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Consider the quotient space obtained by identifying the north and south pole of $S^2$. I think the fundamental group should be infinite cyclic, but I do not know how to prove this.

If it is infinite cyclic, would this and $S^1$ be an example of two spaces which have isomorphic fundamental groups but are not of the same homotopy type?

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A sphere with two points identified is homotopy equivalent to the wedge of a sphere with a circle (this is proved in Hatcher's book, on page 11 of chapter 0). Thus, Van Kampen's theorem implies that the fundamental group is infinite cyclic. However, the second homology group is also infinite cyclic, so it's not homotopy equivalent to the circle.

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Another way to see this is to use the theory of covering spaces. Consider the subspace $Y$ of $\mathbf R^3$ obtained by placing a copy of $S^2$ centered at $(n, 0, 0)$ for each even integer $n$ (I'll try to remember to add a picture later). The group $\mathbf Z$ acts on this space by translation, and the quotient $\mathbf Z\backslash Y$ is the space in question.

enter image description here

Since $Y$ is simply connected (can you prove this?), it follows from Proposition 1.40c of Hatcher that $\pi_1(\mathbf Z\backslash Y) \cong \mathbf Z$. I don't see a nice way of proving that the spaces aren't homotopy equivalent without using homology, as Chris does.

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    $\begingroup$ Lots of negations in that last sentence. $\endgroup$ – Dylan Moreland Mar 31 '12 at 17:24
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    $\begingroup$ I added a simple image for your $Y$. $\endgroup$ – Mariano Suárez-Álvarez Mar 31 '12 at 17:48
  • $\begingroup$ @MarianoSuárez-Alvarez Neat. Thanks! $\endgroup$ – Dylan Moreland Mar 31 '12 at 17:52
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    $\begingroup$ One can show that $\pi_2(X)$ is non-zero, too: it is the same as $\pi_2(Y)$. Let $f:S^2\to Y$ be an "inclusion" onto one of the spheres of $Y$, and let us pick as basepoint some point in the image of $f$. If $f$ were homotopic rel. the basepoint to a constant map, then composing that homotopy with the map $Y\to S^2$ which collapses all the other spheres to one of the two points of contact of $f(S^2)$ to its neighbors would give us an homotopy from the identity of $S^2$ to the constant map. $\endgroup$ – Mariano Suárez-Álvarez Mar 31 '12 at 17:52
  • $\begingroup$ I just want to say that this is a beautiful idea. I'm not sure why Y is simply connected though. $\endgroup$ – ThinkConnect Sep 13 '15 at 1:14
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Additionally, Hatchers book provides another way of answering this problem. We can realize $X$ (the space in question) as a CW complex, with $0$-skeleton a point, $1$-skeleton $S^1 \vee S^1$, say with fundamental group free on $a$ and $b$, and attach $2$ $2$-cells along $ab$, (or any other pair, depending on direction of paths and such). This leads to an isomorphism of the fundamental group of the $0$-skeleton $\mod (ab)$ with the fundamental group of $X$. Despite choice of how to glue the $2$-cells, we get $\langle a, b \mid ab \rangle$ which is infinite cyclic.

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