group is $\langle x, y, z \mid x^2 = y^2 = z^2\rangle$.
Because the inclusion map $f_{n+1}$ induces multiplication-by-two on $\pi_1$, we may inductively see that $$\pi_1(X_n) = \langle x_1, \cdots, x_n \mid x_1^2 = x_2^2 = \cdots = x_n^2\rangle,$$ with $[f] = x_n^2$. Notice that when $n = 1$ the relation says nothing.
To see that these are all different you can abelianize. We have $H_1(X_n) = \pi_1(X_n)^{ab} = \Bbb Z^n/\langle 2x_1 - 2x_2, \cdots, 2x_{n-1} - 2x_n\rangle$. A change of coordinates takes this to $\Bbb Z^n/\langle 2y_2, \cdots, 2y_n\rangle$. One finds that $H_1(X_n) = \Bbb Z \oplus (\Bbb Z/2)^{n-1}$ for all $n \geq 1$, which are all pairwise non-isomorphic.
